Báo cáo toán học: "Global cross sections of unitary and similarity orbits of Hilbert space operators " ppsx

GLOBAL CROSS SECTIONS OF UNITARY AND SIMILARITY ORBITS OF HILBERT SPACE OPERATORS MARTA PECUCH HERRERO INTRODUCTION In [2] D.. Fialkow characterized the operators on a sepa- rable Hilb

GLOBAL CROSS SECTIONS OF UNITARY

AND SIMILARITY ORBITS OF HILBERT

SPACE OPERATORS

MARTA PECUCH HERRERO

INTRODUCTION

In [2] D Deckard and L A Fialkow characterized the operators on a sepa- rable Hilbert space which have local] unitary cross sections These are the operators

of the form A ® B@ 1, where A and B are operators on finite dimensional spaces and 1 is the identity operator on a complex separable infinite dimensional Hilbert space #

The purpose of this article is to characterize the operators which have global unitary cross sections or global similarity cross sections To make these statements more precise, it is necessary to introduce some standard notation, which will be used throughout the paper: Y(#) denotes the algebra of all bounded linear ope- rators on #, &(#’) denotes the group of unitary operators in YH), UT) =

= {U*TU: Ue U(#)} is the unitary orbit of an operator T in ⁄() and +:9() + U(T) is the (norm continuous) function defined by 1(U) = U*TU

A local unitary cross section for t is a pair (y, #) such that @ is a relatively open subset of (7) which contains T and »: #@ + Y(#) is a norm continuous function such that tom =lg and o(T)=1; @ is a global unitary cross section when B=UT)

In Section 3 it is shown that t has a global unitary cross section if and only

if T has the form B® 1, where B is an operator on a finite dimensional space Let T denote the canonical image of an operator T in the Calkin algebra

If T= A4 @ B@ 1 (A, Bacting on finite dimensional spaces), then 7 admits a global unitary cross section only in the trivial case when T is a multiple of the identity As

a corollary, it is shown that if T= B @ 1 has a global unitary cross section », but

T is not a multiple of the identity, then it is impossible to construct @ so that @(7,) —

— @(T,) is a compact operator whenever 7,, T,¢ @(T) and 7, — T, is compact

The second part of the paper is devoted to the study of similarity cross sec- tions Let G(#) denote the group of invertible operators in @(#) and let A(T) =: -= {W-1TW :W eG(#)} be the similarity orbit of an operator T in ¥Y(#) The (norm continuous) function s:¢(9) + F(T) is defined by s(W):-: M-1TM,

L A Fialkow and D A Herrero ({5], [6]) proved that s has a local cross section if and only if Tis similar to a nice Jordan operator, that is, an operator of the form

n ay (a, )

@® Aly, + @® q„” >

where 4q, 2s, ., 2„ are distinct complex numbers, g, is the k x & nilpotent Jordan cell, and for each ?, kj, < kip << < Kim, and a;; = co for at most one value of /

In Theorem 17 it is shown that s has a global cross section if and only if 7

is similar to a very nice Jordan operator, that is, a nice Jordan operator of the form

„

® Gz + a)

ie

For operators on finite dimensional spaces and for the Calkin algebra, it is shown that there are no global cross sections except in the trivial case when the operator

T, or respectively its image T in the Calkin algebra, is a multiple of the identity As

in the case of the unitary orbits, it follows from the Calkin algebra result that if 7 has a global similarity cross section ¢, then ¢ cannot be constructed so that ¢(7,) —

~— ¢(1,) is compact whenever 7,, T,¢ (T) and 7, — T, is compact, except, of course, in the trivial case when T is a multiple of the identity operator

[14] and [15] are used as standard references for Algebraic Topology throughout the paper

The author wishes to thank Professors N Salinas and D Voiculescu for pro- viding several helpful references, and to Professor D A Herrero for suggesting the problem and for encouragement

PART 1

1 A CHARACTERIZATION OF THE UNITARY

ORBIT AS A HOMOGENEOUS SPACE

Let T be an operator in Y(#) and let »/'(T)= {Ae £(#): AT: TA}

be the commutant of T

ProposiTION 1 Jf T= A@®B@1, where A and B are operators on finite dimensional spaces, then the unitary orbit U(T) (norm topology) of T is homeomorphic with the homogeneous space of right cosets UA YU(A) 0 f(T) (quotient topology)

