LINEAR MAPS ON OPERATOR ALGEBRAS THAT PRESERVE ELEMENTS ANNIHILATED BY A ...

13 Pages·2002·0.21 MB·English

PROCEEDINGSOFTHE
AMERICANMATHEMATICALSOCIETY
Volume130,Number8,Pages2383{2395
S0002-9939(02)06362-1
ArticleelectronicallypublishedonFebruary12,2002
LINEAR MAPS ON OPERATOR ALGEBRAS THAT PRESERVE
ELEMENTS ANNIHILATED BY A POLYNOMIAL
JINCHUAN HOU AND SHENGZHAO HOU
(Communicated byDavidR.Larson)
Abstract. In this paper some purely algebraic results are given concerning
linearmapsonalgebraswhichpreserveelements annihilatedbyapolynomial
of degree greater than 1 and with no repeated roots and applied to linear
mapsonoperator algebrassuchasstandardoperator algebras,vonNeumann
algebras and Banach algebras. Several results are obtained that characterize
such linear maps in terms of homomorphisms, anti-homomorphisms, or, at
least,Jordanhomomorphisms.
1. Introduction
Let f(x) be a complex polynomial of degree greater than 1. We say that an
element a in an algebra is annihilated by f(x) if f(a) = 0: We say that a linear
map (cid:30) from one algebra into another preserves elements annihilated by f(x) if
f((cid:30)(a)) = 0 whenever f(a) = 0, and we say that (cid:30) preserves elements annihilated
by f(x) in both directions if f((cid:30)(a)) = 0 , f(a) = 0. In [8], R. Howard gives a
characterization of invertible linear maps which preserve matrices annihilated by
a polynomial. For the in(cid:12)nite dimensional case, let H be a Hilbert space and f a
complexpolynomialofdegreegreaterthan1. Semrl[16]provedthatif(cid:30)isaunital
(that is, (cid:30)(I) = I) surjective linear map from B(H) onto itself and if (cid:30) preserves
operatorsannihilatedbyf(x) inbothdirections,then (cid:30)is eitheranautomorphism
or an anti-automorphism. It is then interesting to ask if we can get similar results
undertheweakerassumptionthat(cid:30)preserveselementsannihilatedbyapolynomial
in one direction only and without the (cid:30)(I) = I assumption. It is also interesting
to ask what we can say for Banach spaces and for general operator algebras. It is
the aim of this paper to discuss such questions.
The questions considered here are in fact among the so-called linear preserver
problems (LPP), that is, the problems of characterizing linear maps on opera-
tor algebras that leave invariant certain properties of operators such as spectrum,
invertibility, solutions of an equation, subsets, relations, etc. The (cid:12)rst papers con-
cerning these problems discussed the spectrum-preserving linear maps on matrix
algebras(i.e., the (cid:12)nite dimensional case) and date back to the nineteenth century
[5, 10]. A greatdeal of e(cid:11)ort has been devoted to the study of LPP in the last few
ReceivedbytheeditorsJune23,2000and,inrevisedform,March23,2001.
2000 Mathematics Subject Classi(cid:12)cation. Primary47B48,47L10, 46L05.
Key words and phrases. Operatoralgebras,linearpreservers,homomorphisms.
TheprojectispartiallysupportedbyNNSFCandPNSFS.
(cid:13)c2002 American Mathematical Society
2383
2384 JINCHUAN HOU AND SHENGZHAO HOU
decades for operatoralgebrasoverboth (cid:12)nite dimensional andin(cid:12)nite dimensional
spaces (for example, see the surveys [11]-[13] as well as [1]-[4], [6]-[9], [14]-[16]).
Many results have revealed the algebraic properties of linear preservers in terms
of homomorphisms,anti-homomorphisms,or at least Jordanhomomorphisms, and
conversely,givenewcharacterizationsofhomomorphisms. Recallthatalinearmap
(cid:30)fromanalgebraAintoanotheriscalledaJordanhomomorphismif(cid:30)(a2)=(cid:30)(a)2
for every a2A.
In the present paper we (cid:12)rst consider, in Section 2, the linear maps which pre-
serve elements annihilated by a polynomial with no repeated roots and of degree
greater than 1 in the purely algebraic case. We show that if such a linear map
(cid:30) is unital, then it is idempotent-preserving (Thm. 2.2). Generally, under the as-
sumption f(0)=0, we get a description of the relationship between (cid:30)(1) and (cid:30)(a)
for idempotents a (Thm. 2.3). These allow us to give a characterization of alge-
braichomomorphisms in terms of preservingelements annihilated by a polynomial
(Corollary 2.5). Then, in Section 3, we apply the results of Section 2 to linear
maps on operator algebras. Let P be the set of all complex polynomials of degree
0
greater than 1 and with no repeated roots such that f(0) = 0. Let (cid:30) : A!B
be a linear map which preserves elements annihilated by a polynomial f(x) in P .
