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[email protected],v([AI)la1.2a3a4)
=
c
~X,'(R)a,,([x])lalaza3a4).
P
The application of an element R of S4 to the oscillation state ~ a ~ a 2 a 3 a 4is) defined as follows. For the permutation R,
R=
( 1 2 3 4 ) = (" T1 72 73 T4
1
we have
Recall that the application of a permutation R to the state only changes the order of four oscillatory quantum numbers a j , but does not change their values, In fact, there exist 24 linearly independent symmetric bases belonging to the inequivalent irreducible representations if four oscillatory quantum numbers aj all are different. The number of the linearly independent bases will decrease if some oscillatory quantum numbers aj are equal to each other. For example, when two and only two quantum numbers coincide with each other, there exist 12 linearly independent symmetric bases belonging to five irreducible representations ([4] @ [2,2] @ 2 [3,1]@ [2,1,1]). 26. Calculate the representation matrices of the generators (1 2) and (1 2 3 4 5 ) of the S5 group in the irreducible representation [2,2, I]
by the tabular method. Solution. From Problem 15, the orthogonal standard Young operators for the Young pattern [2,2,1] of S5 are J+ [E (2 5)] and y p ,2 5 A!, 5 5. Now we will calculate the representation matrices of the elements (1 2) and (1 2 3 4 5) in this representation [2,2,1] by the tabular method.
232
Problems and Solutions in Group Theory
From the tables, the representation matrices for the generators (1 2) and (1 2 3 4 5) in the representation [2,2,1] respectively are
[g fl
1 0  1 0
; [; ; 1
0 0 0 01
111
,
0 0
p1;
0 0
.
1 0 0
27. Respectively calculate the real orthogonal representation matrices of the transpositions Pa of the neighbored objects in two equivalent irreducible representations [23] 21 [16] x [3,3] of SS, and find the similarity transformation matrix X between them. Solution. The standard Young tableaux for the Young pattern [23] and the Young pattern [3,3] are listed from the smallest to the largest, respectively
233
Permutation Groups
12 12 13 13 14 34 35 24 25 25 56 46 56 46 36
123 124 125 134 135 456 356 346 256 246
Obviously, each standard Young tableau in [23] is the transpose of the corresponding one in 3,3], but the larger in [23] becomes the smaller in [3,3]. According to Eq. 6.12), the representation matrices of the transpositions P, for the neighbored objects in the representations p3]and [16] x [3,3] respectively are
t
representat ion [23]
representation [16] x [3,3]
[;H ;Al i ) ,
PI :
1 0
0
0
0 0
0
01
o & o
1
P2+
0
; a; o; ) ,
0 1
0
0
O A 0
P3:
0
0
0

3 0 0 0  3 0
0 0
0 0
g [o0
0
4
0
0
i( i 1
1
0

0
0a 1 0 0 0
1
a 0
 d o
i
0
1
0 0 0 0
i3:3:) 0 0 3
[;;;!l ;) ['P,H i i] 0 0 1 0
:
 2 0 01
d 1
0 3 0 01 4 )0>
0 0
P5
2
+[;a
0
,
0 0 0 0 0  1
0
0 010 0 0 0 1
Each matrix in [16] x [3,3] can be obtained from the corresponding one in [23] by a similarity transformation X = Y Z . Y is a diagonal matrix diag{  1, 1, 1,  1 , l ) . Since the nondiagonal matrix entries of P, in the representation [23] appear only in 12, 13, 24, 34 and 45 rowcolumns, Y changes the signs of all nondiagonal matrix entries of P, in [23]. The
234
Problems and Solutions in Group Theory
matrix entries in 2 are all vanishing except for those on the minor diagonal line which are 1. 2 transpose the matrix in [23] with respect to the minor diagonal line. Namely,
0 0 0 01 0 0 0 1 0
X1[23]X = [16] x [3,3],
X=
This method is suitable for other pair of the representations with the associated Young patterns. 28. Calculate the character of each class of the permutation group
S5
in
the representation [2,2,1] by the graphic method.
Solution. The characters of each class of are calculated in the following table.
class Regular application
,
(I5> neglected
Application parity
[(e>l
5
P3J> 14 24 3 1
(1J2> 13 23 2 1
1
1
S5
in the representation [2,2,1]
(12,3> 12 33 3 1 1
(2,3) 11 22 2 1
(L4) 12 22 2 1
(5)
1
1
0
no
29. Calculate the characters of each class of the permutation group
$3
in the representations [3,2,1], [3,3], and [23] by the graphic method, respectively.
Solution. We calculate the characters in three representations of SS by the graphic method in the following table, where we also list the permitted regular application. The regular application of the identity (1") is neglected. The Young pattern [23]is the one associated with the Young pattern [3,3]. The regular applications for [Z3] are neglected because they can be obtained from those for [3,3] by transpose. The Young pattern [3,2,1] is a selfassociated Young pattern, so the characters of the classes composed of the odd permutations are vanishing.
Permutation Groups
235
Characters in the ii reducible representations of s6 Regular application neglected 1 2 3 1 2 4 1 3 4 1 2 5 135 455 355 255 345 245 124 133 134 334 244 234 113 122 123 2 2 3 133 1 2 3 123 144 123 124 134 44 24 444 344 244 4 3 122 133 122 33 2 3 333 3 2 111 1 2 2 111 1 1 2 22 12 222 122
I (3113)
I
2 0
(3,2,1)
l
1
l
1
2
(4,2)
1
0
1
1
123 333 112 222
1
122 22 2
30. Calculate and fill in the character tables of S3, Sq, Sg, SS and S7 by the graphic method. Solution. Here we will not give explicitly the regular applications, but only list the calculation results in the table. The character of each class in the antisymmetry representation [In] shows whether the permutation in the class is even or odd. n ( l ) denotes the number of elements in the class ( l ) .Instead of showing the pair of associated representations, only one of them is shown for simplicity.
Problems and Solutions in Group Theory
236
(3.12) (3,2j (4,l) (5) \
I
20 20 30 24
I
1 1 1 1
1 1 0 1
1 1 1 1 0
2 0 0 0 1
Permutation Groups
6.5
237
The Inner and Outer Products of Representations
*
The direct product of two irreducible representations of a permutation group S, is called the inner product of two representations for S,, which is reducible in general. There is no special graphic rule for the reduction of the inner product of representations for the permutation group. The CG series and the CG coefficients for S, can be calculated by the character method just like the usual finite group:
Since the irreducible representations of S, are real, the multiplicity [cJ]) is totally symmetric with respect to three representations. This symmetry leads to some common property for the CG series of S,. Letting [A] and [?;I be two associated Young patterns, we have .([A],
[TI,
(6.17)
In the reduction of the inner product of two irreducible representations of S,, there is one identical representation [n] if and only if two representations are equivalent, and there is one antisymmetric representation [ln]if and only if the Young patterns of two irreducible representations are associated.
*
We are going to apply the concepts of the subduced representation and the induced representation, discussed in $3.3, to the permutation group. The subgroup H = S, x S, of the permutation group G = Sn+, is the direct product of two subgroups S, and S,, where S, consists of the permutations among the first n objects in the (n m) objects and S, consists of the permutations among the last r n objects. The order of G is g = (n+rn)!. The order of the subgroup H is h = n!m!and its index is n = g / h . Denote by R j H , 2 5 j 5 n, the left cosets of H in G where the element Rj is a permutation transforming the first n objects to n different positions in all (n rn) objects. Rj is not unique, but we assume that Rj has been chosen. For convenience, we denote the subgroup H by R1H with R1 = E .
+
+
238
Problems and Solutions in Group Theory
Let L be the group algebra of G. Denote by e[W]the primitive idempotent in L,and by Le[W]the minimal left ideal generated by &", where the Young pattern [w] with ( n m) boxes describes the irreducible representation of G. Let P" c LC be the group algebra of the subgroup H . Denote by e[A]e[p] the primitive idempotent in Lnm,and by Ln"e[A]e[p]the minimal left ideal generated by e[A]e[p] , corresponding to the irreducible representation of H described by the direct product of two Young patterns [A] x [ p ] ,where the numbers of the boxes in two Young patterns [A] and [p] are n and m, respectively.
+
*
A subduced representation from an irreducible representation [w]of G with respect to the subgroup H is generally reducible and can be reduced with respect to the irreducible representations [A] x [p] of H , (6.18) From the viewpoint of the group algebra, Eq. (6.18) shows that there are ayp linear independent vectors e[A]e[pltae[w] , which map the left ideal Lce[W] onto the minimal left ideals for H , respectively,
Equation (6.19) gives the linear expansion of the new basis in the minimal left ideal in terms of the bases in L e [ W ] The . coefficients in the expansion constitute the columnmatrices in the similarity transformation matrix X.
*
The primitive idempotent e[']e[p]in Lnm is also an idempotent in L, but generally not primitive. The left ideal Ce[A]e[p] generated by e[A]e[p] in C corresponds to a reducible representation of G, which is the induced representation from the representation [A] x [p] of H with respect to G. For the permutation group, the induced representation is called the outer product of two representations, denoted by [A] 8 [ p ] . This representation can be reduced with respect to the irreducible representations [u] of G,
From the viewpoint of the group algebra, Eq. (6.20) shows that there are b'fD linear independent vectors e [ w ] t ~ e [ x ] which e [ ~ ] , map the left ideal
Permutation Groups
239
Le[x]e[p]onto the minimal left ideals for G, respectively,
(6.21) Equation (6.21) gives the linear expansion of the new basis in the minimal left ideal in terms of the bases in Le[’]e[p]. The coefficients in the expansion constitute the columnmatrices in the similarity transformation matrix Y . Due to the symmetry between left and rightideals, the numbers of the linear independent vectors e[Xle[plt,e[W] and e[w]t~e[x]e[pl are equal, a& = b!p. This is an example of the Frobenius theorem (see Chapter 3).
*
There is a graphic rule, called the LittlewoodRichardson rule, for calculating the multiplicity b z p in the reduction of the outer product of two irreducible representations for the permutation group, although it can be calculated by the character method as usual. The calculation method is as follows. Arbitrarily choose one of the Young patterns in an outer product [A] @ [ p ] ,say [ p ] . Usually, choose the Young pattern with the less boxes. Assign each box in the Young pattern [p] with its row number, namely assign the box in the j t h row with j . Now, attach the boxes assigned with 1 to another Young pattern [A] from its right or from its bottom, and then attach the boxes assigned with 2, and so on, in all possible way subject to the following rule. After attaching the boxes assigned with each digit, say j , the resultant diagram satisfies the following conditions: 1) The diagram constitutes a Young pattern, namely, the diagram is lined up on the top and on the left without unconnected boxes in each row such that the number of boxes in the upper row of the diagram is not less than that in the lower row, and the number of boxes in the left column is not less than that in the right column. 2) No two j appear in the same column of the diagram. 3) We read the attached boxes from right to left in the first row, then in the second row, etc., such that in each step of the reading process, the number of boxes assigned with the smaller digit must not be less than that assigned with the larger digit.
In each way subject to the above rule, the resultant diagram [w] after attaching all boxes assigned with digits denotes an irreducible representation [w] of Sn+m which appears in the reduction of the outer product [A] @ [ p ] .
Problems and Solutions an Group Theory
240
If the same Young pattern [w] is obtained in different ways, the number of the ways is the multiplicity of the irreducible representation [w] in the reduction.
*
Due to the Fkobenius theorem, the LittlewoodRichardson rule can also be used for calculating the reduction of the subduced representation from the irreducible representation [w] of Sn+, with respect to the subgroup S, x S,. In the calculation, all the Young patterns [A] with n boxes and all the Young patterns [p] with rn boxes are taken to constitute the irreducible representation [w] of Sn+m with the multiplicity b;p, if [w] appears in the reduction of the outer product [A] 8 [p]according to the LittlewoodRichardson rule with the same multiplicity. 31. Calculate the ClebschGordan series of the inner product of each pair of the irreducible representations for the permutation groups S3, S4, S5, s6 and S7 by the character method. Solution. We can calculate the CG series for the reduction of the inner product of two representations by Eq. (6.16). We only list some results for the CG series, and neglect the remaining CG series which can be obtained from them by Eq. (6.17).
For the S3 group: For the S4 group:
[2, I] x [’&I]N [3]@ [I3]@ [2, I]. [3,1] x [3,1]= [4] @ [3,1]@ [[email protected] [2, 121, [3,1] x [Z2]
= [3,[email protected] [2, 121,
p21
[41CB [221 @ [141.
p”3
For the S5 group:
[4,1] x [4,13 = [5] @ [4,13 @ [3,21 @ [3,121, [4, 11 x [3,2] = [4,1] @ [3,2] @ [3,[email protected] P2,11, [4, I]
[ 3 , 1 2 1 = [4, 13 @ [3,21 @ ~ 3121 , CB [22,[email protected][2, 131,
[3,21 [3,21 = [51 p, [email protected] [3,21 (3,[email protected] [22,11 @ [2,131, p, 21 [3,121 p, 13 @ p, 21 2[3,[email protected] p,13 @ p, 131, [ 3 , q [3,121 [5]@ [4, 11 @ 2[3,[email protected][ 3 , q @ 2[22,11 @ [2,[email protected]~ 5 1 .
=
For the
s6
group:
Pe rrnu t at ion Group
For the
S7
group:
241
242
Problems and Solutions in Group Theory
Permutation Groups
243
32. Calculate the ClebschGordan coefficients for the reduction of the selfproduct of the irreducible representation [3,2] of S g . Solution. This is a typical example for calculating the reduction of a reducible representation with a higher dimension for a finite group. Usually in the reduction of a reducible representation, one may try to find one or a few commutable elements in the group such that their eigenvalues can designate the bases for each irreducible representation without degeneration. This method has been used in the calculation for the ClebschGordan coefficients of the groups T, 0 and I in Problems 2527 of Chapter 3. For the permutation group S,, when n > 5, the dimension of the irreducible representation is quite high. It is hard to find the eigenvalues of some commutable elements to designate the bases of each irreducible representation without degeneration. For example, one can designate the bases by the eigenvalues of two commutable elements (1 2) and (3 4 5), where the number of the different pairs of the eigenvalues is six. However, there is a 16dimensional representation [3,2,1] in Sg. In the present Problem we are going to introduce another method for calculating the CG coefficients,
244
Problems and Solutions in Group Theory
which will be suitable for the representations with higher dimension. From Problem 31 we have
We take the orthogonal bases where the representation matrix of each transposition Pa of two neighbored objects is known. The basis state before reduction is denoted by Ip)lv),where p and Y both run from 1 to 5, respectively. Denote by I ][A], p ) the basis state after reduction, where p runs from 1 to the dimension of [A]. There is a characteristic in this example that two representations in the direct product are the same. The expansion of the new basis I ][A], p ) can be chosen to be symmetric or antisymmetric in the permutation of two states in the product. For simplicity, we use the short notation for the basis state with the given symmetry in the permutation: I P , v ) S = IP)IY) k IY)IP) and I P , V ) A = IP)Iv)  Iv)IP) When P = v, only symmetric state exists, and we just use the basis state 1p)lp) without the subscript S. The method is outlined as follows. For a given representation, the representation matrix of Pi is diagonal. The eigenvalue of PI is used to classify the basis states. The basis state with the eigenvalue 1 of Pi is called the Atype basis, which has the following expansion
The basis state with the eigenvalue —1 of P1 is ccalled the Btype basis, which has the following expansion
The action of the transposition Pa on the basis state is
where the representation matrix of Pa in [A] is known. The representation matrices of Pa in the representation [3,2]are
Permutation Groups
1 0 0 0 0 PI:
1 0 10 001  10 0o
2
),
p2:;[:
0 0 0 0  1 1Jsoo 0 P32[ 0 01 03 0 0 1 ,
Js 0 0
0 0 3 0 0 0 0  3
245
ooool
 1 o f i o 0  1 0 4 4 0 1 0 0 ' 02 0 4 0 0 0 1 0
0
0 0  1 4 0 0 & 1 0 0 2 0 0 0 la too 0 4 1
P42
For each irreducible representation [A], we first choose a basis state which is the common eigenstate of as many transpositions P a as possible, and write its expansion according to its eigenvalue of PI. Then, in terms of the representation matrices of Pa and the orthonormal condition we can determine the coefficients in the expansion. From the Schur theorem, two basis states belonging to two inequivalent irreducible representations or belonging to the different rows of a unitary irreducible representation are orthogonal. If the multiplicity of [A] in the reduction is one, the expansion must have a definite symmetry in the permutation. If it is larger than one, we can choose the expansions having definite symmetry. From the first basis state we are able to calculate the remaining basis states by Eq. ( c ) , which have the same symmetry in the permutation as that the first basis state has. The orthonormal condition and the symmetry in permutation can be used for check. First, we calculate the basis state in the representation [5], which is the identical representation with dimension one. The basis state II[5]) is given in Eq. (a). All representation matrices of Pa in [5] are 1. According to Eqs. (c) and (d), we have from the action of P2
Problems and Solutions in Group Theory
246
Comparing the coefficient of each term, especially the coefficients of the terms which do not appear in 11[5]), we obtain a2 = a3 = a5 = a6 = 0, a4 = ale, a7 = all, a8 = a12 and a9 = ~ 1 3 namely ,
Applying
P3
to 11[5]),we have
Applying
P4
to 11[5]),we have
Thus, a1 = ag. After normalization we obtain
Second, for the fourdimensional representation [4,1], we obtain the representation matrices of Pa from Eq. (12)
Permutation Groups
O1 0 O0 O 0 ) ) 0 0 1 0 0 0 0 1 3 0 0 0 p 3 ' 31 0O Jg l f i 1O 0 )
0 0
0 3
:) ; ;:).
p 2 : 1 (2o02 O0
P1 :
(
247
2
7
.:(aj
0 0  1 &
oofi 1 a
1
0
0
0
0 1
The solutionm is
The following calculation is similar. We only list the method and the results. Applying P3 to [4,1], 2} gives
Thus,
Problems and Solutions in Group Theory
248
Applying P2 to 1J[4,1],3),we have
Thus,
Note that 11[4,1],4) is a Btype basis state. Third, we calculate the basis states in the representation [3,2] where the representation matrices of Pa were given in Eq. (d). 11[3,2],1) is an Atype basis state. Applying PI and P2 to 11[3,2],1) one by one leads to a formula similar to Eq. (e):
Applying
P4
to 11[3,2], 1)) we have
a1
+ 2a4 + 2a9 = 0,
Thus, a1 = 2a4 and
Applying
Thus,
P3
a9
2al
+ 4a4  6 ~ =9 0.
= 0. The normalization condition leads to
to 11[3,2],1),we have
Permutation Groups
249
Thus,
Thus,
Thus,
Fourth, we calculated the basis strates in the sixdimensional representation [3,1], where the representation matrices of Pa are
PI :
P3 :
2 0 0 01 0
100 0 0 0 010 0 0 0 001 0 0 0 0001 0 0 0 0 0 0 1 0 0 0 0 0 0 1
1
3
6 0 0 0
1 0 0 0 0 30 0
0 0 0 0 3 0 0 0 0 0  1 6
1 2
P2 : 
0 0 1
0 0
ofio 1 o o f i o 0 0 0 0  4 0 0 0  1 6
'
0
0 0 0 0 0 1 0 0  1
&i0
0
0
1
'
0
0
0
0
1
p4:
4
0 0 0 0 0O a 0 l
11[3,12], 1) is an Atype basis. Applying PI and leads to a formula similar to Eq. (e):
P2
0 o 0
 1 6 0 a 1 0 0O 0 O 04 1
*
to 11[3, 12],1) one by one
Problems and Solutions an Group Theory
250
Applying
P4
to 11[3,1’3, 1) changes its sign
+
a8)/2. The orthogonal condition Thus, a1 = 0 and a4 = a9 = &(a7 between 11[3,12],1)and 11[22,1],1) leads to a4 = a9 = 0. Then, a7 = a8. The basis state 11[3,12],1) is antisymmetric in the permutation. After normalization we have
Applying
P 3
to 11[3,12],1))we have
P4
to “[3,1’3, 2), we have
P 2
to 11[3, 12],2)) we have
Thus,
Applying
Thus,
Applying
Thus,
11[3,1’1, 4) is a Btype basis state. Applying
P 4
to 11[3, 1’1, 4), we have
25 1
Permutation Groups
Thus,
Applying
P3
to 11[3, 12],5), we have
Thus,
Fifth, we calculate the basis states in the fivedimensional representation [22, 11, where the representation matrices of Paare
1o&io 00 0 00 0 3 0 0  3
0 0
1
0 0 1 0 0 0 0
0 0
0
o&io
1
0
0  2 0 0
0
0
1Ao
0
0  1 6
The basis state 11[22, 1],5) is the common eigenstate of PI, P 2 and P 4 . We begin the calculation with this state. 11[22, 13,5) is a Btype basis with the form giver p 2
252
Problems and Solutions an Group Theory
The solution is bl = b2 = b3 = b4 = 0, b5 =  b 6 , b l l = b12, namely
Applying
P4
b7
= blo, b8 =  b g and
to 11[2” 1],5), we have
The solution is b5 = bll = f i ( b 7  b8)/2. The orthogonal condition between 11[22,1],5) and 11[3,12], 6) leads to b5 = b l l = 0 and b7 = b g . After normalization we obtain the symmetric expansion
The solution is
Thus,
Permutation Groups
253
Thus,
Thus,
{ IW4  13)13)  1414) + 15)15) + & l V ) s }
IP2,11,1)= (2&)'

