The Kadison Similarity Problem

Similarity Problem

10. The Kadison

Summary: This chapter concentrates on the (still open) Kadison Similarity problem (= Problem 0.2). We show that this problem is equivalent to several seemingly different questions which appeared in other contexts: Arveson's hyperreflexivity problem and Dixmier's invariant opemtor range problem. Finally, we describe the author's recent work on the notion of "length" of an opemtor algebm, which also leads to one more reformulation of Kadison's question.

general form, the Kadison similarity problem asks whether any bounded homomorphism u from a C*-algebra A into B(H) is similar to a homomorphism (= "orthogonalizable"). This is the same as Problem 0.2 from the introduction. Indeed, this can be In its most

reduced to the unital

Recall first how ideal in

A

and if

A and A in C

follows.

C*-algebra

is the unit in

e

we

case as

a

A is embedded into its unitization

A

we

A + Ce,

Ila + AeJJX


1

if

Also since

a

preceding notation,

since A has

u: 0, + Ae

moreover

A is

for any

a

an

in

have

max

Thus,

have

A:

hence -

an

lliill

A is

a

we

have

Ilall.

be extended to

u(a)

+ AL

a

bounded

Clearly,

if ii is

is.

Ilull


H, i.e. we have V T(H). Typically we will be dealing with subspaces which are not closed, and Foias introduced the term "paraclosed subspaces" to designate operator ranges. In the early 70's, this subject was studied by several authors, among whom Fbias [Foi2], Fillmore and Williams [FiW] and Voiculescu [Vo]. The following problem, due to Dixmier, (whose 1949 thesis is devoted to a related topic) has circulated for a long time: This shows that if

=

invariant operator range problem.) Let V C H be an B(H) be a unital C*-algebra. Assume that V is

(The

Problem 10.4.

operator range and let A C invariant under

A'

(=

Dixmier, whom in

a

aV C V for any

A, meaning

the commutant of A) such that I

consulted, kindly

letter addressed to

von

find another operator

T

wrote

me

in A. Is there

an

operator T in

V?

back that he asked this question

Neumann in 1950.

T(H)

Remark. Of course, if V problem is to show that if V =

T(H)

a

=

=

and

T(H)

in A' with the

that V is not assumed closed

T

E

A' then V is invariant under A. The

is invariant with T E

B(H),

then

we can

range. Again, we should emphasize the solution is trivial: we can take

same

(otherwise,

T the orthogonal projection onto V which clearly is in A' if V is closed and invariant). Following an idea of Foias ([Foi2]) we will show that this problem is equivalent

for

to the Kadison

one.

10. The Kadison

172

Proof: a

problem (= Problem 10.4) has a similarity problem (= Problem 10.2) also does.

The invariant operator range

Theorem 10.5.

solution iff the Kadison

positive

Problem

Similarity

Following [Foi2], we will associate to an invariant operator homomorphism. u: A -- B(H). So we assume V

bounded unital

C*-subalgebra. injective. Fix

Let K

A and k

a

E

T(H)

with

C

=

=

Ilk. By the

that aTk

=

V for any a in A where A c B(H) is a unital V and of course TIK is ker(T)J-. Note that T(K)

such that aV

B(H)

T E

range V C H

aT(K)

K. Since closed

c

T(K)

a unique k in K such correspondence k --+ i defines

there is

the

graph theorem,

bounded linear operator which we denote by u(a). It is easy to check that u(a) e B(K) is a unital homomorphism. Moreover, one more application of the closed graph theorem shows that a --* u(a) is bounded. Thus we obtain the a

a

--+

announced

homomorphism. associated

u(a))

Tu(a),

aT

=

hence if 0

to V. Note that

K

TIK:

=

u(a)

VaE,A but in this formula

Then there is

an

0-1 is we

H

--+

Now

S-'ir(a)S

=

(by

have

definition of

0-'aO

=

K such that

have 0-'aO

we

have

we

priori unbounded.

invertible S: K Thus

homomorphism.

a

--+

assume u -+

a

7r(a)

orthogonalizable.

