A Collection of Problems on Spectrally Bounded Operators
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A COLLECTION OF PROBLEMS ONSPECTRALLY BOUNDED OPERATORS ∗ MARTIN MATHIEU
Department of Pure Mathematics, Queen’s University Belfast,Belfast BT7 1NN, Northern [email protected] to Professor Rainer Nagel on the occasion of his retirement.
We discuss several open problems on spectrally bounded operators, some new, some old,adding in a few new insights.
Keywords : Banach algebras, C* -algebras, derivations, Jordan homomorphisms, Lie ho-momorphisms, spectrally bounded operators, spectral isometries, Glimm ideals, Haus-dorff spectrum.AMS Subject Classification: Primary 47A65; Secondary 17C65, 46H10, 46L05, 47A10,47B48.
1. Introduction
Let A be a unital complex Banach algebra. A linear mapping T : E → B from asubspace E ⊆ A into another unital complex Banach algebra B is called spectrallybounded if there is a constant M ≥ r ( T x ) ≤ M r ( x ) for all x ∈ E .Here, and in what follows, r ( x ) stands for the spectral radius of a Banach algebraelement x .This concept evolved in Banach algebra theory, and especially automatic conti-nuity, over time in the 1970’s and 1980’s but the terminology was only introducedin [16], together with its companions spectrally infinitesimal : M = 0; spectrally con-tractive : M = 1; and spectrally isometric : r ( T x ) = r ( x ) for all x . It follows from [1],see also [4], Lemma A, that the separating space of every surjective spectrallybounded operator T on a closed subspace E is contained in the radical of B ; thus T is bounded if B is semisimple. This was used by Aupetit in [1] to give a new proof of ∗ This paper is an expanded version of a talk given at the conference
Jordan Structures: Nonasso-ciative Analysis and Geometry on 6 September 2008 at Queen Mary College, London.1
Martin Mathieu
Johnson’s uniqueness-of-the-complete-norm-topology theorem; cf. also [25]. It wasfurther exploited in [4] to investigate continuity properties of Lie epimorphisms.A systematic study of spectrally bounded operators was begun in [26], with itsmain results published in [22] and [23]. Since then, the interest in the topic hassteadily grown and by now there is a sizeable literature on a variety of aspects. Thepresent paper’s aim is to discuss several of the open problems on spectrally boundedoperators along with some new results that should make it clear why these questionsare natural and important.Section 2 is devoted to a recapitulation of probably the most important anddeepest problem on spectrally bounded operators, the non-commutative Singer–Wermer conjecture. No substantial progress seems to have been made on this overthe last few years. A standard method to reduce a more sophisticated problem inBanach algebra theory to a simpler one is to use quotients. When dealing withoperators between Banach algebras, we, of course, need invariant ideals to per-form this. In Section 3 we discuss the interplay between properties of a (spectrallybounded) operator and the operator it induces on suitable quotient Banach alge-bras in a fairly general setting and recover recent results for operators preservingthe essential spectral radius on B ( H ) in [5].The identity element plays a distinguished role for spectrally bounded operators.Suppose that T : A → B is a surjective spectral isometry between unital C* -algebras A and B . Since T restricted to the centre Z ( A ) of A is a *-isomorphism onto thecentre Z ( B ) of B ([19], Proposition 2.3), T B . Therefore,whenever we study a spectral isometry T , we can without loss of generality assumethat T is unital (i.e., T T
1. As it turns out, if the domainalgebra is sufficiently ‘infinite’, then T B .As a consequence, we are able to extend the main results in [23] and [15] to thenon-unital setting (Theorem 4.3 below). The special case A = B ( H ), B = B ( K )for Hilbert spaces H and K was studied previously in [9].In [19], we propose a non-selfadjoint version of Kadison’s classical theorem stat-ing that every unital surjective isometry between unital C* -algebras is a Jordan*-isomorphism. In many cases, this conjecture (see Problem 5 below) has been con-firmed, however almost always under some strong assumption of ‘infiniteness’ on thedomain C* -algebra. In Section 5, we employ a reduction method developed in [24]and [20] to obtain a new result of a similar ilk (Theorem 5.2). More importantly,perhaps, we propose a new route based on a description of spectrally bounded op-erators in the presence of a trace to solve this open question at least in the settingof II factors (Corollary 5.6). Collection of Problems on Spectrally Bounded Operators
2. Automatic Continuity
The separating space S ( T ) of an operator T between normed spaces E and F isdefined by S ( T ) = { y ∈ F | y = lim n T x n for some ( x n ) n ∈ N ⊆ E, x n → } and measures the degree of discontinuity of T . If E and F are Banach spaces then T is bounded if and only if S ( T ) = { } , by the closed graph theorem. For a detaileddiscussion of the separating space see [10].It follows from [1], see also [25] and [4], that S ( T ) ∩ T E consists of quasinilpotentelements whenever T : E → B is a spectrally bounded operator on a subspace E of a Banach algebra. In fact, r ( T x ) ≤ r ( T x − y ) for each x ∈ E and y ∈ S ( T ).This, together with Zem´anek’s characterisation of the Jacobson radical rad( A ) of A ,entails the following result. Theorem 2.1 (Aupetit).