Finally, we calculate the basis states in the fourdimensional representation [2, 13], where the representation matrices of Pa are
p1:
(
1 0 0 0 010 0 0 0  1 0 0 0 01 3 0 0
P 3 4 ; \ O
2: 0
,
p 2 : 1(
2
'0) ,
03
1ao 0 d l 0 0 0 0  2 0 0 0 02 4 0 0
p4:;(
0 40
0 o
)
'
').
01 4 5 a
1
The basis state )/[2,13], 4) is the common eigenstate of P I , P2 and P3. We start the calculation with this state. 11[2, 13],4) is a Btype basis. Applying P 2 to 11[2, 13],4), we obtain a similar formula to Eq. (g):
Applying
P 3
to 11[2, 13],4), we have
Thus, b5 = b7 = b8 = 0. After normalization we obtain the antisymmetric expansion
Applying P4 to I1[2, 13],4), we have
254
Problems and Solutions in Group Theory
The solution is
Thus,
Thus,
33. Calculate the reduction of the following outer products of the representations in the permutation groups by the LittlewoodRichardson rule: (1) [3,2,118 ~31, (2) [3,218 ~ 211, , (3) P, 11 8 [4, z31. Solution. (1)
The dimensions of the representations on both sides are
9! 6! 3!
x
16 x 1 = 1344 = 105
+ 162 + 120 + 189 + 168 + 216 + 216 + 168.
Permutation Groups
255
The dimensions of the representations on both sides are 8!
5! 3!
(3)
~5~2=560=28+64+14+2~70+56+90+42+56+70.
256
Problems and Solutions an Group Theory
The dimensions of the representations on both sides are
+ +
13! x 2 x 300 = 171600 = 12012 + 9009 + 12870 11583 3 2 x 21450 3! lo! 5005 10296 8580 12870 15015 8580 17160 5720.
+
+
+
+
+
+
+
34. Calculate the reduction of the subduced representations from the fol
lowing irreducible representations of
S6
with respect to the subgroup
s36 x 3 3 :
Solution. (1)
[4,2]
2
[3] x [3] @ [3] x [2,1] @ [2,1] x [3] @ [2,1] x [2,1].
The dimensions of the representations on both sides are 9=1x1+1x2+2x1+2x2. (2)
[22,121 fi [2, 1 3 x [2, 11CD [2, 11 x
[i31 CB [PI x
[2, [email protected] [i31 x
The dimensions of the representations on both sides are 9 = 2 x 2 + 2 x 1+1x 2 + 1 x 1.
p31.
Permutation Groups
257
(3)
The dimensions of the representations on both sides are 5 = 1x 1
+ 2 x 2.
35. Calculate the similarity transformation matrix in the reduction of the
subduced representation from the irreducible representation [3,3] of SS with respect to the subgroup S3BS3.
Solution. The reduction of the subduced representation from [3,3] of SS with respect to the subgroup S36& was calculated in Problem 34:
The standard Young tableaux for the Young pattern [3,3] were given in Problem 18. Denote by Y1 the Young operator for the standard Young tableau
The standard bases in the representation space C of [3,3] are
From the Fock condition (6.5) we have
{ E + (1 4) { E + (1 5) { E (1 6) {E (3 4)
+ +
+ (2 4) + (3 4)) Yl = 0, + (2 5) + (3 5)) Yi = 0, + (2 6 ) + (3 6)) Yi = 0, + (3 5) + (3 6)) Yl = 0.
The idempotent of the onedimensional representation [3] x [3] of the subgroup S 3 8 S 3 is
258
Problems and Solutions an Group Theory
From Eq. (6.19), the basis state for the representation [3] x [3] is
The idempotent of the fourdimensional representation [2,1] x [2,1] of the subgroup S38S3 is
Since y3y1 = 0, from Eq. (6.19), we choose the basis state for the representation [2,1] x [2,1] as = Y3(3 4)Yl = (3 4) {E (1 2)  (1 4)  (2 1 4)) {E
'$1
+
= (3 4) {2E  (1 4)  (4 2)) { E
+ (3 5)  (3 6)  (5 3 6)) YI
+ (3 5)  2(3 6)} Y1
+ 2(4 3 5)  4(4 3 6)  (3 4 1)  (4 1 3 5) + 2(4 1 3 6)  (3 4 2)  (4 2 3 5) + 2(4 2 3 6)} Y1 = 2b2 + 2b3  4(3 6)bl  (1 4)bl  (1 5)bl + 2(6 l ) b l  b4  b5 + 2(6 2)bi = 2b2 + 2b3  b4  b5 + [bl + b2 + b4] + [bl + b3 + b5]  2bl + 6 [bi + b2 + b3] = 6b1 + 9b2 + 9b3.
= {2(3 4)
Then, we have
Thus, we obtain the similarity transformation matrix X:
The generators of
S38S3
36 0
6 9
0
0
3 0
are P I , P 2 ,
6 0
3 0
9 P4,
and
P5.
Their representation
Permutation Groups
259
matrices in [3,3] can be calculated by the tabular method:
1 0 0  1 1 0 1 0  1 0
1 0 0 0 0 0 0 0 1 0
0 0 0 0 1 0 0 0 1 0
0 0 0 11 0 0 0 01
Their representation matrices in [3] x [3] @ [2,1] x [2,1] are
1 0 0 0 0 0 1 0  1 0 ~ ( ~ l ) =0 [email protected] ( ~  ~ ) ~ ( ~ ~ ) = ~ o o 0 0 01  01 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 , 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 D ( P 4 ) = 1 @ ~ ~ ; ) x ( ~ 0 ~1) ( 0 0 0 10 1 01 001  0 o1 ) ,
j,
1 0 0 0 0
0 0 0 0 0 1 1 0 0 0 0 0
1 1
0 01 0 0 0 0 0 0 . 0 1 1 0
Problems and Solutions an Group Theory
260
It is easy to check that X satisfies the similarity transformation
36. Calculate the representation matrices of the generators of S4 in the induced representation from twodimensional irreducible representation [2,1] of S 3 with respect to Sq.
Solution. For simplicity, let A and B respectively be the representation matrices of the generators of S 3 in the twodimensional representation [2, I]:
Denoting by we have
&,
p = 1 and 2 , the state bases in the representation space, 2
2
u=l
u=l
This twodimensional space is invariant for the S 3 group, but not for Sq. The S 3 group is a subgroup of S 4 with index 4 . Extend the twodimensional space:
The eightdimensional space is invariant for S4. In fact,
26 1
Permutation Groups
Therefore, the representation matrices of the generators (1 2) and (1 2 3 4) of Sq in the induced representation are:
D(12)=
(
A 0 0 0 A O), O A O O
D(1234)=
O O O A
(
0 B O 0
0 O B O 0 0 ABOO O A O
)
*
It is a reducible representation of Sq, which can be reduced into the direct sum of the representations [3,1],[2,2] and 12, 12].
37. Calculate the similarity transformation matrix in the reduction of the outer product representation [2, I] 8 (21 of the permutation group.
Solution. The outer product representation [2, I] @ [2] is the induced representation from [2,1] x [2] of the subgroup Ss xS2 with respect to the group S5. Its reduction can be calculated by the LittlewoodRichardson rule: [2,1] 63 [2] = [4,13 @ [3,2] 43 [3, 12]@ p2,11 120 ~2~1=20=4+5+6+5. 6x2
The idempotent of the twodimensional representation [2,1] x [2] of the subgroup [email protected] is
The standard bases in the representation space of [2,13 x [2] are
From the Fock condition (6.5) we have
and the useful formula for the later calculation,
[E + (1 2)] [E  (1 3)] bl = [2E  (1 3)
 (1
2)(1 3)] bl = 3 h 
Problems and Solutions in Group Theory
262
Extending the twodimensional space, we obtain the bases b3
= ( 1 4)bl,
b5
= (2 4 ) h ,
= ( 1 4)bz = ( 1 4 ) ( 2 3)bl, be = ( 2 4)bz = (4 2 3)bl, b4
The representation matrices of the generators P1 = (1 2) and W = (12345) of S5 in the out product representation [2,1] []2 for the bases are
Through a direct calculation we have
After the similarity transformation X, the out product representation [2, I] 8 [2] becomes a direct sum of four irreducible representations:
263
Permutation Groups
where R = Pi and W . By the tabular method we obtain 0[4qp1) =
(
1 0 0  1
’) ,
 1 1 0 0
0[4111(W) =
0 0 1  1 0 0 0  1
(’
’
 1 0 0 1O )  1 0 0 0
111  1 0 0 0 10  1 0 0 0 10
10 0 1 0 0 1 0 1 0
111 0 0 0 1 0 01 1 0 0 101 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0
[ 1o 0l  01 01 11) D[22Jl(q)= 0 0 1 0 0 0 0 0  1 0 0 0 0 01
D[22Jl(w)=
i
,
[;m;mcl) 11 1 1 0 111 0 0 0 0 1 0 0
.
Now, we calculate the similarity transformation matrix X by Eq. (6.21). The idempotent of the fourdimensional representation [4,1] of S5 is
From Eq. (6.21) we obtain the expansions of the basis states [email protected]”’ in the extending space: 20
C
$ ~ ”=]
bvXv3
= [ E T t l 4)
= Y2Y1
+ (2 4) + (1 5) + (2 5) + (1 4)(2 5)]
+
x [E (1 211 [E + (4 511 [E  (1 311 bi = 6 [ E (1 4) (2 4) (1 5) (2 5) (1 4)(2 5)] bi
+ + + + = 6 [bi + + + b g + b i i + b3
b5
b15]
+
,
of [4,1]
264
Problems and Solutions in Group Theory
The idempotent of the fivedimensional representation [3,2] of S5 is
From Eq. (6.21) we obtain the expansions of the basis states in the extending space:
$F’21of [3,2]
Permutation Groups
265
20
=
’2I !$
C buXu7= (4 5)$F’21 u=l
= 3 [bl
+ b9 + b l l + 2b4  2b5  2b6 + b7 + 2b16  b17
20
=
I”!+
C b,X,g
= (2 3)(4 5)$~!’~]
The idempotent of the sixdimensional representation [3,12]of
S5
is
From Eq. (6.21) we obtain the expansions of the basis states $ J ~ ’ ~of’ ’[3,12] in the extending space: 20
Problems and Solutions in Group Theory
266
$,F912]=
20
bvXu13 = ( 2 3)q!Jf3”21
The idempotent of the fivedimensional representation [22,11 of
Y5
S5
is
=
From Eq. (6.21) we obtain the expansions of the basis states +F2’l1of [22, 11 in the extending space:
Permutation Groups
$f2’11=
20
bvXv19 = ( 2 3)(4 5)llfz2’11
26 7
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Chapter 7
LIE GROUPS AND LIE ALGEBRAS
7.1
Classification of Semisimple Lie Algebras
*
An element R in a Lie group G can be described by a set of independent real parameters ( T I ,r2,. . .). Those parameters vary continuously in a given Euclidean region, called the group space. The number g of the parameters, which is equal to the dimension of the group space, is called the order of the group G. In this chapter we only discuss the Lie group whose group space is simplyconnected. A Lie group is said to be compact if its group space is a closed region in a Euclidean space. A Lie group is said to be simple if it does not contain any nontrivial invariant Lie subgroup. A Lie group is said to be semisimple if it does not contain any Abelian invariant Lie subgroup including the whole group. Any Lie group of order one is a simple Lie group, but not a semisimple Lie group. A simple Lie group whose order is more than one must be a semisimple Lie group.
*
The generators I A in a representation D ( R ) of a Lie group G are defined
as
A
The generators I A in any representation D ( R ) of G satisfy the common commutation relation: [ I A , 1131
=iC
cA,”ID,
(7.2)
D
where the real parameters CA,”, independent of the representation, are called the structure constants of the Lie group. 269
270
Problems and Solutions in Group Theory
*
The real linear space spanned by the bases ( 4 1 ~is)called the real Lie algebra if the product of two vectors X and Y in the space is defined by the Lie product:
If the linear space is complex, the algebra is called the complex Lie algebra, or briefly, the Lie algebra. A real Lie algebra of a compact Lie group is compact. A Lie algebra of a simple or a semisimple Lie group is simple or semisimple, respectively. In fact, a simple Lie algebra does not contain any nontrivial ideal, and a semisimple Lie algebra does not contain any Abelian ideal, including the whole algebra. Therefore, a onedimensional Lie algebra is simple, but not semisimple. In the rest of this chapter, if without special indication, “a simple Lie algebra with the dimension larger than one” will be simply called as “a simple Lie algebra” for simplicity. It is proved that a semisimple Lie algebra can be decomposed into the direct sum of some simple Lie algebras.
*
The Killing form QAB of a Lie group or its Lie algebra is defined from its structure constants cA::
is a real symmetric matrix of dimension 9 . When the parameters of a given Lie group are changed, the generators transform like a vector, and the structure constants as well as the Killing form transform like a tensor: QAB
*
The Cartan Criteria says that a Lie algebra is semisimple if and only if its Killing form is nonsingular, and a real semisimple Lie algebra is compact if and only if its Killing form is negative definite.
271
Lie Groups and Lie Algebras
*
If { H i , 1 5 j 5 !} is the largest set of the mutually commutable generators in a simple Lie algebra L, the set constitutes the Cartan subalgebra and e is called the rank of the Lie algebra L. It is proved that the group parameters in a simple Lie algebra L can be chosen such that the generators H j and the remaining generators E, satisfy
[ H j , E a ] = @a,
+p
( N,,pE,+p
when a'
lo
the remaining cases,
is a root,
(7.6)
and the Killing form becomes 9jk =  s j k ,
Sap = S(CY)p,
gj,
= gaj = 0.
(7.7)
The real vector d in the !dimensional space is called the root in L, and the space is called the root space. The space spanned by the roots in a simple Lie algebra of rank l is &dimensional. The roots kG appear in pair. Except for the pair of roots, any two roots are not collinear. There is a onetoone correspondence between the generator E, and the root a'. Hj and E, are the CartanWeyl bases for the generators in the simple Lie algebra L.
*
In a semisimple Lie algebra, we can constitute a root chain from a root a' by adding or subtracting another root several times. Without loss of generality, if d+np, qag 5 n 5 pap, are the roots, and a'+ (pap and a'  (qap 1)pare not the roots, where pap and qap are both nonnegative integer, it is proved that
p
+
When d =
*
+ 1)p
p,the root chain consists of a', 0 and 6.
A root is called positive if its first nonvanishing component is positive, otherwise it is negative. Sometimes, Hj is called the generator corresponding to the zero root, which is l degeneracy. A positive root is called a simple root if it cannot be expressed as the nonnegative integral combination of other positive roots. Therefore, any positive root is equal to a nonnegative integral combination of the simple roots, and any negative root is equal to
Problems and Solutions an Group Theory
272
a nonpositive integral combination of the simple roots. The difference of two simple roots is not a root. From Eq. (7.8), the simple roots are linearly independent. Thus, the number of the simple roots in a trank simple Lie algebra is equal to t . The angle between two simple roots has to be equal to 5.n/6, 3n/4, 2n/3 or n/2, and correspondingly, the ratio of the length squares of the two simple roots is 3, 2, 1, or no restriction, respectively.
*
There are four series of the classical simple Lie algebras (At, Be, Ce and De) and five exceptional simple Lie algebras ( G 2 , Fq, Eg, E7 and Eg), where the Lie group of At is SU(t + l),the Lie group of Be is S 0 ( 2 t + 1), the Lie group of Ce is Sp(2t), and the Lie group of De is SO(2l). The Dynkin diagram for a simple Lie algebra is drawn by the following rule. There are one or two different lengths among the simple roots in any simple Lie algebra. Denote each longer simple root by a white circle, and denote each shorter simple root, if it exists, by a black circle. Two circles denoting two simple roots are connected by a single link, a double link, or a triple link depending upon their angle to be 2n/3, 3n/4 or 57r/6, respectively. Two circles are not connected if two simple roots are orthogonal and the ratio of their lengths is not restricted. The Dynkin diagrams for the simple Lie algebras are listed in Fig. 7.1. G2
Ap, t
1
Bp,
F4
2 1 
t2
3
Dp, .f 2 4
3
c2e1
e
3
e2e1
c
E6
OGO1
ce,e 2 2
2
2
L2.L o==e 1
2
1
2
2
3
e2e1
1
2
3
e3c2
o<'~
4
5
&
tt++~==~E7 1 1
3
e
e
&3
1
2
3
4
5
6
2
3
4
5
6
7
Fig. 7.1 The Dynkin diagrams of the simple Lie algebras.
*
The simple Lie algebra can also be described by the Cartan matrix. The Cartan matrix for a simple Lie algebra with t simple roots r, is an &dimensional matrix A:
273
Lie Groups and Lie Algebras
The diagonal element of A is always equal to 2, and its nondiagonal element can be equal to 0, 1, 2 or 3. In comparison with the Dynkin diagram, if two simple roots r, and r, are disconnected, A,, = A,, = 0, and if they are connected by a single, double or triple link, and the length of r, is not less than the length of r,, A,, = 1 and A,, = 1, 2 or 3, respectively. The Dynkin diagram or the Cartan matrix gives all the information of the simple Lie algebra, such as the angles and the lengths of the simple roots as well as all roots. The roots can be calculated recursively as follows. Any positive root can be expressed as a sum of the simple roots: e
C C,r,,
a'=
C, 2 0,
(7.10)
C, Cvis called the level of the root d. Any simple root is a root of level one. If all positive roots with the level less than n have been found, we want to judge whether a' r, is a root where 6 is a root of level (n  1). C alculat e e e q  p = r (a'/.,) = A&,, if a'= Cur,. (7.11)
+
u=l
u=l
Since q is known, one can calculate p from Eq. (7.11) to determine whether a' rP is a root or not.
+
+ +
1. Prove the number pap qap 1 of roots in a root chain a'+ nB, qap n 5 pap, in a simple Lie algebra is less than five.
5
Solution. Prove this problem by reduction to absurdity. If the length of the root chain is larger than four, one can redefine the root a' such that the following five vectors are all nonzero roots:
a'  2&
a'
6,
a',
a'+
/?,
a'+ 28.
However, the following four vectors are not roots: (a' f 2B) + a' = 2(a' f
B),
(a' f 28)  a' = *2B.
Thus,
o = r[(a'f2 f i ) / a ' ] =
di1(a'* 2B) a' = 2 f 2d;l
r
p . a' .
)
Adding two equations leads to contradiction. This completes the proof. Sin:e qap +pap 1 5 4, both qap and pap are less than four. Thus, I r ( w > IL 3.
+
274
Problems and Solutions in Group Theory
2. Prove that the difference of two simple roots is not a root, and there are
l simple roots in a simple Lie algebra of rank l . Solution. We first prove the difference of two simple roots rcl and r, not to be a root by reduction to absurdity. Suppose rp  r, = a'. If a' is a positive root, rp = r, + a' is the sum of two positive roos. If a' is a negative root, r, = r P + (a') is the sum of two positive roos. It contradicts with the definition of a simple root. Thus,
The angle of two simple roots is not an acute angle. It is obvious that the number of the simple roots is not less than l?, because any root can be expressed as a linear combination of the simple roots. If the simple roots are linearly independent to each other, the number of the simple roots must be l . Now, we prove it by reduction to absurdity. If there is a linear relation among simple roots, we separate the terms with the negative coefficients from those with the positive ones:
P
v
where cp and d, are all positive, and the values of p and u are different. Since a # 0,
0 < a2 = Cc,d,r,
'r,
5 0.
This completes the proof.
3. Prove that the angle between two simple roots has to be equal to 57r/6, 37r/4, 2 ~ / 3or 7r/2, and correspondingly, the ratio of the length squares of the two simple roots is 3, 2, 1, or no restriction, respectively. Solution. Calculate the square of cosine of the angle between two simple roots:
Since rP and r, is not collinear, 1 cos2 81 < 1, and the integer can only be equal to 3, 2, 1 or 0. Namely, the angle 9 can be 57r/6, 37r/4, 27r/3 or 7r/2. Without loss of generality, let the length of rp be not less than that of r,,
Lie Groups and Lie Algebras
275
Due to the restriction for their product, r'(r,/rp) has to be equal to 1 or 0, and I'(r,/r,) is equal to 3, 2, 1 or 0. The ratio of those two number is
depending on the angle of two simple roots to be 5 ~ / 6 , 3 n / 4 , 2 ~ /or3 ~ / 2 . 4. Calculate the Cartan matrix of the Lie algebra E6.
Solution. The Cartan matrix of E6 is a sixdimensional matrix with the diagonal elements to be 2. According to the enumeration for the simple roots in its Dynkin diagram, the rowcolumn numbers of the positions of the nondiagonal elements, which are 1, are 12, 23, 34, 36 and 45, or vice versa,
A = I j l 1 0l o 0 0 1 2 1 0 0  1 2 0 0 0  1 0 0  1 0
0 0 ) 0 1 1 0 2 0 0 2
'
5 . Draw the Dynkin diagram of a simple Lie algebra where its Cartan
matrix is as follows, and indicate the enumeration for the simple roots.
Solution. The Dynkin diagram consists of four circles. Since A32 = 2, the circle designated by 2 is white corresponding to a longer simple root, and the circle 3 is black corresponding to a shorter simple root. The circles 2 and 3 are connected by a double link. Since A12 = A21 = A34 = A43 = 1, the circles 1 and 2 are connected by a single link, so be the circles 3 and 4. Thus, the circle 1 is white and the circle 4 is black.
r\ 1
2
3
4
276
Problems and Solutions an Group Theory
6. Calculate all positive roots in the C2 Lie algebra.
Solution. There are two simple roots in
C2
where r2 is the longer simple root, and r1 is the shorter. Their angle is
3n/4. Usually, the length square of the longer root is taken to be 2. The Cartan matrix of
C2
is
2 2 A=(12). First, we calculate pl in the root chain rl have
+ plr2. From Eq. (7.11)) we
p1 = I'(rl/r2) = A21 = 1.
+
+
+
Thus, rl r2 = (el e2) is a root, but rl 2r2 is not a root. Then, we calculate p2 in the root chain r2 p2r1. Since
+
+
+
+
in addition to r1 r2, we know 211 r2 = fie1 is a root. 3rl 1'2 is not a root. rl + r2 is the only root of level two, and 211 + r2 is the only root of level three. Since (2rl 212) is twice of rl r2, it is obviously not a root. Therefore, the C2 Lie algebra contains four positive roots and four negative roots. The rank of C2 is 2, and its order is ten. This result can be generalized to the Ce Lie algebra. Ce contains 2C2 roots: f f i e j , & m ( e j ek) and m ( e j  ek). The rank of Ce is C and its order is C(2C+ 1).
+
+
+
7. Calculate all positive roots in the B3 Lie algebra. Solution. There are three simple roots in B3: rl = el  e2,
r2 = e2  e3,
r3 = e3,
where rl and r2 are the longer roots with the length square 2, and r3 is the shorter root. The angle between rl and r2 is 2 ~ / 3 The . angle between r2 and r3 is 3n/4. rl is orthogonal to r3. The Cartan matrix of B3 is
Lie Groups and Lie Algebras
277
We first calculate pl, p2 and p3 in the root chains
r2
+ p3r3. From Eq. (7.11)) we have
rl
+ plr2, r1 + p2r3,
Thus, there are two roots of level two: rl + r 2 and r 2 +r3. We also know that 2r3 is a root of level three, and rl 2 1 2 and r2 3313 ' are not the roots. Then, we calculate p4, p5, p6 and p7 in the root chains (rl r 2 ) p4r1, (ri r2) p5r3, (r2 13) p6rl and (12 r3) p7r2. Since
r2
+
+
+ +
+ +
+
+
+ +
+ + + + + + +
+
+
in addition to r 2 2r3, there is another root of level three: rl r 2 + 13. Besides, rl r2 2r3 is a root of level four. Adding r 2 or r3 to the root r2 21'3 is not a root. w e are going to calculate p8 in the root chain (rl r 2 13) ~ 8 1  2 . Since rl + r3 is not a root, we have
Thus, rl + r 2 +2r3 is the only root of level four. Adding r3 to it is not a root. We want to calculate pg and plo in the root chains (rl r 2 2r3) pgrl and (rl r 2 2r3) ~ 1 0 1  2 . Since
+ +
+ +
+
+ +
+
+ +
we obtain the only root of level five: rl 2r2 213. Besides, rl 312 213 is not a root. 2rl 2r2 2r3 is twice of a known root, so it is not a root. Now, we are going to calculate pll in the root chain (rl 2 1 2 213) +pllr3. Since
+
+
+ +
there is no root of level six. In summary, the B3 Lie algebra contains three roots of level one (three simple roots: rl = el  e 2 , r 2 = e2  e3 and r3 = e3), two roots of level two (rl + r 2 = el e3 and r 2 +r3 = e 2 ) , two roots of level three (rl + r 2 + r 3 = el and r 2 2r3 = e2 e3), one root of level four (rl r 2 213 = el e3), and one root of level five (rl 2r2 + 21'3 = el + e 2 ) . The rank of B3 is 3, and its order is 21. This result can be generalized to the Bc Lie algebra. Bc contains 212 roots: f e j , f ( e j + e k ) and (ej  e k ) . The rank of Bc is l , and its order is l ( 2 l + 1).
+
+
+
+ +
+
Problems and Solutions in Group Theory
278
8. Calculate all positive roots in the
D4
Lie algebra.
Solution. There are four simple roots in rl = el  e2,
r2
= e 2  e3,
D4:
r3 = e3  e 4 ,
r4
= e3
+
e4,
where the length squares of four simple roots are all 2. The angle between r 2 and any other simple root ra, a # 2, is 27r/3, and the other pair of simple roots is orthogonal to each other. The Cartan matrix of D4 is
Lie Groups and Lie Algebras
r3
279
+ 14) + p20r1 and ( r 2 + r3 + r 4 ) + p21r2. Since
+ + +
+
there is only one root of level four rl r 2 r3 r 4 = el e3. Adding rl, r3 or 14 to the root of level four is not a root. We calculate p 2 2 in the root chain (rl 12 r3 14) ~ 2 2 1  2 . Since
+ + + +
+
+ +
+
we obtain the only root of level five: rl 2r2 r3 r 4 = el e2. rl + 13 14 is not a root. It is easy to know those vectors by adding r l , r 4 to the root of level five is not a root, because
+ +
3r2 r3 or
Therefore, the D4 Lie algebra contains four roots of level one, three roots of level two, three roots of level three, one root of level four, and one root of level five. The rank of D4 is 4, and its order is 28. This result can be generalized to the Dc Lie algebra. Dc contains 2 l ( l  1) roots: f ( e j e k ) and (ej  e k ) . The rank of Dc is l , and its order is l ( 2 l  1).
+
7.2 Irreducible Representations and the Chevalley Bases
*
Discuss an irreducible representation of a simple Lie algebra where the representation matrices of Hj are diagonal:
where Im) is the basis state in the representation space, and the vector m in an ldimensional space is called the weight of Im). The &dimensional space is called the weight space. A weight m is said to be multiple if the number of the basis states Im) satisfying Eq. (7.12) in the representation space is larger than one. Otherwise, it is a single weight. From Eq. (7.6) we have
280
Problems and Solutions in Group Theory
Therefore, E, is called the raising operator and E, the lowering operator, if a' is a positive root.
*
In a given irreducible representation, one can construct a weight chain from a given weight m and a root d:
+ nd,  q 5 n 5 p , are weights, p 2 0, q 2 0, m + ( p + 1)G and m  ( q + 1)G are not the weights. m
(7.13)
It can be proved that
I? (m/Z) = q  p = integer.
(7.14)
The weight m' = m  GI' (m/d) is related to the weight m by a reflection with respect to the plane through the origin and orthogonal to the root G. m and m' have the same multiplicity and are called the equivalent weights. This reflection is called the Weyl reflection. All the mutually equivalent weights constitute the Weyl orbit, and their number is called the Weyl orbital size.
*
The Chevalley bases H p , E p and F p , 1 5 p 5 l , are another set of bases for the generators in a simple Lie algebra, which can be expressed as the linear combinations of the CartanWeyl bases Hj and E,:
They satisfy the following commutation relations: (7.16)
and the Serre relations:
(7.17)
The generator E, corresponding to a nonsimple root G can be calculated by the commutator (7.6). The main merit of the Chevalley bases is that three Chevalley bases H p , E, and Fp with the same subscript p span a subalgebra, denoted by d p ,which is isomorphic onto the Lie algebra A1
281
Lie Groups and Lie Algebras
(the SU(2) group). In fact, they satisfy the same commutation relations as those satisfied by the generators 22'3, T+ and 2' in A1:
*
In the weight space, define a set of new bases w,, called the fundamental dominant weight, satisfying
r (w,/r,)
(wp  r,) = (2w,  r,)
= d;'
/ (r,  r,)
= b,,.
(7.18)
In comparison with Eq. (7.9), we can relate the fundamental dominant weight wp with the simple root r, by the Cartan matrix L
e
p=l
u=l
In the bases w,, all components of the roots and the weights are integers:
(7.20)
The weight m' equivalent to m with respect to a simple root r, can be expressed as m' = m  r,r (m/rp) = m  mpr,.
(7.21)
*
Rewrite the weight chain in Eq. (7.13) with respect to a simple root rp where m + p r p is redefined as m: m, m  r p ,
. . . ,m  qr,,
q = r(m/rp) = m,.
(7.22)
The states corresponding to the weight chain constitute a (q+l)multiplet of the subalgebra d,. The basis states in the irreducible representation space constitute some multiplets of the different subalgebras d p . If all weights in the weight chain (7.22) are single, the representation matrix entry of Fp for the multiplet is
The representation matrix of E, is the transpose of that of F,. If the multiple weight exists, the state in the multiplet of A, is a suitable com
282
Problems and Solutions in Group Theory
bination of the basis states, calculated by the commutators (7.16) and the orthonor ma1 condition.
*
For an irreducible representation with a finite dimension, there is a state JM)with the highest weight M:
The highest weight M in an irreducible representation must be a single weight. Usually, an irreducible representation can be denoted by its highest weight M. A representation is called fundamental if its highest weight is the fundamental dominant weight. Any basis state Im) with the weight m in the irreducible representation space can be obtained from the highest weight state by the actions of the lowering operators. Thus,
Cm is called the height of the basis state Im). The height of the highest weight state is zero.
*
Two irreducible representations are conjugate with each other if the lowest weight of one representation multiplied by 1 is equal to the highest weight of another representation. The irreducible representations is selfconjugate if its lowest weight multiplied by 1 is equal to its highest weight.
*
A weight is called a dominant weight if its components are nonnegative integers. Due to Eq. (7.14), the highest weight must be a dominant weight, and any dominant weight can be the highest weight in one irreducible representation. However, for a given irreducible representation, there may exist a few dominant weights with multiplicities, among them only one dominant weight is the highest weight of this representation. The dimension of an irreducible representation is equal to the sum of the multiplicities of the dominant weights multiplied by their Weyl orbital sizes. Therefore, the dominant weights play a very important role in the calculation for the basis states and the representation matrices of the generators.
*
The method of the block weight diagram is effective for calculating the representation matrices of the generators in an irreducible representation. In a block weight diagram for an irreducible representation, each basis state corresponds to a block filled with its weight. For convenience, the negative component of the weight is denoted by a digit with a bar over it: Fi = m. The blocks corresponding to the multiple weight are enumerated. The block
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283
for the highest weight is located on the first row. The remaining blocks are arranged downwards as the height increases. The blocks for the weights with the same height are put in the same row. Beginning with the highest weight, from each positive component my > 0, we draw m, blocks downwards. They constitute a weight chain of length (m, 1) with respect to the simple root r,. The states corresponding to the weight chain constitute a multiplet of d,. Two neighbored blocks are related by a link denoting the application of the lowering operator F,. Usually, in the block weight diagram we connect two blocks by different links if their basis states are related by the different lowering operators. It is better to indicate the corresponding representation matrix entry of the lowering operator behind the link. Usually, the matrix entry is neglected when it is equal t o one. If all weights in a weight chain are single, the representation matrix entries of the lowering operator F, between two basis states corresponding to the neighbored weights can be calculated by Eq. (7.23). If a multiple weight appears in the weight chain, the state in the multiplet may be a combination of the basis states. The combination as well as the representation matrix entries of F ’ can be determined by Eq. (7.16) and the orthonormal condition. When a dominant weight appears, we determine its multiplicity which is equal to the number of paths along which the dominant weight is obtained from the highest weight. If the representation matrix entry in a path is calculated to be zero, the multiplicity decreases. All weights equivalent to the dominant weight has the same multiplicity as the dominant weight has. Continue this method until all components of the weight are not positive. In the completed block weight diagram, the number of blocks with the same height first increases and then decreases as the height increases, symmetric up and down like a spindle.
+
*
For a tworank Lie algebra the weights can be drawn in a planar diagram, called the planar weight diagram. The planar weight diagram of an irreducible representation is the inversion of that of its conjugate representation with respect to the origin. The planar weight diagram of a selfconjugate irreducible representations is symmetric with respect t o the inversion. Note that the simple roots as well as the fundamental dominant weights are not required to be along the coordinate axes.
*
The representation in whose space the basis is the generator of the Lie algebra is the adjoint representation. Therefore, the root coincides with
284
Problems and Solutions in Group Theory
the weight in the adjoint representation, and the root space coincides with the weight space. To draw the block weight diagram or the planar weight diagram for the adjoint representation is another method for calculating all roots in a simple Lie algebra. 9. Draw the block weight diagrams and the planar weight diagrams of the representations (1,0), (0, l ) ,(1, l ) ,(3,0), and (0,3) of the A2 Lie algebra (the SU(3) group).
Solution. The Cartan matrix A and its inverse A' of A2 are 2 1
A  1 = 1( 2 1 3 1 2
) *
The relations between the simple roots and the fundamental dominant weights are rl = 2wl  w2, w1
= (1/3) (2rl
r2 = w1+ 2w2,
+ ra) ,
w2
= (1/3) (rl
+h).
The Weyl orbits of the dominant weights ( l , O ) , ( O , l ) , ( l , l ) ,(3,O) and (0,3) respectively are: (1,o) rl, ( i , i )
rZ, ( o , i ) ,
(0,1) r2, ( 1 , i ) %
(i,o),
(3,O) rl, (3,3) % (0,3), (0,3) 3 ( 3 , s ) %
(5,O).
The block weight diagrams of the representations ( l , O ) , (0, I), (1, l),(3,0), and (0,3) are given in Fig. 7.2. In the diagrams, the single link means the action of Fl and the double link means that of F2. First, we discuss the fundamental representation (1,O). From the highest weight (1,O) we obtain a doublet of A1 with ( i , l ) . Then, from the weight (i,1) we obtain a doublet of .Ap with ( 0 , i ) . The weight (0,T) contains no positive component. It is the lowest weight in the representation (1,O). The dimension of the representation (1,O) is three, where three weights are equivalent to each other. The representation matrix entries of
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285
the lowering operators are
Second, we discuss the fundamental representation (0,l). From the highest weight ( 0 , l ) we obtain a doublet of d 2 with (1,i). Then, from the weight (1,i) we obtain a doublet of ,A1 with (T,O). The weight (T,O) contains no positive component. It is the lowest weight in the representation ( 0 , l ) . The dimension of the representation (0,l) is three, where three weights are equivalent to each other. The representation matrix entries of the lowering operators are
It is worthy to notice that the block weight diagram of ( 0 , l ) can be obtained from that of (1,O) by turning upside down and changing the signs of all weights. In fact, the lowest weight (0,i) in the representation ( l , O ) , times by 1, is the highest weight in the representation (0,l). Two representations are conjugate to each other.
0, i
Fig. 7.2 The block weight diagrams of some representations of A2
286
Problems and Solutions in Group Theory
Third, we discuss the adjoint representation (1,l) of A2. From the highest weight (1,l)we obtain a doublet of A1 with (i,2)and a doublet of A2 with ( 2 , i ) . Then, from the weight ( 2 , i ) we obtain a triplet of A1 with (0,O) and (2, l), and from the weight (i, 2) we obtain a triplet of A2 with (0,O) and (1,2). There are two paths to the dominant weight (0, 0), which may be a double weight. We define two orthonormal basis states (0,O)l and (0, O),, where (0,O)lbelongs to the triplet of A1 with (2, i)and (2,l), and (0, O), is a singlet of 41. The state (0,O) in the triplet of A2 with (i,2) and (1,2) has to be a combination of (0, O), and (0,0)2:
Applying El F2 = F2E1to I(1,1), (T,2)), we have
m.
Thus, a = Choosing the phase of 1(1,1),(0,0)2) such that b is a positive real number, we have b = = Let
Applying
E2F2 = F2E2
d
m
m.
+ H 2 to l(1, l), (0,0)1),we have
Choosing the phase of l(1, l), (1,g)) such that c is a positive real number, we have c = @, and then, d = @.From the weight (1,T)we obtain a doublet of A1 with (i,i),and from the weight (2,l) we obtain a doublet of A2 with (1,i). Since the weight (i,i)is equivalent to (1,I), it is a single weight and is the lowest weight in the representation (1,l). The representation (1,l) is selfconjugate. The dimension of the representation (1,l)is eight containing six equivalent single weights and one double weight (0,O). The representation matrix entries of the lowering operators are
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287
We see from the block weight diagram of the adjoint representation (1,l) that the A2 Lie algebra contains three positive roots:
2 w l  w2 = r l ,
 w1+2w2 = 12,
w1+
w2 = r1 +r2.
Fourth, we discuss the representation (3,O). From the highest weight
(3,O) we obtain a quartet of A1 with (1,l), ( i , 2 ) and (3,3), where the
a, a,
representation matrix entries of F1 are 2, and respectively. The weight ( 1 , l ) is a dominant weight, which is a single weight because there is only one path to ( 1 , l ) lowered from the highest weight. All weights equivalent to (3,O) or ( 1 , l ) are single, too. Then, from the weight (1,l) we obtain a doublet of A2 with (2, i). From the weight (T, 2) we obtain a triplet of d 2 with (0,O) and ( l , z ) , and from the weight ( 2 , i ) we obtain a triplet of d1 with (0,O) and (2,l). There are two paths to the dominant weight ( O , O ) , which may be a double weight. We define two orthonormal basis states (0, O), and (0,0)2, where (0,O)l belongs to the triplet of d1 with ( 2 , i ) and (2, l),and (0, O), is a singlet of d1. The state (0,O) belonging the triplet of A2 with ( i , 2 ) and (1,2) has to be a combination of (0, O), and (0,0)2:
F21(3,0),(T,2))=a1(3,0),(0,0)1) +b1(3,0),(0,0)2),
a2 + b 2 = 2 .
Applying E1F2 = F2E1to 1(3,0),(i, 2)), we have
~ i 1 ( 3 , 0 )(T,2)) , = d N 3 , o), ( 2 , i ) ) = F2Ei1(3,0), ( i , 2 ) ) = 2F21(3,0), (1,1)) = 21(3,0),(2,T))Thus, a = fi and b = 0. It means that the weight (0,O) is single. The basis state ((3,0),(0,0)2) does not exist, and ((3,0),(0, O),) can be denoted by i(3, o), (0,o)). From the weight (1,2) we obtain a doublet of A1 with (i,i). From the weight (3,3) we obtain a quartet of d2 with (2, 1), (7,T), and (0, Z), where the representation matrix entries of F2 are 2, and respectively. The weights ( 2 , l ) and (T, 1)are single because they are equivalent to ( 1 , l ) . Thus, the state ( 2 , l ) coincides with the state in the triplet of A1 with (2,T) and (0, 0), and the state (T, T) coincides with the state in the doublet
a, a,
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Problems and Solutions in Group Theory
of d1 with (1,2). The block weight diagram of the representation (3,O) is complete. The dimension of the representation (3,O) is ten, containing three single dominant weights (3,0), ( 1 , l ) and (0, 0), with the Weyl orbital sizes 3, 6 and 0, respectively. The representation matrix entries of the lowering operators are
The lowest weight (0,z) in the representation (3,0), multiplied by 1, is the highest weight of the representation (0,3). Therefore, the representation (0,3) is conjugate with the representation (3,0), and its block weight diagram is upside down of that of (3,0), where all weights change their signs.
t"'
(i,i)
Fig. 7.3 The planar weight diagrams of some representations of
A2.
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289
In the rectangular coordinate frame, the simple roots and the fundamental dominant weights of A2 are
Physically, people prefer to put el along the ordinate axis, and e2 along the abscissa axis for A2. The planar weight diagrams for the representations ( l , O ) , (0,1), (1,1),(3,0), and (0,3) are given in Fig. 7.3. Generally, the planar weight diagram for the representation ( 0 , M ) of A2 is a straight regular triangle where all weights are single, and that of ( M ,0) is a regular triangle upside down. The planar weight diagram of ( M , M ) of A2 is a regular hexagon, where the weights on the boundary are single, and the multiplicity increases as the position of the weight goes to the origin. 10. Draw the block weight diagrams and the planar weight diagrams of two fundamental representations and the adjoint representation (2,O) of the C2 Lie algebra.
Solution. The Cartan matrix A and its inverse A' in 2 2
A  l =  1( 2 2 2 1 2
)
C2
is