0
We show that (cid:30) has nice algebraic properties in many situations. For example,
we prove that there exist an invertible operator T and a complex number (cid:21) with
(cid:21)k = 1 such that either (cid:30)(A) = (cid:21)TAT(cid:0)1 for all A 2 A or (cid:30)(A) = (cid:21)TA(cid:3)T(cid:0)1 for
all A 2 A if (1) A and B are standard operator algebras acting on Banach spaces
X and Y, respectively, and (cid:30) is surjective and weakly continuous (Thm. 3.1), or,
if (2) A = B(X), B = B(Y), where X has in(cid:12)nite multiplicity (for example, if
X =c , l , L [0;1];1(cid:20)p(cid:20)1, or any in(cid:12)nite dimensional Hilbert space) and (cid:30) is
0 p p
surjective with (cid:30)(F)6=0 for some (cid:12)nite rank operator F (Thm. 3.6). If A is a von
Neumann algebra and B is any unital Banach algebra, and if (cid:30) is bounded, then
(cid:30) is in fact a Jordanhomomorphism multiplied by a k-potent element (Thm. 3.2).
Our results generalize some known results in [2, 15, 16] even for the Hilbert space
case. In the case when A is a Banachalgebra, we show that (cid:30) is a homomorphism
multiplied by a k-potent element if and only if (cid:30) preserves elements annihilated
2
by a polynomial in P (Thm. 3.8).
0
Note that, in our results,we do not assume that (cid:30) is unital nor that(cid:30) preserves
elements annihilated by f in both directions. And in many cases, we even do not
assume surjectivity and any continuity.
2. Linear maps on algebras
In this section we give some results about the characterization of linear maps
betweenalgebraswhichpreservetheelementsannihilatedbyapolynomialofdegree
greater than 1 and with no repeated roots.
Iff(x)isapolynomial,weletZ(f)denotethesetofzeroesoff andG(f)denote
the set of (cid:21)2C such that (cid:21)Z(f)=Z(f).
Theorem 2.1. SupposeA;B areunitalcomplex algebras and(cid:30):A!B is aunital
linear map. The following are equivalent.
(1) There is a polynomial f(x) with no repeated roots and deg(f)(cid:21)2 such that
f((cid:30)(a))=0 whenever a2A and f(a)=0:
(2) (cid:30) is idempotent-preserving.
LINEAR MAPS ON OPERATOR ALGEBRAS 2385
(3) For every polynomial g(x) with no repeated roots and degg (cid:21)2, g((cid:30)(a))=0
whenever a2A and g(a)=0.
Proof. The implication (3)=)(1) is obvious.
(2)=)(3). Suppose(cid:21) ;::: ;(cid:21) aretherootsofg:Standardlinearalgebraargu-
1 n
ments show that g(a)=0 if and only if Pthere is an algebrPaically orthogonalfamily
e ;::: ;e ofidempotentssuchthat1= n e anda= s (cid:21) e :However,if(cid:30)
1 n j=1 j j=1 j j
is idempotent-preserving,then(cid:30) alsopreservesorthogonalityofidempotents, since
idempotents e and f are orthogonal (i.e., ef = fe = 0) if and only if e+f is an
idempotent.
(1)=)(2):Suppose(1)istrue. Byreplacingf(x)withf((cid:11)+((cid:12)(cid:0)(cid:11))x),wemay
assume f(0) = f(1) = 0. A nonzero complex number (cid:21) such that (cid:21)Z(f) (cid:26) Z(f)
must be a root of unity (since (cid:21);(cid:21)2;(cid:21)3;::: must be in Z(f)). Let G = f0 6=
(cid:21) 2 C : (cid:21)Z(f) (cid:26) Z(f)g: It follows from the (cid:12)niteness of Z(f) that G is a (cid:12)nite
multiplicative subgroup of the unit circle, and thus, for some positive integer k;
G = f(cid:21) 2 C : (cid:21)k = 1g = f(cid:21) 2 C : (cid:21)Z(f) = Z(f)g = G(f): It follows that
g(x)=xk+1(cid:0)x divides f(x) and G is the set ofall nonzeroroots of g(x). If e2A
is an idempotent and (cid:11) 2 Z(f); then f((cid:11)e) = f((cid:11)(1 (cid:0)e)) = 0; which implies
f((cid:11)(cid:30)(e))=f((cid:11)(1(cid:0)(cid:30)(e)))=0:Assume that (cid:21) is a rootofthe minimal polynomial
of (cid:30)(e): Since (cid:21)Z(f) (cid:26)Z(f) and (1(cid:0)(cid:21))Z(f)(cid:26)Z(f); we know both (cid:21) and 1(cid:0)(cid:21)
are in G. It follows that the minimal polynomials of both (cid:30)(e) and 1(cid:0)(cid:30)(e) divide
g(x). Hence we have g((cid:30)(e))=g(1(cid:0)(cid:30)(e))=0. Suppose, for a contradiction, that
the minimal polynomial of (cid:30)(e) has a root (cid:21) with 0 6= (cid:21) 6= 1. Then (cid:21)k = 1 and
(1(cid:0)(cid:21))k =1. Itfollowsthateither(cid:21)=e(cid:25)3i or(cid:21)=e(cid:0)(cid:25)3i . Notethat,forany(cid:21)1 and
(cid:21) inGandany(cid:11)inZ(f),onehasf((cid:11)((cid:21) (1(cid:0)e)+(cid:21) e))=f((cid:11)(cid:21) )(1(cid:0)e)+f((cid:11)(cid:21) )e=
2 1 2 1 2
0. Thereforeeverynonzerorootofthe minimalpolynomialof(cid:21) (1(cid:0)(cid:30)(e))+(cid:21) (cid:30)(e)
1 2
is in G. In particular, taking (cid:21) =1 and (cid:21) =(cid:21)(cid:0)1, one gets 2(cid:0)(cid:21)2G, which is a
1 2
contradiction.