Let T : E → B be a spectrally bounded operator on aclosed subspace E of a Banach algebra onto a Banach algebra B . Then S ( T ) ⊆ rad( B ) . In particular, if B is semisimple, then T is bounded. The above estimate on the spectral radii is used in [4] to describe the continuityof surjective Lie homomorphisms as follows.
Theorem 2.2 (Aupetit–Mathieu).
The separating space S ( θ ) of a Lie epimor-phism θ between two Banach algebras A and B is contained in Z ( B ) , the centremodulo the radical. The centre modulo the radical, Z ( B ), is the pre-image of the centre of ˆ B = B/ rad( B ) under the canonical map B → ˆ B and turns out to be the largest quasinil-potent Lie ideal of B . As a consequence, it is invariant under every Lie epimorphismand thus the above result yields the following automatic continuity statement ([4],Corollary). Corollary 2.3.
Let θ : A → B be a Lie epimorphism between the Banach algebras A and B . The induced Lie epimorphism ˆ θ : ˆ A = A/ rad( A ) → ˆ B between the BanachLie algebras ˆ A and ˆ B is continuous. The centre modulo the radical also plays an important role in the automaticcontinuity of derivations on Banach algebras. It is an open question whether theseparating space of each of the iterates δ n , n ∈ N of a derivation δ : A → A on aBanach algebra A is contained in the radical of A . If this is the case, the followinglong-standing problem would have a positive answer. Problem 1 (Noncommutative Singer–Wermer Conjecture).
Does [ x, δx ] ∈Z ( A ) for all x ∈ A imply that δA ⊆ rad( A )?Here, [ x, y ] of course stands for the commutator xy − yx . If Z ( A ) is replaced bythe proper centre Z ( A ) of A , then Problem 1 has an affirmative answer, due to a Martin Mathieu reduction technique developed in [21] and Marc Thomas’ famous theorem [28]. Theconnection to spectrally bounded operators is provided by the next result [7].
Theorem 2.4 (Breˇsar–Mathieu).
A derivation on a Banach algebra A is spec-trally bounded if and only if it maps into rad( A ) . Hence, Problem 1 is equivalent to the following question.
Problem 1 ′ . Does [ x, δx ] ∈ Z ( A ) for all x ∈ A imply that δ is spectrally bounded?It appears that no progress on this question has been made in the last decade.For a fuller discussion see [16] and [10].
3. Quotients
In the previous section, we exploited the concept of spectrally bounded operatorsto study linear mappings satisfying some additional algebraic conditions, such asLie homomorphisms, derivations, etc. Another main direction of research has beenon the question which algebraic properties can be derived from the assumption ofspectral boundedness. Maybe the most prominent of these problems is the followingone from [14]. By a
Jordan epimorphism we will, of course, understand a surjectivelinear mapping preserving the Jordan product x ◦ y = ( xy + yx ). Problem 2 (Kaplansky).
Let A and B be semisimple unital Banach algebras.Suppose T : A → B is unital, surjective and invertibility-preserving. Is T necessarilya Jordan epimorphism?Note that a Jordan epimorphism necessarily has all the properties assumed inProblem 2 and hence is bounded, by Theorem 2.1. Many contributions on Kaplan-sky’s problem have been made over the past decades but so far it eludes a finalanswer. From Aupetit’s substantial work on the question we only quote the follow-ing beautiful result in [3]; see also the references therein. Theorem 3.1 (Aupetit).