The relations between the simple roots and the fundamental dominant weights are rl = 2 w l 
w2,
r2
=  2 w l + 2w2,
The Weyl orbits of the dominant weights (1, 0)) (0, 1))and (2,O) respectively are: (1,o) rl,(i,1)
3(1,i) % (i,o),
(0, 1) r2,( 2 , i ) rl,(2,i) 3(o,i),
(2,O) rl,(2,2)
3( 2 3 I'l, (2,O).
Their Weyl orbital sizes are all 4. The block weight diagrams of the representations (1,0), (0, l),and (2,O) are given in Fig. 7.4. In the diagrams, the single link means the action of F1 and the double link means that of F2.
Problems and Solutions in Group Theory
290
Fig. 7.4
The block weight diagrams of some representations of C2.
First, we discuss the fundamental representation (1,O). From the highest weight (1,O) we obtain a doublet of d1 with (i,1). Then, from the weight (7,l) we obtain a doublet of A2 with (1,i). From the weight (1,i) we obtain a doublet of dl with ( i , O ) , which contains no positive component. The dimension of the representation ( 1 , O ) is four, where four weights are equivalent to each other. The representation matrix entries of the lowering operators are
Second, we discuss the fundamental representation (0,l). From the highest weight ( 0 , l ) we obtain a doublet of da with ( 2 , i ) . Then, from the weight (2,T) we obtain a triplet of d1 with (0,O) and (2,l). The representation matrix entries of F1 in the triplet are both The weight (0,O) is a dominant weight. There is only one path to (0,O) lowered from the highest weight, so (0,O) is single. From the weight (2,l) we obtain a doublet of A2 with ( O , i ) , which contains no positive component. The fivedimensional representation ( 0 , l ) contains two single dominant weights ( 0 , l ) and (0,O) with the Weyl orbital sizes four and one, respectively. The representation matrix entries of the lowering operators are
a.
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291
At last, we discuss the adjoint representation (2,O). From the highest weight (2,O) we obtain a triplet of A1 with (0,1> and (2,2), where the The dominant weight representation matrix entries of F1 are both ( 0 , l ) is single because there is only one path to (0,l) lowered from the highest weight. From the weight ( 0 , l ) we obtain a doublet of A2 with (2,i). From the weight (2,2) we obtain a triplet of A2 with (0,O) and (2,2), and from the weight (2, i) we obtain a triplet of A1 with (0,O) and (2,l). There are two paths to the dominant weight (O,O), which may be a double weight. We define two orthonormal basis states (0, O), and (0, O),, where (0,O)l belongs to the triplet of A1 with (2,i) and (2, l),and (0,0)2 is a singlet of 41. The state (0,O) in the triplet of A2 with (2,2) and (2,T) has to be a combination of (0,O)l and (0,0)2:
a.
Thus, a = 1. Choosing the phase of ((2,0),(0,0)2) such that b is a positive real number, we have b = \/2i" = 1. Applying E2F2 = F2E2 H2 to P,01, (O,O)l), we have
+
Choosing the phase of 1(2,0), (2,2)) such that c is a positive real number, we have c = 1, and then, d = 1. All weights in the representation (2,O) are single except for (0,O). From the weight (2,2) we obtain a triplet of d1 with (0,T) and (z,O), where the representation matrix entries of F1 are both From the weight (2,l) we obtain a doublet of A2 with (0,T). Since the weight (0,T) is single, it also has to belong to the triplet of 41 with (2,2) and (3,O). Thus, the block weight diagram of the representation (2,O) is complete. The representation (2,O) contains two single dominant weights (2,O) and (0,1), and one double dominant weight (0,O). The dimension of the representation (2,O) is 1 x 4 1 x 4 2 x 1 = 10. The representation matrix entries of the lowering operators are
a.
+
+
292
Problems and Solutions in Group Theory
We see from the block weight diagram of the adjoint representation (2,O) that the C2 Lie algebra contains four positive roots: 2 w l  w2 = r1, w2 = rl
+ 12,
 2 w l + 2w2 = 1'2, 2wl = 2rl
+r2.
In three representations (1,0), (0, 1), and (2,O) of C2, the lowest weights are equal to the highest weight times by 1, respectively. It means that three representations are all selfconjugate. In the rectangular coordinate frame, the simple roots (see Problem 6) and the fundamental dominant weights of C2 are
Thus, we obtain the planar weight diagrams for the representations ( l , O ) , ( 0 , l ) and (2,O) as follows.
11. Draw the block weight diagrams and the planar weight diagrams of three representations ( O , l ) , ( l , O ) , and (0,2) of the exceptional Lie algebra G2.
Lie Groups and Lie Algebras
293
Solution. The Cartan matrix A and its inverse Al in G2 are
i).
'),
A = ( 3 2
The relations between the simple roots and the fundamental dominant weights are rl = 2wl  3w2,
1'2
= w1+ 2w2,
The Weyl orbits of the dominant weights (0, l ) , (1,0), and (0,2) respectively are: ( o , i ) 3 ( i , i ) %( i , 2 ) r2,( i , 2 )
(1,o)
rl,( i , i ) 3 (o,i),
3( i , 3 ) % (2,s) 3(2,3) r2, (1,s)rl,(i,o),
(0,2) rz, ( 2 3 1p1, (2,4) r2,(2,;I) rl,(2,2) 3 (0)Z). Their Weyl orbital sizes are all 6. First, we discuss the fundamental representation ( 0 , l ) . From the highest weight ( 0 , l ) we obtain a doublet of d 2 with ( 1 , i ) . Then, from the weight (1,i ) we obtain a doublet of d1with ( i , 2 ) . From the weight (i, 2) we obtain a triplet of d 2 with (0,O) and (1)z).The representation matrix entries of F 2 in the triplet are both The weight (0,O) is a dominant weight. There is only one path to (0,O) lowered from the highest weight, so (0,O) is single. From the weight (1,Z) we obtain a doublet of A1 with (1,l). From the weight (1,l) we obtain a doublet of d 2 with (O,i), which contains no positive component. The dimension of the representation ( 0 , l ) is seven, where six weights are equivalent to each other and (0,O) is equivalent only to itself. The representation matrix entries of the lowering operators are
a.
F21(0,
(071)) = KO, 11,( l , i > > ,FlI(0, 11,( l , i ) >= KO, 11,
m),
F21(0,1>,(i, 2)) = JZI(0, I), ( O , O ) ) , FZl(0,1), (0,O)) = JZl(O,l>, (1,2)>, F11(0,1>,(1,W = KO, 11,( i , 1 ) > ,
F2I(O,
11, (i,1))= I(O,l),(O,i)).
Second, we discuss the fundamental representation (1,O). From the highest weight (1,O) we obtain a doublet of d1 with ( i , 3 ) . Then, from the weight ( i , 3 ) we obtain a quartet of d 2 with (0, l), ( 1 , i ) and (2,s). The representation matrix entries of F1 in the quartet are 2, and respectively. The weight ( 0 , l ) is a dominant weight. There is only one path to ( 0 , l ) lowered from the highest weight, so ( 0 , l ) is single. All weights
a, a
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Problems and Solutions in Group Theory
equivalent to (1,O) or (0,l) are single, too. From the weight (1,T) we obtain a doublet of A1 with (i, 2). From the weight (2,z) we obtain a triplet of A1 with (0,O) and (2,3), and from the weight (i, 2) we obtain a triplet of A2 with (0,O) and (1,2). There are two paths to the dominant weight (O,O), which may be a double weight. We define two orthonormal basis states (0,O)l and (0,0)2, where (0,O)lbelongs to the triplet of A1 with (2,s)and (2,3),and (0,0)2is a singlet of d1. The state (0,O) in the triplet of A2 with (i,2)and (1,g)has to be a combination of (0,O)land (0,0)2:
m.
Thus, a = Choosing the phase of I(l,O), (0,0)2)such that b is a positive real number, we have b = = Applying E 2 F 2 = F2E2 H 2 to 1(1,0),(0,0)1)we , have
+
d
m
m.
Choosing the phase of 1(1,0),(1,2))such that c is a positive real number, we have c = and then, d = The remaining part of the block weight diagram is easy to calculate. We only list the results in the block diagram and the representation matrix entries of the lowering operators in the following. The representation (1,O) contains two single dominant weights (1,O) and (0,l) with the Weyl orbital size 6 and one double dominant weight (0,O). The dimension of (1,O) is 1 x 6 + 1 x 6 + 2 x 1 = 14.
m,
m.
Lie Groups and Lie Algebras
295
The representation ( 1 , O ) is the adjoint representation of G2. F'rom the block weight diagram we obtain all positive roots of G2 as follows:
At last, we discuss the representation (0,2). From the highest weight (0,2) we obtain a triplet of d 2 with (1,O) and (2,2), where the representation matrix entries of F 2 are both The weight ( 1 , O ) is a dominant weight. There is only one path to ( 1 , O ) lowered from the highest weight, so ( 1 , O ) is single. From the weight ( 1 , O ) we obtain a doublet of d1 with (i,3). F'rom the weight (2,2) we obtain a triplet of A1 with ( 0 , l ) and (2,4), and from the weight (i,3) we obtain a quartet of d 2 with (0, l),(1,i),and (2,s).There are two paths to the dominant weight (0,1), which may be a double weight. We define two orthonormal basis states (0,1)1 and (0,1)2, where ( 0 , l ) l belongs to the triplet of d1 with (2,2) and (2,4), and (0,1)2 is a singlet of d1. The state ( 0 , l ) in the quartet of d2, composed of (T, 3), ( O , l ) , (l,i'),and (2,3), has to be a combination of (0,O)l and (0,0)2:
a.
Applying
E1F2
= F2E1 to I(0,2), (T,3)), we have
Thus, a1 = 1. Choosing the phase of [(O, 2)) (0,1)2) such that a2 is a positive real number, we have a2 = d m= Since the weight (1,i) is equivalent to the dominant weight (0,1), it is a double weight. Define two orthonormal basis states I(0,2), (1,i)l)and I(0,2), ( l , i ) 2 ) , where I(0,2), (1,i)l)is proportional to F21(0,2), (0,1)1) with a positive real coefficient bl:
a.
Applying
E2F2
= F2E2
+ H2 to [(O, 2), (0,1)1) and I(0,2), (0,1)2), we have
296
Problems and Solutions in Group Theory
F21(0,2), ( 1 ) i ) i ) = b41(0,2), (2,3)), F21(0,2), ( l , i ) 2 ) = b51(0,2), (2,Z)). Applying
E2F2
= F2E2
+ H2 to 1(0,2),( 1 , i j l ) , we have
~%F21(0,2),( 1 , i ) i ) = bz1(0,2), ( 1 , i ) i ) = (F2E2 + H2) I(0,2), ( 1 , i ) l )
hb51(0,2), (1,i)z)
+
= (2 + 1  l > l ( O , 2), (1,i)l) JZI(O,2), (hQ2).
Choosing the phase of I(0,2), (2,s)) such that b4 is a positive real number, we have b4 = fi and b5 = 1. F'rom the weight (2,4)we obtain a quintet of d 2 with ( i , 2 ) , (O,O), (l,T), and (2,4), where the representation matrix entries of F 2 are 2, &, fi,and 2, respectively. Since the weights (T, 2) and (1,T) are equivalent to the dominant weight (0, l ) , they are the double weights. We define the 2)2, (1,2)1, and (1,2)2, where (i,2)l orthonormal basis states (T, 2)1, (i, and (1,2)1 belong to the quintet of A2, and ( i , 2)2 and (1,2), belong to a triplet of A2 with (0,O). Let FlI(0, 21, ( 1 J ) l )= c11(0,2), ( i , 2)l) + c21(0,2), (i, 2)2),
c;
+ c;
= 1,
F11(o,2)7(17i)2)= C3I(o72),(i,2)i)+c41(0,2),(i,2)2),
Ci + C i
= 1
Applying
E2F1
= F1E2 to /(0,2),( 1 , i ) l ) and 1(0,2), (1,i)2), we have
E 2 J N m (1,i)l)= 2c11(0,2),(2,4))
Thus, c1 = 1 and c2 = c3 = 0. Choosing the phase of I(0,2), (i, 2)2) such that c4 is a positive real number, we have c4 = 1.
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Lie Groups and Lie Algebras
In addition to a quintet and a triplet of d 2 , which both contain the weight (O,O), there is another triplet of d1 containing (0,O). The triplet of A1 consists of (2,3),(O,O), (2,3). There are three paths to the dominant weight (0,O) lowered from the highest weight, so (0,O) may be a triple weight. We define three orthonormal basis states (0, O),, (0, O),, and (0, O),, where (0,O)l belongs to the quintet of A2 composed of (2,4), (T,2)1, (0,0)1, (1,2)1 and (2,3), (0,0)2 belongs to the triplet of A2 composed of ( i , 2 ) 2 , (0,0)2 and and ( O , O ) 3 is a singlet of A2. Thus, we have
(l,a),,
Let
Fl
Fl Fl Applying
E2F1
= FIE2 to 1(0,2),(2,3)),we have
m.
and d2 = Choosing the phase of I(0,2), (0,0)3) Thus, dl = such that d3 is a positive real number, we have d3 = Applying EIFI = FlEl HI to I(0,2), (0, 0)3), we have
+
m.
Choosing the phase of I(0,2), (2,3)) such that d6 is a positive real number, and then d4 = and d5 = It is easy to we have d6 = calculate the remaining part of the block weight diagram.
m,
m.
298
Problems and Solutions an Group Theory
q 0 ,i
Fig. 7.5.
Ji
Ji
lT,oI The block weight diagrams of some representations of G2
.
0
0
The results for the representation (0,2) as well as (1,O) and ( 0 , l ) are given in the block weight diagrams (Fig. 7.5). In the diagrams, the single link means the action of Fl and the double link means that of F2. The planar weight diagrams are also given, where the simple roots and the fundamental dominant weights of GZ in the rectangular coordinate frame
Lie Groups and Lie Algebras
299
are
7.3 Reduction of the Direct Product of Representations
*
The ClebschGordan (CG) series for the reduction of the direct product M1 x M2 of two irreducible representations of a simple Lie algebra can be calculated by the dominant weight diagrams. The main point is to list all dominant weights N contained in the product space, to count their multiplicities n(M1 x M2,N) in the space, to calculate their Weyl orbital sizes OS(N), and to calculate the multiplicities n(M,N) of those dominant weights N in the representations M contained in the CG series, respectively. First, we list OS(N) and n(M1 x M2,N) of all dominant weights N contained in the product space at the first column of the diagram. Let M1 M2  N = C j cjrj, Cjc j is called the height of the dominant weight N. Arrange the dominant weights N downwards as the height increases. The height of M1 +M2 is zero and is listed in the first row. The representation MI+ M2 is contained in the CG series with multiplicity one. Then, the second column of the diagram lists the dominant weights N in the representation M1 M2 and their multiplicities n(M1 M2, N) (as the superscript). The difference n(M1 x M2,M’)  n(M1 M2,M‘) for the dominant weight M’ with the next smallest height is equal to the multiplicity of the representation M‘ contained in the CG series. If it is not zero, the third column of the diagram lists the dominant weights N in the representation M‘ and their multiplicities n(M’,N) (as the superscript). Continue the calculation to determine the multiplicities of the representations N in the CG series by comparing n(M1 Ma, N) with the sum of the multiplicities of N in the representations, which we have known to be contained in the CG series, until all dominant weights N are calculated. At last, calculate the dimensions d(M‘) of the representations M’ contained in the CG series by
+
+ +
+
+
d(M’) =
OS(N)n(M’,N),
(7.26)
N
to see whether their sum is equal t o the dimension of the direct product
300
Problems and Solutions an Group Theory
space:
d(M1 x M2) =
C OS(N)n(M1 x M2,N) = C d(M‘). N
(7.27)
M‘
*
Denote by JM1,ml)JM2,ma) the basis state before reduction, and by llM,m) the basis state after reduction. Usually, the basis state IM1, ml)IM2,ma) is briefly written as Iml)lm2) for simplicity.
are the ClebschGordan (CG) coefficients. Applying the where CmM,lmM,2,Mm lowering operator Fp to the basis state, we have:
The main steps of the calculation are as follows. First, the highest weight state in the representation M1 + M2 is
In terms of Eq. (7.29) we are able to calculate the expansions of all basis states in the representation MI+ M2. Then, by the orthonormal condition or by Eq. (7.24) we calculate the highest weight state in the representation M’ contained in the CG series. At last, we calculate the expansions of all basis states in the representation M’ by Eq. (7.29). When M1 = M2, the expansion of each basis state can be symmetric or antisymmetric with respect to the permutation between two basis states Iml)lm2) and Im2)lml). For simplicity, we denote by (sym. terms) or (antisym. terms) the remaining terms obtained from the preceding terms by the permutation, multiplied by 1 or 1, respectively.
12. Calculate the ClebschGordan series and the ClebschGordan coefficients for the direct product representation ( 1 , O ) x ( 1 , O ) of the C2 Lie algebra. Solution. In Problem 10 we have calculated the block weight diagrams of the representations ( l , O ) , ( 0 , l ) and (2,O) of C2. The representation (1,O) contains only one single dominant weight (1,O). The representation ( 0 , l )
Lie Groups and Lie Algebras
30 1
contains two single dominant weights (0,l) and (0,O). The representation ( 2 , O ) contains two single dominant weights (2,O) and (0, 1),and one double dominant weight (0,O). The Weyl orbital sizes of the weights (2,0), (0, l), and (0,O) are 4,4, and 1, respectively. The linearly independent basis states for the dominant weights in the product space are Dominant Weight
Basis State
No. of States
Thus, we draw the diagram of the dominant weights as follows.
0s x n 4x2 1x4 16
=
1
0
+
5
+
1
Now, we calculate the expansions of the basis states. First, for the representation (2,0), from Eq. (7.30) we have
From the block weight diagram of (2,O) of C2 (see Fig. 7.4) and Eq. (7.29),
we have
According to the orthonormal condition, we obtain the highest weight state in the representation (0,l)
302
Problems and Solutions in Group Theory
Applying the lowering operators to the basis states continuously, we obtain
For the basis states in the representation ( 0 , l ) we have
The expansion of the basis state in the representation (0,O) can be calculated by the orthonormal condition. But now, we prefer to calculate it by the condition (7.24) for the highest weight state. Let
Lie Groups and Lie Algebras
303
Applying El and E2 to it, we have
Thus, a = b = c = d. After normalization we obtain
It can be checked that four basis states with the weight (0,O) are orthogonal to each other. In the permutation between lml)and Im2), the representation (2,O) is symmetric, but ( 0 , l ) and (0,O) are antisymmetric. 13. Calculate the ClebschGordan series and the ClebschGordan coefficients for the direct product representation ( 0 , l ) x ( 0 , l ) of the C2 Lie algebra. Solution. One may solve this problem following the way used in the preceding problem. But now, we prefer to calculate the block weight diagram and the CG coefficients with another method, where the CG coefficients for (0,l) x (0,l) and the block weight diagram for (1,l) are calculated at the same time. From the block weight diagram of ( 0 , l ) given in Fig 7.4, we know that the linearly independent basis states for the dominant weights in the product space are as follows, where we neglect the basis state which can be obtained by the permutation of two states Iml)and Im2).
We have calculated the block weight diagrams of the representations ( 0 , l ) and (2,O). However, we did not calculate the block weight diagram of the representation (0,2). The representation ( 0 , l ) contains two single dominant weights ( 0 , l ) and (0,O). The representation (2,O) contains two single dominant weights (2,O) and (0, l ) , and one double dominant weight (0,O). The Weyl orbital sizes of the weights (2,0), (0, l),and (0,O) are 4,
304
Problems and Solutions an Group Theory
4,and 1, respectively. The Weyl orbital of the weights (0,2) contains four weights: (0,2), (4,?),(3,2),and (O,?). F'rom Eq. (7.30) we have the expansion of the highest weight state in the representation (0,2)
From the highest weight (0,2) we have a triplet of d2:
The dominant weight (2,O) is single in the representation (0,2) because there is only one path to it lowered from the highest weight. However, the multiplicity of the weight (2,O) in the product space is 2. It means that one representation (2,O) is contained in the CG series for ( 0 , l ) x (0,l). From the weight (2,O) we have a triplet of A1,
The dominant weight ( 0 , l ) is single in the representation (0,2) because there is only one path to it lowered from the highest weight. The representation (2,O) contains a single dominant weight (0, l), and the product space contains two basis states with (0,l). It means that the representation ( 0 , l ) is not contained in the CG series for ( 0 , l ) x ( 0 , l ) . F'rom the weight (4,2) we have a quintet of d1 with ( 2 , i ) , (O,O), (2, l), and (4,2), and from the weight (?,2) we have a triplet of d 2 with (0,O) and (2,?). There are two paths to the dominant weight (0,O) lowered from the highest weight, so (0,O) may be a double dominant weight in the representation (0,2). We define two orthonormal basis states for the weight (0,O). "(0,2), (0,O)l) belongs to the quintet of d1,and 11(0,2),(0,0)2) has to belong to a singlet of St1 because the weight (2,T) is a single weight.
Lie Groups and Lie Algebras
305
We obtain the inner product between this basis state and Il(0,2), (0,O)l) to be Thus,
m.
m.
After normalization we have a = In fact, a can also be calculated by the property of a triplet: u2 + 1/3 = 2. Hence,
It is easy to calculate the remaining expansions of the basis states in the representation (0,2). We list the results as follows.
306
Problems and Solutions an Group Theory
In addition, we have
From the above calculation, we are able to draw the dominant weight diagram for ( 0 , l ) x ( 0 , l ) of Cz and the block weight diagram and the planar weight diagram of the representation (0,2) as follows.
0s x
n
4 x 1 4x2 4x2 1x5 25
=
1
4
+
1
0
+
1
Now, we calculate the expansions of the basis states in the representation (2,O). Since 11(2,0), (2,O)) is orthogonal to Il(0,2), (2,0)), we have
The remaining basis states are
Lie Groups and Lie Algebras
307
We may calculate the basis state Il(O,O), (0,O)) by the orthonormal condition to the preceding four basis states with the weight (0,O). Here we calculate it by Eq. (7.24). Let
308
Problems and Solutions an Group Theory
From Eq. (7.29) we have
Thus, a = b = c = d = e. After normalization, we have
It can be checked that five basis states with the weight (0,O) are orthogonal to each other. In the reduction of the direct product representation ( 0 , l ) x ( 0 , l ) of C2, the representations (0,2) and (0,O) are symmetric with respect to the permutation between Iml) and Imz), and the representation (2,O) is antisymmetric.
14. Calculate the ClebschGordan series for the direct product representation (1,O) x ( 0 , l ) of the Ca Lie algebra and the expansion for the highest weight state of each irreducible representation in the CG series. Solution. From the block weight diagrams for the representations ( 1 , O ) and ( 0 , l ) given in Fig 7.4, we know that the linearly independent basis states for the dominant weights in the product space are as follows.
There are a few methods to determine the dominant weights and their multiplicities contained in an irreducible representation. One method is to draw the upper part of the block weight diagram until all dominant weights in the representation appear. Another method is to look over the table book, for example, the table book [Bremner et al. (1985)] contains the information for the simple Lie algebras with rank less than 13. Anyway, we know that the representation (1,l) contains one single dominant weight (1,l) and one double dominant weight (1,O). The representation (1,O) contains only one single dominant weight (1,O). The Weyl orbital size of the weight (1,O) is 4. The Weyl orbital of the weights ( 1 , l ) contains eight weights: (1,l ) , (1,2)) ( 3 , i ) ) (3,2), (3,2), (3, l ) , (1,2) and (i,i).NOW,we
309
Lie Groups and Lie Algebms
are able to draw the diagram of the dominant weights for the direct product representation (1,O) x (0,I) of Ca.
0s x n 8 x 1 4x3 20
1 6 + 4
=
Fig. 7.6. The diagram of the dominant weights for (1,O) x ( 0 , l ) of C2.
The highest weight states of the representations (1,l) and (1,O) can be calculated by Eqs. (7.30) and (7.24). The results are
15. Calculate the CG series for the following direct product representations of the Ga Lie algebra and the expansion for the highest weight state of each irreducible representation in the CG series:
where the dimensions d(M) of some representations M, the Weyl orbital sizes OS(M) of M, and the multiplicities of the dominant weights in the representation M are listed in the following table.
6
1
1
6
2
1
6
3
2
12
4
6
5
6
5
4 4 3
1 1 2
1 2
3
2
3
2
1 1 1
1
1
1
Solution. In Problem 11 we have studied the block weight diagrams of the
310
Problems and Solutions an Group Theory
representations ( 0 , l ) and (1,O) of G2. The representation ( 0 , l ) contains seven single weights
The representation ( 1 , O ) contains 12 single weights and one double weight:
In the following, we first list the linearly independent basis states Iml)Im2)for each dominant weight in the product representation space, where we neglect the basis state which can be obtained by the permutation of two states Iml)and Im2). Then, comparing the number of the basis states of the dominant weight in the product representation space with the multiplicities of the dominant weight in the relevant representations, we are able to calculate the ClebschGordan series. The expansion for the highest weight state of each irreducible representation in the CG series can be calculated by Eq. (7.24). The results are listed in the following. a) The product representation ( 0 , l ) x ( 0 , l ) . Dominant Weight
Basis State
No. of States
The expansions for the highest weight states of the irreducible representations in the CG series are
In the permutation between m1) and m2), the representation (0,2)and (0,0) are symmetric, but (1,0) and (0,1 are antisymmetric.
Lie Groups and Lie Algebras
311
OSxn
49
= 2 7
(0,1) x (0, I)= (0,2)
+
14
+
7
+
1
+
(170)
+
(0,1)
+
(0,O)
Fig. 7.7. The dominant weight diagram for ( 0 , l ) x ( 0 , l ) of G2.
b) The product representation (0,l) x (1,O). Dominant Weight
Basis State
No. of States
The expansions for thje highest weight states of the irreducible representations in the CG series are
312
Problems and Solutions in Group Theory
Fig. 7.8. T h e dominant weight diagram for ( 0 , l ) x ( 1 , O ) of G2.
c) The product representation (1,O) x (1,O).
The expansions for the highest weight states of the irreducible representations in the CG series are
Lie Groups and Lie Algebras
313
In the permutation between Iml)and Im2), the representation (2,0),(0,2) and (0,O) are symmetric, but (0,3) and (1,O) are antisymmetric.
Fig. 7.9. The dominant weight diagram for ( 1 , O ) x ( 1 , O ) of G2.
16. Calculate the ClebschGordan series for the direct product representation ( O , O , 0 , l ) x (O,O, 0 , l ) of the F4 Lie algebra and the expansion for the highest weight state of each irreducible representation in the ClebschGordan series, where the dimensions d( M) of some representations M, the Weyl orbital sizes OS(M) of M, and the multiplicities of the dominant weights in the representation M are listed in the following table.
0s 1 24 24 96 24
Solution. F’rom the Cartan matrix of F4 we obtain the relations between the simple roots rj and the fundamental dominant weights wj
The representation (0, 0, 0 , l ) of F4 contains 24 equivalent single weights
Problems and Solutions in Group Theory
314
and one doublet weight (O,O, 0,O):
First, we list the linearly independent basis states Im1)lmz)for each dominant weight in the product representation space (0, 0, 0 , l ) x (0, 0, 0, l ) , where we neglect the basis state which can be obtained by the permutation of two states Iml)and Im2). Dominant Weight
Basis States
No. of States
Then, comparing the number of the basis states of the dominant weight in the product representation space with the multiplicities of the dominant weight in the relevant representations, we are able to calculate the ClebschGordan series, as given in the dominant weight diagram. The expansion for the highest weight state of each irreducible representation in the CG series can be calculated by Eq. (7.24). The results are listed in the following.
Lie Groups and Lie Algebras
315
In the permutation between m1) and m2), the representation (0,0,0,2), (0,0,0,1) and (0,0,0,0) are symmettric, but (0,0,1,0) and (1,0,0,0) are antisymmetric.
0s x n 24 x 1 96 x 2 24 x 6
Fig.7.10…The dominant weight diagram for (0,0,0,1)´(0,0,0,1) of F4.
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Chapter 8
UNITARY GROUPS
8.1
The S U ( N ) Group and Its Lie Algebra
*
The set of all Ndimensional unimodular unitary matrices, in the multiplication rule of matrices, constitutes the SU(N) group. It is a simplyconnected compact Lie group with order ( N 2  1) and rank ( N  1). The generators in its selfrepresentation are traceless hermitian matrices:
6cd
&d
0
(8.1)
when c < a,
[2a(a  1)]1’2
[(a 1 ) / ( 2 ~ ) ] ” ~ when c = a, when c > a,
2
L a L N,
where the subscripts a and b are the ordinal indices of the generator, while c and d are the row and column indices of the matrix. The matrix
Ti:)
is symmetric with respect to both ab and cd, but Tit) is antisymmetric. TA3)is diagonal. Usually, three types of the generators can be enumerated uniformly in the following order 2’1 = Ti:), T2 = Ti;), T3 = T i 3 ) T , 4 = Ti:), T5 = Ti:),Ts = T J i ) 27 , = T23 (2) , Tg = Ti3),and so on. The generators TA Satisfy the normalization condition: TAT') = 6,4B/2. Any element u in SU(N) can be expressed in the exponential matrix form: N21
N
317
318
*
Problems and Solutions an Group Theory
Let N
E
l + 1, and define ( 1 + 1)vectors V
a
in the !dimensional space:
v1=
vz = v 3
=
1
1
/=,
d2 l ( l + 1 ) ’ @(cq’
.a={
””
1
v N = {  d X
)
2 ( l + 1)’
0,
2a
0,
...,
‘
‘
O,
....*..........
0,
. .. . .. . .. . .. .. .
8
)
0
1
,
The CartanWeyl bases for the generators in the selfrepresentation of SU(N) can be expressed as
Hj = &T,(!\+,,
Esab= Tab (1) + iTit),
0341
when l!  j
1 [Hj,
Ezob when a
+ 2 = a,
< l  j + 2 < b,
Ea,,l =
(e  j
+ 2)(! j + 1) when l  j + 2 < a ,
where a
< b. Namely,
1j+2 when l  j
+ 2 = b,
or l  j + 2 > b ,
Unatary Groups
319
.
I
When a < b, Ezabcorresponds to the positive root Gab = &{V,  vb}= rP. Therefore, the Lie algebra of the S U ( l + 1) group is Ae. The rP = w1 we, which is the highest weight largest root in At is c3 = 1 of the adjoint representation of A t .
c/.L=l
+
1. Calculate the anticommutation relations for the generators TA in the
selfrepresentation of SU ( N ).
Solution. The anticommutator for two generators TAand TB is hermitian, but not traceless. Let
P=l
Taking the trace, we have cN = ~ T ~ ( T A T =B BAB. ) Thus, N21
'I
2. Calculate the structure constants and d A B D in Eq. (8.7) for the
su(3)
group.
Solution. Write the commutation relations for the generators TA in the selfrepresentation of SU(N):
where CAj$'are the structure constants. Multiplying Eqs. (8.7) and (8.8) by To and taking the trace, we have
Problems and Solutions in Group Theory
320
CA,” are totally antisymmetric with respect to three indices, but ~ A B D are totally symmetric. Substituting the generators TA into Eq. (8.9), we obtain the nonvanishing structure constants CA,” and the coefficients ~ A B D of SU(3) as listed in Table 8.1. Table 8.1 The nonvanishing structure constants CAf and the coefficients ~ A B Dof SU(3). ABD
ABD
123
147
118
I A B D I 355
156
146
366
246
157
377
257
228
448
345
247
558
367
256
668
458
338
778
678
344
888
I
3. Calculate the Killing form for SU(N) group. Solution. The generator Iidjin the adjoint representation is related to the structure constant CA,Dby (8.10) Thus, the Killing form g A B is
(8.11) Being an ( N 2 1)dimensional matrix, g is commutable with any generator Iidjin the adjoint representation of SU(N):
Unitary Groups
321
Due to the Schur theorem, g A B is a constant matrix, the constant c for the SU(3) group by taking A = 3:
= C ~ A B Calculate .
Note that T4 = Ti:) and TG= Ti:). When generalizing the result to the SU(N) group, the subscript 3 in Ti:) and Ti;) becomes a, 3 5 a 5 N , so that the number in the curve brackets becomes 1 2 ( N  2)/4. Therefore,
+
gAB
=N~AB,
for SU(N).
(8.12)
Irreducible Tensor Representations of SU(N )
8.2
*
The Chevalley bases for the generators in the selfrepresentation of SU(N) are
(EP)cd
=
(Erp)cd
1 5p
= 'W6(p+l)d,
(Fp)Cd
=
(E'p)cd
= d(p+I)cdpd,
5 e = N  1.
(8.13) Note that for a given p, three generators H,, E p and Fp constitute the subalgebra Ap,which is isomorphic onto the A 1 algebra.
*
as
A covariant tensor
T a l . , . a n of
rank n with respect to SU(N) transforms
Problems and Solutions an Group Theory
322
Since the matrix entries ?&idi in Eqs. (8.14) and (8.15) are commutable, the symmetry among the indices of a tensor keeps invariant in the SU(N) transformation. Therefore, the tensor space is reducible, and can be reduced into the direct sum of some irreducible tensor subspaces with the definite permutation symmetry among the indices.
*
In order to describe the symmetry of the indices of a tensor, we define the permutation operators R for the tensor. R is a linear operator by which a tensor T is transformed into a new tensor T'. The product of two permutation operators should satisfy the multiplication rule of elements in the permutation group. Let us begin with a simple example. If
R=
(ii;) (iii) =
=(123)=(12)(23),
then a tensor transforms as
On the other hand, a tensor basis transforms as
They transform in the different but compatible ways. In fact, expanding a tensor T in the tensor basis @&, we have
In general, if
we have
Unitary Groups
323
It is easy to show that the product order of an SU(N) transformation 0, and a permutation R among indices of a tensor are commutable
ROUT= 0,RT.
(8.17)
This property is called the Weyl reciprocity. It is the theoretical foundation for decomposing a tensor space by the Young operators.
*
In general, the tensor space ‘T is reducible. The minimal tensor subspace
ex]corresponding to an irreducible representation [A] of SU(N) can be
obtained by the projection of a Young operator, = y117.Two tensor subspaces denoted by different Young patterns are inequivalent. Two tensor subspaces denoted by different standard Young tableaux of the same Young pattern are equivalent, but are linearly independent.
*
can be An arbitrary tensor basis [email protected] in a tensor subspace denoted by a tensor Young tableau with the Young pattern [A], where the box filled with j in the standard Young tableau yF1is now filled with the subscript a j . For example, for the Young operator $”’ corresponding to the tensor Young tableau of the tensor the standard Young tableau
wl,
basis
$’[email protected]
is F
l
. For convenience we write (8.18)
Note that the same tensor Young tableaux in different tensor subspaces denote the different tensor bases. For example, if the Young operator $”I corresponds to the standard Young tableau E l , the tensor Young
l denotes the different tensors in two tensor subspaces: tableau F
Due to the symmetry and the Fock condition (6.5) satisfied by the Young operator, the tensor Young tableaux in a given tensor subspace satisfy some linear relations. For example, for the Young operator y?’l1we have
324
Problems and Solutions an Group Theory
E The relations similar to Eq. (8.19) hold for the general cases. Those relations obviously depend upon the Young pattern, but are independent of the Young operators with the same Young pattern as well as the tensor subspaces. The first equality says that the tensor Young tableau is antisymmetric with respect to any two digits filled in the same column. The tensor Young tableau is equal to zero if there are two equal digits filled in its same column. We may rearrange the digits in each column of the tensor Young tableau by the antisymmetric relation such that the digit increases downwards. The second equality is also called the Fock condition.
*
Any tensor in a tensor subspace $I’ is a combination of the tensor can be expressed as bases yF1Oa1...,,. yp I’[
Since yFIRFY are the standard bases in the minimal right ideal of the permutation group produced by the Young operator yI1,the following tensor bases YFIR,[email protected] , >
bl
5 b2 5    5 bn,
(8.20)
constitute a complete set of the tensor bases in the tensor subspace $’I, where R,, is the permutation transforming the standard Young tableau y;’] to the standard Young tableau We are going to discuss the characteristic of the basis (8.20). Since
ypl.
, , tensor the tensor Young tableau for the tensor basis Y ~ l R p , @ ~ l .in. . the is the same as that for the tensor basis y~’]@b,...,,, in the tensubspace $I’ sor subspace In fact, if R permutates ~j to j, R ~ ~ o ~ ~=. ~ . . ,b,...bp,. ,
d’].
yrl
Therefore, the box filled with j in the standard Young tableau is filled with T j in the standard Young tableau y;’], and is filled with bj both in the tensor Young tableau y ~ l R p , @ b l . . . and b , in the tensor Young tableau
yuI’[
@b1
*A
...b , .
tensor Young tableau is called standard if the digit in each column of the tableau increases downwards and in each row does not decrease right
325
Unitary Groups
wards. Since YLxl is a standard Young tableau and bl 5 b2 5 . . . 5 b,, the tensor Young tableau [email protected],,,..,n as well as the tensor Young tableau yP [XI [email protected], is standard. The standard tensor Young tableaux constitute a complete set of the tensor bases in the tensor subspace The standard tensor Young tableau is the eigenstate of HP where, due to Eq. (8.13), the eigenvalue is equal to the number of the digit p filled in the tableau, subtracting the number of the digit ( p 1). Therefore, the representation matrix of the Chevalley basis HP is diagonal in the tensor bases. The action of F, on the standard tensor Young tableau is equal to the sum of all possible tensor Young tableaux, each of which is obtained from the original one by replacing one filled digit p with the digit ( p 1). If without special indication, we use the notation for the inner product of tensors that the bases @al...a, are orthonormal. Two standard tensor Young tableaux with the different sets of the filled digits are orthogonal, but may not be normalized to the same number. Two standard tensor Young tableaux with the equal set of the filled digits but different filling order may not be orthogonal. To find a set of the orthonormal bases, one need to make suitable combination and normalization from the standard tensor Young tableaux.
+
+
*
Since RpvYLx1are the standard bases in the minimal left ideal of the permutation group corresponding to the irreducible representation [A], Eq. (8.21) also shows that the same tensor Young tableaux in all tensor subspace 7;Ix1with the same Young pattern [A] constitutes a complete set of the standard bases in the irreducible representation space of the permutation group, denoted by [A]. Therefore, in the decomposition of the tensor space
the tensor subspace corresponds to the irreducible representation [A] of SU(N), and for a given Young pattern [A], the tensor bases with the same span the irreducible tensor Young tableau in the d[xl(S,) subspaces I$' representation space of the permutation group S, denoted by [A].
*
The dimension d[xl(SU(N)) of the representation [A] of SU(N) can be calculated by the hook rule. In this rule the dimension is expressed by a quotient, where the numerator is the product of ( N m i j ) , and the
+
326
Problems and Solutions in Group Theory
denominator is the product of hij. mij = j  i is called the content of the box located in the j t h column of the ith row of the Young pattern [A]. hij is called the hook number of that box in the Young pattern, which is equal to the number of the boxes on its right in the ith row of the Young pattern, plus the number of the boxes below it in the j t h column, and plus one. (8.23)
where the symbol Yyl is a tableau obtained from the Young pattern [A] by filling ( N mij) into the box located in its ith row and j t h column, and the symbol YLX1is a tableau obtained from the Young pattern [A] by filling hij into that box. The symbol Y y l means the product of the filled digits in it, so does Yh [XI .
+
*
The onerow Young pattern [n]denotes the totally symmetric tensor representation. The onecolumn Young pattern [In]denotes the totally antisymmetric tensor representation. The totally antisymmetric tensor space of rank N , denoted by [IN],contains only one standard tensor Young tableau, corresponds to the tensor basis (8.24)
The tensor E is an invariant tensor in the SU(N) transformation, so [l”] denotes the identical representation of SU(N ) . The tensor subspace described by a Young pattern with more than N rows is a null space. The reduction of the direct product of two irreducible representations described by two Young patterns [A] and [p] can be calculated by the LittlewoodRichardson rule. However, the Young patterns with more than N rows contained in the reduction should be removed. The representation denoted by the Young pattern [A] with N rows is equivalent to the representation denoted by the Young pattern [A’] obtained from [A] by removing its first column.
*A
contravariant tensor Tal...anof rank n with respect to SU(N) transforms as
=
C
d l ...d,
Tdl...dn ( u  ~. .). ( u~  ’~) ~~, ~,~,
u E SU(N).
(8.25)
Unitary Groups
327
(8.26)
d l ...d , d l ...d ,
The contravariant tensor space T can also be decomposed by the projection of the Young operator, T = @$T = @$"*. The tensor subspace corresponds to the representation [A]* of SU(N), which is the conjugate one of the representation [A]. The basis in the tensor subis denoted by the standard tensor Young tableau with star, for space example $ " I '
$"*
(8.27)
*A
mixed tensor T$;;t; of rank (n,m) with respect to SU(N) transforms
as dl
dn
( o ~ T )= ~ ~ ~ * . ~u a l~c l ~' ' * U a n d n T c l : : : c n ( U  l ) d l b l Cl...Cn
' ' '
(u')dnbn
'
d l ...d n
(8.28) Obviously, the mixed tensor (D): = 6; is an invariant tensor of rank (1,l) in the SU(N) transformation. A mixed tensor is said to be made a contraction (or the trace) of its one pair of indices, one covariant and one contravariant, if to take the pair of indices to be equal and then to sum over the indices. The trace tensor of a mixed tensor of rank (n,m)is a tensor of rank ( n  1,m  1). A mixed tensor can be decomposed into the sum of a series of traceless tensors with different ranks. For example,
} ;[ c l ~ d c: +dt
Ti]7
d
+D
1
(8.29)
T:]
7
d
where the tensor in the curve brackets is a traceless mixed tensor of rank (1,l), and the tensor in the square brackets is a scalar [the tensor of rank (0, O)]. Therefore, a mixed tensor space can be decomposed into the direct
328
Problems and Solutions in Group Theory
sum of a series of traceless tensor subspaces with different ranks, each of which is invariant in the SU(N) transformation. The traceless tensor space can be decomposed into the direct sum of some minimal tensor subspace by the projection of the Young operator YF1 on the covariant indices and the projection of the Young operator yF1 on the contravariant indices, where the Young patterns [A] and [TI contain n and m boxes, respectively. The minimal tensor subspace corresponds to an irreducible representation of SU(N), denoted by [A]\[T]*. It can be shown that the tensor subspace corresponding to [A]\[T]* is null space if the sum of the numbers of rows of [A] and [TI is larger than N , and the representation [A]\[T]* of S U ( N ) is equivalent to the representation [A']\[T']* if [TI contains t rows, [ T ' ] is obtained from [TI by removing its first column, and [A'] is obtained from [A] by attaching a new column with ( N  t ) boxes from its left. Thus, the representation [TI* of SU(N) is equivalent to the representation [A] if by rotating the Young pattern upside down and then by attaching it to the Young pattern [A] from below, it just forms a rectangle with N row.
[TI
4. Calculate the dimensions of the irreducible representations denoted by
the following Young patterns for the SU(3) group and for the SU(6) group, respectively:
Solution.
34
67
1 345
1 678
329
Unitary Groups
5 . Calculate the ClebschGordan series for the following direct product representations, and compare their dimensions by Eq. (8.23) for the
SU(3) group and for the SU(6) group, respectively:
o o l l l $ o o l l $ o0 0 1 1
Solution. a): O O 8 1 1 1 z 0
01
0
1
001 $01 , 1
+ 27 + 10 + 8, SU(6) : 70 x 56 = 3920 = 1050 + 1134 + 840 + 896.
SU(3) : 8 x 10 = 80 = 35
00011$0001$000 b): 0 0 0 ~ 1 1 1 ~ 0 0 0 1 1 1 $ 1 11 1 1 1’
+ 27 + 10, SU(6) : 56 x 56 = 3136 = 462 + 1050 + 1134 + 490.
S U ( 3 ) : 10 x 10 = 100 = 28 + 35
4:
00011 0001 000 000 ooolll$ooo $000 $000, l l ~ @ o o o  o o o 1 11 111 N
+
+
SU(3) : 10 x 10 = 100 = 64 27+ 8 1, SU(6) : 56 x 490 = 27440 = 9240 + 11340 5880
d):
[email protected]
00
2
N
o o o o l l $ oooo o o l l
002
r)
L
+
+ 980.
00001 @0012
00001 00001 00001 $00 $001 $002 2 1 12
Problems and Solutions in Group Theory
330
00001 O0 @ I 2
0000
ool
0000 0000 @[email protected] 2 1 2
2
0000 00 @11
’
2
SU(3) : 27 x 8 = 216 = 6 4 + 35 + 35 + 2 x 27+ 10 O + 1 0 + 8 + 0 + 0, SU(6) : 1134 x 70 = 79380 = 9240 + 11550 + 5880 2 x 11340 6000 5670 4704 5880 4536 3240.
+
+
+
+
+ +
+
+
7;(2’11= $’ll‘T of SU(3), and prove that the standard tensor Young tableaux constitute the complete bases in this subspace.
6 . Write all tensor Young tableaux in the tensor subspace
Solution. The standard Young tableau
$’11
is F
l
. Thus,
If there is a pair of the filled digits to be equal in the tensor Young tableau, from Eq. (8.19) we have
where the left tensor Young tableau in each equality is standard. If all three filled digits in the tensor Young tableau are different, from Eq. (8.19) we
Unitary Groups
331
have
where the left tensor Young tableaux in the first two equalities are standard. The dimension of the subspace 7;[2’11 is eight. There are eight standard tensor Young tableaux. The first six standard tensor Young tableaux are normalized to 6 , but the last two to 4. In addition, the last two standard tensor Young tableaux are not orthogonal to each other.
7.Try to express each nonzero tensor Young tableau for the irreducible representation [3,1] of SU(3) as the linear combination of the standard tensor Young tableaux.
Solution. For the SU(3) group, the digits filled in the tensor Young tableaux are 1, 2 and 3. The equal digits cannot be filled in the same column. There are three types of nonzero tensor Young tableaux. A tensor Young tableau in the first type contains three equal digits. From Eq. (8.19) we have
When a < b the left tensor Young tableaux is standard, while when a > b the right one is standard. There are 3 x 2 = 6 standard tensor Young tableaux in the first type. A tensor Young tableau in the second type contains two pairs of equal digits. From Eq. (8.19) we have
When a < b, the first tensor Young tableau is standard, while when a > b the fourth one is standard. There are 3 standard tensor Young tableaux in the second type. A tensor Young tableau in the third type contains one pair of equal digits. There are six standard tensor Young tableaux in the third
Problems and Solutions an Group Theory
332
type. The other tensor Young tableaux can be expanded in the standard tensor Young tableaux by Eq. (8.19),
U
The dimension of the irreducible representation [3,1] of SU(3) is 15. There are 15 standard tensor Young tableaux. 8. Write the explicit expansion of each standard tensor Young tableau in the tensor subspace JfP1]7, where 5' is the tensor space of rank four for the standard Young tableau of the Young operator

