Additive preserving rank one maps on Hilbert $C^\ast$-modules

Abstract

In this paper, we characterize a class of additive maps on Hilbert $C^\ast$-modules which maps a "rank one" adjointable operators to another rank one operators.

Full text

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arXiv:0705.0640v1 [math.OA] 4 May 2007
ADDITIVE PRESERVING RANK ONE MAPS ON HILBERT
C∗-MODULES
BIN MENG
Abstract. In this paper, we characterize a class of additive maps on Hilbert
C∗-modules which maps a ”rank one” adjointable operators to another rank
one operators.
1. Introduction and preliminaries
The problem of determining linear maps Φ on B(X) preserving certain properties
has attracted attention of many mathematicians in resent decades. They have been
devoted to the study of linear maps preserving spectrum, rank, nilpotency, etc.
Rank preserving problem is a basic problem in the study of linear preserver
problem. Rank preserving linear maps have been studied intensively by Hou in [2].
In [4, 9] the authors used very elegant arguments to completely describe additive
mappings preserving (or decreasing) rank one.
In our study of free probability theory, we find that module maps on Hilbert
C∗−module preserving certain properties are also important [6, 7], and thus the
study of the modules preserving problem becomes attractive. In [8], a complete
description of modules maps which preserving rank one modular operators is given.
Naturally in the present paper we consider additive maps on modules preserving
rank one, which much more complicated than the module maps case.
A Hilbert C∗−module overa C∗−module overa C∗−algebraA is a left A−module
M equipped with an A−valued inner product ?,? which is A−linear in the first and
conjugate A−linear in the second variable such that M is a Banach space with the
norm ?v?M= ??v,v??
of Kaplansky [3], who used them to prove that derivations of type I AW∗−algebras
are inner. Now a good text book about Hilbert C∗−module is [5]. In this paper
we mainly consider Hilbert A−module H ⊗ A (or denoted by HA), where H is
a separable infinite dimentional Hilbert space and A is a unital C∗−algebra. HA
play a special role in the theory of Hilbert C∗−modules (see [5]). Obviously, HA
is countably generated and possesses orthonormal basis {ei⊗ 1}, which {ei} is an
orthonormal basis in H.
We introduce a class of module maps which is analogues to rank one operators
on a Hilbert space. For all x,y ∈ HA, define θx,y: HA→ HAby θx,y(ξ) := ?ξ,y?x,
for every ξ ∈ HA. Note that θx,yis quite different from rank one linear operators
on Hilbert space. For instance, we cannot infer x = 0 or y = 0 from θx,y= 0. We
n ?
2000 Mathematics Subject Classification. 47B49; 46B28.
Key words and phrases. Hilbert C∗−module; coordinate inverse; Additive preservers .
1
2, ∀v ∈ M. Hilbert C∗−modules first appeared in the work
denote {
i=1αiθxi,yi,∀n ∈ N} by F(HA).
1

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2 BIN MENG
In the present paper, we will describe the additive map Φ : F(HA) → F(HA)
which maps θx,yto some θs,t. The methods are analogues to those in [2, 4, 9] but
much more complicated.
2. Definitions and lemmas
In this section we mainly introduce some definitions and prove some lemmas.
Definition 1. [8] Let ε be a orthonormal basis in HA. x ?= 0 in HAwill be called
coordinatly invertible if for all e ∈ ε, ?e,x? is invertible in A unless ?e,x? = 0.
We denote the set of all the coordinatly invertible elements in HAby CI(HAor
CI for short.
Coordinatly invertible elements in Hilbert C∗−module are analogues to elements
in linear space to some extents.
Lemma 2. [8] Supposing y ∈ CI and θx,y= 0 then x = 0.
Lemma 3. [8] Let M be Hilbert A−modules, where A is a unital C∗−algebra, and
let φ,σ : M → A be A−linear operators. Suppose that σ vanishes on the kenel of
φ. Then there exists a b ∈ A such that σ = φ · b.
Corollary 4. [8]Let g1,g2∈ M. If for all x ∈ M, ?x,g1? = 0 implying ?x,g2? = 0,
then there is a a ∈ A such that g2= ag1.
The following Lemma 5, which will be used frequently, has been obtained in [8].
Nevertheless we give its proof for the sake of completeness.
Lemma 5. Let A be a unital C∗−algebra, x1,x2 ∈ M, g1,g2 ∈ CI satisfying
θx1,g1+ θx2,g2= θx3,g3. Then at least one of the following is true:
(i)there exists a invertible α1∈ A such that g1= α1g2;
(ii) there are β1,β2∈ A such that x1= β1x3, x2= β2x3.
Proof. We will complete the proof by considering the following four cases.
Case 1. For all ξ ∈ HA, ?ξ,g2? = 0 implying ?ξ,g1? = 0. From Corollary 4, there
exists α1∈ A such that g1= α1g2. Furthermore since g −1,g2∈ CI, we infer that
α1∈ A is invertible.
Case 2. For all ξ ∈ HA, ?ξ,g1? = 0 implying ?ξ,g2? = 0. Still from Corollary 4,
there is a α2∈ A such that g2= α2g1and α2is invertible.
Case 3. There exists a ξ0 ∈ HA such that ?ξ0,g2? = 0 but ?ξ0,g1? ?= 0. We
can find e ∈ ε such that ?e,g2? = 0 but ?e,g1? ?= 0.
?e,g2?x2 = ?e,g3?x3, it follows ?e,g1?x1 = ?e,g3?x3.
x1= ?e,g1?−1?e,g3?x3. We put β1= ?e,g1?−1?e,g3? and get θβx3,g1+θx2,g2= θx3,g3.
Thus θx2,g2= θx3,g3−β∗
β∗
β∗
1g1? then we obtain (ii).
Case 4. There exists ξ0∈ HAsuch that ?ξ0,g1? = 0 but ?ξ0,g2? ?= 0. Similar to
Case 3, we get (ii) again.
Then from ?e,g1?x1+
Since g1 ∈ CI, we have
1g1. Now choosing a e′∈ ε, we get ?e′,g2?x2 = ?e′,g3−
1g1?x3and thus x2= ?e′,g2?−1?e′,g3− β∗
1g1?x3. Putting β2= ?e′,g2?−1?e′,g3−
?
Corollary 6. With the notations in the above lemma, suppose g1?= αg2, for all
α ∈ A and g3 ∈ CI. Then there exist β1,β2 ∈ A which are invertible such that
x1= β1x3, x2= β2x3.