Let A and B be von Neumann algebras. Then everyunital surjective invertibility-preserving linear mapping T : A → B is a Jordan epi-morphism. To the best of our knowledge, Problem 2 is still open even in the case when bothdomain and codomain are C* -algebras; cf. also [11]. The structural investigation ofspectrally bounded operators is partly motivated by the question to what extentthe hypotheses in Kaplansky’s problem can be relaxed while retaining the samegoal, to show that the mapping is a Jordan homomorphism. The relevance of thevalue at 1 will be discussed in the next section. When considering the possibility ofreplacing “invertibility-preserving” by “spectrally bounded” we must keep in mindthat (a) every bounded linear mapping defined on a commutative C* -algebra isspectrally bounded (since spectral radius and norm coincide in spaces of the form C ( X )); (b) a finite trace on a C* -algebra is a spectral contraction, hence, alreadyon the n × n matrices there is a unital spectrally bounded operator onto C which Collection of Problems on Spectrally Bounded Operators is not a Jordan epimorphism. Notwithstanding this there are satisfactory results inthe setting of ‘very infinite’ C* -algebras; see Sections 4 and 5.Many examples are known illustrating the fact that no strong results can beexpected for non-surjective spectrally bounded operators in general; see, e.g., [11].We shall now study the situation when a spectrally bounded operator is merelysurjective ‘up to’ or ‘modulo’ an ideal.Suppose I ⊆ A and J ⊆ B are proper closed ideals in the unital Banach algebras A and B , respectively. We say that a linear mapping T : A → B is essentiallyspectrally bounded (more precisely, I - J - essentially spectrally bounded ) if there is aconstant M ≥ r ( T x + J ) ≤ M r ( x + I ) for all x ∈ A . We call T surjectivemodulo J if, for every y ∈ B , there is x ∈ A such such y − T x ∈ J . If T I ⊆ J ,we can define the induced linear mapping ˆ T : A/I → B/J by ˆ T ( x + I ) = T x + J , x ∈ A . The following proposition relates the properties of ˆ T to those of T . Proposition 3.2.
Let A and B be unital Banach algebras. Suppose that I and J are proper closed ideals of A and B , respectively, such that B/J is semisimple. Fora linear mapping T : A → B the following conditions are equivalent. (a) T is essentially spectrally bounded and surjective modulo J ; (b) T I ⊆ J and ˆ T is spectrally bounded and surjective. Proof. (b) ⇒ (a) Under the assumption T I ⊆ J , ˆ T is a well-defined linearmapping which is spectrally bounded if and only if T is essentially spectrallybounded, by definition. Moreover, ˆ T surjective precisely means that, for each y ∈ B , y + J = ˆ T ( x + I ) = T x + J for some x ∈ A ; that is, T is surjective modulo J .(a) ⇒ (b) By the first paragraph in this proof, it suffices to show that T I ⊆ J . Take x ∈ I . By hypothesis, for each y ∈ B , there is x ′ ∈ A such that y − T x ′ ∈ J . Let λ ∈ C . We have r ( λ ( T x + J ) + y + J ) = r ( λ T x + y + J )= r ( T ( λ x + x ′ ) + J ) ≤ M r ( λ x + x ′ + I )= M r ( x ′ + I ) , for some M ≥
0. Consequently, the subharmonic function λ r ( λ ( T x + J ) + y + J )is bounded on C , hence constant. It follows that r ( T x + J + y + J ) = r ( y + J )for all y ∈ B from which we conclude that T x + J ∈ rad( B/J ), by Zem´anek’scharacterisation of the radical. As
B/J is semisimple, we obtain that
T x ∈ J asdesired.As an immediate consequence we have the following variant of the main resultin [15]. Corollary 3.3.
Let T : A → B be a unital linear mapping from a unital purelyinfinite C* -algebra A with real rank zero into a unital Banach algebra B . Let J ⊆ B Martin Mathieu be a proper closed ideal in B such that B/J is semisimple and suppose that, forsome proper closed ideal I ⊆ A of A , the operator T is I - J -essentially spectrallybounded and surjective modulo J . Then T is a Jordan epimorphism modulo J . Proof.
By Proposition 3.2,
T I ⊆ J and the induced mapping ˆ T : A/I → B/J isunital, spectrally bounded and surjective. Since
A/I is purely infinite and has realrank zero, Corollary 2.5 in [15] implies that ˆ T is a Jordan epimorphism; hence, T is a Jordan epimorphism modulo J .In particular, if T is even a spectral isometry modulo J in the above situation,then ˆ T provides a Jordan isomorphism between the quotients A/I and
B/J . Thiswas obtained in the special case A = B = B ( H ) and I = J = K ( H ) for an infinitedimensional Hilbert space H in [5], Theorem 3.1.In order to make use of quotients, invariant ideals are needed. In the settingof C* -algebras, so-called Glimm ideals offer themselves as good candidates, seeSection 5 below. However, so far their invariance has only been established underadditional hypotheses, for instance for spectral isometries on von Neumann alge-bras [24]. Problem 3.