Solution. First, we write the explicit formula for the application of the
Unitary Groups
333
Similar to Problem 7, there are three types of standard tensor Young tableaux. In the first type there are six standard tensor Young tableaux, each of which contains three equal digits.
PI
P I = [3 11
= Y2 ’ Oaaba = [email protected]  2 0 b a a a  2 0 a b a a  2 0 a a a b .
When a < b the left tensor Young tableau is standard, and when a > b the right one is standard. The module square of the standard tensor Young tableaux in the first type is 48. In the second type there are three standard tensor Young tableaux, each of which contains two pairs of equal digits:
The module square of the standard tensor Young tableau in the second type is 24. In the third type there are six standard tensor Young tableaux, each of which contains one pair of equal digits:
334
Problems and Solutions in Group Theory
where a and b are respectively taken 2 and 3. The module square of the first form, which includes two standard tensor Young tableaux, is 24, and that of the remaining standard tensor Young tableaux is 18. Two standard tensor Young tableaux with the equal set of the filled digits are not orthogonal.
9. Transform the following traceless mixed tensor representations of the SU(6) group into the covariant tensor representations, respectively, and calculate their dimensions:
Solution. (1) [3,2,I]* N [i3]\[2,I]* N [23,1]\[1]* 21 [33,2,1], 6 5 4 3
678 56
d [ 6 ,5,4,3](su(6) =
6 5 4 3
7 6 5 4 9 7 5 3
8 9 10 11 78 9 67 5 8 6 4 2
7531 531 31 1
= 147840.
78 67 56 4
Unitary Groups
d[7,6,4,3,1](SU(6))
=
6 5 4 3 2
7 6 5 4
335
8 9 10 11 12 7 8 9 10 67 5
11
= 612500.
976421 6431 421 1
10. Prove the identity:
are the generators in the selfrepresentation of the SU(N) where (TA),, group.
Solution. Define a mixed tensor T of rank (2,2) with respect to SU(N)
It is easy to show that T is invariant in the SU(N) transformation,
Therefore, T can be expressed as the product of the invariant mixed tensors d: of SU(N),
336
Problems and Solutions in Group Theory
Since ’T is a traceless matrix, 0 =  N p . AS
C , TE,d = ( N p + q) 6:,
we have q =
when A # N 2  1 when A = N 2  1, we have N1 2N

Tgg = p + q =  ( N
 1)p.
The solution is p =  ( 2 N )  ’ and q = 1 / 2 . This completes the proof.
8.3
Orthonormal Bases for Irreducible Representations
*
Recall that the standard tensor Young tableau is the eigenstate of the Chevalley basis H p where the eigenvalue is equal to the number of the digit p filled in the tableau, subtracting the number of the digit ( p + 1 ) . The action of the lowering operator Fp on the standard tensor Young tableau is equal to the sum of all possible tensor Young tableaux, each of which is obtained from the original one by replacing one filled digit p with the digit ( p + 1 ) . Two standard tensor Young tableaux with different sets of the filled digits are orthogonal, but may not be normalized to the same number. Two standard tensor Young tableaux with the equal set of the filled digits but different filling order may not be orthogonal. The merit of the method of Young operators is that the complete tensor bases in an irreducible representation space of SU(N) have been explicitly known. The shortcoming is that the tensor bases are not orthonormal, and the obtained representation is not unitary. On the other hand, the method of the block weight diagram gives the representaion matrices of the Chevalley bases of generators in the unitary irreducible representation, but the orthonormal bases are given only symbolically. To combine two methods will offset one’s shortcoming by the other’s strong point.
*
In the standard tensor Young tableau corresponding to the dominant weight, the number of each digit p filled in the tableau must not be less than the number of the digit (p+ 1)filled there. It is the criterion for determining which standard tensor Young tableau corresponds to the dominant weight in the representation. The states for the multiple weight correspond to the
Unitary Groups
337
standard tensor Young tableaux with the equal set of the filled digits but in different filling order. In the standard tensor Young tableau corresponding to the highest weight, each box located in the j t h row of the tableau is filled with the digit j , because each raising operator E, eliminates it. Therefore, the component MP of the highest weight M in the representation [A] is (8.30)
M , = A,  &+I.
*
Gelfand presented the symbols for the orthonormal basis states of an irreducible representation [A] of SU(N), usually called the Gelfand bases, and calculated the analytic formulas for the representation matrix entries of the generators (the Chevalley bases) of SU(N) in the Gelfand bases. For a multiple weight, he chose the orthonormal basis states first according to the multiplets of the subalgebra d1,then, according to the multiplets of the subalgebra A2 if there still exist the multiple states, and so on. Gelfand described the basis state with the weight m in the representation [A] of SU(N) by N ( N 1)/2 parameters &b, 1 5 a 5 b 5 N , arranged as a regular triangle upside down:
+
w1N
lwab)
... ... w(N1)N ... ... ...
W2N W1(Nl)
=
W ( N  1) ( N  1)
w22
w12 (J11
where W
~ N =
A,,
WNN
= AN = 0, and
(8.31) i
The representation entries of the generators are
338
Problems and Solutions an Group Theory
A,,
(Wab)
=
{

P1
(&(@I)
t=l
.(
 Wvp
t
+ v  1)
n
(Wp(p+l)
 WV,
p
p=l
,
d#v,d=l
n
P+l
(Wdp
 Wv,  d
+ v)
(Wd,
 Wv,  d
+ v  1)
+ ).
r2
} I 2
(8.32)
Namely, the eigenvalue of H, is equal to twice the sum of the parameters in the pth row from the bottom, minus the sum of the parameters in the ( p 1)th row and the ( p + l ) t h row from the bottom. The application of E, (F,) to a Gelfand basis leads to a linear combination of the possible basis states, each of which is obtained by increasing (decreasing) one parameter wap in the pth row from the bottom under the condition (8.31). The parameters of the highest weight state in the Gelfand bases satisfy
11. Expand the Gelfand bases in the irreducible representation [3,0]of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.
Solution. The tensor subspace described by the representation [A] = [3,0] consists of the completely symmetric tensors of rank three. There is no multiple weight in the representation. Its highest weight is M = (3,0), which corresponds to the standard tensor Young tableau and the Gelfand basis are
There are three types of the standard tensor Young tableaux which are normalized to 6,12 and 36,respectively:
where a, b and c are all different. In the application of the lowering operator F, , all possible terms, which may be nonstandard tensor Young tableaux,
Unitary Groups
need to be considered. For example
Now, the orthogonal bases can be calculated as follows.
2 02 0 0 ) )
2 0 0 0 ) )
339
Problems and Solutions in Group Theory
340
The expansions of the orthonormal basis states with respect to the standard tensor Young tableaux give the wave functions of the hadrons. Substituting the expansions into the planar weight diagram of (3,O)given in Fig. 7.3, we obtained the wave functions for the baryon decuplet. In the planar weight diagram, the ordinate axis and the abscissa axis are the eigenvalues of T3 and 2'8, respectively (see the end of Problem 14). 12. Expand the Gelfand bases in the irreducible representation [3,3] of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.
Solution. The representation [A] = [3,3] of SU(3) is the conjugate representation of [3,0], [3,3] N [3,0]*. This problem can be solved in the same way as that used in Problem 11. However, in the present problem we prefer to calculate the standard tensor Young tableaux of the contravariant tensor instead of those of the covariant tensor. Both results for two kinds of the standard tensor Young tableaux are given here. Note that the action of the generator on the basis state of the conjugate representation is different from that on the basis state of the original representation. Denote by D ( I A ) and D'(IA) the representation matrices of the generator I A in the representations D ( R ) and those in its conjugate one D ( R ) * ,respectively:
D(R) = 1  i
C D(IA)wA,
+ix A
D(R)* = 1
D(IA)*wA
A
=1
ix
D'(IA)wA.
A
Thus, D ' ( I A )=  D ( I A ) * . For the Chevalley bases we have
Namely, the standard tensor Young tableau with star is the eigenstate of the Chevalley basis H p where the eigenvalue is equal to the number of the digit ( p 1) filled in the tableau, subtracting the number of the digit p. The action of the lowering operator Fp on the standard tensor Young tableau with star is equal to the sum multiplied by 1, of all possible tensor Young tableaux with star, each of which is obtained from the original one by replacing one filled digit ( p 1) with digit p. The order of the filled
+
+
Unitary Groups
341
digits in the standard tensor Young tableau changed to 3, 2, 1 instead of 1, 2) 3. The highest weight of the representation [3,3] is M = (0,3), which corresponds to the standard tensor Young tableau and the Gelfand basis are
=wl= l3 :
=m]*
1(0,3),(0,3))
1 1 1 2 2 2
3
3 O).
The normalization factor for the standard tensor Young tableau with star is similar to that for the standard tensor Young tableau. Now, the orthogonal bases can be calculated as follows.
342
Problems and Solutions in Group Theory
13. Expand the Gelfand bases in the irreducible representation [2,1] of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.
Solution. The representation [A] = [2,1] is the adjoint representation of SU(3). For the tensor subspace 7;12’11 = yi2’117, the standard Young operator is $’11 = E + (1 2)  (1 3)  (2 1 3), corresponding to the Young tableau
Fl
. There are six single weight and one double weight (0,O)
in the representation [2,1]. The typical standard tensor Young tableaux and their normalization factors are
Due to the Fock condition (6.5), we have
In particle physics, a hadron is composed of three quarks (baryon) or a pair of one quark and one antiquark (meson). The expansions of the orthonormal basis states in [2,1] with respect to the standard tensor Young tableaux give the flavor wave functions for the baryon octet or for the meson octet. For the meson octet, the standard tensor Young tableau should be
Unitary Groups
343
transformed into the traceless mixed tensor by the totally antisymmetric tensor &bc:
(8.35)
The highest weight in [2,1] is M = (1,l), which corresponds to the standard tensor Young tableau and the Gelfand basis are
The remaining orthonormal bases can be calculated from the block weight diagram given in Fig. 7.2:
U
U
I
I
I
344
Problems and Solutions in Group Theory
+ F V
P I =I
dm *
2
2 0l o 0 ) ,
where we use the formula calculated from Eq. (8.32):
F2I22:l0)=&I 3
2 l l1 l o )
+Ll 1
2 , l 1O O ) .
Each basis state in the representation [2,1] is normalized to 6 . Especially,
Unitary Groups
345
the basis state
14. Express each Gelfand basis in the irreducible representation [4,0] of
the SU(3) group by the standard tensor Young tableau and calculate the nonvanishing matrix entries for the lowering operators Fp. Draw the block weight diagram and the planar weight diagram for the representation [4, O] of SU(3).
Solution. The tensor subspace described by the representation [A] = [4,0] consists of the completely symmetric tensors of rank four. There are four single dominant weights in the representation:
The highest weight of the representation [4,0] is M = (4,0), which corresponds to the standard tensor Young tableau and the Gelfand basis:
=
)(4,0),(4,0))
M
I
=
l4
4
0
").
There are four types of the standard tensor Young tableaux which normalize to 48, 96, 144, and 576, respectively. The normalization factors for them are &,2, and I, respectively.
m,
Obcac
346
Problems and Solutions in Group Theory
where a , b, and c all are different. Now, we calculate the remaining basis states. From the highest weight [A] = [4,0] we have a quintet of d1:
Unitary Groups
=
dZ[11212131=
347
4 3 : 0 0 ) , 1/2)F2
{ dZ pppp1)
=
2ml=/4 ,
=
m l = 1 o 4 .
O)
0
O)
It is easy to check that
The block weight diagram and the planar weight diagram are now drawn as follows. From Eqs. (8.3) and (8.6), the simple roots and the fundamental dominant weights of the SU(3) group in the rectangular coordinate frame
348
Problems and Solutions in Group Theory
are
From Eq. (8.4) we know that the first and the second components of the weight m in the rectangular coordinate frame are related to the eigenvalues of T8 = Ti3) and T3 = Ti3) in SU(3), respectively. Namely, the ordinate axis and the abscissa axis are the eigenvalues of T3 and T8, respectively.
15. Express each Gelfand basis in the irreducible representation [3,1] of the SU(3) group by the standard tensor Young tableau and calculate the nonvanishing matrix entries for the lowering operators F’. Draw the block weight diagram and the planar weight diagram for the representation [3,1] of SU(3).
Solution. There are two single dominant weights ( 2 , l ) and (0,2), and one double dominant weight (1,O) in the representation [A] = [3,1] of SU(3):
Unitary Groups
349
, The highest weight of the representation [3,1]is M = (2,1), which corresponds to the standard tensor Young tableau and the Gelfand basis:
P
the standard Young tableau is In the tensor subspace 7;[3”1= 3/1[3’117, I , and the action of the Young operator on the tensor basis
Note that
where a may be filled in the second column or the third column. There are four types of standard tensor Young tableaux which normalize to 18, 24, 24, and 48, respectively. The normalization factors for them are and 1, respectively.
m,a,a,
J“(J1/12 ( 3 G a a a d  O d a a a  O a d a a  e a a d a ) } 
350
Problems and Solutions in Group Theory
From the highest weight state ( 2 , l ) we have a triplet of A1 with (0,2) and (2,3), and a doublet of A2 with ( 3 , i ) .