Remark 2.1. Theorem 2.1 cannot be generalize(cid:20)d to the(cid:21)case of a polynomial with
(cid:11) (cid:12)
repeated roots. For example, if we let A = f : (cid:11);(cid:12) 2 Cg and let (cid:30) :
(cid:20) (cid:21) (cid:20) 0 (cid:11)(cid:21)
(cid:11) (cid:12) (cid:11)+(cid:12) 0
A!M (C) be de(cid:12)ned by 7! , then (cid:30) is obviously unital
2 0 (cid:11) 0 (cid:11)
and preserves id2empotents, 3while it does not preserve elements annihilated by x2.
(cid:11) 0
If we let A = f4 (cid:12) 5 : (cid:11), (cid:12), (cid:13) 2 Cg and let (cid:30) : A!M (C) be de(cid:12)ned by
3
2 3 02 (cid:13) 3
(cid:11) 0 (cid:11)+(cid:12)(cid:0)(cid:13) 0
4 (cid:12) 5 7! 4 (cid:12) 5, then (cid:30) is unital and preserves elements
0 (cid:13) 0 (cid:13)
annihilated by x2. But (cid:30) does not preserve idempotents.
The following lemma is technical.
Lemma 2.2. Suppose k (cid:21)2 is a positive integer, B is a unital algebra over C; and
u;v2B are such that vk+1 =v, (u(cid:0)v)k+1 =u(cid:0)v and
(cid:16) (cid:17)
k+1
e2(cid:25)kisu+(e2(cid:25)kit (cid:0)e2(cid:25)kis)v =e2(cid:25)kisu+(e2(cid:25)kit (cid:0)e2(cid:25)kis)v
for 1(cid:20)s;t(cid:20)k: Then uv =vu, and u=0 implies v =0:
2386 JINCHUAN HOU AND SHENGZHAO HOU
Proof. Letting s = t = k, we see that uk+1 = u: Note that when m 2 N we have
(cid:16) (cid:17) (cid:16) (cid:17)
P m P t
kt=1 e2(cid:25)kit = kt=1 e2(cid:25)kim equals0 if k does not divide m; andequals k if k
divides m: It is clear that if u=0; we have
(e2(cid:25)kit (cid:0)e2(cid:25)kis)k+1v =(e2(cid:25)kit (cid:0)e2(cid:25)kis)v
for 1(cid:20)s;t(cid:20)k: Since k (cid:21)2; this means v =0:
Let w = u(cid:0)v: Since e2(cid:25)kisu+(e2(cid:25)kit (cid:0)e2(cid:25)kis)v =e2(cid:25)kisw+e2(cid:25)kitv; letting s =k;
we have
(cid:16) (cid:17)
k+1
w+e2(cid:25)kitv =w+e2(cid:25)kitv
(cid:16)for 1 (cid:20)(cid:17) t (cid:20) k: If we multiply the left side we get wk+1 plus terms of the form
m
e2(cid:25)kit qm, where qm is the sum of all products containing m factors of v and
(cid:16) (cid:17)
P k+1
k+1(cid:0)m(cid:16)factor(cid:17)s of w: If we take the sum kt=1 w+e2(cid:25)kitv , we see that all
m
the terms e2(cid:25)kit qm disappear except when m=k: This gives us
Xk (cid:16) (cid:17)
k+1
w+e2(cid:25)kitv =kwk+1+kqk =kw+kqk:
t=1
(cid:16) (cid:17)
P
Ontheotherhand,thesumoftherighthandside,i.e., kt=1 w+e2(cid:25)kitv ,isequal
tokw:Itfollowsthatq =0:However,asimplecomputation(usingvk+1 =v)shows
k
that
0=vq (cid:0)q v =vw(cid:0)wv:
k k
Hence we have vw =wv; from which it follows that uv=vu:
Recall that P is the set of all polynomials f with no repeated roots and of
0
degree greater than 1 such that f(0)=0.