Does every unital spectral isometry from a unital C* -algebra ontoanother one leave each Glimm ideal invariant?The setting of C* -algebras is favourable also because a spectral isometry inducesa spectral isometry on every quotient ([24], Proposition 9). In a more general settingit is easy to see that a spectrally bounded operator induces a spectrally boundedquotient operator provided the domain algebra is an SR -algebra; cf. [20].
4. The Relevance of the Value at 1
A surjective Jordan homomorphism between unital algebras attains the value 1 at 1.In this section, we shall discuss the behaviour of an arbitrary spectrally boundedoperator at 1 to see how this affects the possible difference from a Jordan homomor-phism. Remembering that every bounded operator from a space C ( X ) is spectrallybounded, we have to be careful not to expect too much in a general setting.The following observation follows directly from our earlier results. Proposition 4.1.
Let T : A → B be a spectrally bounded operator from a unital C* -algebra A onto a unital semisimple Banach algebra B . For every pair p, q ofmutually orthogonal properly infinite projections in A we have ( T a ) (
T b ) + (
T b ) (
T a ) = 0 ( a ∈ pAp, b ∈ qAq ) . (4.1) Proof.
By [23], Corollary 3.2, T preserves elements of square zero. By [15], Propo-sition 2.1, every element in the corners pAp and qAq can be written as a finite sumof elements of square zero (in pAp and qAq , respectively), since both p and q areproperly infinite. The claim thus follows from Lemma 3.3 in [23]. Collection of Problems on Spectrally Bounded Operators Corollary 4.2.
Let T : A → B be a spectrally bounded operator from a unital C* -algebra A with real rank zero onto a unital semisimple Banach algebra B . Supposethat every non-zero projection in A is properly infinite. Then T is an invertibleelement in the centre of B . Proof.
The basic idea of the argument has been used before in some special cases,see, e.g., [9]. Let p ∈ A be a non-trivial projection. Applying the identity in Propo-sition 4.1 to a = p and b = q = 1 − p we obtain( T p ) T (1 − p ) + T (1 − p ) ( T p ) = 0 , that is, ( T p ) ( T
1) + ( T
1) (
T p ) = 2 (
T p ) . Upon multiplying this identity first on theleft, then on the right by T p and subtracting the resulting two identities we obtain(
T p ) ( T
1) = ( T
1) (
T p ) for every (non-trivial) projection p .Let { p , . . . , p n } be an orthogonal family of projections in A . By applying (4.1)inductively we find that (cid:0) T (cid:0)P i p i (cid:1)(cid:1) = P i (cid:0) T p i (cid:1) . Since A has real rank zero, itfollows that ( T
1) (
T a ) = ( T a ) ( T
1) for all a in a dense subset of A sa and since T is bounded, thus for all a ∈ A sa . As ab + ba = ( a + b ) − a − b and ( a + ib ) = a − b + i ( ab + ba )for all selfadjoint a, b ∈ A , we conclude that T T x ) for every x ∈ A . The surjectivity of T entails that T B but since 2 y = (1 + y ) − − y for each y ∈ B , we finally obtain that T B .Going back to the first paragraph of the proof we therefore have T ( p ) ( T
1) =(
T p ) ( T
1) = (
T p ) for every projection p ∈ A . Using the same argumentationas above and the fact that A has real rank zero another time we conclude that T ( x ) ( T
1) = (
T x ) for all x ∈ A . As T is surjective, it follows that T T is a spectrally bounded operator such that T T defined by ˜ T x = ( T − T x , x ∈ A is a unital spectrally bounded operator; thus we can apply the results knownin this situation. Theorem 4.3.
Let T : A → B be a spectrally bounded operator from a unital C* -algebra A onto a unital semisimple Banach algebra B . Suppose that (i) A is a properly infinite von Neumann algebra, or (ii) A is a purely infinite C* -algebra with real rank zero.Then there is a unique Jordan epimorphism J : A → B such that T x = ( T Jx for all x ∈ A . Moreover, T is a central invertible element in B . Martin Mathieu
Proof.