Proof Let p:U(H#) ¬ ()/()n/'(T) be the canonical projection Since «71 ({T}) = UH) 0 f(T), there is a bijective continuous map ¢ : U(#)/ [U(#) 0 A(T) > UT) which makes the following diagram commutative:

2() ——> #()/2() n.#'(T)

UIT) °

If T satisfies the hypothesis, then t is an open mapping [2], [4], and it follows easily that g is indeed a homeomorphism

Remark Since %(T) is pathwise connected, W(H#)/WH) 1 x'(T) has the

same property

Consider the commutative diagram (1.1) In [2] it was shown that t has a local cross section; this implies that p also has a local cross section via the homeo- morphism ¿ Since 2()n Z (7) 1s a closed subgroup of %(#), it follows [1S,

p 57) that

2(#)n #'(T) ——> %() —”—> #()J2() n (1)

is a fibre bundle (¡: 2(⁄)n.Z{(T) > UA) is the inclusion map.) Via g, the existence of a global (norm continuous) cross section for t is equivalent to the existence of a global cross section for p

From now on we will only consider the latter question, that is, we will work with the above mentioned fibre bundle

First of all we need to determine the homotopy type of #(3/)n '(T) This

is done in the next section

2 CHARACTERIZATION OF 2(⁄)n (7)

PROPOSITION 2 If A = N° ~ N@1,, for some irreducible operator N in L(C"), then the unitary operators which commute with A are the operators of the jorm 1,@ U with Ue @(C”)

Let A=4,@4,@ @A, and B= BOB @ @OB,, where 8; and A, are irreducible operators for each i,j, 1 <igm, 1 <j<n lt BFA, for every i,j, we say that A and B are disjoint In this case, it follows from [3, p 8] and a simple calculation that if U(A ® B) = (A ® B)U for some unitary operator

U, then U = U, ® U,, where U, commutes with A and U, commutes with B

Let T=A@BO@1 with A=A OAV @ @APM ELC), B=

= Bi? @ B® ® Ba” « LCD, where 4,e.Z(C) and B,e.Z(C”) are

irreducible operators, s = Ÿ_s;&, and / = Mi,

We can assume that A and B are disjoint: if A; ~ B; for some pair i,j, then

we can write

(s,)

T= A? ® OAL OA @ OAM © BY O10 OB @1@

©{4?°@ B/” @1) © B7 @ 1@ @ BH” @1~

jy)

(That is, all the common irreducible terms can be “absorbed” in B) Also notice

that BO1~B@1@ @B,@1, because Be @1~ B® 1, @1

~ B; @ 1 Hence, we can directly assume that ¢; = 1 forall j, 1 <j < m

From Proposition 2, the above remarks and the fact that m(#(C®) == Z for all k > 1, and 2(#) is contractible [11], we obtain:

PROPOSITION 3 For an operator T= A ® B@ 1 of the above form,

UH) 0 f'(T)={1,,® U,@® @®1, @U,@1,@W;@ @1, @W„:

:U,e(C”), V,e#(,}

and

m(UH) 1 A(T) =n (WUC) Ox (WC ))@

Om (UH) On (UH,))=Z

3 THE EXISTENCE OF GLOBAL CROSS SECTIONS

THEOREM 4 Let T= B@ 1, where B is an operator on a finite dimensional space Then the fibre bundle

(3.1) UH) 0) tT) —> UH) ——> U(#)|UH) 0 f(T)

has a global (continuous) cross section

J=1

Proof By Proposition 3, &(#)n#(T) = lê 1, @ E;,: Vị;e#(#,) je24,2, , my It follows that %#(3/) n f’(T) is isomorphic with a (finite) pro- duct of copies of #(#) and therefore it is contractible [11]

We will prove that the base space U(H#)/U@(#) 1 '(T) is also contractible Since a fibre bundle with contractible base is equivalent to a product fibre bundle

(which obviously has global cross sections), it will follow that p has a global cross section

Consider the exact homotopy sequence

—>m(0#) n #'(1)) “> (UG) A> (UA YUH) 1 f'(T)) ~>

(3.2)

2 (UH) 0 #'(1)) <> > (UH) ND)

corresponding to the fibre bundle (3.1) Since x„(2(2)) = 0 for all z and x„(()n nZ(T))=0 for all ø, we obtain z„(()/2(⁄)n ZŒT)) =0 for all n > 1 Since UH UH) Nn '(T) ~ X(T) is pathwise connected, we also have 2,(U(#)/ [U(H) 0 L'(T)) = 0