21 1 ° ) , I
From the weight (3,i) we have a quartet of d1 with ( l , O ) , (i, l), and (3,2), where the dominant weight ( 1 , O ) and its equivalent weight (i,1) are the double weights. We define one basis state for ( 1 , O ) or for ( i , l )belongs to the quartet of d1, and the other belongs to the doublet.
Unitary Groups
351
From the weight (0,2) we have a triplet of A2 with (1,O) and (2,2), where the basis state with the weight (1,O) is a combination of 1(1,0)1) and 1(1,0)2). Letting
we have
m,
m.Applying
Thus, a1 = and a2 = [email protected] = H2 to the basis state ((2, 1), (l,O)l), we have
E2F2
= F2E2
+
Problems and Solutions in Group Theory
352
f i o m the weight (2,2) we have a triplet of A1 with ( 0 , i ) and (2,0), where the weight (0, i)is a double weight. We define one state basis I(0, i),) belongs to the triplet of A1, and the other 1 ( O 7 i ) 2 ) belongs to the singlet.
From the weight (2,3) we have a quartet of A2 with (T, 1))(0,T) and
(l,s),where the states with the weights (i,1) and ( 0 , i ) are the combinations of the. basis states. Letting
we have
Unitary Groups
353
m, = m,ci = m,and
Thus, bl = E2F2 = F2E2
b2
c3
=
m. Applying
+ H2 to the basis state l ( 2 , 1 ) , ( i , l ) 2 ) , we have
Choosing the phase of the basis state )(2,1),(0, i ) 2 ) such that c4 is real positive, we obtain c4 = fi,and then c2 = 0. Applying E2F2 = F2E2+ H2 to the basis state )(2,1),( O , i ) 2 ) , we have
P,11,( O , i ) 2 ) = dld2 1(2,1), ( 0 , i ) l ) + di; l(2,1), ( O , i ) 2 ) = (F2E2 + H2) ((2,1), ( O , i ) 2 ) = Jz [email protected], 11,( 0 , i ) l ) + (3  1) 1(2,1), (0,1)2).
E2F2
(1,s))
Choosing the phase of the basis state l(2, l), such that d2 is real and then dl = 1. From the above calculation, positive, we obtain d2 = we have
a,
l(27 I>,( O A 2 ) = r n ( F 2 1 ( 2 , 1 > , ( i , 1 ) 2 )  r n I ( 2 , 1 > ,( O J ) l ) }
{ m p !
= m F 2
 J . j ( 2 P l = + p =1 1 = 3 1 33
;
P I } 0) ,
l(2, 1))( L a ) = l / W 2 1 ( 2 , 1 > , ( O , i ) 2 ) = F21(2,1), ( 0 , i ) l )
354
Problems and Solutions in Group Theory
Note that those representation matrix entries of F 2 related to the basis states with the multiple weights can also be calculated with the formula of Gelfand.
From Eq. (8.32) we have
Unitary Groups
355
The block weight diagram and the planar weight diagram are given in the figure. The planar weight diagram of the representation [A,, A,] of SU(3) is a hexagon with the lengths of two neighbored sides to be (A1  A,) (i.e. M I ) and A2 (i.e. M z ) . The weights on the boundary is single, and the multiplicity increases as the position of the weight in the diagram goes inside until the hexagon reduces to a triangle.
16. Calculate the ClebschGordan series for the direct product representation [2,1] x [2,1] of the SU(3) group, and expand the highest weight state of each irreducible representation in the CG series with respect to the standard tensor Young tableaux. Solution. The representation [2,13 is the adjoint representation of SU(3) and was studied in Problem 13. The highest weight of [2,1] of SU(3) is ( 1 , l ) .
Problems and Solutions in Group Theory
356
Usually, the ClebschGordan series for the direct product representation of the SU(N) group is best calculated with the LittlewoodRichardson rule, as used in Problem 5. For [2,1] x [2,1] of SU(3) we have
However, the method of the dominant weight diagram discussed in Chapter 7 is also a good method for it. In this method we need to know the dominant weights and their multiplicities in the relevant representations, which can be obtained, say, by the method of the block weight diagram, or by consulting the table book. They can also be obtained directly from the standard tensor Young tableaux. This is the aim of this problem. Dominant weight
Standard tensor Young tableaux
No. of states 1 2
X
2
X
6
10
Unitary Groups
357
We have known that the weight for a standard tensor Young tableau is dominant if and only if the number of each digit p filled in the tableau is not less than the number of the digit ( p 1) filled there. We can easily count by this criterion which and how many standard tensor Young tableaux correspond to the dominant weights in the direct product representation space and in the relevant representation spaces. First, we list the standard tensor Young tableaux with the dominant weights in the direct product representation space [2,1] x [2,13, where only one standard tensor Young tableau in the pair related by a permutation of two product factors is listed for simplicity. Then, we calculate the standard tensor Young tableaux with the dominant weights in the representation (2,2) which is contained once in the CG series because the weight (2,2) is the highest weight in the product space.
+
Dominant weight
Standard tensor Young tableaux 111111111
No. of states
1111
12121 By comparing respectively the multiplicities of the dominant weights (3,O) and (0,3) in the product space with those in the representation (2,2), 2 1 = 1,we conclude that there is one representation (3,O) and one representation (0,3) contained in the CG series. Third, we calculate the standard tensor Young tableaux with the dominant weights in the representations (3,O) and (0,3). Comparing the numbers of the standard tensor Young tableaux with the dominant weight (1,l) in the product space and in the three representations (2,2), (3,O) and (0,3), 6  2  1  1 = 2, we conclude that the representation (1,l)is contained in the CG series twice.
358
Problems and Solutions in Group Theory ~~
Dominant weight
Standard tensor Young tableaux in (3,O) in ?0,3)
No. of states in (3,O) in (0,3)
TTTTTq
Fourth, we calculate the tensor Young tableaux with the dominant weights in the representation (1,l). Comparing the numbers of the standard tensor Young tableaux with the dominant weight (0,O) in the product space and in the representations (2,2), (3,0), (0,3), and two ( l , l ) , 10  3  1  1  2 x 2 = 1, we conclude that the representation (0,O) is contained once in the CG series. Therefore, we complete the diagram of the dominant weights for the direct product representation [2,1] x [2, I], where the subscript S or A denotes the symmetric or antisymmetric combination in the permutation of two factors in the product, respectively.
1
Standard tensor Young tableaux
No. of states
0s x n 6x1 3x2 3x2 6x6 1 x 10
R
2
0,o
At last, we calculate the expansion for the highest weight state of each
Unitary Groups
359
representation contained in the CG series by Eq. (7.24), namely, each raising operator E, annihilates the highest weight state. The results are

Fl Fl+
(antisym. terms),
x
X
X
terms)
U
U
U
In terms of the orthonormal basis states we have
U
360
Problems and Solutions in Group Theory
17. A neutron is composed of one u quark and two d quarks. Construct the wave function of a neutron with spin S3 = 1/2, satisfying the correct permutation symmetry among the identical particles. Solution. In particle physics, a hadron is composed of three quarks (baryon) or a pair of one quark and one antiquark (meson). The strong interaction among quarks is the gauge interaction of the colored SU(3) group. In addition to the color, a quark brings the flavor quantum number. The u and d quarks span the selfrepresentation space of the flavor SU(2) group, the isospin symmetry group. The isospin is conserved in the strong interaction. The next quark is s quark, which is heavier than the u and d quarks. u, d and s quarks span the selfrepresentation space of the flavor SU(3) group, which describes an approximate symmetry, but played an important role in the history of particle physics. Even now, it is still useful in the rough estimation. The expansions of the orthonormal basis states with respect to the standard tensor Young tableaux give the flavor wave functions of the hadrons. The flavor SU(3) group is a compact simple Lie group with order 8 and rank 2. The two commutable generators have important physical meanings. In the selfrepresentation, T3 describes the third component of the isospin of a quark, and T8 is proportional to the supercharge Y of a quark:
The basis states in an irreducible representation space of the flavor SU(3) group can be described in the planar weight diagram with the coordinate axes T3 and Y , as shown in Fig. 7.3 and Problems 14 and 15 in this Chapter. Since the planar weight diagram is intuitive, it has widely applied in particle physics. A neutron consists of three quarks. It is a system of three identical fermions. According to the Fermi statistics, the wave function has to be totally antisymmetric in the permutation of quarks. The internal orbital angular momentum of quarks is assumed to be zero. The total wave function is composed of three parts: the color part, the flavor part, and the spinor part. Since the neutron has to be the color singlet, its wave function
Unitary Groups
361
of the color part is totally antisymmetric in the permutation. Thus, the product of the wave functions of the flavor part and the spinor part should be combined into the totally symmetric state in the permutation. The neutron belongs to a flavor octet with T = T3 = 1/2 and Y = 1, and to a spinor doublet. In this problem we discuss the state with S3 = 1/2. The Young patterns corresponding to two parts both are [2,1]. Choosing the standard Young tableau
F l , we obtain two basis states of the flavor
wave function by the standard tensor Young tableaux:
F l= udd + dud  2ddu,
(23)
F l= udd + ddu  2dud .
There are also two basis states of the spinor wave function:
F l= (+  ) + ( + )

 2(  +) ,
(23)
 +)
E l= (+  ) + (
 2( f
) .
In this set of the standard bases, the representation matrices of the generators of the permutation group S3 can be calculated by the tabular method (see Problem 20 in Chap. 6),
In the direct product representation,
362
Problems and Solutions in Group Theory
Their eigenvectors for the eigenvalue 1 respectively
The totally symmetric state corresponds to the common eigenvector with eigenvalue 1, which is ( 2 1 1 2 ) T . Thus, the total wave function of the flavor part and the spinor part for the neutron with S3 =  1 / 2 is
r
+
= {udd dud  2 d d ~ } 3 {(+  )  (  +)} {udd ddu  2 d ~ d ) 3{(+  )  ())
+
*
+
+
*
= 3{2~+ddd+d~d+~d~dd++2dd~+
 dud+
 ud+d
 dd+u
+2d~+d}
For normalization, one may replace the factor 3 with
8.4
.
dm.
Subduced Representations
*
Now, we discuss the reduction problem of the subduced representation of an irreducible representation of the SU(N) group with respect to its subgroup. There are three kinds of important subgroups for the SU(N) group. One is SU(N 1) cSU(N), the second is SU(M)xSU(N/M) cSU(N), and the third is SU(M)xSU(N  M) cSU(N).
*
The reduction of the subduced representation from an irreducible representation [w] of SU(N) with respect to the subgroup SU(N  1 ) can be solved directly by the standard tensor Young tableau. The tensor indices taken the value N keep invariant in the subgroup SU(N  1). In a standard tensor Young tableau, the box filled with N can only be located in the lowest position of each column, where the neighbored right position has to be empty or a box filled with N . Removing the boxes filled with N in a standard tensor Young tableau of SU(N), we obtain a basis state of an
Unitary Groups
363
irreducible representation of the subgroup SU(N  1). Thus, by removing several (including zero) boxes located in the lowest positions in some columns from the Young pattern [A], we obtain the Young pattern [p],denoting an irreducible representation of SU(N  1) in the ClebschGordan series for the reduction of the subduced representation:
(8.37) This method may be applied continuously to the reduction for the subgroups in the group chain SU(N) 3 SU(N  1) 3
. ‘ . 3 SU(3) 3 SU(2) .
Thus, the subduced representation of an irreducible representation [A] of SU(N) with respect to its subgroup SU(N’) with N’< N can be reduced.
*
For the SU(NM) group, containing the subgroup SU(N) x SU(M), each tensor index a consists of two indices (a;),where a is the tensor index of the subgroup SU(N), and i is the tensor index of the subgroup SU(M). In the reduction of a subduced representation from an irreducible representation [w] of SU(NM) with respect to the irreducible representations [A] 8 [p] of the subgroup SU(N)xSU(M), the numbers of the boxes in three Young patterns are the same n. The three Young patterns respectively describe the permutation symmetry among n indices a , n indices a , and n indices a. When n indices i are fixed, the n indices a have the permutation symmetry denoted by [A]. When n indices a are fixed, the n indices i have the permutation symmetry denoted by [ p ] . When both sets of indices are transformed, n indices a! have the permutation symmetry denoted by [w]. Therefore, the representation [w] of the permutation group S, should be contained in the reduction of the inner product of two representations [A] and [ p ] :
Now, being an irreducible representation of SU(NM), the number of rows of the Young pattern [w] cannot be larger than N M . Correspondingly, the number of rows of the Young pattern [A] for SU(N) cannot be larger than N, and that of the Young pattern [p] for SU(M) cannot be larger than M .
*
+
For the SU(N M ) group, containing the subgroup SU(N)x SU(M), each tensor index a! takes two possible kinds of values, either the tensor in
364
Problems and Solutions in Group Theory
dex a of the subgroup SU(N),or the tensor index i of the subgroup SU(M). In the reduction of the subduced representation from the irreducible representation [w] of the SU(N IM) group with respect to the irreducible representation [A] 8 [p] of the subgroup SU(N)xSU(M), the number n1 of boxes in the Young pattern [A] and the number 722 of boxes in the Young pattern [p] are not fixed, but the sum nl + n2 of their numbers of boxes is fixed and equal to the number n of boxes in the Young pattern [w]. In the permutation among the tensor indices, all indices are divided into two sets. The n1 indices a in the first set have the permutation symmetry denoted by [A], which contains n1 boxes, and the 722 indices i in the second set have the permutation symmetry denoted by [ p ] ,which contains 122 boxes. Altogether, they have the permutation symmetry denoted by [w], which contains n = n1 n2 boxes. Therefore, the representation [w] of the permutation group S, should be contained in the reduction of the outer product of two representations [A] and [p] of the permutation group:
+
+
[A] 8 [p] = a:, [w] (3
*
 
The reduction of the subduced representation can be calculated by the LittlewoodRichardson rule.
+
Being an irreducible representation of SU(N M ) , the number of rows of the Young pattern [w] cannot be larger than N M . Correspondingly, the number of rows of the Young pattern [A] for SU(N) cannot be larger than N , and that of the Young pattern [p] for SU(M) cannot be larger than M .
+
18. Reduce the subduced representation from [3,1] of the SU(4) group with respect to the subgroup SU(3) and list the standard tensor Young tableaux as the basis states in each irreducible representation space of SU(3). Solution. The dimension of the representation [3,1] of SU(4) is
Unitary Groups
365
Its subduced representation with respect to SU(3) is reduced into the direct sum of six irreducible representations of SU(3). They and their basis states are listed as follows. 1). The representation [3,1] of SU(3) is 15dimensional with the following standard tensor Young tableaux as the basis states 111 111 1 1 2 112 113 2 ' 3 ' 2 ' 3 l 2 ' 113 122 122 123 123 3 ' 2 ' 3 l 2 l 3 ' 133 133 222 223 233 2 l 3 ' 3 j 3 ' 3 *
2). The representation [3,0] of SU(3) is 10dimensional with the following standard tensor Young tableaux as the basis states
111 1 1 2 113 122 123 ' 4 ' 4 ' 4 ' 4 ' 4 133 222 223 233 333 4 ' 4 l 4 l 4 ' 4 3). The representation [2,1] of SU(3) is 8dimensional with the following standard tensor Young tableaux as the basis states 114 2 ' 134 2 l
114 3 l 134 3 l
124 124 2 ' 3 224 234 3 l 3
*
4). The representation [2,0] of SU(3) is 6dimensional with the following standard tensor Young tableaux as the basis states 114 4 '
124 4 '
134 4 '
224 4 '
234 4 '
334 4

5 ) . The representation [1,1]of SU(3) is 3dimensional with the following standard tensor Young tableaux as the basis states 144 2 '
144 3 '
244 3
*
6). The representation [l,01 of SU(3) is 3dimensional with the following
Problems and Solutions in Group Theory
366
standard tensor Young tableaux as the basis states
144 4 )
244 4 '
344 4

19. Reduce the subduced representations from the following irreducible representations of the SU(6) group with respect to the subgroup SU(3)@SU(2),respectively:
Solution. In particle physics, a representation of the flavor SU(3) group describes a flavor multiplet, and a representation of the spinor SU(2) group describes a spinor multiplet. All those representations are denoted by the state numbers contained in the multiplets.
6
E
(3, 2)
15 N (3*, 3) @ (6, 1)
21 21 (3*, 1) @ (6, 3)
56 N (10, 4) @ (8, 2)
Unitary Groups
E P
367
n
70 "
20
21
35 N Notice the reductions of the subduced representations from [3]and [2, 14]2 [1]\[1]*. The former gives the observed multiplets of the lowenergy baryons in the sixties of the 20th century, namely, a decuplet with spin 3/2 and an octet with spin 1/2. Since they belong to the same representation of SU(6), their parities have to be the same (to be positive relative to the parity of a proton). The latter gives the observed multiplets of the lowenergy mesons in that years, namely, a singlet with spin 1, an octet with spin 0, and an octet with spin 1, whose parities are the same (to be negative from experiments). It is the success of the SU(6) symmetry theory for hadrons. In the SU(6) symmetry theory, the spin is described by the SU(2) group, so that this theory is only a nonrelativitic theory. This theory cannot explain many phenomena in high energy particle physics. 2 0 . Reduce the subduced representations from the following irreducible
representations of the SU(5) group with respect to the subgroup SU(3) @SU(2), respectively:
Problems and Solutions in Group Theory
368
Solution. In the grand unified model proposed in 1974, one needs to study the reduction of the subduced representations from some irreducible representations of the SU(5) group with respect to the subgroup SU(3)@SU(2). There is a U(1) subgroup in the SU(5) group with the generator Y =diag{1/3,  1/3,  1/3, 1/2, 1/2}, which is commutable with all elements in the subgroup SU(3)@SU(2). However, since there are five common elements in two subgroups in addition to the identity, the SU(5) group is not the direct product of two subgroups SU(3)@SU(2) and U(1). The color subgroup SU(3) describes the strong interaction of quarks, and the subgroup SU(2)@U(1) describes the weakelectric interaction. Each representation in the reduction is denoted by two multiplets of SU(3) and SU(2) and by the eigenvalue of Y as the subscript. In the SU(5) grand unified model, the electric charge Q is equal to T 3 Y , where T 3 is the third generator of the subgroup SU(2):
+
Q=
7'3
+ Y = diag (0, 0, 0, 1/2,
+ diag{1/3, = diag{1/3,

1/2}
 1/3,  1/3, 1/2, 1/2} 
1/3,
 1/3,
1, 0} .
Unitary Groups
369
In the SU(5) grand unified model, the state of a particle is divided into the lefthand state and the righthand state. They are filled into the 5dimensional and 10dimensional representations, respectively. The first generation of particles are filled as follows:
where the superscript c denotes the charge conjugate state, and the subscript R and L denote the righthand state and lefthand state, respectively. The 10dimensional represent ation is the antisymmetric tensor represent ation, so only the upper half matrix is needed to be filled with the particle states. The 24dimensional representation is the adjoint representation, describing the gauge particles, which is not listed here.
8.5
*
Casimir Invariants of S U ( N )
The generators TA in the selfrepresentation of SU(N) satisfy the commutation relations and the anticornmutation relationship
Problems and Solutions in Group Theory
370
Due to the normalization condition, ” ~ ( T A T B =)6,4B/2, the coefficients c A iand d A B D can be expressed by T’ as given in Eq. (8.9)
Recall that the adjoint representation Dadj(u)of SU(N) satisfy N21
C
D [ ’ ~ ( u )  ~ I ~ ~ D [= ’ ~ ( u ) D i 2 1 ( u ) I $ 1 , u E SU(N),
(8.38)
A‘=l
It1
where D[’] ( u )and are the representation matrices of the element u and the generator I A in the representation [A] of SU(N), respectively. When [A] is the selfrepresentation [l]of SU(N), D[’](u) = u and I?] = TA. Let CA,” and ~ A B Dbe the totally symmetric and antisymmetric tensors of rank three with respect to the adjoint representation. We are going to show that they are the invariant tensors. Prove it with CA; as example. CA,D
+
C
Di$iD$$iD$DiCAiD,:
A’B’D‘
= 2in
{ U  ~ T A U U  ~ T B U U  ~ T DUu  ~ T ~ u u  ~ T ~ u u 
= CA!.
(8.39) According to the LittlewoodRichardson rule, except for N = 2, the CG series for the direct product of two adjoint representations of SU(N) contains, in addition to the other representations, only one identical representation and two adjoint representations. Therefore, the identical representation only appears twice in the reduction of the direct product of three adjoint representations of SU(N). Equation (8.39) says that CA: and ~ A B Dare the only invariant tensors of rank three, up to a constant factor, with respect to the adjoint representation of SU(N). For the SU(2) group, since there is only one adjoint representation contained in the CG series for the direct product of two adjoint representations, and then, there is only one identical representation contained in the CG series for the direct product of
Unitary Groups
371
three adjoint representations. On the other hand, d ~ = g0 for~SU(2), and the invariant tensor of rank three with respect to the adjoint representation of SU(2) has to be proportional to CA,D. Now, we replace TA on the right hand side of Eq. (8.9) with They are still the symmetric or antisymmetric invariant tensors of rank three with respect to the adjoint representation. Thus, they are proportional to cA[or dAgg, respectively
It1.
From the first equality we obtain
Tz([X]) and A([X]) are respectively related to the Casimir operators of rank two and rank three of SU(N), and sometimes are simply called the Casimir invariants of rank two and rank three in the literature. Both Tz([X]) and A( [A]) are independent of the similarity transformation of the representation. It is easy to show from Eq. (8.40) that
(8.42)
In the calculation of T ! ( [ X ]with ) Eq. (8.41), we take A = D = 3 such that both generators contained in Eq. (8.41) belong to the subgroup SU(2). Denote T2([A]) in SU(2) by T2(')([X]),which is calculated to be T 2 ( O ) ( [ X ] ) = Tr
{
=
1 12 X(X + 1)(X + 2) .
(8.43)
Calculate T2( [A]) in the subduced representation from the irreducible representation [A] of SU(N) with respect to the subgroup SU(N  2)@SU(2), which is equivalent to a direct sum of some irreducible representation [v] @I [p] of the subgroup, (8.44)
Problems and Solutions in Group Theory
372
In the calculation of A([X]) with Eq. (8.40), we take A = B = 3 and D = 8 such that three generators are contained in Eq. (8.40) all belong to the subgroup SU(3), where d338 = Denote A([X]) in SU(3) by
m.
d o"XI), )
In the planar weight diagram, the states located in the same horizontal level have the same eigenvalues of Tix1,and the sum of (Tix1)2can be calculated by Eq. (8.43). Then, we sum up the contribution for the states with different TiX1.For a onerow Young pattern, we have x
A(')([A,O]) = 2
C
(A  3m)Tz("'([A m ] )
m=O
= (1/6)
Cx m=O
= (1/120)X(A
(A  3m)(X  m)(X  m
+ 1)(X  m + 2)
(8.45)
+ 1)(X + 2)(X + 3)(2X + 3).
Generally,
A(O)([A + A', A']) = A(O)([A]CZI [A']*)  A(O)([A  11CZI [A' = dpll ( s W ) ) A ( O )
([XI)