Case (ii) is immediate from Corollary 4.2, since every non-zero projectionin a purely infinite C* -algebra is properly infinite. Thus we can define J by Jx =( T − T x , x ∈ A , which is a Jordan epimorphism by [15], Corollary 2.5.The case (i) will not only need the result in the unital case, [23], Theorem 3.6,in the same manner as just explained but also an elaboration of the projectiontechniques used in the main technical result to obtain the unital case, which is [23],Proposition 3.4. Note that, by the proof of Corollary 4.2, it suffices to show that T T p ) for every projection p ∈ A ; once this is verified, T p and 1 − p are properly infinite. Then the assertionfollows immediately from identity (4.1).Next suppose that p is properly infinite but 1 − p is not. By the Halving Lemma,there is a subprojection f of p such that p ∼ f ∼ p − f , where ∼ denotes Murray–von Neumann equivalence. Hence, all projections f , 1 − f , p − f and 1 − p + f areproperly infinite, see the proof of [23], Proposition 3.4. By our first step we thushave ( T
1) (
T f ) = ( T f ) ( T
1) and ( T T ( p − f ) = T ( p − f ) ( T . Applying (4.1) to f and p − f we have ( T p ) = T ( p − f ) + ( T f ) , wherefore T T p ) .Suppose now that p is infinite but not properly infinite. Then there is a non-trivial central projection z in A such that zp is properly infinite and (1 − z ) p isfinite. We need the following preliminary observation. Suppose that q is a properlyinfinite projection in A but 1 − q is not. Choosing the subprojection f of q as in thelast paragraph we have( T q ) ( T
1) + ( T
1) (
T q ) = (
T f ) ( T
1) + ( T
1) (
T f ) + T ( q − f ) ( T
1) + ( T T ( q − f )= 2 ( T f ) + 2 T ( q − f ) = 2 ( T q ) , where the last two equality signs come from identity (4.1). Multiplying this identityfirst on the left, then on the right by T T ( T q ) − ( T q ) ( T = 2 (cid:0) ( T
1) (
T q ) − ( T q ) ( T (cid:1) = 0 , by the previous paragraph.To simplify our calculations we will now use the usual commutator notation[ x, y ] = xy − yx . Since zp is a properly infinite projection, the preliminary observa-tion yields [( T , T ( zp )] = 0. By the proof of [23], Proposition 3.4, the projection q = (1 − z )(1 − p ) is properly infinite as well; therefore, [( T , T ((1 − z )(1 − p ))] = 0,too. As a result,[( T , T ((1 − z ) p )] = [( T , T p ] − [( T , T ( zp )] = − [( T , T (1 − p )]= − [( T , T ( z (1 − p ))] − [( T , T ((1 − z )(1 − p ))]= − [( T , T z ] + [( T , T ( zp )] = 0 , Collection of Problems on Spectrally Bounded Operators since every non-zero central projection in A , in particular z , is properly infinite andthus [( T , T z ] = 0 as well.From identity (4.1) we have T ( za ) T ((1 − z ) b ) + T ((1 − z ) b ) T ( za ) = 0 ( a, b ∈ A ) , (4.2)since 1 − z is non-zero, hence properly infinite. In particular,( T z ) T ((1 − z ) p ) + T ((1 − z ) p ) ( T z ) = 0 . It follows that(
T q ) = T ((1 − z )(1 − p )) = (cid:0) T (1 − z ) − T ((1 − z ) p ) (cid:1) = T (1 − z ) + T ((1 − z ) p ) − ( T (1 − z ) T ((1 − z ) p ) + T ((1 − z ) p ) T (1 − z ))= T (1 − z ) + T ((1 − z ) p ) − (( T T ((1 − z ) p ) + T ((1 − z ) p ) ( T . Combining this with[( T , T w ] = T T , T w ] + [ T , T w ] T T , T T w + T w T
1] ( w ∈ A )we find that0 = [ T , ( T q ) ]= [ T , T (1 − z ) ] + [ T , T ((1 − z ) p ) ] − [ T , ( T T ((1 − z ) p ) + T ((1 − z ) p ) ( T T , T ((1 − z ) p ) ] − [( T , T ((1 − z ) p )] , that is, [ T , T ((1 − z ) p ) ] = [( T , T ((1 − z ) p ]. The commutator on the right handside is zero, as we saw above. Therefore,[ T , ( T p ) ] = [ T , T ( zp ) ] + [ T , T ((1 − z ) p ) ] = 0 , where we used (4.2) once again.Finally suppose that p is finite. Then p ′ = 1 − p is infinite. Letting z ′ be thecentral projection such that z ′ p ′ is properly infinite and (1 − z ′ ) p ′ is finite, werecollect the necessary information from the arguments above.(i) [ T , ( T p ′ ) ] = 0;(ii) [( T , T ((1 − z ′ ) p ′ ] = [ T , T ((1 − z ′ ) p ′ ) ] = 0;(iii) [( T , T ( z ′ p ′ )] = [ T , T ( z ′ p ′ ) ] = 0.Consequently,[( T , T p ′ ] = [( T , T ( z ′ p ′ )] + [( T , T ((1 − z ′ ) p ′ )] = 0 . As [ T , T (1 − p ) ] = [ T , ( T ] + [ T , ( T p ) ] − [ T , ( T
1) (
T p ) + (
T p ) ( T T , ( T p ) ] − [( T , T p ] = [ T , ( T p ) ] + [( T , T (1 − p )] , we conclude that [ T , ( T p ) ] = [ T , ( T p ′ ) ] − [( T , T p ′ ] = 0 . Martin Mathieu
This completes the argument that [ T , ( T p ) ] = 0 for every projection p ∈ A and, as explained above, this is sufficient to prove the result. Remark 4.4.