Su(a)S-'

=

and if we let

x

=

is

a

OS-': K

x7r(a) ax for any a in A. But since A is a C*-algebra and 7r(a*) x7r(a)* x7r(a*) a*x, hence taking adjoints lr(a)x* x*a, and 7r(a)*, xx*a. But we can also write finally (after left multiplication by x) x7r(a)x* H

we

have

=

have

we

=

=

=

=

x7r(a)x* Thus

the x

we

have

(x7r(a))x*

xx*: H

proved that

(xx*)'I'.

for

same

=

=

(ax)x*

=

axx*.

A, which implies polar decomposition of

H is in the commutant

--+

A moment of

thought

on

the

shows that

T(H) and since

x

=

OS-' with S invertible

x(K) Thus

we

We have thus shown that lem 10.4.

Conversely,

u:

A

-+

a

=

=

have

T(H).

.

.

.

be

a

bounded unital

notation. Let

(6,,)

B(H,,)

-+

be

u(A) j

T

c

A'.

a

one

to Prob-

+

homomorphism. Assume H separable H, and let H,, C H

dense sequence in +

u(A) ,.

defined by u.,,(a) hence, by Theorem 7. 11,

B(H.,,) set

,

able. Whence S,,: H,, -4

with

positive answer to Problem 0.2 implies positive answer to Problem 10.4.

homomorphism u,,: A ,, as cyclic 1,

admits

A

O(K)

T(H)

H,,

7r,,:

clearly

assume a

B(H)

simplicity of be defined by

for

The

we

x(K)

conclude V

Let

=

=

H,, invertible (and bounded) and

such that

u,,(.)

=

S,-,1,7r,,(-)S.,,.

u(a)IH,,, obviously

=

u,, a

is orthogonaliz*-homomorphism

We may

as

well

assume

10. The Kadison

JIS,,11

1. Let

B(ft)

ii: A

Hi 6) H2 (D

be defined

We have

(a)

is

1,

=

S-' (a),

:

F1

Problem

`R: A

B

--

173

(ft)

and

by

EDS,

=

and let

...

Similarity

(a)

injective,

where here

ax,,(a)

=

its

-1

and

i (a)

u,,(a).

=

r( )

is invariant under

image

and

have

we

priori unbounded. Suppose for a moment that we can show that i is orthogonalizable. Then it is easy to see that u itself must be orthogonalizable. Indeed, we have 11U11cb IP16 hence SUP JIU,,Ilcb u is orthogonalizable by Corollary 4.4 (one can also prove it directly). Thus to complete the proof it suffices to show that any homomorphism u: A B(H) of the form u(a) H bounded, injective and with 0-17r(a)0 with 0: H V O(H) invariant under 7r(A), is orthogonalizable. If we assume a positive solution to Problem 10.4, this is easy: we can find 0 in 7r(A)' such that O(H) V. Let Ho (ker0)_L. Note that Ho is invariant =

is

a

=

=

--+

--+

=

=

_

=

=

7r(A). 7r(a)Oo

under

Let

have

=

Oo: Ho

--+

0o7r(a)jH,,

H be the restriction of 0 to

for any

a

7r(a) But V

=

O(H)

=

0-10o7r(a)IH000

W(H)

10.

But

in A

=

hence,

on

Ho which is injective. We 00 we have

the range of

Oo7r(a)IHOOO 1.

is also the range of 00. Hence we have now 00 10: H -* Ho and 0-100: Ho

u(a) ---

=

H

0-17F(a)O

are

bounded

by the closed graph theorem (note here that, of course, Ho is closed), hence u is orthogonalizable. This completes the proof that "yes" to Problem 10.4 implies 11 44yes" to Problem 0.2. Remark. Note that

we

may

as

well

assume

in Problem 10.4 that A is

a von

already know this for Problem 0.2 (by passing bidual). See [On] for more details on this particular point. mann

algebra,

since

The results in for the

(SP)

Chapter 4, and notably Theorem 4.3 provide ample

Neu-

to the

motivation

following:

Generalized the

we

Similarity

Problem.

following property (SP) For any H, every bounded

Which unital operator

algebras A have

?

unital

homomorphism

u:

A

--+

B (H) is

c.

b.