Theorem 2.3. SupposeA;B are unitalcomplex algebras and(cid:30):A!B is a linear
map, and suppose f(x) is a polynomial in P such that f((cid:30)(a)) = 0 whenever
0
a 2 A and f(a) = 0: Then there exist a positive integer k; a polynomial h(x) and
an orthogonal family p ;::: ;p of idempotent in B such that
1 k
(1) f(x)=xh(xk);
(2) if (cid:21) is any root of xk(cid:0)1; then (cid:21)Z(f)=Z(f);
(3) g((cid:30)(a))=0 for every idempotent a2A; where g(x)=xk+1(cid:0)x;
P
(4) (cid:30)(1)= ks=1e2(cid:25)ksips;
(5) p commutes with (cid:30)(a) for 1(cid:20)s(cid:20)k and for every idempotent a2A;
s P
(6) (cid:30)(a)= k (cid:30)(a)p for every idempotent a2A:
s=1 s
Proof. Let G = G(f) = f0 6= (cid:21) : (cid:21)Z(f) = Z(f)g. G is a (cid:12)nite multiplicative
subgroup of the unit circle, and thus, for some positive integer k; G = f(cid:21) 2 C :
(cid:21)k = 1g: It follows that (1) and (2) must be true. As in the proof of Theorem
2.1, we see that if a is an idempotent in A; and (cid:21) 6= 0 is a root of the minimal
polynomial of (cid:30)(a); then (cid:21)2G: Hence (3) is true. Since g((cid:30)(1)) =0; (4) must be
true, with possibly p =0 for some values of s:
s
Ifk =1;itfollowsfrom(3)that(cid:30)isidempotent-preserving,and,thus,forevery
idempotenta2A;(cid:30)(a)and(cid:30)(1(cid:0)a)mustbeorthogonalidempotentswhosesumis
LINEAR MAPS ON OPERATOR ALGEBRAS 2387
(cid:30)(1): This implies that, for every idempotent a 2 A; (cid:30)(a)(cid:30)(1) = (cid:30)(1)(cid:30)(a) = (cid:30)(a),
which yields both (5) and (6).
On the other hand, suppose k (cid:21) 2: It follows, forhevery idempotent a 2 Ai; for
every(cid:11)2Z(f)andallintegers1(cid:20)s;t(cid:20)k;thatf((cid:11) e2(cid:25)kis1+(e2(cid:25)kit (cid:0)e2(cid:25)kis)a )=
0;soany nonzerorootof the minimal polynomialofe2(cid:25)kis(cid:30)(1)+(e2(cid:25)kit (cid:0)e2(cid:25)kis)(cid:30)(a)
must be in G: Hence the hypothesis of Lemma 2.2 holds, with u = (cid:30)(1) and
v =(cid:30)(a). Thus (5) and (6) follow from Lemma 2.2.
For an algebraA overC we denote by A(cid:10)M (C) the algebraof n(cid:2)n matrices
n
with entries in A. If (cid:30) : A ! B is a linear map, we denote by (cid:30) the linear map
n
fromA(cid:10)Mn(C)intoB(cid:10)Mn(C)whichmapseverymatrix(aij)n(cid:2)n to((cid:30)(aij))n(cid:2)n.
By use of Theorems 2.1 and 2.3 we can get a characterization of homomorphisms
between algebras.
Corollary 2.4. Let A and B be unital complex algebras. Let (cid:30):A!B be a linear
map. If (cid:30) preserves the elements annihilated by a polynomial f(x) 2 P , then
2 0
there is a Jordan homomorphism : A!B such that (c) (c(cid:0)1ac) = (a) (c)
holds for any a and invertible element c in A, and such that (cid:30)=(cid:30)(1) = (cid:30)(1).
Proof. Obviously, (cid:30) preserves elements annihilated by f, too. Take the orthog-
onal family of idempotents p ;:::;p in B as that in Theorem 2.3. Then (cid:30)(1) =
P P 1 k
ks=1e2(cid:25)ksips, p = ks=1ps = (cid:30)(1)k, and for any idempotent e in A we have
(cid:30)(1)(cid:30)(e) = (cid:30)(e)(cid:30)(1) and (cid:30)(e) = p(cid:30)(e) = (cid:30)(e)p. We claim that these hold(cid:20)true fo(cid:21)r
1 a
all elements of A. In fact, for any a 2 A, if we take an idempotent A=
(cid:20) (cid:21) 0 0
(cid:30)(1) 0
and write (cid:30) (1)= , we get
2 0 (cid:30)(1)
(cid:20) (cid:21) (cid:20) (cid:21)
(cid:30)(1)2 (cid:30)(1)(cid:30)(a) (cid:30)(1)2 (cid:30)(a)(cid:30)(1)
=(cid:30) (1)(cid:30) (A)=(cid:30) (A)(cid:30) (1)=
0 0 2 2 2 2 0 0
and
(cid:20) (cid:21) (cid:20) (cid:21)
(cid:30)(1) p(cid:30)(a) (cid:30)(1) (cid:30)(a)
=(cid:30) (A)k+1 =(cid:30) (A)= ;
0 0 2 2 0 0
(cid:20)fromwh(cid:21)ichitfollowsthat(cid:30)(1)(cid:30)(a)=(cid:30)(a)(cid:30)(1) and(cid:30)(a)=p(cid:30)(a). IfonetakesA=
1 0
, one get (cid:30)(a) = (cid:30)(a)p. Hence the range of (cid:30) is contained in B = pBp,
a 0 1
whichis a subalgebraofB with unit p: Moreover,everyp commutes with (cid:30)(a) for
P s
all a 2 A. Let b0 = ks=1e(cid:0)2(cid:25)ksips and = b0(cid:30). Then (cid:30) = (cid:30)(1) . Regard (cid:30) as a
linear map from A into B . It follows from Theorem 2.3 (2) that both and
1 2
are unital and still preserve elements annihilated by f, since e(cid:0)2(cid:25)ksiZ(f) = Z(f).