Evidently, all we need to assume on T in the proof of the abovetheorem is that T is surjective, bounded and preserves elements of square zero.Thus, the main result in [15], Theorem 2.4, extends appropriately to the non-unitalsetting.As pointed out above, we cannot expect a result like Theorem 4.3 for arbitrary C* -algebras of real rank zero or even von Neumann algebras. However, in a morerestricted setting the situation might change. Problem 4.
Let T be a spectrally bounded operator defined on a finite von Neu-mann factor (onto a semisimple unital Banach algebra B ). Is T B ?
5. Spectral Isometries
Generalising the classical Banach–Stone theorem, Kadison showed in [12] that everyunital surjective isometry T between two unital C* -algebras must be a Jordan *-isomorphism. In fact, he showed a stronger result in [13], Theorem 2, namely that itsuffices that T maps the selfadjoint part A sa of A isometrically onto the selfadjointpart B sa of B . There are many interesting consequences of these results, all relatingisometric properties to selfadjointness. As an example, we state and prove a variantof a theorem of Chan ([8], Theorem 3).Recall that the numerical radius ν ( x ) of an element x in a unital C* -algebra A is defined by ν ( x ) = sup {| ϕ ( x ) | | ϕ a state of A } . Theorem 5.1.
Every unital surjective numerical radius-preserving operator T be-tween two unital C* -algebras A and B is a Jordan *-isomorphism. Proof.
Since ν is a norm, T is injective and it is clear that thus T − is numericalradius-preserving as well. As ν ( x ) = k x k for all x ∈ A sa , by Kadison’s theorem itsuffices to show that T A sa ⊆ B sa . Let a ∈ A sa , k a k = 1. Write T a = b + ic with b, c ∈ B sa . Suppose c = 0; then there is 0 = γ ∈ σ ( c ) and we may assume that γ >
0. For each state ϕ on B and each n ∈ N , we have | ϕ ( c + n ) | ≤ | ϕ ( b ) | + | ϕ ( c + n ) | = | ϕ ( b ) + iϕ ( c + n ) | = | ϕ ( T a + in ) | . Hence, for large n , ν ( a + in ) = 1 + n < ( γ + n ) ≤ ν ( c + n ) ≤ ν ( T a + in ) which is impossible as T is numerical radius-preserving. Therefore, c = 0 and T a isselfadjoint.Kadison’s theorem in one direction and an application of the Russo–Dye theoremin the other direction show that a unital surjective spectral isometry is selfadjoint
Collection of Problems on Spectrally Bounded Operators (i.e., maps selfadjoint elements to selfadjoint elements) if and only if it is an isometry,see [19], Proposition 2.4. These results and others, and the analogy to Kaplansky’squestion (Problem 2 above), made us ask the following question which we indeedstate as a conjecture in [22]. Problem 5.
Is every unital surjective spectral isometry between unital C* -algebrasa Jordan isomorphism?By now, there is a fair number of results affirming this conjecture in reasonable,though not full generality. Combining the methods of [18], [20] and [24] we can coveranother new case below. Theorem 5.2.