1]*)
4 x 1(sw~))A(O)([X'I)
+ A')(A + A' + 2)(A + 2X' + 3)(2A + A' + 3).
 ~~~,~I(SU(~))A  (1'3))( [ X ~ ~ ~  ~ I ( S U ( ~ ) ) A( 11) O)([X'
1
=  (A 120
+ 1)(A' + 1)(A

(8.46) Calculate A( [A]) in the subduced representation from the irreducible representation [A] of SU(N) with respect to the subgroup SU(N  3)@SU(3), which is equivalent to a direct sum of some irreducible representation [v]@ [p] of the subgroup,
21. Calculate the Casimir invariants T2([1']) and A([1']) in the representation denoted by a onecolumn Young pattern ([l']) of the SU(N) group.
373
Unitary Groups
Solution. From Eqs. (8.44) and (8.47), we obtain
(
1 N2 Tz([l‘]) = d[lr~l(SU(N 2)>T4O’([1])= 2 r1
)’
A([lr]) = {d[,r1](SU(N 3))  d[lp21(SU(N 3))) A‘O’([1]) (N  3)! (N  3)! (r 2)!(N  r  l)! (r  l)!(N  r  2)!  (N  3)! {(N  r  1)  (T  1)) (r  l)!(N  r  l)!

22. Calculate the Casimir invariants T2([3])and A([3]) in the irreducible
representation [3] of the SU(6) group.
Solution. From Eq. (8.43) we have T4O’([3])= 5,
Ti0’([2]) = 2,
TiO’([l])= 1/2
TiO’([O]) = 0.
From the reduction of the subduced representation from [3] of SU(6) with respect to the subgroup SU(4)@SU(2)
we obtain
From Eq. (8.45) we have A(’)([3]) = 27,
A(’)([2]) = 7,
A(’)([l]) = 1
A(’)([O])= 0.
From the reduction of the subduced representation from [3] of SU(6) with respect to the subgroup SU(3)@SU(3)
Problems and Solutions in Group Theory
374
we obtain
23. Calculate the Casimir invariants T2([X]) and A([X]) in the representation ([A] = [A, O,O, 01) of the SU(5) group. Solution. The calculation method is the same as Problem 22. Here we only list the result.
+ 1)(X + 2)/12, A(O)([X]) = X(X + 1)(X + 2)(X + 3)(2X + 3)/120, d[A](SU(3)) = (A + 2)(X + 1)/2, d[A](SU(2)) = + 1. T2(O)([X]) = X(X
Hence
Generalizing this result to the SU(N) group, we obtain N n=O
 X(X+l).(X+N) 2 ( N + l)! N
X(X
)
+ 1)    (A + N ) ( U + N ) ( N + 2)!
Chapter 9
REAL ORTHOGONAL GROUPS
9.1
Tensor Representations of S O ( N )
*
The set of all Ndimensional unimodular real orthogonal matrices, in the multiplication rule of matrices, constitutes S O ( N ) group. In this chapter we discuss only the case of N 2 3, because S O ( 2 ) group is an Abelian group which is isomorphic onto the U(1) group. The S O ( N ) group is a doublyconnected compact Lie group with order N(N  1 ) / 2 . S O ( N ) is a subgroup of SU(N). The generators in the selfrepresentation of S O ( N ) is usually taken as the second type of generators in the selfrepresentation of SU(N) multiplied by two
(Tab),, = 2 (TLt))
rs
Tr ( T a b T c d ) = 2SacSbd
+ ibasdbr,
= idarSbs
a
< b, (94
 26adSbc,
where the indices a and b are the ordinal indices of the generator, while T and s are the row and column indices of the matrix. N(N  1 ) / 2 matrices Tab all are antisymmetric, and they are also antisymmetric with respect to the ordinal indices. Any matrix R in S O ( N ) can be expressed in the form of exponential matrix function: R=exp
{ iC a
WabTab
1
.
(9.2)
j , The S O ( N ) group with N = 2 l or N = 2 l + 1 contains l mutually commutable generators,
375
Problems and Solutions in Group Theory
376
which span the Cartan subalgebra of the Lie algebra of S O ( N ) . Thus, the ranks of both SO(2t) and S0(2C+ 1) are C. We want to find the common eigenvectors of H j , [ H j , Ea] = ajEa. When N = 2 4 the eigenvectors are
1 [T(2a)(2bl) + iT(2a1)(2bl) E,,( 2 )  5
+ i T ( 2 a ) ( 2 b ) T ( 2 a  1 ) ( 2 b ) ]
(3) Eab
+ i T ( 2 a ) ( 2 b )+ T ( 2 a  l ) ( 2 b ) ]
=
5 [T(2a)(2b1)
 iT(2a1)(2bl)
)
(9.4) 7
+
+
with the eigenvalues { e ,  e b } j , { e, eb}j,{ e , e b } j ,and { e,  e b } j , respectively, where a < b, and { e , } j = 6,j. When N = 2 t + 1, in addition to Eq. (9.4), there are another two types of eigenvectors:
=
8
+i~(2al)(2C+l~l
[T(2,)(2l+l)
(9.5)
3
with the eigenvalues { e , } j and  { e , } j , respectively. The simple roots for the S 0 ( 2 l + 1) group are
The remaining positive roots are
where a < b. The largest root, which is the highest weight of the adjoint representation of SO(2C + l),is
v=2
Therefore, the Lie algebra of S0(2C+ 1) is Bl,and L3 = w2. The simple roots for the SO(2t) group are
The remaining positive root are
377
Real Orthogonal Groups
where a < b. The largest root, which is the highest weight of the adjoint representation of S 0 ( 2 l ) , is
Therefore, the Lie algebra of S O ( 2 l ) is De,and c3 = w 2 .
*
The Chevalley bases in the selfrepresentation of the S 0 ( 2 l + 1) group (the Be Lie algebra) are
Hp = Ep =
Fp = He = Ee = Fe =
q2e)(2e+l)
+ iq2el)(2e+l),
(9.10) where 1 5 p < l. The Chevalley bases in the selfrepresentation of the SO(2l) group (the DCLie algebra) are the same as those of the S 0 ( 2 l + 1) group except for those when p = l , which are
He = =
Fe =
*
T(2[3)(2&2)
+ 7'(2ei)(2e),
$ { T(212)(2el)  iT(213)(2ll) + iT(212)(21) + T(2e3)(21!)} $ {~(2e2)(2e1) + i ~ ( 2 e  3 ) p e  i )  i~(2[2)(2e) + T(ze3)pe)) 
7
(9.11)
The definition for a tensor with respect to the S O ( N ) group is the same as that to the SU(N) group, except that the transformation matrix u ESU(N) is replaced with R & O ( N ) . The Weyl reciprocity still holds for the tensors of S O ( N ) . However, this replacement makes some new characteristics for the tensors of S O ( N ) . Since the transformation matrix R E S O ( N ) is real, there is no difference between the covariant tensor and the contravariant tensor for the S O ( N ) group. As a result, first, the tensor space of S O ( N ) group can be decomposed into a direct sum of a series of the traceless tensor subspaces 7 ,which are invariant in S O ( N ) . Each 7can be projected by the Young operators into the subspace = $?IT.
378
Problems and Solutions in Group Theory
Second, since the tensor is traceless, the tensor subspace is a null space if the sum of the numbers of boxes in the first and the second columns of the Young pattern [A] is larger than N. Third, since there is no difference between the covariant tensor and the contravariant tensor, the representation described by a Young pattern [A] with the row number m larger than N / 2 is equivalent to that by [A'], where [A'] is obtained from [A] by replacing its first column with a column containing ( N  m) boxes. Fourth, when the row number of [A] is equal to N / 2 , the representation space '7$^] is reduced into two minimal invariant subspaces with the same dimension. They are the selfdual and the antiselfdual tensor subspaces, whose representations are denoted by [+A] and [A], respectively.
*
A tensor of rank two for S O ( N ) is easy to be decomposed into the sum of the traceless tensors:
where the tensor in the curve bracket is a traceless tensor of rank two, and the tensor in the square bracket is a tensor of rank zero. However, generally speaking, the decomposition of a tensor of rank n into a sum of some traceless tensors of different ranks is straightforward, but tedious. As discussed in Chap. 8, the irreducible tensor bases for SU(N) are obtained explicitly. In fact, they are the standard tensor Young tableaux. The bases can be orthonormalized by making use of the block weight diagram. However, due to the traceless condition, some standard tensor Young tableaux for S O ( N ) become linearly dependent. It becomes a new problem how to find the complete set of the independent irreducible tensor bases for S O ( N ) explicitly. Another problem is that the tensor basis Oal...,, is not the common eigenstate of the Chevalley bases H,. Since a tensor basis is a direct product of n vector bases O,, ..O,,, we only need to find the new vector bases such that H p are diagonal. Namely, we want to diagonalize H p in the selfrepresentation of S O ( N ) by a similarity transformation. Both problems are well solved for the SO(3) group. Now, we generalize the solution to the S O ( N ) group.
*
+
In the selfrepresentation space of SO(2C 1) we defined a set of new orthonormal bases, called the spherical harmonic bases,
379
Real Orthogonal Groups
i
(l)ea+l
+
i02,)
when 1 5 a
5 l,
when c t = l + l , when ! 2 5 a 5 2 l + 1. ( @ 4 ~  2 , + 3  [email protected]+4) (9.12) The tensor basis @ , l . . . ( y , of rank n for S 0 ( 2 l + 1) is the direct product of n vector bases a,    @ a n . In the spherical harmonic bases a, the nonvanishing matrix entries of the Chevalley bases are @a
=
@2e+1
+
(9.13)
where 1 5 p 5 l  1. In the spherical harmonic bases, the standard tensor Young tableau $l+,,...(y,, in the representation [A] of S0(2!+ 1) corresponds to the highest weight if each box in its j t h row is filled with the digit j , because each raising operator E p annihilates it. Therefore, the relation between the highest weight M and the Young pattern [A] for S 0 ( 2 l + 1) is
Although the standard tensor Young tableaux is generally not traceless, the standard tensor Young tableau with the highest weight in [A] of S 0 ( 2 l + 1) must be traceless, because it only contains with a < l + 1. For example, is traceless. The remaining the tensor basis 0 1 0 1 is not traceless, but traceless tensor bases in an irreducible representation space [A] of S 0 ( 2 l + 1) can be calculated from the basis with the highest weight by the lowering operators Fp. In the calculation the block weight diagram is helpful.
*
In the selfrepresentation space of SO(21) we defined a set of new orthonormal bases, called the spherical harmonic bases,
380
Problems and Solutions an Group Theory
The tensor basis @)(Y1...CYnof rank n for SO(2t) is the direct product of n vector bases aal    a , . In the spherical harmonic bases a,, the nonvanishing matrix entries of the Chevalley bases are
(9.16)
where 1 5 p 5 C  1. In the spherical harmonic bases, the standard tensor [XI @al...Qn in the representation [A] of SO(2t) corresponds Young tableau yp to the highest weight if each box in its j t h row, j < t, is filled with the digit j and if all boxes located in its t t h row, if [A] contains k' row, are filled with the same digit l or k ' + 1, depending on whether the representation is selfdual [+A] or antiselfdual [A]. Therefore, the relation between the highest weight M and the Young pattern [A] for SO(2t) is
Since the standard tensor Young tableau with the highest weight contains with a < C, it is a traceless tensor. either @e or @!+I, in addition to The remaining traceless tensor bases in an irreducible representation space [A] of SO(2l) can be calculated from the basis with the highest weight by the lowering operators Fp.
*
The dimension d[xl(SO(N))of the representation [A] of S O ( N ) can be calculated by the hook rule. In this rule the dimension is expressed as a quotient, where the numerator and the denominator are denoted by the
Real Orthogonal Groups
381
symbols Y g l and YL”],respectively:
Let us explain the meaning of two symbols YL”]and Y P ] . The hook path (i, j) in the Young pattern [A] is defined to be a path which enters the Young pattern at the rightist of the ith row, goes leftwards in the i row, turns downward at the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j) is the same path as the hook path (i, j) but with the opposite direction. The number of boxes contained in the path (i, j ) , as well as in its inverse, is the hook number hij. YL”’ is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the hook number hij. Define a series of the tableaux Y g l recursively by the rule given below. Y p l is a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Y g l in the series. The symbol Y g l means the product of the filled digits in it, so does the symbol YL”]. The tableaux Y g l are defined by the following rule: (a) Y g l is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the digit ( N j  i).
+
(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recursively the Young pattern [A(”)] by removing the first row and the first column of the Young pattern until [A(”)] contains less than two columns. (c) If [A(”)] contains more than one column, define Y g l to be a tableau of the Young pattern [A] where the boxes in the first ( a  1) row and in the first ( a  1) column are filled with 0, and the remaining part of the Young pattern is nothing but [A(”)]. Let [A(“)] have r rows. Fill the first r boxes along the hook path (1, 1) of the Young pattern [A(”)], beginning with the box on the rightmost, with the
382
Problems and Solutions in Group Theory
digits (Xi"'  l ) , (A?)  l ) , ", (A?)  l),box by box, and fill the first (Xi"'  1) boxes in each inverse path (i, 1) of the Young pattern [A(")], 1 5 i 5 T , with 1. The remaining boxes are filled with 0. If a few 1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern Ygl is zero. 1. Calculate the dimensions of the irreducible representations of the SO(8) group denoted by the following Young patterns: ( a ) [4,2], (b) [3,2], (c) ~ 4 ~ 4 (1d, ) [+3,2,1,11, (el [+3,3,1,11.
Solution. (a).
d[4,2~(S0(8))=
8910
891011 78

781114 58 5421 2 1
= 4312.
2 1
11 891011 78 910+21 d[4,4](S0(8))
0
71012
2 1
(c).
2
5421 2 1
112
.
1  1 1 3 +
=
3
3 0 0 0 0'01 1
1
0 7 8 1 3 14 2  5 6 7 12

5432 4321
5432 4321
= 8918. 0 1 2 8 9 10 00 78 6 '1
81012 78 5
4 1 2x 2 1
8 9 10
(el.
d[+3,3,1,1](S0(6))
=
789 6 5
41 2x 2 1
0 0 0 0  1 1
0 2 2 0 0 0 +  2 2
+ o 0 632
521 2x 2 1
81112 7 7 1 0 4  3 6 3 2 = 4312. 2 x
521 2 1
2. Calculate the ClebschGordan series for the subduced representations
of the following irreducible representations of the SU(N) group with respect to the subgroup S O ( N ) , and then check the results by their dimensions for N = 7:
Real Orthogonal Groups
383
Solution. The following results for the reductions of the subduced representations of SU(N) with respect to the subgroup S O ( N ) are independent of N , except for the Young patterns in the CG series with the row number not less than N/2, or with the digit number in the first two column larger than N.
210 = 189
+ E P
+ 77 + 105 + 7. 1008 = 819
@
+ 21.
Problems and Solutions an Group Theory
384
882 = 693
+ 77 + 105 + 7.
'3
756 = 616
490 = 378
t
224 = 189
924 = 714+ 182
+ 27+
+ 105 + 35.
+ 105 + 7.
+ 35.
1.
[email protected] 2646 = 1911
2100 = 1560
1176 = 825
+ 182 + 168 + 330 + 2 x 27 + 1.
+ 189 + 330 + 21.
+ 330 + 21.
Real Orthogonal Groups
2352 = 1617
385
[email protected]@P
+ 189 + 330 + 168 + 27 + 21.
840 = 616
+ 189 + 35.
490 = 294+ 168+ 27+ 1.
= 378
+ 189 + 21.
140 = 105 t.35.
3. Calculate the ClebschGordan series in the reductions of the following direct products of the irreducible tensor representations, and check the results by their dimensions for the SO(7) group:
(1)
PI €3 PI,
(2)
PI €3 [I,11,
(3) [31 €3 [2,11*
Solution. (1):
27 x 27 = 729 = 182
+ 330 + 168 + 27 + 21 + 1. m
27 x 21 = 567 = 330 + 189
+ 27 + 21.
Problems and Solutions in Group Theory
386
E F + + + + + + +
77 x 105 = 8085 = 1750 1911 1560 1617 182 330 168 330 189 27 21.
+
+
+
4. Calculate the orthonormal bases in the irreducible representation space
[2,2] of SO(5) by the method of the block weight diagram, and then express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank four for SO(5).
Solution. The Lie algebra of SO(5) is B2. The relations between the simple roots rp and the fundamental dominant weights wp are rl = 2wl  2w2,
r2
= wl
+ 2w2,
The highest weight of the representation [2,2] of SO(5) is (0,4), and its dimension is
56
11
21
67
21
From the highest weight state (0,4) we have a quartet of A2 with (1,2), (2,0), (3,g) and (4, where the matrix entries of F 2 are 2, &,&,and 2, respectively. The weights (1,2) and (2,O) are the single dominant weights. From (1,2) we have a doublet of A1 with (T, 4). From the weight (2,O) we have a triplet of d1 with (0,2) and (z,4), and from the weight (T,4) we have a quintet of A2 with (0,2), (l,O), (2,2) and (3,a). The weight (0,2) may
z),
Real Orthogonal Groups
387
be a double dominant weight. Define the first basis state I(0,2)1) belonging to the triplet of A1 and the other I(0,2)2) belonging to a singlet of CAI. Let
Applying El F 2 = F2El to the basis state
l(i,4))) we have
Thus, a1 = &, and then, a2 = d m = 1. From the weight (3,2) we have a quartet of d1 with (1,0), (i,2)) and (3,4).On the other hand, applying F 2 to two states )(0,2)1) and 1(0,2)2),we obtain another two states with the weight (1,O). Thus, the dominant weight (1,O) may be a triple weight. We define the first basis state (1, O), belonging to the quartet of dl and the other two belonging to two doublets of d1. The action of F 2 on 1(0,2)1) is the combination of I ( 1 , O ) l ) and I(l,0)2). The weight (2,2) is the double weight because it is equivalent to (0,2). The weight ( i , 2 ) is equivalent to (1,O). We have
Let
Applying
E1F2
= F2El to the basis state )(0,2)1) and 1(0,2)2), we have
388
Problems and Solutions an Group Theory
Choosing the phases of the basis states 1(1,0)2) and 1(1,O)3) such that a4 and a7 are real positive, we have a4 = 1, a6 = &, and a7 = 0. It means that the dominant weight (1,O) as well as its equivalent weight (T,2) in [2,2] are the double weights, not triple. The states 1(1,O)3) and I(T, 2)3) do not exist in [2,2]. From the weight (4, q) we have a quintet of 41 with the weights (2,2)1, (0,0)1, (2,2)1 and (3,4),where the matrix entries of F1 are 2, &, & and 2, respectively. Another basis state 1(2,2)2) belongs to a triplet of 41 with (0, O), and (2,2)2, where both matrix entries of F1 are Let
Jz.
Applying
E1F2
= F2El to the basis state I(l,O)1) and 1(1,0)2), we have
F2E2
+ H2 to the basis
Choosing the phase of the basis state )(2,2)2) such that we have a9 = 1 and all = 2. Applying E2F2 = F2E2 state 1(2,2)1), we have
+ H2 to the basis
Thus, a8 = 4 and a10 = 0. Applying state I ( l , O ) 1 ) , we have
E2F2
=
a9
is real positive,
Real Orthogonal Groups
389
Choosing the phase of the basis state 1(3,;?>)such that a12 is real positive, we have a12 = 1 and a13 = 4. Applying F2 to the states with the double weight (T,2), we have two basis states with the weight (O,O), in addition to the state bases I(0,O)l) in the quintet and I(O,O)2) in the triplet of 41. Thus, the dominant weight (0,O) may be a quadruple weight. Both I ( O , O ) 3 ) and I(0, O),) are singlet of A1. Let
2)1) and [(i) 2 ) 2 ) , we Applying ElF2 = F2E1 to the basis states 1(2, 4))) I(i, have
Thus, we have bl = b2 = b3 = b4 = b7 = fi and b6 = 0. Applying = F2E2 H2 to the state basis [(i, 2)1), we have
E2F2
+
Thus, b5 = 0, and the basis state 1(0,0)4) does not exist. So, b g = 0. Applying E2F2 = F2E2 H2 to the basis state l(i, 2 ) 2 ) , we have
+
Choosing the phase of the basis state 1(0,0)3) such that b8 is real positive, we have bs = fi. The remaining basis states and the matrix entries of the lowering operators Fp can be calculated similarly. The results are given in Fig. 9.1.
390
Problems and Solutions in Group Theory
Fig. 9.1 The block weight diagram for [2,2] of SO(5).
Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau
yi2’21 is
Fl
. F’rom the Fock
condition (8.19) we have
The normalization factors for the typical tensor Young tableaux are
Real Orthogonal Groups
391
From Problem 2 , we know that the number of linearly independent trace is 50  35 = 15. tensors in the tensor space of rank four projected by yPy2] In fact, they belong to the representation [2,0] and [O,O] of SO(5). We can calculate the trace tensors from the highest weight condition (7.24) and in terms of the lowering operators. In the following we only list the trace tensors with the dominant weights.
392
Problems and Solutions in Group Theory
+ 2 p 2+ q5
2
Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (9.13) and the block weight diagram. Due to the WignerEckart theorem, the state basis belonging to the representation [2,2] of SO(5) is orthogonal to the basis states belonging to [2, O] and [O,O], so it is a traceless tensor. For simplicity we neglect the index (0,4) for the highest weight in the basis state I(0,4), (ml ,ma)).
Real Orthogonal Groups
393
’
394
Problems and Solutions in Group Theory
5 . Calculate the orthonormal bases in the irreducible representation space
[2,0,0,0] of SO(8) by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless symmetric tensor space of rank two for SO(8).
Solution. The Lie algebra of SO(8) is D4. The relations between the simple roots rp and the fundamental dominant weights w p are rl = 2 w l  w2,
12
= w1+ 2w2  W Q  w4,
The highest weight of the representation [2,0,0,0] of SO(8) is (2,0,0,0), and its dimension is
From the highest weight state (2,0,0,0) we have a triplet of A1 with
Real Orthogonal Groups
395
a.
(0, l , O , 0), and (2,2,0,0), where both matrix entries of F1 are The weight (0, l , O , 0) is a single dominant weight. The weights equivalent to two dominant weights are
Therefore, the representation [2,0,0,0] contains two simple dominant weights (2,0,0,0) and (0, 1,0,O) and one triple dominant weight (O,O, 0,O). The block weight diagram for the representation [2,0,0,0] is very easy except for the part related to the triple weight (O,O,O,O), which will be explained below. In Fig. 9.2 we give the block weight diagram for the representation [2,0,0,0] where only those matrix entries which are not equal to 1 are indicated. Let 1(2,%,0,0)),1(0,0,0,0)1) and I(?, l , O , O ) ) constitute a triplet of A1, and the other two basis states I(O,O, 0,0)2)and I(O,O, O , O ) 3 ) are the singlets of C A I . We assume
where we neglect the index (2,0,0,0) for the highest weight in the basis state I(2,0,0,0), ( m l ,m2, m3, m4)) for simplicity. Applying El F2 = F2E1 to the basis state I (i, 2, i,i)), El F3 = F3El to the basis state I(0, i,2, 0)), and ElF4 = F4E1 to the basis state I ( O , i , O , Z ) ) , we have
E1F2 ( ( i , Z , i , 1))= J Z C l ((2, i,0,O)) = F ~ ((i, E ~2 , i , i ) ) = F~ ((I,i,i,i))= ](2,i,o,o)),
396
Problems and Solutions an Group Theory
Thus, c1 = d1/2and a1 = bl = 0. Choosing the phase of the basis state I ( O , O , O , O ) 2 ) such that a2 is real positive, we have a2 = Applying E3F2 = F2E3 to the basis state I(i, 2 , i , i ) ) , and E3F4 = F4E3 to the basis state I(0, T, 0,2)), we have
a.
Thus, c2 = and b2 = 4. Then, b3 = 0. Choosing the phase of the basis state 1(0,0,0,0)3)such that c3 is real positive, we have c3 = 1. Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau yi2’o’0’01 is . The normalization factors for two typical tensor Young tableaux are
ml
There is only one trace tensor in the tensor space of rank two projected by
y/2~0~0~01 . It belongs to the representation [O,O, O,O] of SO(8):
Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (9.16) and the block weight diagram. Due to the WignerEckart theorem, the basis state belonging to the representation [2,0,0,0] of SO(8) is orthogonal to the basis state belonging to [O,O, O,O], so it is a traceless tensor.
Real Orthogonal Groups
397
I2, ?,o(01 Iii;,i,iI
+ 27 B = 36 + 45 C=18 + 27 + 36 A = 18
(?,O,O,Ol
 45
Fig. 9.2 The block weight diagram for [2,0,0,0] of SO(8).
398
Problems and Solutions in Group Theory
Real Orthogonal Groups
6. Calculate the spherical harmonic functions space.
399
Yglin
an Ndimensional
Solution. The relations between the rectangular coordinates xa and the spherical coordinates T and e b in an Ndimensional space are
(9.19)
a=l
The unit vector along x is usually denoted by j;: = x / r . The volume element of the configuration space is N1
N
The orbital angular momentum operators Lab are the generators of the transformation operators PR for the scalar function, R ESO(N),
F'rom the second Lie theorem, Lab satisfy the same commutation relations as those of the generators Tab in the self representation of SO(N). The Chevalley bases H,(L), E,(L) and F,(L) can also be expressed by Eqs. (9.10) and (9.11), where Tab are replaced with Lab. Because N
N
c=l
c=l
where J a b are the generators of OR. The common eigenfunctions X, of H,(L) can be obtained from @a given in Eq. (9.12) for S 0 ( 2 l + 1) and in Eq. (9.15) for SO(2l) by replacing the vector basis 0, with the rectangular coordinate z a / T , where the factor T  ~ stands for removing the dimension of length.
Problems and Solutions in Group Theory
400
The spherical harmonic function Ykl(ii) is the eigenfunction of the orbital angular momentum for a single particle. Since there is only one coordinate vector x, the representation [A] for the spherical harmonic function Ykl(P) has to be the totally symmetric representation denoted by the onerow Young pattern [A] = [A, 0 , . . . ,O]. X,, multiplied by a normalization factor, is nothing but the spherical harmonic function Ykl(ii)in the selfrepresentation [A] = [l]. The component mp of the weight m is the eigenvalue of H p ( L ) in the eigenfunction X,. Generally, for the highest weight state M = (A, 0,. . . , 0 ) we have
Y i l ( k ) = cjvxx: = CN,*
{ (V
+ is,)}*
(51 T d 5
7
(9.22)
where t is equal to l when N = 2 l + 1 and to ( l  1) when N = 21. The remaining spherical harmonic function Y k l ( i ) with the weight m can be calculated by the lowering operators F p ( L ) ,just like we have done in Problems 4 and 5. Here we have to calculate the normalization factor C N , and the eigenvalue of the angular momentum square L2. Both problems can be solved from the highest weight state Y&](P). Since z1 + ix2 = reiel sin 02 . . . sin ODM1and
An
in
1 . 3 . 5 ... ( n  1) I,.
de (sir
('*
1 . 3 . 5 . : . (4
'
when n is odd,
we have
For SO(2l
+ 1) group, we have 2 4  6  (2A + 2 1  2) (z ...() A+l1 ) 2  1*3.5..*(2A+2ll)
 2 ' 4 4 7 1 ) A+1 x+2 2 * + 2 W A ! ( A + l  l)! (2X + 2l  l)! '
*
7r
* *
*
(9.23)
Real Orthogonal Groups
401
For SO(2l) group, we have

21x7r
('>
( 1 ) ... x + l x+2
(X + l  l ) 7r
T'X!
= 2Xl(X+&
I)!. (9.24)
Assuming both b and d are not equal to 1 and 2, we have
Thus, the eigenvalue of L2 in the spherical harmonic function k'(k'+ N  2).
Ykl(Ec) is
7.Calculate the complete set of the independent bases for the eigenfunctions of angular momentum in a threebody system which is spherically symmetric in an Ndimensional space. Solution. After separating the motion of the center of mass, there are two Jacobi coordinate vectors in a threebody system, denoted by R1 = x and R 2 = y. Thus, the eigenfunction of angular momentum in the system belongs to an irreducible representation, denoted by a Young pattern with only one or two rows, [XI, X 2 , 0 , . . . ,O] = [XI, A,]. In a threedimensional space, the eigenfunction of angular momentum is described by the representation D ' and the magnetic quantum number m. But, in the Ndimensional space, N > 3, the eigenfunction of angular momentum is described by the representation [A,, A,] and the weight m. Due to the spherical symmetry, we only need to calculate the eigenfunction of angular momentum with the highest weight, namely, m = d in the threedimensional space and m = M = (A1  X2, X 2 , 0 , . . . ,0) in the Ndimensional space. The remaining eigenfunctions can be calculated by the lowering operators Fp( L ) ,just like we have done in Problems 4 and 5. Therefore, the eigenfunction of angular momentum with the highest weight in the Ndimensional space is described by two parameters A1 and X2. Denote by Q;lX (x,2 y) the independent bases of the eigenfunctions of angular momentum with the highest weight, where q is understood as an ordinal parameter. What the complete set means is that any eigenfunction (x,y) with the angular momentum [XI, A,] and the highest weight M can be expanded with respect to Q;IX2(x, y),
$Jk
(9.25)
402
Problems and Solutions in Group Theory
+q(c)
where the coefficients only depend upon the internal invariants t. In the threebody system there are three internal variables: = xx, 62 = y  y and 6 3 = x  y. What the independent bases means is that any basis cannot be expanded with respect to the other bases like Eq. (9.25). Obviously, a basis is not a new independent basis if it is a product of another basis and a function of the internal variables. Consider the product of two spherical harmonic functions Ykl(ii) and (y). It belongs to the direct product [q] x [p3 of two irreducible representations, which is usually a reducible one. Its reduction can be calculated by the LittlewoodRichardson rule and the contraction of the components of x and y. In the contraction the factor of the internal variables appears. Therefore, the independent bases of angular momentum come from the calculation result by the LittlewoodRichardson rule.
Yk)
The representations with t > 0 correspond to the bases which are not independent. Conversely, any independent basis of angular momentum [XI, A,] can be calculated by the product (k)Y$](y) where X2 5 q 5 X1 and p = X1 X2  q. Note that the highest weight state corresponds to the standard tensor Young tableau where each box in the first row is filled with 1 and each box in the second row is filled with 2 , and the tensor indices in the same column of the tensor Young tableau are antisymmetric. Now, the digit 1 means X1 = c(z1 i 2 2 ) or Y1 = c(y1 + iy2) and the digit 2 means X2 = C(Q izd) or Y2 =  c ( y ~ iy4), where c is a constant factor. Up to the normalization factor, the independent basis state Q?lX2(x,y) is expressed as a product of three parts:
Ykl
+
+
+
+
For SO(4) group the irreducible representation denoted by a tworow Young pattern can be selfdual or antiselfdual. The standard tensor Young tableau for the highest weight state of the selfdual representation is still in the form ( 9 . 2 6 ) ,but that of the antiselfdual representation is different. In fact, each box in the second row of that standard tensor Young tableau for the highest weight state of the antiselfdual representation is filled with 3,
Real Orthogonal Groups
403
not 2. In this case, the independent basis state Q;IX2 (x,y) is expressed as
For SO(3) group, due to the traceless condition, the sum of the numbers of boxes in the first two columns of the Young pattern describing the irreducible representation is not larger than 3, namely, A2 = 0 or 1. On the other hand, the representation [ A , l ] for SO(3) is equivalent to the representation [A, 01. But, the bases of angular momentum for [A, 01 and [A, 11 have different parities.
9.2
*
Spinor Representations of S O ( N )
Let N matrices yu satisfy the anticommutation relations,
The set of all their products constitutes a finite group, denoted by r N . I’N is a matrix group. Usually, the matrix “(a is defined by its representation matrix in a faithful unitary irreducible representation of r N . The matrices yu, called the irreducible yu matrices, are unitary and hermitian. When N = 2 l is even, define
yf = (i)”2y1yz.. .yp/*
(9.29)
yf is also unitary and hermitian, and yf is anticommutable with any matrix ya. Thus, the matrix yj and N matrices yu satisfy the anticommutation relations (9.28). They are regarded as the definition of yu in that
rN+l.
Note
The lefthand side of Eq. (9.30) changes its sign if one transposes arbitrary two matrices ya and yb,or to change the sign of arbitrary one matrix yu. When N = 4rn I, the righthand side of Eq. (9.30) is not a new element, the group r4m+1is isomorphic onto the group r4m.When N = 4m  1, the righthand side of Eq. (9.30) is a new element, the set r 4 m  2 and its
+
404
Problems and Solutions an Group Theory
product with i is isomorphic onto the group
l?4m1.
r4m+l M r4m, r4m1 = (r4m2,
ir4m2).
(9.31)
When N = 21, there are 22'+1 elements in For any element of rN, except for the constant matrix, there exists at least one element which is anticommutable with the given element. Thus, the trace of the element is zero. The dimension d2')of the irreducible ya matrices which satisfy Eq. (9.28) can be calculated to be 2' by the character formula. The matrices in r2', neglecting their signs, are linear independent, and form a complete set of bases for the 2'dimensional matrices. The equivalent theorem says that any two sets of irreducible Ta matrices which respectively satisfy Eq. (9.28) are equivalent to each other. Namely, they can be related onetoone by a similarity transformation X : = XlyaX, 1 5 a 5 21. (9.32)