So,by Theorem2.1,both and are idempotent-preserving. We provethat is
2
indeed a Jordan homomorphism. (cid:20) (cid:21)
a c
For any element a and invertible element c in A, is
(cid:20) c(cid:0)1(a(cid:0)a(cid:21)2) 1(cid:0)c(cid:0)1ac
(a) (c)
anidempotentinA(cid:10)M . So, isanidempotent.
2 (c(cid:0)1(a(cid:0)a2)) 1(cid:0) (c(cid:0)1ac)
Hence we have
(2.1) (a)2+ (c) (c(cid:0)1(a(cid:0)a2))= (a)
2388 JINCHUAN HOU AND SHENGZHAO HOU
and
(2.2) (a) (c)+ (c)(1(cid:0) (c(cid:0)1ac))= (c):
Let c=1 in (2.1). We get
(a2)= (a)2;
that is, is a Jordan homomorphism. It follows from (2.2) that
(2.3) (c) (c(cid:0)1ac)= (a) (c);
completing the proof.
Corollary 2.5. Let A and B be unital complex algebras. Let (cid:30) : A ! B be a
linear map and f(x) a polynomial in P . The following statements are equivalent.
0
(1) (cid:30) preserves elements annihilated by f(x).
3
(2) There exist a homomorphism :A!B and an n-potent element b for some
integer n(cid:21)2 such that the nonzero roots of a minimal polynomial of b are in G(f)
and (cid:30)=b = b.
Proof. (2)=)(1). Write (cid:30) = b , where is homomorphism and b is an n-potent,
i.e., bn =b. Let f(cid:21) ;:::;(cid:21) g be the set of nonzero roots of the minimal polynomial
1 l
of b. Then there exists a set of mutually orthogonal idempotents fe ;:::;e g such
P 1 l
thatb= l (cid:21) e ande (a)= (a)e foralla2A. Ifa2Aandf(a)=0,then
s=P1 s s s P s
f((cid:30)(a)) = f( l (cid:21) e (a)) = l (f((cid:21) a))e = 0, since is homomorphic
s=1 s s s=1 s s
and f((cid:21) a) = 0; by the assumption that (cid:21) Z(f) = Z(f) for 1 (cid:20) s (cid:20) l: As is
s s 3
still a homomorphism, one easily checks that (cid:30) preserves elements annihilated by
3
f, too.
(1)=)(2). Since (cid:30) also preserves the elements annihilated by f, by Corollary
2
2.4,thereexista(k+1)-potentelementbandaJordanhomomorphism satisfying
equation (2.3) such that (cid:30) = b = b, where k is the degree of the group G(f).
We prove that is indeed a homomorphism.
Without loss ofgeneralitywe may assume is unital(otherwise,replace B with
B =pBp as in the proof of Corollary2.4). Then is idempotent-preserving. For
1 3
any elements a;w and invertible c in A,
2 3
(a) (c) 0
4 (c(cid:0)1(a(cid:0)a2)) 1(cid:0) (c(cid:0)1ac) 0 5
(wa(cid:0)w) (wc) 1
is idempotent since
2 3
a c 0
4 c(cid:0)1(a(cid:0)a2) 1(cid:0)c(cid:0)1ac 0 5
wa(cid:0)w wc 1
is. Thus we have
(wa(cid:0)w) (c)+ (!c)(1(cid:0) (c(cid:0)1ac))+ (wc)= (wc);
that is,
(2.4) (wa) (c)(cid:0) (w) (c)+ (wc)(cid:0) (wc) (c(cid:0)1ac)=0:
LINEAR MAPS ON OPERATOR ALGEBRAS 2389
It is well known that a unital Jordan homomorphism is invertibility-preserving
and (c(cid:0)1)= (c)(cid:0)1. So, by (2.3) we have (c(cid:0)1ac)= (c)(cid:0)1 (a) (c), and (2.4)
becomes
(2.5) (wa) (c)(cid:0) (w) (c)+ (wc)(cid:0) (wc) (c)(cid:0)1 (a) (c)=0:
Taking w =c(cid:0)1 in (2.5), we see that
(c(cid:0)1a)= (c)(cid:0)1 (a)
for all a and invertible c. Using (2.5) again, we get (wa) = (w) (a), that is,
is a homomorphism.
3. Linear maps on operator algebras
Inthissection,weapplytheresultsinSection2tothelinearmapsfromaunital
operatoralgebrainto anotherwhichpreserveelementsannihilatedbyapolynomial
in P , that is, a polynomial of degree greater than 1 and with no repeated roots
0
such that f(0) = 0. The question we consider here is mainly to determine when
the structure of such linear preservers can be described by homomorphisms, anti-
homomorphisms, or, at least, Jordan homomorphisms.
Thefollowingtheoremgivesacharacterizationofallweaklycontinuoussurjective
linear maps on standard operator algebras which preserve elements annihilated by
apolynomialinP . RecallthataclosedsubalgebraAinB(X)iscalledastandard
0
operator algebra if it contains the identity I and the set F(X) of all (cid:12)nite rank
operators. Of course B(X) itself is a standard operator algebra. We say that a
linear map (cid:30) : A!B commutes with a polynomial f(x) if f((cid:30)(a)) = (cid:30)(f(a)) for
all elements a2A.