Let T : A → B be a unital spectral isometry from a unital C* -algebra A with real rank zero and without tracial states such that Prim ( A ) is Hausdorff andtotally disconnected onto a unital C* -algebra B . Then T is a Jordan isomorphism. The above statement includes in particular the simple case but we will reducethe more general situation to this one. To this end, we need the notion of a Glimmideal in a C* -algebra.There are (at least) five structure spaces associated with a unital C* -algebra A ,which we will briefly discuss. Their relation can be depicted as follows. Irr ( A ) −→ Spec ( A ) −→ Prim ( A ) β −→ Glimm ( A ) ∼ = −→ Spec ( Z ( A ))The set Irr ( A ) of all irreducible representations of A maps onto the set Spec ( A ) of allunitary equivalence classes of such representations. Since equivalent representationshave the same kernel, there is a canonical surjection from Spec ( A ) onto Prim ( A ), theset of all primitive ideals of A . The latter carries a natural topology, the Jacobsontopology, which can be pulled back to Spec ( A ), and then to Irr ( A ), to turn theminto topological spaces in a canonical way. Since Z ( A ) is a commutative unital C* -algebra, Spec ( Z ( A )) allows several equivalent descriptions of which we choose theone via maximal ideals. For each maximal ideal M of Z ( A ), the closed ideal AM = n n X j =1 x j z j | x j ∈ A, z j ∈ M, n ∈ N o is called a Glimm ideal of A . For each I ∈ Glimm ( A ), M = I ∩ Z ( A ) gives thegenerating maximal ideal in Z ( A ) back, wherefore there is a bijection between thetwo sets. Transporting the natural topology of Spec ( Z ( A )) over to Glimm ( A ) thusturns the latter into a compact Hausdorff space. From the other direction, we canapply the complete regularisation map β : Prim ( A ) → Glimm ( A ) defined by P, Q ∈ Prim ( A ) : P ∼ Q if f ( P ) = f ( Q ) (cid:0) f ∈ C b ( Prim ( A )) (cid:1) and β ( P ) is the equivalence class with respect to this relation. If I is the Glimmideal given by I = A ( P ∩ Z ( A )) then β ( P ) can be identified with T Q ⊇ I Q . Among Martin Mathieu the consequences of this is T Glimm ( A ) = T Prim ( A ) = { } , that is, the Glimmideals separate the points of A .It follows from the definition of the Jacobson’s topology that Prim ( A ) is T ifand only if every primitive ideal is maximal. In a similar vein, Prim ( A ) is T , i.e.,Hausdorff, if and only if every Glimm ideal is maximal ([20], Lemma 9). Proof of Theorem 5.2.
By hypothesis,
Prim ( A ) is Hausdorff and thus the map-ping β is a homeomorphism. Consequently, C ( Prim ( A )) = C ( Glimm ( A )) = Z ( A )has real rank zero; Lemma 8 in [20] therefore yields that J = T I is a Glimm ideal in B for every I ∈ Glimm ( A ). The induced unital surjective operator ˆ T : A/I → B/J given by ˆ T ( x + I ) = T x + J , x ∈ A is a spectral isometry by [24], Proposition 9.Since Prim ( A ) is Hausdorff, every Glimm ideal is maximal so that A/I is simplefor each I ∈ Glimm ( A ). Moreover, A/I has real rank zero and no tracial states.Theorem 3.1 in [18] therefore entails that ˆ T is a Jordan isomorphism, and since thisholds for every I , we obtain that T itself is a Jordan isomorphism. (cid:3) In contrast to results like Theorem 5.2, very little is known about the behaviourof spectral isometries on C* -algebras carrying a trace. In the remainder of this paperwe suggest a possible new route to tackle this situation.The following example is mentioned with less detail in [27]. Example 5.3.