+
When N = 21 1, the dimension of matrix ^la is still 2[. The equivalent condition for two sets of the irreducible ya matrices should include, in addition to Eq. (9.28), that two products y1 . . . y~ are equal to each other.
*
The equivalent theorem of ya plays an important role in the spinor theory of S O ( N ) . The charge conjugate matrix C and the spacetime inversion matrix B in particle physics can be defined based on it. When N = 21, clyac =  ( y a y , detC = 1,
CtC = 1, CT = (_l)W+1)/2C,
BlyaB = (%IT detB = 1,
B~B = 1, B T = (1)!?('1)/2B.
,
Due to the condition (9.30), only matrix C can be defined when N = 4m1, and only matrix B can be defined when N = 4m 1. The fundamental spinor representation D["]( R )of SO( N ) , which is simply called the spinor representation in literature, is also defined based on the equivalent theorem:
+
Real Orthogonal Groups
405
Its generators Sab = iYaYb/2, a < b, are called the spinor operators. The fundamental spinor representation, being a group, is homomorphic onto S O ( N ) by 2 1 correspondence. When N is odd, the fundamental spinor representation D['I(R),or denoted by [s], is irreducible. [s] is real when N = 8k f 1, but selfconjugate when N = 8k f 3. Its dimension is di,] = 2(N1)/2. The fundamental spinor representation [s] is reducible when N is even. It can be reduced into two inequivalent representations, denoted by D[*"](R)or [fs],[s]N [+s] @ [s], with the same dimension d [ k S ]= 2(N/2)1. [fs] are conjugate to each other when N = 4k + 2. [ f s ] are real when N = 8k and selfconjugate when N = 8k + 4.
* In S O ( N ) transformation, Jr is called the fundamental spinor of S O ( N )
if it transforms by the fundamental spinor representation D[']( R ) :
where Q is a column matrix with d[,] components. When N is even, @ can be decomposed into two parts by the project operators Pk = (1/2)(1 fyf), Q* = P*Q.
*
The spintensor is a spinor with the tensor subscripts,
The spintensor space is reducible. In its minimal invariant subspace, the spintensor has to satisfy the usual traceless condition and the second traceless condition:
and is projected by the Young operator. The Young pattern for the Young operator has the row number not larger than N / 2 , otherwise the subspace corresponds to the null space. The representation corresponding to the minimal invariant subspace is called the irreducible spinortensor representation or simply the spinor representation, which are described by a Young diagram with a symbol s: [s,X] = [ s , X ~ , X ~ ,    for ] an odd N or [fs,A] = [ f s ,XI, X2, .] for an even N . The spinor representation is doublevalued one of SO ( N ).

*
The dimension of a spinor representation [s,A] of S 0 ( 2 l + 1) or [fs,A] of SO(2l) can be calculated by the hook rule. In this rule the dimension
406
Problems and Solutions in Group Theory
is expressed as a quotient, where the numerator and the denominator are denoted by the symbols Ygland YiA1,respectively:
(9.35)
We still use the concept of the hook path (i, j ) in the Young pattern [A], which enters the Young pattern at the rightmost of the ith row, goes leftwards in the i row, turns downwards at the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j ) is the same as the hook path (i, j ) except for the opposite direction. The number of boxes contained in the hook path (i, j ) is the hook number hij of the box in the j t h column of the ith row. YiA1is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the hook number hij. Define a series of the tableaux Yitlrecursively by the rule given below. Yklis a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Yg' in the series. The symbol Yplmeans the product of the filled digits in it, so does the symbol YiA1. The tableaux YE1are defined by the following rule: (a) Y t ' is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the digit ( N  1 + j  i).
(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recursively the Young pattern [A(")] by removing the first row and the first column of the Young pattern [A("')] until [A(")] contains less than two rows. (c) If [A(")] contains more than one row, define YE1to be a tableau of the Young pattern [A] where the boxes in the first ( a  1) row and in the first ( a  1) column are filled with 0, and the remaining part of the Young pattern is nothing but [A(")]. Let [A(")] be T row. Fill the first (T  1) boxes along the hook path (1, 1) of the Young pattern [A(")], beginning with the box on the rightmost, with the digits A?',
At),
Real Orthogonal Groups
407
   A?), , box by box, and fill the first A!)' boxes in each inverse path (i, 1) of the Young pattern [A(")], 2 5 i 5 r , with 1. The remaining boxes are filled with 0. If a few 1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern Y x l is zero. 8. Calculate the dimension of the irreducible spinor representation of the SO(7) group denoted by the following Young patterns:
Solution.
5431 321 678 5 $ 4
0 1 1 0 2
521 2 1
=8x
678 0 2 2 0 0 0 5 6 +I 0 $0 1 45 21 01 541 32 21
689 5 2 521 2 1
= 1728.
6 9 10 47
= 8~
L 541 } = 3024. 32 21
9. Calculate the highest weight of the fundamental spinor representations
of the S O ( N ) Solution. Fkom the second Lie theorem, the spinor operators Sub satisfy the same commutation relations as those of the generators Tab in the self representation of S O ( N ) . The Chevalley bases H,(S), E,(S) and F,(S) in
Problems and Solutions an Group Theory
408
the fundamental spinor representation can also be expressed by Eqs. (9.10) and (9.11)) where T a b are replaced with S a b . It is convenient to choose an explicit forms of the ya matrices for calculating the highest weight of the fundamental spinor representations, although the calculated result is independent of the chosen forms. Since the dimension of ya for the SO(2l) group is 2e, we may define the ya matrices of l?2e to be the direct product of l Pauli matrices such that Eq. (9.28) is satisfied:
 = 1x
y2P1
... x
1x01 x
03
x
x
03
x
P1
x
;x02
P1
yf =
03
x
03
,
03
,
1P
t x .;
=
y2P
... x
... x
(9.36)
eP
... x 0 3 . e
In fact, they are also ya matrices of r 2 0 + 1 with 7 2 e + 1 = yf. Thus, the Chevalley bases in the fundamental spinor representation of SO(2l + 1)) whose Lie algebra is Be, are
HP= 1x . . x 1xdiag(0, 1,  1, 0) x 1 x .. x 1 )
ePi
P1
E P = IX X ~
X ( O + X O  } X ~ X.
.xl
1
, " P1
ePi
F P = IX x ~ x { ~  x o + } x ~x xl , P1
He=
IX~ 1 x 0 3 ,
e
Ee =
(9.37)
ePi
03
x
1
. * *
e
x
0 3 XQ+
)
Fe =
03
x *..x
e
1
0 3 XCT
)
1
where 1 5 p < l . The Chevalley bases in the fundamental spinor representation of the SO(2l) group, whose Lie algebra is De,are the same as those of the S 0 ( 2 l + 1) group except for those when p = !, which are

He = 1x
x l x d i a g ( 1 , 0, 0, 1)
)
e2
Ee=
 l x
... x l x { o + x a + } ,
e2
Fe=  1~. . . x I x { a _ x a  } . e2
(9.38)
Real Orthogonal Groups
Let a and
p be two eigenfunctions of 03, a=(;), ff3a= a ,
asp = p,
409
respectively,
P=(;),
ff+a= 0,
a+p = a ,
f f  a = p, aP = 0.
Denote by x(m) the basis states with the weight m in the fundamental spinor representation [s]of S 0 ( 2 & +1) and by x*(m) the basis states with the weight m in the fundamental spinor representation [ks] of SO(2l). From the condition that each raising operator annihilates the highest weight state, we obtain the basis state x(M) for the highest weight for [s] of S0(2k‘+ 1) and the basis state x,t(M) for the highest weight for [ks]of SO(2C) as follows:
x(M) =
CXX.XQI
1’
e
M = (0, 0, ..., O , l ) , x+(M) =
C Y X .  . X ~
’ e M = (0, 0, ..., O , l ) , x(M) =
for [s] of SO(2&+I),

(9.39) for [+s] of S0(2&),
C Y X .  . X C X X ~ ,
e
1
M = (0, 0, ..., l,O),
for [s] of SO(2l).
10. Calculate the highest weight of the spinor representation [s,A] of
S0(2&+1),and the highest weights of the spinor representations [fs,A] of SO(2t) Solution. The spintensor of S O ( N ) corresponds to the irreducible representation [s,A] when N = 2 & + l or to [ks,A] when N = 2&if the spintensor satisfy the usual traceless condition and the second traceless condition, given in Eq. (9.34). The generators in the irreducible spinor representation are the total angular momentum operators Jab = L a b Sub,a < b. The orbital angular momentum operators Lab and the spinor angular momentum operators S,b are respectively applied to the tensor part and the spinor part of the spintensor. The total angular momentum operators Jab satisfy the same commutation relations as those of L a b and Sab. The Chevalley bases H,(J), E,(J) and F,(J) in the irreducible representation [s,A] can also be expressed by Eqs. (9.10) and (9.11), where Tubare replaced with
+
410
Problems and Solutions in Group Theory
Jab. Therefore, the weight of a spintensor basis is the sum of the weights of the tensor part and the spinor part. Let [A] = [A,, Aa, . . . , A,]. The highest weight M of the spinor representation [s,A] of S0(2e+ 1) is
The highest weight M of the spinor representation [+s,A] of SO(2e) is
The highest weight M of the spinor representation [s,A] of SO(2e) is
11. Calculate the ClebschGordan series for the direct product of the tensor representation [A] and the fundamental spinor representation [s] of S0(2C+ 1) or [fs] of SO(24, where [A] is a single row Young pattern or a single column Young pattern.
Solution. If the tensor part of a spintensor qal...,,,belongs to the tensor representation [A] but it does not satisfy the second traceless condition, namely, its tensor part satisfies the traceless condition for each pair of the subscripts ai and aj and is projected by a Young operator, then, the spintensor space corresponds to the direct product of the tensor representation (A] and the fundamental spinor representation. The method for reducing the direct product representation is to decompose the spintensor space into a series of subspaces satisfying the second traceless condition (9.34). In fact, xb TbQal...b...a,, is a new spintensor where the tensor rank decreases by one. If [A] is a single row Young pattern, the tensor part of the spintensor is a totally symmetric traceless tensor such that twice application of the Tb~c*al...b...c...a, = 0 due to Eq. second traceless condition is trivial, (9.28). If [A] is a single column Young pattern, the tensor part of the spintensor is a totally antisymmetric tensor. Note that the antisymmetric tensor space of rank n of S O ( N ) is equivalent to that of rank N  n. Thus, the reductions of the direct product representations for SO(2.t 1) are
zbc
+
[A,O,.
..,O]
[PIx
[s] 21 [s, I”] CEI [s, l n   l ]
x [s] 2 [s,A,O,.
and those for SO(2l) are
. ,O] c3 [s, (A *
CB .. .
[s],
 l),O,
* * *
n 5 .t
,O],
,
(9.40)
411
Real Orthogonal Groups
[A)O,.
. . ,O]
x,o, . . . ,O] a3 [ys,(A  l ) , O , . . . ,O],
x [h] N [As,
where P* = (1 f y f ) / 2 . The sign in f s changes because P k y , = Those formulas can be shown with the method of the dominant weight diagram. The reader is encouraged to check them for the groups SO(7) and SO(8). Equations (9.40) and (9.41) can be partly checked by their dimensions. From Eqs. (9.18) and (9.35), the dimensions of [A] and [s,X] of S O ( 2 t + l ) ,for example, are ( 2 t ) ( 2 t + 1).. . ( 2 t + x  2 ) ( 2 t + 2 x  1) qx,o, ...,01 = A! ( 2 t + x 2 ) ! ( 2 t + 2 x  1) > (21  l)!A! ( 2 l f l)! ' when n 5 t , d ~ l n=~ ( 2 t  n 1 ) ! n !7 (2!+ (2!)(2t 1).. . ( 2 t  1) =2 +,x,o ,...,01 = 2' A!
+
+
'
+x
x  l)!
Then, for Eq. (9.40) we have 2e
(2t+
x  2 ) ! ( 2 t + 2 x  1) = 2 '( 2 t + A  l)!
+
( 2 t  l)!X! ( 2 t  l)!X! ( 2 t + l)! ( 2 t + 1 ) ! ( 2 t  2a + 2 ) 2e =2 e x ( 2 t  a 2)!a! (21 1  n)!n! a=O
+
+
2t
+
( 2 t A  2)! ( 2 t  1)!(A  l ) ! )
)
when n
5 t.
12. Calculate the basis states in the fundamental spinor representation [s]
of SO(7).
Solution. The Lie algebra of the SO(7) group is B3. The highest weight of the fundamental spinor representation [s]of SO(7) is (0, 0 , l ) . Its dimension is 8. Its block weight diagram is as follows. All nonvanishing matrix entries of the lowering operators FD are 1.
412
Problems and Solutions an Group Theory
By the notation used in Problem 9, we have
i , i ,1
x ( i 7 i 1) 7=
I JI1,i , i
x p x a,
x(i,o,i) = 
x
p x p,
x(o,i,i)=  p x p x a , ~ ( oo, i ) =  p x p x p. Fig. 9.3 T h e block weight diagram of [s] of B3.
13. Calculate the basis states in the fundamental spinor representations [AS]of SO(8).
Solution. The Lie algebra of SO(8) group is Dq.The highest weights of the fundamental spinor representations [+s] and [s] of SO(8) are (O,O, 0 , l ) and (0, 0,1,0), respectively. The dimensions of both representations are 8. Their block weight diagrams are as follows. All nonvanishing matrix entries of the lowering operators Fp are 1. By the notation used in Problem 9, we have
x a x p,
x + ( O , O , O , 1) = a x a x a x a ,
x(O,O,l,O)
x+(o,i,o,i) = 
x(o,i,i,o) = a x a x p x
p x p, x + ( i , i , i , o ) =  x p x x p, x + ( i , o , i , o ) =  p x a x x p, x
x
x+(i,o,i,o) =  a x p x p x a,
=a x
0,
x(i,i,o,i) = a x p x a x a , x(i,o,o,i) = p x x x a , x(i,o,o,i) =  a x p x p x P ,
x + ( i , i , i , o ) =  p ~ ~ ~ px ~( i ~, i , o , i ) =  p x c ~ x p x p ,
x+(o,i,o,i)=  p x p x a x a , X+(o,o,o,i) = p x p x p x p,
x(o,i,i,o) =  p x p x x(o,o,i,o) =  p x
x p,
p x p x a.
Real Orthogonal Groups
a) [+sl Fig. 9.4
b)
413
[Sl
The block weight diagrams of [fs]of Dq.
14. Expand the eigenfunction of the total angular momentum with the highest weight in terms of the product of the spherical harmonic function Y:’(EE) and the spinor basis x(m). Solution. In Problem 11 the ClebschGordan series for the direct product representation [X,O, . . . ,O] x [s] is given. For the SO(2t 1) group, we express the representations in Eq. (9.40) by their highest weights,
+
(A,O..
.,O)
x ( 0 . . . ) O , 1) 21 (A,O..
. ) O ) 1)@ (A  1 , o . . . , O ) 1).
Let (J) = (A,O.,.)0,1), (L)= ( A , O , . , , O ) , @ + I ) = ( A + 1 , 0 ...,O), and (S) = ( 0 . . . , O , 1). There are two ways to obtain the total angular momentum state ( J ) : ( L ) x (S) and ( L 1) x (S). The highest weight state can be calculated by the condition that it should be annihilated by each raising operator. The highest weight state for the first case is easy to calculate:
+
where C(z~+l),x is given in Eq. (9.23). The highest weight state for the second case is
414
Problems and Solutions in Group Theory
For the SO(2C) group, we express the representations in Eq. (9.41) by their highest weights,
(A, 0 . . . , O ) x ( 0 . . . ,o, 1) 2 (A, 0 . . . 0 , l ) @ (A  1)0 . . . ,o, 1)O), )
(A,O
* *
 ,O)
x (0.  ) O , 1 , O ) *
N
(A,O.. . , O ) 1,O) 63 (A  1 ) O . . . ,o, 1).
+
Let ( J + ) = (A$. . . , O , l),( J  ) = (A$. . . , O , 1,0), ( L ) = (A$. . . , O ) , ( L 1) = (A 1 , O . . . , O ) , (+S) = ( 0 . . . , 0 , l),and (S) = ( 0 . . . ,O,l,O). There are two ways to obtain the total angular momentum state (J*): ( L )x (fS) and ( L 1) x ( r S ) . It is easy to calculate the highest weight state for the first case:
+
+
where C ( 2 1 )is, ~given in Eq. (9.24). The highest weight state for the second case is
Real Orthogonal Groups
415
+ (z3+ i ~ &  [ ( i , i ,0 , . . . ,o, i)] + (21 + iz2)x[(i, 0, ,0, l)]}* * *
9.3
SO(4) Group and the Lorentz Group
*
The inequivalent irreducible representations of the Lorentz group can be obtained from those of SO(4) group. This is the general method for studying the inequivalent irreducible representations of a noncompact group.
*
Find new bases of generators of SO(4) from six generators Tab,
416
Problems and Solutions an Group Theory
The new bases are separated into two sets. Two generators in different sets are commutable. Express them in the form of the direct product of the Pauli matrices:
By the similarity transformation N , we have
10 0 i 0 0 0 1 1 0 ii
N =  1(
Jz
1 i 0 0
)

Therefore, any element R of SO(4) can be expressed in the direct product of two matrices
(
\
4
R = exp i
WubTab a
)
1
= expli
I
where 3
Thus, matrix R is explicitly expressed in the product u1 x 212 of twodimensional unimodular unitary matrices. If two matrices u1 and u 2 change their signs at the same time, R keeps invariant. Therefore, Eq. (9.43) gives
Real Orthogonal Groups
417
a onetotwo correspondence between an element R in SO(4) and two elements u1 and u2 in SU(2). Furthermore, this correspondence keeps invariant in the multiplication of the group elements. Thus,
SO(4)

SU(2) x SU(2)'.
(9.44)
We restrict the varied area of the parameters of the SO(4) group such that there is a onetoone correspondence between a set of parameters and a group element, at least in the region with nonzero measure: (9.45) where O(*) and v(*)are the polar angle and the azimuthal angle of the direction n(*). Since the group space of the second SU(2)' group is reduced to that similar to the group space of the SO(3) group, namely, the two ends of a diameter in the group space correspond to the same element, the group space of SO(4) is doublyconnected. Any irreducible representation of SO(4) can be expressed as the direct product of two irreducible representations of two SU(2) groups, denoted by Djk : Djk
= ~ (fi(+),J+)) . i
(fi(+),u(+);~(),u())
~k
(&),J)).
(9.46) The row (column) index of Djk is denoted by two indices ( p v ) . The dimension of D j k is ( 2 j + 1)(2k 1). Its generator I:"*) is expressed by the generator I: of Su(2):
+
(9.47)
Djk
with different superscripts jIc are inequivalent to each other. When
+ Ic) is an integer, D j k is a singlevalued representation of S0(4), i.e., the irreducible tensor representation. When ( j + k  1/2) is an integer, D j k is a (j
doublevalued representation of S0(4), i.e., the spinor representation. The correspondence between D j k and the irreducible representation denoted by
418
Problems and Solutions in Group Theory
+ k is an integer, we have
a Young pattern [X,, A,] is as follows. When j
Djj
when j = k ,
[ 2 j , 01,
Djk N [ + ( j Djk 21 [  ( j
+ k ) ,( j  k ) ] + k),(k  j ) ]
when j when j
+ k  1/2 is an integer, we have Djk 21 [+s, ( j + k  1/2), ( j  k  1/2)], Djk 21 [s, ( j + k  1/2), (k  j  1/2)],
> k, < k,
and when j
when j > k, when j
< k.
Note that [+X,,Xz] is the selfdual representation, and [X,,Xz] is the antiselfdual one. DOo is the identical representation, Dqo and Do* respectively are two fundamental spinor representations [+s] and [s]) and D f 3 is equivalent to the selfrepresentation.
*
The Lorentz transformation matrix A between two inertial frames is a
4 x 4 orthogonal matrix:
A*A = A A = ~ I.
(9.48)
The matrix entries of A satisfy the following condition: Aab and A44 are real Aa4
a and b = 1, 2, 3.
and A4a are imaginary
This condition keeps invariant in the product of two matrices A. The set of all such orthogonal matrices A, in the multiplication rule of matrices, constitutes the homogeneous Lorentz group, denoted by O(3,l) or Lh. The orthogonal condition (9.48) gives detA = f l ,
Ai4 = 1 +
3
IA,4I2 2 1. a=l
These two discontinuous constraints separate the group space of O ( 3 , l ) into four unconnected sections. The element in the section where the identity belongs to satisfies detA = 1,
A44 2 1.
(9.49)
Those elements satisfying Eq. (9.49) constitute a simple Lie subgroup, called the proper Lorentz group, denoted by L,. Since there is no upper limit of A44, the group space of L, is a n open area in the Euclidean space. Thus, L, is a noncompact Lie group. The representative elements in three
Real Orthogonal Grozlps
419
cosets of L, are usually chosen to be the space inversion sion r and the whole inversion p: o = diag(1,

the time inver
1,  1, 1),
r = diag(1, 1, 1,  1), p = diag(1,  1,  1,
rt Let A be an infinitesemal
CT,
 1).
element of L,:
A = I iaX,
1 = ATA = 1  ia ( X
+XT),
1 = d e t A = 1 i a T r X ,
AT = 1  i a X T , XT = X, Trx = 0.
Thus, X is a traceless antisymmetric matrix. Expand X with respect to the generators in the selfrepresentation of SO(4): 3
3
a
a=l
(9.50)
a=l
where
dab
is real,
wa4
is imaginary, and 3
(9.51) b
Except for the imaginary parameters, the generators in the selfrepresentations of SO(4) and L, are completely the same, so are the generators in the corresponding irreducible representations of two groups. The finitedimensional inequivalent irreducible representations of L, are also denoted by Djk(L,), whose generators are given in Eq. (9.47). Since the parameters for SO(4) and L, are different, the global properties of the two groups are very different. Any finitedimensional irreducible representation D j k of L,, except for the identical representation, is not unitary. There exist infinitedimensional unitary representations for L,.
*
The transformation generated by T a b is obviously a pure rotation, belonging to the subgroup SO(3). Let us study the transformation generated
420
by
Problems and Solutions in Group Theory
T34
with the parameter
w34
= iw:
A(G3,iw) rexp{i(iw)T34} =
("
0
0
0 0 coshw isinhw 0 0 isinhw coshw
w = ctanhw,
) (9.52)
coshw = (1  w2/c2)1'2,
sinh w = (w/c) (1  'u2/c2)1'2
.
It describes the Lorentz boost along the direction of x axis with the relative velocity w = c tanh w. Any Lorentz transformation A satisfying Eqs. (9.48) and (9.49) can be decomposed into the product of some rotational transformation and the Lorentz boost along the x direction. Let A 4 4 = coshw, from which 13 is determined. Extracting a factor i sinh w from Aa4, we obtain a unit vector n(8, cp) in the threedimensional space, whose rectangular coordinates are (i/ sinhw)(A14rA24,A34): = COShw, iA24/ sinh w = sin 8 sin cp, A44
iA14/sinhw = sinOcoscp,
(9.53)
iA34/ sinh w = cos 8.
Extending the rotational matrix into a 4 x 4 matrix, we have
R(g3 ,p)R(s2)0) =
coscpcos8  sincp coscpsin8 0 sincpcoso coscp sincpsin8 0 0
0
0
0
1
Leftmultiplying to A the inverse matrix of R(G3, cp)R(&,8)A(G3)i w ) ) we obtain a rotational matrix R(a,p, T),
from which the Euler angles a , P and y can be calculated. Thus,
The geometrical meaning of the decomposition is evident. Two rotations in twosides of A(&, i w ) rotate two z axes in two frames before and after the Lorentz transformation A to the direction of the relative motion, and the remaining axes to be parallel to each other, respectively. After two rotations, the Lorentz transformation A is simplified into A(&, '1'13).
Real Orthogonal Groups
42 1
*
Since all inequivalent irreducible representations of L, have been known, the inequivalent irreducible representations of the homogeneous Lorentz group O(3,l) can be obtained by determining the representation matrices of r and p. The representation matrix of o can be calculated by the formula o = ~ p .Since p is commutable with any element in O(3,l) and the p2 is equal to the identity, the representation matrix of p in an irreducible representation of O(3,l) is a constant matrix, where the constant is fl in a singlevalued representation. The property of the representation matrix of r can be determined by the relations in the selfrepresentation
Let V ( x ) X, = 1, 2, 3 and 4, be four inequivalent irreducible representations of the fourorder inversion group V4. When j = k , The irreducible .. representation D J J of L, induces four inequivalent irreducible representation Ajjx of O(3,l):
AjjX(A)= D j j ( A ) , A E L,, (7) V(’) (T)S,,~ 1 5 X 5 4. Ajjx(p) = V(’)(p)1,
(9.55)
SvcLj,
+
When j # k , while j k is an integer, the representation D j k @ Dkj of L, induces two inequivalent irreducible representation Ajk* of O(3,l):
where a and P are used to identify two representation spaces of D j k and D k j . Since Ajk*(r) has no diagonal entries, changing its sign leads to an equivalent representation. When j k is a halfodd number, we only discuss the Dirac spinor representation D(O(3,l)) here. For definiteness, the index p runs from 1 to 4, while the index a runs from 1 to 3. Let T~ be four anticommutable matrices and C be the chargeconjugate transformation matrix:
+
cly,c = y;,
C+C= 1,
CT = c,
d e t C = 1.
422
Problems and Solutions in Group Theory
Thus, D(A) satisfies 4
D(A)'ypD(A) =
Apuyu,
detD(A) = 1, (9.57)
u=l
ClD(A)C =
IA44 I
{D(Al)}*.
The additional factor A44/IA441 comes from the physical reason. Its generators are
The representation matrices of the representative elements in the cosets of L, are
D ( a ) = fiy4,
D ( 7 ) = fy4y5,
D ( p ) = fiy5.
(9.59)
15. Discuss the classes in the SO(4) group and calculate their characters
in the irreducible representation
Djk.
Solution. Any element R of SO(4) can be transformed into the standard form by a real orthogonal similarity transformation X:
where Eq. (9.43) is used. Therefore, the class of SO(4) is described by two parameters cpl and y2. Its character in the representation D j k ( R )is
Real Orthogonal Groups
423
16. Calculate six parameters of the following proper Lorentz transformation A, and write its representation matrix in the irreducible representation Djk ( A )of the proper Lorentz group L,:
A(cp7 "w 7
"
/1 0 0 0 \ 0 &/2 (cosh w ) / 2 i(sinh w ) / 2 = 0  1 / 2 &(cosh w ) / 2 i&(sinh w ) / 2 \O 0 isinhw coshw 1
Solution. From the fourth column of A(cp,8, w , a ,P,r),the boost parameter is w , and 6' = 7r/6 and cp = 7r/2 because sin 6' cos cp = 0, sin0 sin cp = 1 / 2 and cos 0 = &/2. The matrix form of the rotation R(7r/2,7r/6,0) is 0100 1 0 00 R(7r'2,n/6,0)=( 0 0 l o ) 0 0 01
/
(
fi/2 0
0 &/2
;/2 0
:
;)
1/2 0
0
&2 0
1/2
o\
R(n/2,7r/6,0)' =
Thus,
From this we obtain representation matrix of A in
and
Thus, we obtain the
424
Problems and Solutions in Group Theory
17. Prove that the Dirac spinor representation satisfy
Solution. The Dirac spinor representation is not unitary. The aim of this problem is to study the property of its conjugate representation. From the definition (9.57) of the Dirac spinor representation we have:
To make the formulas uniformly, we make a similarity transformation
74:
From the equivalent theorem of 7a, we obtain
Substituting Eq. (9.59) into it, we determine the constant c :
This is the reason why the matrix 7 4 should be inserted between two spinors $t and 1(, to construct an invariant quantity of the Lorentz transformation in the quantum field theory:
18. Discuss the classes of the proper Lorentz group L,.
Solution. In the selfrepresentation of L, we write the pure rotation R ( MP, , 7 ) and the Lorentz boost A(&, i w ) with the relative velocity along the z axis in terms of Eq. (9.43)
Real Orthogonal Groups
425
Thus, any element A(cp,8, w , a , P, r) in L, can be expressed as
where
The matrix M is a twodimensional matrix with determinant +1, which belongs to the twodimensional unimodular complex matrix group SL(2 ,C). We are going to show the homomorphism between SL(2,C) and L, with a 2:l correspondence. First, we have known from Eq. (9.60) that an arbitrary Lorentz transformation A = A(cp,8,w , a , P, 7) corresponds to a matrix M E SL(2,C). Conversely, any M given in Eq. (9.61) determines a Lorentz transformation A. We will show that any element P in SL(2,C) can be expressed in the form (9.61) later. If two matrices M and M' correspond to the same Lorentz transformation A, then
Thus, M  l M ' = c l . Since detM = 1 and detM' = 1, c2 = 1 and c = f l . There is a 2:l correspondence between fM given in Eq. (9.61) and the Lorentz transformation A. This correspondence keeps invariant in the multiplication of elements. In fact, if
we have
AA' = N { M M ' x
[ 0 2 ( M M ' ) * ~ 2N ] } l .
Now, we have proved that the twodimensional unimodular complex matrix group SL(2,C) is homomorphic onto the proper Lorentz group L, with a 2:l correspondence. Thus, we can determine the class of L, by SL(2,C). Recall that two matrices fM eSL(2,C) correspond to the same Lorentz transformation A E L,. Two classes in SL(2,C) different by a sign correspond to the same class in L,. If two eigenvalues of M eSL(2,C) are different, M can be diagonalized through a similarity transformation X eSL(2,C)
426
Problems and Solutions in Group Theory
Thus, we have
A=
(
0 coscp  s i n 9 sincp coscp 0 0 0 coshw 0 0 isinhw coshw
n
I cp I
r 1
The class is described by two parameters w and cp. If two eigenvalues of M are equal, the eigenvalues have to be fl.fl correspond to the identity element of L,, which constitutes one class in L,. If M is not equal to fl, M can be transformed into the Jordan form by a similarity transformation Y eSL(2,C) (see Problem 16 in Chap. 1):
Y%Y=f(,
1>.
12
Since
we obtain 01 = 22, 0 2 = 2, 0 3 = 0. In order to calculate the six parameters of A , we make a transformation
The solutions are cp = and
7r,
1
6' = r/4, coshu = 3, a = T , p = 37r/4, y = 0, 0
2  ei2T31 2Tl4
2 2i
0 1 2i 0  2i 3
It is easy to check that A is a proper Lorentz transformation with all four eigenvalues to be 1. A has only two linearly independent eigenvectors for the eigenvalue 1. A is transformed into the Jordan form by the similarity
Real Orthogonal Groups
427
transformation 2:
0 0 21 040 04io
1 0
ZlAZ=
0 0
1 0
O
O
( ) 1 0 0 0
0 1 0 0
0 1 1 0
0 0 1 1
0 0 0 i/4 l
i
*
Finally, we are going to show that, any twodimensional unimodual complex matrix P ESL(2,C) can be expressed into the form of Eq. (9.61), namely
Solving the equation, we have
where c,g = cos(O/2), = sin(O/2) and cp = cos(P/2) and so on. The direct calculation leads to
+ c$cgew+ sis$e*, IB12 = (sin8 sinp cosa) /2 + cis$ew + sic$ew, ICI2 = (sin8 s i n p cosa) /2 + s;4cgew+ czsge", 1OI2 =  (sin 8 sin p cos a ) /2 + sis$ew + cic$ew, AB = eiCP {  cos p sin 8 + sin p (cieweia + siew ia> > / 2 , AC = eiy {case s i n p + sin8 (c$eweia  s$eweia)} /2, IAI2 =  (sin 8 sin p cos a ) /2
AD = s i s $ +tic$  secespcp (eweia  e+eia),
+
[A12 IBI2
+ [C12+
+
= ew ew,
428
Problems and Solutions in Group Theory
p can be respectively calculated from the last three formulas. Then, a is determined by the formula for AD. At last, y and 9 can be calculated from the formulas for AC and AB. Here, the calculated method is not unique due to d e t P = 1. This completes the proof. For example, we calculate the parameters of a matrix P :
w , 8 and
P = 1 ( 4
[a 1 i (a+ l)] a+ 1 i 3 (a+ 1) a+1 i 3 ( A 1) d [ A +1 + i ( A I ) ]
4
coshw = 3,


sinhw =
a,
ew/2=
case = 112, e = 7r/3, Then, from A D = 3(1  i&)/S,
Jz+ 1,
C O S = ~
1/2,
ew/2 =
Jz 1
p = 21~/3.
we obtain
+ eweiff = 6 c o s a  i 4 a s i n a = The solution is a = 7r/2. Finally, from
AC = a [  1  2i(3  &)]  eiy 8
we have ezv = i and last, we have
eeiY
= 1, and determine cp = 7r/2 and y = 7r. At
19. Express an arbitrary element in the proper Lorentz group L, in the form of exponential matrix function.
Solution. In Problem 18, we have shown that L,

SL(2,C) with the
Real Orthogonal Groups
429
correspondence (9.60):
An arbitrary element A in L, can be written in the form of exponential matrix function if its corresponding elements f M in SL(2,C) can. Since detM = 1, its two eigenvalues have the same sign. If its eigenvalues are positive, M can be transformed into the form of exponential matrix function by a unimodular similarity transformation. If its eigenvalues are negative,  M can be transformed into the form of exponential matrix function by a unimodular similarity transformation. Ignoring the irrelevant sign, we can express an arbitrary element M in SL(2, C) as
M
= fexp ( 4 6  $12)
,
02
M*u2 = fexp (46*. $12)
.
By Eq. (9.60), the proper Lorentz transformation A corresponding to can be expressed as
M
a=l
a
where Ti*), Tab and Ta4 are the generators in the selfrepresentation of S0(4), and the parameters Wab and wc4 are the real part and the imaginary part of O,, respectively: 3
1 Wab = 2 c=
&bc 1
(0,
+ n:),
Wc4
1 2i
=  (a,  n:).
430
Problems and Solutions in Group Theory