Theorem 3.1. Let A and B be standard operator algebras on Banach spaces X
and Y, respectively. Suppose (cid:30):A!B is a linear surjective map which is contin-
uous in the weak operator topology and f(x) is a polynomial with no repeated roots
and degf (cid:21)2 such that f(0)=0. Then the following are equivalent.
(1) (cid:30) commutes with f(x).
(2) (cid:30) preserves elements annihilated by f(x).
(3) There exists a scalar (cid:21) which is a root of 1 satisfying (cid:21)Z(f) = Z(f), and
either there exists an invertible operator T 2 B(X;Y) such that (cid:30)(A) = (cid:21)TAT(cid:0)1
for all A 2 B(X), or there exists an invertible operator T 2 B(X(cid:3);Y) such that
(cid:30)(A)=(cid:21)TA(cid:3)T(cid:0)1 for all A2B(X). In the last case both X and Y are re(cid:13)exive.
Proof. (1)=)(2). Obvious.
(3)=)(1). Suppose(3). LetkbethedegreeofthegroupG(f)andletdegf =n.
The condition (cid:21)Z(f) = Z(f) implies that k divides n(cid:0) 1, since f(0) = 0. So
f((cid:21)x) = (cid:21)f(x). If (cid:30)(A) = (cid:21)TAT(cid:0)1, then f((cid:30)(A)) = Tf((cid:21)A)T(cid:0)1 = (cid:30)(f(A)), that
is, (1) is true.
(2)=)(3). Suppose that (cid:30) : A!B is a weakly continuous surjective linear
map which preserves elements annihilated by the polynomial f(x). We claim that
(cid:30)(I) = (cid:21)I for some scalar (cid:21) satisfying (cid:21)Z(f) = Z(f) and (cid:21)k = 1 for some k (cid:21) 1.
Take k the same as in Theorem 2.3. Then, for every idempotent A 2 A one has
(cid:30)(A)k = (cid:30)(A) and (cid:30)(I)(cid:30)(A) = (cid:30)(A)(cid:30)(I). Notice that every rank one operator
can be represented as a linear combination of at most two rank one idempotent
2390 JINCHUAN HOU AND SHENGZHAO HOU
operators. Theweakcontinuityof(cid:30)andtheweakdensityofall(cid:12)niterankoperators
in B(X) together imply that
(cid:30)(I)(cid:30)(S)=(cid:30)(S)(cid:30)(I)
for every S 2 A. Thus (cid:30)(I) is the identity multiplied by a scalar, since F(X) (cid:26)
A. It is clear that (cid:30)(I) 6= 0, for otherwise Theorem 2.3 (6) would imply that
(cid:30)(A) = (cid:30)(A)(cid:30)(I) = 0 for all idempotents A, and consequently (cid:30) = 0. Therefore
we must have (cid:30)(I)=(cid:21)I for some (cid:21)6=0 with (cid:21)k =1. By Theorem 2.3 (2) and (4),
(cid:21)Z(f)=Z(f).
Let = (cid:21)(cid:0)1(cid:30). Then : A!B is a weakly continuous surjective unital linear
map which also preserves elements annihilated by f(x), as (cid:21)(cid:0)1Z(f) = Z(f). By
Theorem 2.1, is idempotent-preserving. It follows from [3, Theorem 3.3] that
is either an isomorphism or an anti-isomorphism. Therefore, either there is an
invertible operator T 2B(X;Y) such that
(cid:30)(A)=(cid:21)TAT(cid:0)1
for every A2A, or there is an invertible operator T 2B(X(cid:3);Y) such that
(cid:30)(A)=(cid:21)TA(cid:3)T(cid:0)1
for every A 2 A, with (cid:21)k = 1. It is clear that in the last case, X and Y are
re(cid:13)exive(for the proof of re(cid:13)exivity,see [3, Prop. 3.1]or [7, Thm. 1.4]). The proof
is (cid:12)nished.
Remark 3.1. Theorem 3.1 remains true if the assumption of weak continuity on
the linear map (cid:30) is replaced by the assumption that (cid:30) is continuous in the strong
operator topology or in the (cid:27)-weak operator topology.
Next we consider linear maps on von Neumann algebras.
Theorem 3.2. Let A be a von Neumann algebra and B a unital Banach algebra,
and let (cid:30) : A!B be a bounded linear map. Let f(x) be a polynomial with no
repeated roots and degf (cid:21)2 such that f(0)=0. Then the following statements are
equivalent.
(1) (cid:30) commutes with f(x).
(2) (cid:30) preserves elements annihilated by f(x).
(3) There exist an r-potent element B 2 B with spectrum contained in f(cid:21) :
(cid:21)Z(f) = Z(f)g and a Jordan homomorphism from A into B such that (cid:30) =
B = B.
Proof. It is obvious that (1)=)(2).