Let T : M n ( C ) → M n ( C ) be a linear mapping. Then T is unital,surjective and spectrally bounded if and only if there are a Jordan automorphism S of M n ( C ) and a non-zero complex number γ such that T x = γ Sx + (1 − γ ) τ ( x ) (cid:0) x ∈ M n ( C ) (cid:1) , (5.1)where τ denotes the normalised centre-valued trace on M n ( C ).Evidently, the formula (5.1) defines a unital mapping which is spectrallybounded, since the images under S and τ commute. Moreover, T is surjective:let y ∈ M n ( C ) and z ∈ M n ( C ) be such that Sz = y . Put x = γ ( z − τ ( z )) and x = τ ( z ). Then x ∈ ker τ . Setting x = x + x we have T x = T ( x + x ) = γSx + γSx + (1 − γ ) τ ( x + x )= Sz − τ ( z ) + γτ ( z ) + (1 − γ ) τ ( z )= Sz = y. Conversely, if T is spectrally bounded, it leaves ker τ invariant, as this space isspanned by the nilpotent matrices and T preserves nilpotency ([23], Lemma 3.1).Assuming that T is surjective, it is in fact bijective and hence remains injectivewhen restricted to ker τ . Since the latter is finite dimensional, T | ker τ is bijectivefrom ker τ to ker τ . By [6], there exist a Jordan automorphism S of M n ( C ) and anon-zero complex number γ such that T | ker τ = γ S | ker τ . Hence, for each x ∈ M n ( C ), T ( x − τ ( x )) = γ S ( x − τ ( x ))which is nothing but identity (5.1), if T is unital. Collection of Problems on Spectrally Bounded Operators Specialising the above description of spectrally bounded operators on matrixalgebras to spectral isometries we recover Aupetit’s result from [2], Proposition 2,which was proved using holomorphic methods.
Example 5.4.
Every unital spectral isometry T from M n ( C ) into itself is a Jordanautomorphism.Since T : M n ( C ) → M n ( C ) is injective, by [22], Proposition 4.2, it is surjective aswell. By the above description (5.1), T | ker τ = γ S | ker τ for a Jordan automorphism S of M n ( C ) and γ ∈ C \ { } . Suppose first that n ≥ y ∈ ker τ with σ ( y ) = { , − } ; for instance, y = e − ( e + e ), where e ij , 1 ≤ i, j ≤ n denotethe standard matrix units. Then σ ( T y ) = γ σ ( Sy ) = γ σ ( y ) = { γ, − γ } . On the other hand, T preserves the peripheral spectrum, by Proposition 4.7 in [22].Therefore γ = 1, and the claim follows from (5.1). In the case n = 2, note thatidentity (5.1) entails that τ ( T x ) = τ ( x ) for all x ∈ M n ( C ). Since T always preservesone eigenvalue (in the peripheral spectrum), it follows that T preserves the entirespectrum and hence must be a Jordan automorphism. The case n = 1 is trivial.Let us now state a problem motivated by these examples. Problem 6.
Let A be a II factor with normalised centre-valued trace τ . Let T : A → A be a unital surjective spectrally bounded operator. Are there a Jor-dan automorphism S of A and a non-zero complex number γ such that T x = γ Sx + (1 − γ ) τ ( x ) (cid:0) x ∈ A (cid:1) ? (5.2) Remarks 5.5.
1. Every Jordan epimorphism S : A → A is either multiplicativeor anti-multiplicative, by a classical result due to Herstein, as A is simple, henceprime. Therefore its kernel is an ideal of A . Since A is simple, S must be injective.2. By the remark just made together with the final part of the argument in Exam-ple 5.3 it suffices to find a Jordan epimorphism S on A such that T | ker τ = γ S | ker τ for some non-zero γ ∈ C .3. The “if”-part in Example 5.3 remains valid in general, so identity (5.2) wouldin fact be a characterization of unital surjective spectrally bounded operators onII factors.4. It is easily seen that the representation (5.2) is unique. Suppose that T = γ Sx + (1 − γ ) τ ( x ) = γ ′ S ′ x + (1 − γ ′ ) τ ( x ) (cid:0) x ∈ A (cid:1) are two representations of the spectrally bounded operator T , where γ, γ ′ ∈ C \ { } and both S , S ′ are Jordan automorphisms of A . It suffices to show that γ = γ ′ .Since γ S | ker τ = γ ′ S ′| ker τ and both S and S ′ are spectral isometries, it follows that | γ | = | γ ′ | . Let p ∈ A be a projection with τ ( p ) = and put x = p − τ ( p ) ∈ ker τ . Martin Mathieu
Then σ ( x ) = { , } − = {− , } . As both S and S ′ preserve the spectrum, itfollows that {− γ , γ } = γ σ ( x ) = γ ′ σ ( x ) = {− γ ′ , γ ′ } and thus γ = γ ′ .Adapting the argument in Example 5.4 to the II factor situation, we can inferthe following result from a positive answer to Problem 6. Corollary 5.6 (Under the assumption that Problem 6 has a positive an-swer).
Every unital surjective spectral isometry T from a II factor A onto itselfis a Jordan automorphism. Proof.
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