This means that @()/W@(#) N x'(T) is weakly homotopy equivalent [14] to

a contractible space To prove that it is actually contractible, it will be enough to show that it is homotopy equivalent to a CW-complex [14] This is done in the following auxiliary result

If Z is a subset of L(H#), then we define Z* = {R*: Re BZ}

Lemma 5 (4(#),G(#) 0 LT), HH) (GA) 0 S(T) Nn [#(⁄) n N'(T))}*) and (U(#), UH) '(T)) are homotopy equivalent relative to CW-complexes

Proof Since G(#) is paracompact and (4) 0 x'(T) is a closed subset of G(#), according to [13] it is enough to prove that the following property is satisfied: there is a neighborhood Q of the diagonal in (#7) x GH), a (continuous) function 2:2x[0, 1] ~ G&A) and an open covering {4,} of YH) such that the following properties are satisfied:

A(V,W,0) = W for every (V,W) in Q,

2(V,W,1)—=W_ for every (V, W) in Q,

ÀA(V,V,!)=V_ for every Vin (7) (0 < t < 1),

AV Wi teGA) nf (T) if VW) is in HH)0'(T) X OP) A(T) (<< i),

Py X Py <Q

and

AG, < Sy X [0,1)) = S, for every ©, in the covering

Let Q= ((V,W)e GH) x WH) = \|\V— WI < min{\| VU, WU -H},

let O, = {We GH): ||V — W|| < 1/3||V-1|)} for each Win #(Z) and let 2: Qx

<[0, 1] #2) be defñned by 2(V, M, £) = 0V +- (1 — OW

It is easy to see that all the required conditions are satisfied This shows that (GA), UH) 1 '(T)) is a relative CW-complex.

To prove the second statement of the lemma, we notice that 2(V, W, t) belongs

to [F(F)n A(T) N{[FA)n we (T) if V and W are in [GH)N WV (TIS

1 [G(H#H) 1 Z/'{T)]*, so that the same proof holds in this case

For the last statement, consider the map r: G() > W@(#) defined by

r{ƒ⁄)= V*V)~!* for V in Ø(#)

If 7: (2) ¬ #() denotes the inclusion map, then ro/ = lacy) Thus r

is a retraction

TẾ Ve[#(⁄Z)n Z(Tjln[Z(⁄)n./'(T))*, then both V and V* commute with T and, a fortiori, the whole W*-algebra W*(V) generated by V is contained

in «/(T) In particular, (V*V)—1? € o'(T) Since V is invertible, so is (V7V)~2?, and therefore r(V) == V(V*V)~¥?2 e€ UMA) 1[FA) 1 A(T) NGM) 0 (TI) =

== U(H) 1 f(T) Hence,

r|[[Z(#)n 4()1n [#(C#)n 01]: :12()n Z')1n1Ø()n

Nn f(T)" > (⁄)n 0)

Therefore r induces a map of pairs

r: (#(#), [2(#)n A(T) 1[HL) 0 L(T)") > (UH), MH) AL"T)), where [Z(⁄)n.Z(7)]n{[2()œ.Z(T)Ị! and ()n ¥'(T) are closed subsets

of G(A#) and U(#), respectively Since a retract of a relative CW-pair is also a rela-

tive CW-pair [12, p 127], [14], the proof is complete

The proof of Theorem 4 is now complete

By using Lemma 5 and [14, p 402] we obtain the following:

CoROLLARY 6 #(#)/Ø()n (T7), Ø(⁄)/0Ø(#)n /(T)1n[2() n.2/(T1]E and (3)/2()n '(T) are homotopy equivalent to CW-complexes

THEOREM 7 Let T= A@®B@1, where A and B are disjoint operators on finite dimensional spaces, and A #0 Then the fibre bundle (3.1) does not have a global cross section

Proof In the exact homotopy sequence (3.2) we have x„(2(2)) = 0 for all

n and m(2()n.Z/'(T)) = Z™ (Proposition 3) for some m2 1 Therefore T„+a(()/()n f'(T)) = 1,(U(H) 1 '(T)) and, in particular,