(3)=)(1). SaythespectrumSp(B)=f(cid:21) ;:::;(cid:21) g. Thereisanorthogonalfamily
1 Pl
P ;:::;P of idempotents in B such that B = l (cid:21) P . As every (cid:21) satis(cid:12)es
1 l s=1 s s s
(cid:21) Z(f) = Z(f), one has Bk+1 = B, where k (cid:21) 1 is the integer as in Theorem
s
2.3. It is clear that P (A) = (A)P . Now if A is any element in A, then,
s s P
since is a Jordan homomorphism, one has f((cid:30)(A)) = l f((cid:21) (A)P ) =
P P s=1 s s
l (cid:21) f( (A))P = l (cid:21) P (f(A))=(cid:30)(f(A)).
s=1 s s s=1 s s
(2)=)(3). Denote the unit by I for both A and B. By Theorem 2.3, there
is an orthogonal familyPP1;:::;Pk(cid:0)1 of idempotents in B commutinPg with every
(cid:30)(A), andP(cid:30)(A) = (cid:30)(A) ks=(cid:0)11Ps whenever A2 = A, while (cid:30)(I) = ks=(cid:0)11e2k(cid:25)(cid:0)s1iPs.
Let P = k(cid:0)1P and B = (cid:30)(I) + (I (cid:0) P). Then P2 = P, Bk(cid:0)1 = I, and
s=1 s 0 0
B (cid:30)(A) = (cid:30)(A)B = (cid:30)(I)(cid:30)(A) if A2 = A. It is well known that the subspace
0 0
LINEAR MAPS ON OPERATOR ALGEBRAS 2391
S of all linear combinations of projections (i.e., self-adjoint idempotents) in a von
Neumannalgebraisadensesubsetunderthenormtopology. Since(cid:30)iscontinuous,
we get (cid:30)(S) = (cid:30)(S)P = P(cid:30)(S) and B (cid:30)(S) = (cid:30)(S)B = (cid:30)(I)(cid:30)(S) for all S in A.
0 0
Let B = PBP and = B(cid:0)1(cid:30). It is clear that : A!B is bounded with
1 0
range contained B , and preserves element annihilated by f, since B is a Banach
1 1
algebra with unit P and (I) = P. By Theorem 2.1 is idempotent-preserving
andhence transformsa setoforthogonalidempotents of Ainto a setoforthogonal
idempotents of B (cid:26)B.
1
Now, if A=A(cid:3) is a self-adjointelement in the vonNeumann algebraA, then A
is the limit of a sequence of linear combinationsof orthogonalprojections andcon-
sequently,bycontinuityof , (A)isthelimitofasequenceoflinearcombinations
of orthogonal idempotents. It turns out that (A2)= (A)2 holds for self-adjoint
elementsA. ReplacingAbyC+D inthisrelation,whereC andD areself-adjoint,
weget (CD+DC)= (C) (D)+ (D) (C). Forany S 2A, letC = 1(S+S(cid:3))
2
and D = 1(S(cid:0)S(cid:3)); then C, D are self-adjoint and S =C+iD. Hence
2i
(S2)= (C2(cid:0)D2+i(CD+DC))
= (C)2(cid:0) (D)2+i( (C) (D)+ (D) (C))= (S)2;
that is, is a Jordan homomorphism and (cid:30)=B =B with B =(cid:30)(I).
0
Corollary 3.3. LetA,B,(cid:30)andf bethesameasinTheorem3.2. If(cid:30)issurjective,
then the following statements are equivalent.
(1) (cid:30) commutes with f(x).
(2) (cid:30) preserves elements annihilated by f(x).
(3) There exists an invertible r-potent element B in the center of B satisfying
Sp(B)(cid:26)((cid:21): (cid:21)Z(f)=Z(f)g, and there exists a Jordan homomorphism from A
onto B such that (cid:30) =B = B. Here the identity element is denoted again by I
for both A and B.
P
Proof. Let B = (cid:30)(I) and P = k(cid:0)1P as in the proof of Theorem 3.2. Since (cid:30)
s=1 s
is surjective, B and P commute with every element in B, i.e., both B and P are
in the center of B. Moreover, (cid:30)(S)= (cid:30)(S)P holds for every element S in A. This
implies that P = I and hence (cid:30)(I) is invertible. Now let = B(cid:0)1(cid:30) and apply
Theorem 3.2.
Notice that, in the above corollary,if B is a factor, then B =(cid:21)I for some scalar
(cid:21) with (cid:21)k(cid:0)1 =1. If B is prime, then the Jordan homomorphism onto B is either a
homomorphismor an anti-homomorphism. So if B is a factor and prime, then (cid:30) is
a homomorphism or anti-homomorphism multiplied by a scalar which is a root of
1. In particular, we have
Corollary 3.4. Let H and K be Hilbert spaces and (cid:30):B(H)!B(K) a bounded
surjective linear map. Let f(x) be a polynomial of degree greater than 1 and with
no repeated roots such that f(0)=0. Then the following are equivalent.
(1) (cid:30) preserves elements annihilated by f(x):
(2) There exist a scalar (cid:21) which is a root of 1 satisfying (cid:21)Z(f)=Z(f) and an
invertible operator T 2B(H;K) such that either (cid:30)(A)=(cid:21)TAT(cid:0)1 for every A, or
(cid:30)(A) = (cid:21)TAtT(cid:0)1 for every A, where At denotes the transpose of A relative to a
basis.
2392 JINCHUAN HOU AND SHENGZHAO HOU
Comparing Theorem 3.1 with Corollary 3.3, we weaken the assumption of (cid:30)
being weakly continuous to that of (cid:30) being bounded for the case A = B(H) and
B =B(K) with H and K being Hilbert spaces. After the next lemma, we can get
a stronger one by omitting the assumption of continuity.