Loosely speaking, this property (SP) could be described as "automatic complete boundedness" in analogy with the field of automatic continuity for homomorphisms between Banach algebras (see [DW]). The Kadison problem is equivalent to showing that every C*-algebra satisfies

(SP). [P9],

In

for

an

show that this property (SP) is closely linked to a notion of "length" operator algebra. Related results appear in [P10-11-12]. The rest of this we

chapter The

is

an

outline of all these results.

"length"

that

we

have in mind is

analogous

to the

following

situation:

a unital semi-group S and a unital generating subset B C S, it is natural to say that B generates S with length < d if any x in S can be written as

consider

a

product

x

=

b1b2

...

bd with each bi in B. Our main idea

can

be illustrated

10. The Kadison

174

in

Similarity Problem

rather transparent way

a

simple model of semi-groups

the above

on

fol-

as

lows. Assume that B generates S with length < d. Then any homomorphism 7r: S -- B(H) (i.e. 7r(st) 7r(s)-7r(t) and 7r(l) 1) which is bounded on B with =

=

sup

11-7r(b) 11

be bounded

< c must

11 7r(s) 11A

the whole of S with sup

on

SES

bEB

Conversely, assume that we know that morphisms ir: S -- B(H) satisfy, for some sup

11 7r (b) 11

:5

c =:>.

for

sup

11 7r (s) I




0, all homo-

following implication:

> 1, the

c

and

> 0

some a

rw'.

SES

bEB

Then it is rather easy to see that B (=integral part of a), so that we can

necessarily generates S with length replace a by [a] and r, by 1.


2, since otherwise B(H) would semi-nuclear, which contradicts the main result in [And]. Hence, we have d(B(H)) > 3. Actually, using the length t(B(H)) instead, we can obtain a very simple proof that d(B(H)) 3, as follows. Proof that (B(H)) < 3: This very direct proof comes from [P12]. Fix n > 1. Let W, and W2 be any two n x n unitary matrices such that be

=

V i,j

JW2(iii)l

JW1(iJ)J

Note that there

are

associated to the

examples

many

of such

(suitably normalized)

n- 1/2

:=

matrices, for. instance the

Fourier transform

on

group of order

there

are

matrix

any commutative

n. Then, for any x in the unit ball of Mn (B (H)) (with dim H diagonal matrices D1, D2, D3 also in the unit ball of Mn (B (H))

=

00)

such

that x

The

proof of this orthogonal ranges

is very so

=

DlWlD2W2D3-

simple. Let Si,

i

1,

=

.

.

.

,

n

be isometries

on

H with

that

V iJ

Si*Si

=

6ii I

-

Then let

D, (i, i) and

=

Si*

and

D3

Sj

moreover

D2 (k, k)

=

n

7 W, (i, k) Sixij Sj* W2 (k, ij

The announced properties

are now

but

a

simple verification.

El

10. The Kadison

Similarity Problem

179

Proposition 10.12. The Kadison similarity problem has a positive answer for all unital C*-algebras A iff there is an integer do such that (A) :! do for any C*-algebra A. By Proposition 10.6, if we have a bound for the length, the Kadison conjecture is immediate. Conversely, if there are C*-algebras Ad with length tending to infinity with d, then clearly the length of their direct sum (in the f,, sense) must be infinite, so this direct sum must fail the Kadison conjecture. El Unfortunately, up to now, the highest known value of t(A) for a C*-algebra is 3, but we conjecture that there are examples of arbitrarily large length. However, in the non-self-adjoint case, we have recently been able to prove the following. Prooh

Theorem 10.13. For any integer d > operator algebra Ad such that

i(Ad)

1, there

=

is

a

(non-self-adjoint)

unital

d.

algebras with arbitrarily large finite length? example with 2 < t(A) < oo is known. However, algebras is proved in [P9] that any proper uniform algebra A must satisfy t(A) > 2. It also unknown whether there are Q-algebras (= quotients of uniform algebras)

Problem. Are there uniform For uniform

with 2