TMH fU(A) 0 Z'(T)) ~ m(2(⁄)n Z'(7)) = ZO,

If there exists a global cross section for p, then the exact homotopy sequence

- —> m(#()n s2) ~> m(Q€)) ?x m(()/@) n2/)) o>

splits [14, p 418], that 1s, 2(W(#)) ~ m;(2() n '(T)) ® m;(()/2()n

fn xf'(T)) But Z cannot be a direct summand of 2,(%(#)) == 0; therefore p does

not have a global cross section

4 THE FINITE DIMENSIONAL CASE

In what follows, the notation U(n) = &(C") will be used Let A be an operator

on C” Proceeding exactly as in Proposition |, we see that %(A) is homeomorphic with U(n)/U(1) 0 (4) (right cosets) and that

(4.1) U0) n z/(4)—> UÚ) ”> U0)JU(@w) n2!)

is a fibre bundle (Indeed, this can also be proved by a direct argument.)

LEMMA 8 Let A= NO ~ N@1,,, where N is an irreducible operator in L(C%), n = qm Then the fibre bundle (4.1) does not have a global cross section unless

q = 1, that is, unless A is a multiple of the identity

Proof By Proposition 2, U(n)n (A) = {l,@ U: Ve U(m)} Hence, m(U(nn.Z{49)) ~ m(U0n)) ~ 2

The last terms of the exact homotopy sequence of the bundle we are consi- dering are

Z > Z =>z/(U(/U0) n '(4)—>0

We will show that /,(z) = gz for every z¢ Z The generator of 2,(U(n)) is the homotopy class of the map y, :{0, |] ~ U(), where y,(t) is the diagonal matrix diag(e?*", 1,1, ., 1) Since io y,(t) = diag(e’*", 1, ., le, 1,0 1, , Ă€?z 1, , , |), where ec? appears g times on the diagonal and the remaining ele- ments on the diagonal are equal to J, itis easy to show that jo y,, and qj, are homo- topic Therefore, i,[y,,] == đ[,}

Since 2,(U(#)) = Z is commutative, if there is a global! cross section, then the exact homotopy sequence will also split at z, But

n,(U(n)/U(n) n f'(A)) ~ 2 (U(n))/ker p., = a (UC) /im i, ~ Z/qZ = Z, Since Z, cannot be a direct summand of Z, there is no global cross section unless g:= 1

Lemma 9 Let 4 = 4! @®Ao* @® @A,", 1 2 2, where the operators

A, € £(C*) are irreducible and disjoint, and ¥) sik; == s Then the fibre bundle

(4.2) U(s) 0 Z⁄(4)—> UG) -> UG)JUG) ñ 3⁄44)

does not have a global cross section

Proof 7(U(s) 0 #(A) = Z@Z@ ®@ Z= Z™ (Proposition 3) and

7, (U(s)) = Z

The last terms of the exact homotopy sequence of the fibre bundle (4.2) are

= Z® ~ Z = m(UG)/UG)n '(4) — 0

Since n > 2, Z™ cannot be a direct summand of Z; therefore the exact homo- topy sequence cannot split This implies that the fibre bundle has no global cross

Combining Lemmas 8 and 9, we obtain the following result

THEOREM 10 An operator acting on a finite dimensional space has a global unitary cross Section if and only if it is a multiple of the identity

5 UNITARY CROSS SECTIONS IN THE CALKIN ALGEBRA

We need to introduce the following notation: % denotes the ideal of compact operators in Z(H), SL = L(A) KH is the Calkin algebra, (4#) denotes the unitary 8roup m Z and 2⁄;(.4Z) 1s the pathwise connected component of iin Uf), where T denotes the image of T ¢ L(#) under the canonical projection onto / Let #(T): =

=: {V*FV: Vin Ut) In this section we will only consider operators of the form T::=A@®B@i1, AC L(C), Be LC’), for which t: U(x) ~ UT) is known to have a local unitary cross section [2] Also, notice that 7 is unitarily equivalent to (B@ 1)”

As before, we have a commutative diagram

%(#) ——> #()/#(#) n.'()

UP) a

where the map @ is continuous and bijective We will prove that if T is not a mul- tiple of the identity, then p does not have a global cross section, in which case t cannot have one either: if €: UT) — ⁄(.Z) is a(norm continuous) mapping such that to ¢ = Lag ¿Œ) =z 1, then @spsŠs@ =tofom = @ Since @ is injective, this implies that p o € og = 1 Therefore &o y is a cross section for p