Thus, an arbitrary Lorentz transformation A can be expressed in the form of exponential matrix function, where the parameters u a b are real, and ma4 are imaginary. The set of parameters up,is mainly for the theoretical study, and the set of parameters (cp,8, u,a , P, 7 )is convenient for calculation. The relation between two sets of parameters are given by Eq. (9.60), namely
20. Let 3
x = ix41+
C
xaga
23
=
x1
a=l
 ix4
 ixa
+ ix2 23  ix4 ) ’ x1
where xa is real, 24 = ict is imaginary, and X is a hermitian matrix. Let M E S L ( ~ , C ) , 3
M X M ~= X I =  i x & l +
C
XLua.
a=l
(a) Calculate Tr X and det X . (b) Prove xh = C , Ap,x,, where A (c) Prove that L, SL(2,C). is a proper Lorentz transformation. (d) Calculate the corresponding matrix M , when A is equal to B(G3,cp), R(&,8), and A ( & ,i w ) , respectively. (e) Calculate the matrix M corresponding to an arbitrary proper Lorentz transformation 4% 6,u,a , P, 7 ) . 4
Solution. (a) TrX = i2x4,
det X = 
x:. p=l
(b) Since XI = M X M t is hermitian, it can be expanded with respect to the Pauli matrices and 1, where the coefficients xb, are real, and x i is imaginary. Further, x; can be expressed as the linear combinations of x p , 4 u=l
where A a b and det X’ = det X ,
A44
are real, and
A4a
and
4
4
p=l
p=l
Aa4
are imaginary. Due to
43 1
Real Orthogonal Groups
Therefore, A is a Lorentz transformation matrix, which is an orthogonal matrix with the matrix entries depending upon M . When M = 1, A is the identity. Since the group space of SL(2,C) is connected, the set of A can be continuously extended from the identity, namely, A is a proper Lorentz transformation matrix. (c) From a given matrix M , we can calculate X' = MXMt, and then, determine a proper Lorentz transformation matrix A uniquely. If the matrices X ' calculated from M and M' are the same, so are the matrices A , then MlM' is commutable with three Pauli matrices ua. Thus, it is a constant matrix. Due to detM=detM' = 1, M' = & M . Namely, there is a 2:l correspondence between & M and A . This mapping is obviously invariant in the multiplication of elements. Therefore, SL(2,C) is homomorphic onto L,, L, SL(2,C). (d) Ignoring the sign in front of the matrix M , we can express the matrix M and A by the same parameters. If A ESO(3) is a pure rotation, M obviously belongs to SU(2),

M(&, cp) = u(e'3, cp) = 1cos (cp/2)  i 0 3 sin (cp/2), M(e'2,O) = u(e'2, 8) = 1cos (8/2)  iaasin (e/2),
For a boost A = A(&,iw), we have M(&,iw)XM(e'3,iw)t = il(ix3 sinhw + 2 4 coshw)
+ 03(x3 coshw  ix4 sinhw) + 01x1 + 02x2.
If only one component xp is nonvanishing, we have
M(G3, iU)OlM(e'3, i w ) t = 01,
x1 = 1,
M(e'3, iw)a2hf(e'3,iw)t = 0 2 , Ad(&, iw)a3M(Z3, iw)t = 1 sinhw
x2 = 1,
M(G3,iw)M(G3,iw)t= 1 coshw
+ 0 3 coshw,
+ 0 3 sinhw,
23
= 1,
24
= i.
Let M(e'3,iw) = 1Mo+alM1 +a2M2+a3M3. Extracting the Pauli matrix from the lefthand side of the first two formulas, and then, moving it to
432
Problems and Solutions an Group Theory
their righthand side, we have
(1Mo + a l M 1  02M2 (1Mo  a& + a2M2
 03M3)
M t = 1,
 a3M3) M += 1.
Adding and subtracting two equations, we obtain MI= M2 = 0, IMo12 lM3I2 = 1, and MOM,* are real. Thus, the last two formulas become the same, and turn to lMOl2 lM3I2 = coshw and 2MoM,* = sinhw. Recall that det M = 1. The solution is MO = coshw/2 and M3 = sinhw/2, namely
+
M(e"3,iw)=
( eow/2) ewoi2
= u(Z3,iu).
(e) For an arbitrary proper Lorentz transformation A(p, 8, u,a, p,?), we have
Chapter 10
THE SYMPLECTIC GROUPS
10.1 The Groups Sp(24 R ) and Usp(21) j r Take the subscript

p, 1 5 p
a of a vector in a (2l)dimensional space to be p or
5 i?, in the following order: a=l,i,2,2,
,e,Z.
(10.1)
Define a (2l)dimensional antisymmetric matrix J : when a = p, b = p when a = p , b = p otherwise, det J = 1 . J = l c x (ia2) = Jl = JT, Jab
=
{
1 1 0
(10.2)
The set of all (2t) x ( 2 4 real matrices R satisfying R T J R = J,
R* = R,
(10.3)
in the multiplication rule of matrices, constitutes the real symplectic group Sp(2l, R ) . Both the inverse R' and the transpose RT of any element R of Sp(2l, R) satisfy Eq. (10.3): R1
= JRTJ,
( ~  1 )J R~ 1
= J,
J (  J R T J ) ~ J (JRTJ) J = RJRT = J.
(10.4)
Sp(2[, R) is not a compact Lie group, because the general form for a diagonal matrix & E Sp(2l, R) contains the parameters up without an upper limit:
& = diag {ewl,
ewl,
ew2, e  W 2 , 433
. . . ,e W t , e  w t } .
(10.5)
434
*
Problems and Solutions in Group Theory
The set of all ( 2 4 x ( 2 4 unitary matrices u satisfying
'
uT J u = J,
ut = u ,
(10.6)
in the multiplication rule of matrices, constitutes the unitary symplectic group USp(2.t). Similarly, both the inverse ul and the transpose uT of any element u of USp(2l) satisfy Eq. (10.6). Since any matrix entry of a unitary matrix is finite, USp(2.t) is a compact Lie group. As an example, the general form of a diagonal matrix uo EUSp(2.t) is:
It can be shown that the determinant of R ~ S p ( 2 . t , R )or u ~ U S p ( 2 t )is +l. The group spaces of both Sp(2.t, R) and USp(2.t) are connected.
*
For the infinitesimal elements R ~ S p ( 2 [R , ) and u E U S P ( ~ ~ ? ) ,
R = 1iaX, u = 1  ipY,
XT =JXJ, YT = J Y J ,
X* = X, Y t = Y.
(10.8)
An imaginary matrix X of 2.t dimensions contains 4.t2 real parameters, so does a hermitian matrix Y . The component form of the condition XT = J X J is
which gives .t2 + .t(.t  1) real constraints. The first equation gives T r X = 0, which means detR = 1. The component form of Y T = J Y J is
yV P =yP"
7
y" P =yP"'
Due to Y t = Y , the first equation contains .t2 real constraints, including T r y = 0, and the second equation contains l ( 1  1) complex constraints. Therefore, there are l ( 2 t 1) real parameters both for R and for u . The order of Sp(2t, R) and the order of USp(2.t) are both t(2.t+ 1).
+
*
We choose the generators TA in the selfrepresentation of USp(2l)
where 7':;)
and T::) are the generators in the selfrepresentation of the
su(1)group, and
(Ti:))
bd
= &d&b.
The generators TA are hermitian and
The Syrnplectic Groups
435
normalized:
Thus, the structure constants of USp(2l) are totally antisymmetric with respect to all three indices. The diagonal generators span the Cartan subalgebra of USp(2E):
H p = Ti;) x
03/&,
15
/L
(10.11)
5 l?.
Their simultaneous eigenvectors in the Lie algebra, [H,, E,] = apE,, are
where a < b and the eigenvalues (roots) are and
a
< b, respectively. The simple roots are
The first ( l  1) simple roots are shorter, d, = (rp  rp)/2 = 1/2, and the last one is longer, de = (re re) /2 = 1. The remaining positive roots are a
b1
/A=,
e
b 1
e
p=,
p=b
1
1
Thus, the Lie algebra of USp(2t) is Ce. The largest root, which is the highest weight of the adjoint representation of Ce,is ei
W‘ = h e 1 = 2
C rp +re = DL=1
2 ~ 1 .
(10.13)
436
Problems and Solutions an Group Theory
* The Chevalley bases of USp(2l) are
(10.14)
*
The generators in the selfrepresentation of the Sp(24 R ) group are pure imaginary, X * =  X . The explicit forms of the generators can be obtained from Eq. (10.9) by replacing 0 a with T a : r1 = iu1,
72
=02,
73
= ia3.
(10.15)
Namely, some generators change by a factor i so that the Sp(2!, R ) group is a noncompact Lie group. USp(2l) and Sp(24 R) have the same complex Lie algebra, but different real Lie algebras. In another viewpoint, one may change the relevant parameters of Sp(2l, R ) to be imaginary such that the generators in the corresponding irreducible representations of both USp(2l) and Sp(2e,R) are the same. 1. Prove that the determinant of R in Sp(2l, R) and the determinant of u in USp(2l) are both +l.
Solution. In the following proof for the determinant of R in Sp(2t, R ) ,we did not use the conjugate operation so that the proof is also effective for u in USp(2t). Let R ~ S p ( 2 t , R be ) the transformation matrix for a vector x a in the 21dimensional real space:
Define a pseudoproduct for two real vectors x and y :
e
=
c
(X,Y,
X D P )
*
(10.17)
p=1
Obviously, the pseudoproduct is invariant in the transformation R: = {Rx, RY1,
*
(10.18)
The Symplectic Groups
437
The selfpseudoproduct of one vector is vanishing, {x, x } = ~ 0. For the totally antisymmetric tensor of rank 2t, Eala..aze, we can prove the following identity by Eq. (10.2)
In fact, the nonvanishing terms in the sum come from the permutations among J's ( l !terms) and the transpositions between two subscripts of any J (2' terms), and all are equal to one. Moving the column index of a 2G dimensional matrix X to be a superscript, we rewrite the determinant of X as
Permutating the order of X:, and changing the summing indices to restore the order of the superscripts, we obtain
Thus, we have det X = (2'l!) = (2el!)I a1
...a 2 ~
Since the righthand side is invariant in the transformation we have det (RX) = det X,
det R = 1 .
In the same reason, det u = 1. Both Sp(2l, R) and USp(2l) are the simplyconnected Lie groups. 2. Count the number of the independent real parameters of R in Sp(2l, R ) and u in USp(21) directly from their definitions (10.3) and (10.6).
Solution. A (2l)dimensional real matrix contains (2t)2 real parameters. Equations (10.3) and (10.17) show that the pseudoproduct of two column
438
Problems and Solutions in Group Theory
matrices of R ESp(2l,R) is equal to
Jab:
It is an identity when a = b. It gives l ( 2 l  1) independent real constraints when a # b, so that the order of Sp(2l, R) is (21)2  l ( 2 l  1) = l ( 2 l + 1). A (2l)dimensional complex matrix contains 2( 21)2 real parameters. From ul = J u T J , we have u * =  J u J ,
Each gives 212 independent real constraints. From the unitary condition Ed U;,Udb = S a b , we have the following constraints. The constraints for a = b = p are the same as those for a = b = 1.. There are l real independent constraints : e V=l
c
e v= I
When p # u, the constraints for a = p and b = u are the same as those for a = V , b = j!Z as well as those for exchanging a and b. There are l ( l  1)/2 complex independent constraints:
When p # u, the constraints for a = p, b = i7 are the same as those for a = u , b = as well as those for exchanging a and b. There are l ( l  1)/2 complex independent constraints:
The Symplectic Groups
439
One complex constraint is equivalent to two real constraints. Altogether there are ( 2 t ) 2 e+ 2t(!  1) = 6e2  i! real constraints. The number of the independent real parameters of u in USp(2l) is 812  (612  l ) = l(2C l), which is the order of the USp(C) group.
+
+
3. Express the simple roots of USp(2C) by the vectors V, given in Eq. (8.3) for the SU(e 1) group, and then, write their CartanWeyl bases of generators in the selfrepresentation of USp(2l)
+
Solution. From the CartanWeyl bases (10.11)) we have
If we choose another set of the normalized bases HL in the Cartan subalgebra:
p=
H: = Ti!:+, x
1
03
(10.20) er+1
HeT+2,
the component (rL)uof the simple root can be calculated from the formula Hh by replacing H p with (dpp  d p ( p + l l ) when p < e and &dpe when p = e.
440
Problems and Solutions in Group Theory
Namely,
(10.21) The remaining positive roots are b 1
p=a
10.2
Irreducible Representations of Sp(2C)
*
Since the generators in the corresponding irreducible representations of USp(2e) and Sp(2l, R ) ,as well as the orthonormal bases in the representation space are the same, we will only discuss the irreducible representations of Sp(2l, R ) . In the following, Sp(2l) stands for both USp(2t) and Sp(24 R) for simplicity.
*
Let R E Sp(2l,R) be the coordinate transformation in a real (2t)dimensional space: (10.22) From Eq. (10.22) we can define the tensors for Sp(2t, R ) transformations:
Similar to the tensors of SU(N) and S O ( N ) ,the Weyl reciprocity holds for the tensors of Sp(2l,R). The tensor space can be reduced by the Young operators. In addition, there are two invariant tensors for Sp(2l,R). One is the antisymmetric tensor Jab of rank two
The Symplectic Groups
The contraction two:
cab
JabTabc...
441
decreases the rank of the tensor
Tabc ...
by
ab ab a'b'c'
=
C C' ...
...
Rot..
.
(z
Ja'b'Ta'b'c'
)
...
*
(10.25) Sometimes, this operation is also called the trace. The trace subspace is invariant in Sp(2&,R). In order to reduce the tensor space, we have to decompose the tensor space into the direct sum of a series of traceless tensor subspaces with decreasing rank two by two. Applying the Young operator yp 1XI to a traceless tensor space 7 ,we obtain the minimal tensor subspace
yrl7, corresponding to an irreducible representation [A] of Sp(2l,
R).
The other invariant tensor of Sp(2l, R) is the totally antisymmetric tensor € al . . . a z t of rank 2&. However, this tensor violates the new traceless condition. It is irrelevant to the reduction of the tensor space for Sp(24 R).
*
The traceless tensor subspace denoted by a Young pattern [A] with the row number larger than & is empty. The irreducible representation of the Sp(2l,R) group is described by a Young pattern with at most l rows. The standard tensor Young tableau is still the common eigenstate of the Chevalley bases H p . The problem is that it is not necessary traceless. Fortunately, the contraction occurs only when a pair subscripts p and F appears. It is convenient to rearrange the bases @ a in the vector space: @,
= @,,
@e+p = @,
15 P
5 e.
(10.26)
From Eq. (10.14) the nonvanishing matrix entries of the Chevalley bases in the bases @ a are
(10.27)
In the bases
@a,
1
5 a 5 2&, the standard tensor Young tableau
Problems and Solutions in Group Theory
442
Y [XI , @ a l . . , a n in the representation [A] of Sp(2l, R ) corresponds to the high
<
est weight if each box in its j t h row, j L, is filled with the digit j. The standard tensor Young tableau with the highest weight is a traceless tensor because all subscripts take the values less than l 1, but the trace occurs only when a pair of subscripts p and 2 l  p + 1 appears. The relation between the highest weight M and the Young pattern [A] for Sp(2l,R) is
+
The remaining traceless tensor bases in an irreducible representation space [A] of Sp(2L,R) can be calculated from the basis with the highest weight by the lowering operators F,.
*
The dimension of an irreducible representation [A] of Sp(2l) can be calculated by the hook rule. In this rule, the dimension is expressed as a quotient, where the numerator and the denominator are denoted by the symbols Y p l and YiA1,respectively: (10.29) We still use the concept of the hook path (i, j ) in the Young pattern [A], which enters the Young pattern at the rightmost of the ith row, goes leftwards in the i row, turns downwards a t the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j) is the same path as the hook path (i, j) except for the opposite direction. The number of boxes contained in the hook path (i, j) is the hook number hij of the box in the j t h column of the ith row. YiA1is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the hook number hij. Define a series of the tableaux Yktl recursively by the rule given below. YF1is a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Y g l in the series. The symbol Ypl means the product of the filled digits in it, so does the symbol YiA1. The tableaux Y g l are defined by the following rule:
(a) Y g l is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the digit (2& j  i).
+
443
The Symplectic Groups
(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recursively the Young pattern [A(")] by removing the first row and the first column of the Young pattern [A("')] until [A(")] contains less than two rows. (c) If [A(")] contains more than one row, define YE1to be a tableau of the Young pattern [A] where the boxes in the first (a  1) rows and in the first (a  1) columns are filled with 0, and the remaining part of the Young pattern is nothing but [A(")]. Let [A(")] have r rows. Fill the first ( r  1) boxes along the hook path (1, 1) of the Young pattern [A(")], beginning with the box on the rightmost, with the digits A?),   ., A?), box by box, and fill the first A:") boxes in each inverse path (i, 1) of the Young pattern [A(")], 2 5 i 5 r , with 1. The remaining boxes are filled with 0. If a few 1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern YAtl is zero. 4. Calculate the dimensions of the irreducible representations of the Sp(6) group denoted by the following Young patterns:
At),
(1) [4,2],
(2) [ 3 J l ,
(3) [4,417
Solution. 0
6789
(1):
d[4,2](SP(6))
=
0
0 2
} { %} 67811
=
t 5 41 2 1 1
56
( 5 ) [4,4, 31.
(4) [3,3,21,
=924.
21 678
(2):
(3):
(4):
0 0 2 1  1
+
d[4,4](SP(6))
=
d[3,3,2](SP(6))
=
6 7 10
6789 0 0 0 4 5 6 7 8 + 514 31 2 1  1 } = {
67813
4321
4321
678 567 45
+
0 2 3 11 0 2  1 542 431 21
+
45 54 63 27 ) = 1274.
0 0 0 0 0 1 01
6 9 11 45 8 542 431 21
444
Problems and Solutions in Group Theory
(5):
d[4,4,3](SP(6)) =
6789 0 0 3 4 0 0 + 0 5678+111 456 211
0 0
6 7 11 13 4 5 6 10 23 4
0 0 0 2
6542 5431 321
6542 5431 321
= 2002.
5 . Calculate the orthonormal bases in the irreducible representation [1,1,0]
of the Sp(6) group by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank two for Sp(6).
Solution. The Lie algebra of Sp(6) is C3. The relations between the simple roots rp and the fundamental dominant weights wp are rl = 2wl  w2,
w1 = rl
1 + + r3, 2 12
r2
= w1+ 2w2  w3,
w2 = rl
+ 2r2 + r3,
r3
= 2w2
w3 = rl
+ 2w3,
3 + 2r2 + r3. 2
The highest weight of the representation [l,1,0] of Sp(6) is M = (0, 1 , O ) . The dimension of [l,1,0] is d[l,l,o~ = (7  4)/(2  1) = 14. The weights equivalent to the highest weight are
Therefore, the representation [1,1,01 contains one simple dominant weight (0, 1 , O ) and one double dominant weight (O,O,O). From the highest weight (0,1,0) we have a doublet of d 2 with (1, T, 1). From (1,i,1) we have a doublet of A1 with ( i , O , 1) and a doublet of d 3 with ( 1 , l ) i ) . From ( l , l , i )we have a doublet of .A1 with ( i , 2 , i ) and a doublet of d2with ( 2 , i , 0). From @,O, 1) we have a doublet of ds with (i, 2, i).Although (i, 2, i)comes from two paths, it is still a single weight because it is equivalent to (0, 1 , O ) . Now, we meet the double weight (O,O, 0) by applying Fl to 1(2,i,O)) and applying F 2 to l ( i , 2 , i ) ) . Letting 1(2,i,O)), I(O,O, O),) and 1(2,1,0))constitute a triplet of d1,and the other basis state I(O,O, 0)2) is a singlet of d1,we have
The Symplectic Groups
445
where we neglect the first index (0, 1,0), which denotes the representation, in the basis state I ( O , l , 0), ( m l ,m2, m3)) for simplicity. Applying ElF2 = F2El to the basis state I ( % , 2, i)),we have
0.
Choosing the phase of the state basis (0,0,0)2)such Thus, u = that b is real positive, we have b = = Applying E2F2 = F2E2 H2 to 1(0,0,0)1)we have
+
d
m
m.
Choosing the phase of the basis state 1(1,2,1))such that c is real positive, we have c = and d = The remaining basis states and the matrix entries of the lowering operators F, can be calculated similarly. The results are given in Fig. 10.1, where only those matrix entries of F, which are not equal to 1 are indicated.
m.
II Fig. 10.1
The block weight diagram and the state bases for [ l ,1,0] of Sp(6).
446
Problems and Solutions in Group Theory
Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau
y[ljljO] is
. A standard tensor
Young tableau is expanded as follows:
There is only one trace tensor in the tensor space of rank two projected by y[ljl*O]. It belongs to the representation [O,O,O] of Sp(6):
Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (10.27) and the block weight diagram. Due to the WignerEckart theorem, the basis state belonging to the representation [1,1,0] of Sp(6) is orthogonal to the basis state belonging to [O,O,O], so it is a traceless tensor.
m{B 1 
+2
44 7
The Symplectic Groups
U
l(o,i,o)) = p21(i,i , i ) )= 6. Calculate the orthonormal bases in the irreducible representation [I,1,1]
of the Sp(6) group by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank two for Sp(6). Solution. The Lie algebra of Sp(6) is C3. The relations between the simple roots rp and the fundamental dominant weights wp was given in Problem 5. The dimension of [l,1,1]is
131
= 14.
The highest weight of the representation [l,1,1]of Sp(6) is M = (O,O, 1). The weights equivalent to the dominant weight (O,O, 1) are
From the highest weight state (0, 0 , l ) we have a doublet of d 3 with (0,2, i), From ( 0 , 2 , i ) we have a triplet of d 2 with (1,0,0) and (2,2,1). Both the matrix entries of F2are A. The weight (1,0,0) is a single dominant weight. The weights equivalent to the dominant weight (1,0,0) are
448
Problems and Solutions in Group Theory
Therefore, all weights in the representation [I,1,1]are single. It is easy to draw the block weight diagram for the representation [l,1,1]. In Fig. 10.2 we give the block weight diagram for the representation [l,1,1] where only those matrix entries of Fp which are not equal to 1 are indicated.
11,090
I
I
IIJZ
Fig. 10.2 The block weight diagram and the state bases for [l,1,1] of Sp(6).
Now, we expand the orthonormal basis states in the standard tensor . Young tableaux. The Young tableau y[lilsllis
. The standard tensor
Young tableau is
The trace tensors in the tensor space of rank three projected by Y['v1i1] belong to the representation [l,0, 01 of Sp(6):
The Symplectic Groups
449
Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (10.27) and the block weight diagram. Due to the WignerEckart theorem, the basis state belonging to the representation [l,1,1]of Sp(6) is orthogonal to the basis state belonging to [l,O,O], so it is a traceless tensor.
450
Problems and Solutions in Group Theory
+i}
m{i
m { I [} 
45 1
The Symplectic Groups
7. Calculate the ClebschGordan series for the reduction of the direct product representation [l,1,0] x [l,l , O ] of the Sp(6) group and the highest weight states for the representations contained in the series by the method of the standard tensor Young tableau.
Solution. There are a few methods to calculate the ClebschGordan series of a reducible representation. The method of the standard tensor Young tableau is one of them. The main point in the calculation is to count the multiplicities of the dominant weights in the relevant representations. The difficulty in Sp(2l) group, as well as S O ( N ) group, is how to remove the trace tensor subspace. In fact, the dimension of the tensor space projected by a Young operator [A] is nothing but the dimension d[A](SU(2l)) of SU(2l) group. Subtracting the dimension 4x1 (Sp(2t)) of Sp(2l) group, we obtain the dimension of the trace tensor subspace. Conversely, we can calculate the multiplicities of the dominant weights in the representation [A] of Sp(2l) by counting the multiplicities both in the tensor space projected by the Young operator Y[’] and in the trace tensor subspace. As far as the highest weight state is concerned, the condition that each raising operator E p annihilates the highest weight state is useful to calculate the basis state. First, we count the multiplicities of the dominant weights in the space of the direct product representation [l,l , O ] x [I,1,0] of the Sp(6) group. Since the block weight diagram for the representation [I, l , O ] of Sp(6) was given in Fig. 10.1 of Problem 5, it is easy to count the multiplicities of the dominant weights. In the direct product space, there are one basis state with the dominant weight (0,2,0), two basis states with the dominant weight ( l , O , l), four basis states with the dominant weight (2,0,0), eight basis states with the dominant weight (0, 1,0), and 16 basis states with the dominant weight (0, 0,O). Second, since there is one basis state with the dominant weight (0,2,0) in the direct product space, one representation [2,2,0] is contained in the ClebschGordan series. Its highest weight state is
The decomposition of the tensor subspace projected by [2,2,0] +
[a, 2,0] CB [I,1 , O ] CB [O,O, 01,
y[2)2)o] is
105 = 90
+ 14 + 1.
The tensor subspace contains the basis states with the dominant weights
Problems and Solutions in Group Theory
452
.as follows Weight
No. State bases
We have to subtract the basis states in the representations [I, l , 01 and [O,O, 01: one state with (0, 1,O) and three states with (O,O, 0). Third, since the difference between the multiplicities of the dominant weight (1,0,1) in the direct product space and in the representation space [2,2,0] is 2  1 = 1, one representation [2,1,1] is contained in the ClebschGordan series. Its highest weight state is
The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. The decomposition of the tensor subspace projected by Y[’i1r1] is [Z, 1,1]
+
[2,1,1]@ [2,0,0]@ [l,1,0],
105 = 70
+ 21 + 14.
The tensor subspace contains the basis states with the dominant weights
The Symplectic Groups
453
as follows.
Weight
No. State bases
1
2
5
9
We have to subtract the basis states in the representations [2,0,0] and [I,1,0]: one state with (2,0,0), two states with (0,1,0) and five states with (0, 0,O). Note that the tensor space for [2,0,0] is traceless and contains the basis states with the dominant weights as follows.
Weight
No. State bases
(2)O)O) :
1
(0,1,0):
1
I1I 1I) I 1 1 2 1,
Fourth, since the difference between the multiplicities of the dominant weight (2,0,0) in the direct product space and in the representation spaces [2,2,0] and [2,1,1) is 4  2  (2  1) = 1, one representation [2,0,0] is
454
Problems and Solutions in Group Theory
contained in the ClebschGordan series. Its highest weight state is
The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. The basis states with the dominant weights contained in the traceless tensor space of [2,0,O] was given. Fifth, since the difference between the multiplicities of the dominant weight (0,1,0) in the direct product space and in the representation spaces [2,2,0], [2,1,1], and [2,0,0] is 8(41)(52)1 = 1, onerepresentation [I,1,0] is contained in the ClebschGordan series. Its highest weight state is
The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. We have known from Fig. 10.1 of Problem 5 that the traceless tensor space for [l,1,0] contains one basis state with the dominant weights ( O , l , 0) and two basis states with the dominant weight (O,O,O). Finally, counting the multiplicities of the dominant weight (0, 0,O) in the direct product space and in the representation spaces [2,2,0], [2,1,1], [2,0,0], and [I,1,0], we have 16  (9  3)  (9  5)  3  2 = 1. One representation [0, 0, 01 is contained in the ClebschGordan series. The expression for its highest weight state is quite long. We write it in the form of the basis states instead of the standard tensor Young tableaux. The
The Symplectic Groups
455
latter form can be calculated from Fig. 10.1.
In summary, the dominant weight diagram for the direct product representation [1,1,0] ´ [1,1,0] is given in Fig. 10.3.
1 4 1~ 4=196=
90
+
70
+
21
+
14
+
1
Fig. 10.3 The dominant weight diagram for [l,1,0] x [l,1,0] of Sp(6).
In the permutation of two factors in the direct product representation space [l,l , O ] x [l,l , O ] , the basis states in the representations [2,2, 01, [l,1,0] and [O,O, 01 are symmetric, and the representations [2,1,1] and [2,0,0] are antisymmetric.
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Index
left, 29 right, 29 crystal, 173 crystal lattice, 173 basis vector of the reciprocal, 175 besis vector, 173 vector, 173 crystal system, 178 crystallographic point group, 174 cycle, 193 cycle structure, 194
l?N matrix group, 403 ya matrix, 403
anticommutator, 319 associative law, 27 basis, 7 orthogonal, 212 standard, 211 block weight diagram, 282, 285, 289, 292, 298, 303, 309, 336, 345, 348, 355, 378, 386, 394, 412, 445, 448 Bravais lattices, 174, 180
determinant, 1 dominant weight diagram, 299, 301, 356 doublevector form, 175 Dynkin diagram, 272
Cartan criteria, 270 Cartan matrix, 272 Cartan subalgebra, 271, 376, 435 CartanWeyl bases, 271, 318, 439 character, 43 graphic method, 213 character table, 48 Chevalley bases, 280, 321, 378, 409, 436, 441 class, 30, 48, 54, 139, 422, 424 selfreciprocal, 54 mutually reciprocal, 30 selfreciprocal, 30 ClebschGordan coefficients, 80, 101, 106, 149, 154, 215, 243 ClebschGordan series, 80, 240, 299 composition functions, 140 coset, 29
eigenequation, 1 eigenvalue, 1 eigenvector, 1 element, 27 conjugate, 30 equivalent theorem, 404 Euler angles, 123, 125, 127 filling parity, 214 Fock condition, 199, 323 F’robenius theorem, 67, 239 Gelfand bases, 337 generator, 27, 269 461
462
Problems and Solutions in Group Theory
gliding plane, 178 gliding vector, 177, 178 group, 27 C4, 32 c8, 35 D2n+1, 68 D2n, 68 D4, 33 D N , 32 v4, 32 direct product, 55 homomorphism, 33 I, 73, 75, 106, 121, 134 0, 59, 70, 85, 101 T, 81 Abelian, 27 c10, 37 C2n, 38 C4h, 35 c6, 32 C N , 65, 68, 69 Dn, 38 S O ( N ) ,375 S0(3), 115 SU(N), 317 SU(2), 117 Sp(2C, R), 433 USp(2C), 434 v2, 34 covering, 117, 140 direct product, 51 improper point, 51 proper point, 51 quaternion, 35 quotient, 31 T, 40 group space, 52, 140, 269 height, 282 homomorphism, 121 kernel of, 33 hook number, 198 hook rule, 198, 325, 380, 405, 442 horizontal operator, 199 ideal, 205
idempotent, 205 equivalent, 205 orthogonal, 205 primitive, 205 identity, 27 infinitesimal element, 141 international notation, 178 inverse, 27 irreducible tensor operator, 146 isomorphic, 27 Jordan forms, 18 Killing form, 270, 320 Lie algebra, 270 compact, 270 rank, 271 simisimple, 270 classical, 272 exceptional, 272 Lie group, 115, 140, 269 compact, 140, 269 order, 269 semisimple, 269 simple, 269 Lie product, 270 Lie Theorem, 141, 142 LittlewoodRichardson rule, 239, 326, 356, 364, 370, 402 Lorentz group, 415 homogeneous, 418 proper, 418 lowering operator, 280 matrix, 2 hermitian, 2, 5 negative definite, 2, 270 positive definite, 2 positive semidefinite 2 real orthogonal, 2, 5 real symmetric, 2, 6 unitary, 2, 5 multiplication rule, 27 multiplication table, 27
Index
multiplicity, 49, 53, 67, 77, 80, 97, 280, 282, 289, 299, 304, 308, 313, 355, 356, 451, 454 order, 140 of the element, 29 of the group, 27 Pauli matrices, 3, 33 period, 29 permutation, 193 even, 194 horizontal, 199 odd, 194 vertical, 199 permutation parity, 194 planar weight diagram, 283, 288, 289, 292, 298, 306, 340, 345, 348, 355, 360, 372 projection operator, 81 raising operator, 280 rank, 2, 142, 146, 317, 321, 326, 327, 376, 379, 380, 440, 441 rearrangement theorem, 27 reduced matrix entry, 147, 149 regular application, 214 representation, 43 regular, 52, 81, 85 subduced, 130 adjoint, 142, 283 conjugate, 43 equivalent, 47 faithful, 43 identical, 43 induced, 66, 237 inner product, 237 irreducible, 48, 123, 211, 279, 440 outer product, 238 real orthogonal, 43 reduced, 47 reducible, 47 subduced, 66, 237, 362 unitary, 43 unitary of infinite dimension, 166 root, 271
463
positive, 271 simple, 271, 319, 435 root chain, 271 rotation, 45, 115 improper, 51 proper, 51 Schur theorem, 47, 143, 245, 321 screw axis, 177 secular equation, 1 Serre relations, 280 similarity transformation, 8 space group, 174, 186 symmorphic, 174 spherical harmonic function, 399 spin tensor, 405 spinor, 405 spinor operators, 407 spinor representation, 404 structure constants, 141, 269, 319 subgroup, 29 cyclic, 29 index, 30 invariant, 30 normal, 30 symmetric center, 177 symmetric operation, 173 symmetric plane, 177 symmetric straight line, 177 symmetry transformation, 44 symplectic group, 433 real, 433 unitary, 434 tabular method, 212 trace, 1 Transformation Operators for a Scalar Function, 43 translation group, 173 transposition, 194 vertical operator, 199 weight, 279 dominant, 282, 336 equivalent, 280, 281
464
Problems and Solutions an Group Theory
fundamental dominant, 281 highest, 282, 337, 379, 442 multiple, 279 single, 279 Weyl orbit, 280 Weyl orbital size, 280 Weyl reciprocity, 323, 377, 440 Weyl reflection, 280 WignerEckart theorem, 51, 147 Young operator, 323, 377, 440 Young pattern, 198, 211, 323, 379, 441, 442 associated, 214 Young tableau, 198 standard, 198 standard tensor, 324, 337, 362, 441 tensor, 323