We say that a Banach space X has in(cid:12)nite multiplicity if X is isomorphic to
X (cid:8) X and to an in(cid:12)nite direct sum X(1) of copies of X in such a way that
arbitrarypermutations of coordinates and transformations of the form A(1) (with
A 2 B(X)) are bounded operators, and such that, for any bounded operators A
andB onX,A(1)(cid:8)B(1) actingonX(1)(cid:8)X(1) issimilarto(A(cid:8)B)(1) actingon
(X (cid:8)X)(1). In(cid:12)nite dimensional c -space, in(cid:12)nite dimensional lp-space, Lp[0;1],
0
1(cid:20)p(cid:20)1, and in(cid:12)nite dimensional Hilbert spaces are examples of Banachspaces
of in(cid:12)nite multiplicity [6].
Lemma 3.5. Let X and Y be complex Banach spaces and (cid:30) : B(X) ! B(Y) a
linear map. Let f(x) be a polynomial of degree greater than 1 and with no repeated
roots such that f(0) = 0. If (cid:30) preserves elements annihilated by f and if X has
in(cid:12)nite multiplicity, then (cid:30)(A)(cid:30)(I) = (cid:30)(I)(cid:30)(A) for every A in B(X), and there
existsanidempotent-preservinglinearmap :B(X)!B(Y)suchthat(cid:30)=(cid:30)(I) .
Proof. It follows from Theorem 2.3 that there is an orthogonal family P1;:::;Pk(cid:0)1
of idempotent operators in B(Y) such that P commutes with every (cid:30)(A) and
P s P
(cid:30)(A)=(cid:30)P(A) ks=(cid:0)11Ps whenever A2 =A, and (cid:30)(I)= ks=(cid:0)11e2k(cid:25)(cid:0)s1iPs. Let B =(cid:30)(I)
andP = k(cid:0)1P . ThenBk =B andP2 =P. SinceX hasin(cid:12)nitemultiplicity,by
s=1 s
[6], every S 2 B(X) is a di(cid:11)erence of sums of idempotents in B(X). This implies
that (cid:30)(S)B = B(cid:30)(S) and (cid:30)(S) = (cid:30)(S)P = P(cid:30)(S) hold for all S 2 B(X). Let
B = B +(I (cid:0)P) and = B(cid:0)1(cid:30). Then preserves elements annihilated by f,
0 0
and (I)=P. IfweletY =ranP,the rangeofP,and(cid:18) :B(X)!B(Y )de(cid:12)ned
0 0
by (cid:18)(S) = (S)j , then (cid:18) is unital, and therefore (cid:18), as well as , is idempotent-
Y0
preserving, by Theorem 2.1. It is clear that (cid:30)=B .
Theorem 3.6. Let X, Y, (cid:30) and f be as in Lemma 3.5. If (cid:30) is surjective and
(cid:30)jF(X) 6= 0, then (cid:30) preserves elements annihilated by f(x) if and only if there is
a scalar (cid:21) which is a root of 1 satisfying (cid:21)Z(f) = Z(f) and either there is an
invertible operator T 2 B(X;Y) such that (cid:30)(A) = (cid:21)TAT(cid:0)1 for all A 2 B(X) or
there exists an invertible operator T 2 B(X(cid:3);Y) such that (cid:30)(A) = (cid:21)TA(cid:3)T(cid:0)1 for
all A2B(X). In the last case both X and Y are re(cid:13)exive.
Proof. By Lemma 3.5 it is obvious that (cid:30)(I) = (cid:21)I for some complex number (cid:21)
with (cid:21)k = 1 and (cid:21)Z(f) = Z(f) such that (cid:30) = (cid:21) , where is surjective and
idempotent-preserving. We have to prove that is isomorphic or anti-isomorphic.
IfA2B(X)isanidempotentofrankone,we(cid:12)rstshowthat (A)alsohasrank
at most one. Let M = AB(X)A, M = AB(X)(I (cid:0)A), M = (I (cid:0)A)B(X)A
1 2 3
and M =(I(cid:0)A)B(X)(I(cid:0)A). Thus B(X)=M (cid:8)M (cid:8)M (cid:8)M . Similarly,
4 1 2 3 4
B(Y)=N (cid:8)N (cid:8)N (cid:8)N whereN = (A)B(Y) (A),N = (A)B(Y)(I(cid:0) (A)),
1 2 3 4 1 2
N =(I(cid:0) (A))B(Y) (A)andN =(I(cid:0) (A))B(Y)(I(cid:0) (A)). AsM =CAwe
3 4 1
have (M )=C (A)(cid:18)N . TakeS 2M :ThenA+(cid:11)S,aswellas (A)+(cid:11) (S),
1 1 2
is idempotent for any (cid:11)2C. Hence we have
(A) (S)+ (S) (A)+(cid:11) (S)2 = (S);
andthisyields (S)2 =0and (S)=(I(cid:0) (A)) (S) (A)+ (A) (S)(I(cid:0) (A)) 2
N (cid:8)N . Therefore, (M ) (cid:18) N (cid:8)N , and similarly, (M ) (cid:18) N (cid:8)N . Let
2 3 2 2 3 3 2 3