THEOREM 11 Let T= B@1, where Be Y(C") The fibre bundle

%()n #'(Ð —> Ø(#) —>#(#)/2(#) n #'()

does not have a global cross section unless T is a multiple of the identity

Proof Consider the following morphism of bundles:

UA) 1 ol'(P) ——» Ut) 0 b'(T)

%()/%4#)n /()—>()/(#)n fF)

where 1, %, i and /, are inclusion mappings, p and py are the canonical projections onto the classes of right cosets and :’’ is induced by the inclusions i and 7’ so that the diagram is commutative Itis easy to see that i’’ is well defined and injective Since U,( xf) is the (pathwise connected) component of I in Ws), we have

ig | Ty(Up( I), 1) ~ 1, (U(x), 1) for n> 1

Observe that the path component of I in &(s/) N (7) is equal to the path component of | in Uf) 0 o'(T), whence

it (Ug) 1 8'(T), Ú) ~ m(@(#)n 7), 1) for n> l

Now consider the exact homotopy sequences of the bundles in (5.1) and the induced homomorphism i,, i, and iy We have the following commutative diagram:

(Aol) 0 ob'(F), 1) 2 rey Uo), 1) P 1( g(t) [Ug ot) ñ 22G), p(Ñ) ->

1"

i

->m(()n 2), 1) ~> m(0(),Ï) —Š m(0(2)/2(2) n f'(P), ply) 2s (5.2)

Since i,, and i, are isomorphisms, the 5-lemma implies that ij’ is also an isomorphism Therefore if one of the exact sequences splits, the same holds for the other one

We will show that the upper sequence in (5.2) does not split unless T isa multiple

of 1 This implies that the exact homotopy sequence of p : U(x) > 2⁄(Z)/2(Z)n

n '(T) does not split either, whence p cannot have a global cross section To this end, first we compute 7,(%)(.), 1) We use the fact that U,(./) is isomorphic with UH) UA) 1 + #), which is a deformation retract of£ ()/2(⁄)n(1 +), where 1+ 4% ={14+ K: Ke x}.

Let r: GH) ¬ U(H#) and h: GH) xX [0, 1] + GH) be defined by

r(Vf)== V(V*V)~1!2 for Vin GH) and

AV, D2 VI — t(V*V)-1E + rlỊ for Vìn Ø(#),0 <t< 1,

respectively It is easy to see (by using the spectral theorem for compact hermitan operators) that A(GM)N(L +X) X (0,1) <— #⁄)n(L-+-Z) and r((#)n a( +#)) < U(H#)N( +) Therefore r and A give maps of pairs

r:(#(#).2(#)n(1 — )) > AH), MH) ACU + )

and

h: (GH) x [0, 1], #(Z)n( + ) x [0, l) ¬ (@(), ##)n(1 ‹- #))

with A(V,0) =r(V) and ;(, l) = V Thus, &@(#) is a deformation retract of GH), UA) AU + #) is a deformation retract of GS/)N (1 + #) and #(#)‡

;{Z)0n (1 ++ 4H) is a deformation retract of #(Z)/2(#)n(1 +2) As a con- sequence,

n((#)/#(#)n( + #)) > m(0(#)/2(#) ñ(L +2), n>0

Since #()/Z()n(1 +2) is homotopy equivalent to the space of Fredhoim operators of index zero [10], which is homotopy equivalent to BU (the classifying space of vector bundles with group U(oo) = lim U()) [1], [10], we obtain

0 #n=2m — 1

[15, p 215]

Case I Let T= B@1, Be L(C") If B=A®@I]1,, where 4EL(C’) is irreducible, then B@ 1 ~ A@1, @ 1 is unitarily equivalent to A @ 1 Therefore

we can assume that B = A is irreducible By using a faithful unital *-representation

of the finite dimensional C*-algebra C*(T), and Proposition 2, one can show that

%(Z)n.Z⁄Œ) = {(„@ U)”:Ue()} Therefore

Ty( No(s) 0 f(T), 1) = m(®()1#()n( +2), n >1

Let [y] be a generator of z;(2()/2()n(1 + )) >~ 7:

y: (0, 1] > {0, 11, 20, 1] x [0, 1) ¬ 2(2/2()n +).