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ADDITIVE PRESERVING RANK ONE MAPS ON HILBERT C∗-MODULES3
Proof. Denote {x|?x,gi?} by kergi, i = 1,2. Since g1 ?= αg2, g2 ?= βg1, we have
kerg1? kerg2, kerg2? kerg1. So there exists e1∈ HA, such that ?1,g1? ?= 0 but
?e1,g2? = 0. Then ?e1,g1?x1+ ?e1,g2?x2= ?e1,g3?x3, i.e. ?e1,g1?x1= ?e1,g3?x3
and thus x1= ?e1,g1?−1?e1,g3?x3. Putting ?e1,g1?−1?e,g3? = β1is invertible.
Similarly there exits a e2∈ HAsuch that ?e2,g1? = 0 but ?e2,g − 2? ?= 0. Then
x2= ?e2,g2?−1?e2,g3?x3. Putting β2= ?e2,g2?−1?e2,g3?.
?
Definition 7. Φ : F(HA) → (HA) is a map. If for any x ∈ HA, y ∈ CI, there are
s ∈ HA, t ∈ CI such that Φ(θx,y) = θs,t. Then we call Φ is rank one decreasing. If
x ?= 0 implying s ?= 0 then Φ will be called rank one preserving.
Definition 8. Φ : F(HA) → F(HA) is an additive map and for arbitrary θx,y,
Φ(λθx,y) = τx,y(λ)Φ(θx,y) where τx,y: A → A is a map. Then Φ will be called a
locally quasi-modular map.
If there exists a τ : A → A is a map such that Φ(λT) = τ(λ)Φ(T), for all
T ∈ F(HA) then Φ will be called a τ− quasi-modular map.
Lemma 9. Let A be a unital commutative C∗−algebra , let A,B be injective
τ−quasi-modular continuous maps on HA with dimA(A(HA)) ≥ 2 and suppose
τ is surjective. There is λx∈ A, such that Bx = λxAx. Then B = λA for some
λ ∈ A.
Proof. There are x1 ∈ CI, x2 ∈ HA such that ?Ax1,Ax2? = 0. From the as-
sumption there exist λ1,λ2,λ3 ∈ A such that Bx1 = λ1Ax1, Bx2 = λ2Ax2 and
B(x1+ x2) = λ3A(x1+ x2). Therefore, (λ1− λ3)Ax1+ (λ2− λ3)Ax2= 0 and
?(λ1− λ3)Ax1+ (λ2− λ3)Ax2,(λ1− λ3)Ax1+ (λ2− λ3)Ax2?
(λ1− λ3)?Ax1,Ax1?(λ1− λ3)∗+ (λ2− λ3)?Ax2,Ax2?(λ2− λ3)∗
0.
=
=
It follows that (λ1− λ3)?Ax1,Ax1?(λ1− λ3)∗= 0 and (λ2− λ3)?Ax2,Ax2?(λ2−
λ3)∗= 0. Furthermore (λ1−λ3)Ax1= (λ2−λ3)Ax2= 0. We infer Bx1= λ1Ax1=
λ3Ax1and Bx2= λ2Ax2= λ3Ax2.
For every x ∈ HA such that Ax,Ax1,Ax2 is an orthogonal set, we claim that
there is a λ ∈ A such that Bx = λAx. In fact, Bx = λxAx = λx+x1Ax and
Bx1 = λ1Ax1 = λx+x1Ax1 = λ3Ax1. It follows that λx+x1x1 = λ3x1 from A is
injective. And then λx+x1= λ3. Thus we obtain Bx = λ3Ax.
A(HA) is submodule of HAsince τ is surjective. Let Ax1,Ax2,··· be a orthog-
onal basis in A(HA). For arbitrary x ∈ HA, Ax = α1Ax1+ α2Ax2+ ···. Since τ
is surjective, there exist λ1,λ2,··· , such that αi= τ(λi),i = 1,2,···. Therefore
Ax = A(λ1x1+ λ2x2+ ···) and x = λ1x1+ λ2x2+ ···. We infer there exists a
λ ∈ A such that
Bx
=
B(λ1x1+ λ2x2+ ···)
α1B(x1) + α2B(x2) + ···
α1λAx1+ α2λAx2+ ···
λA(λ1x1+ λ2x2+ ···)
λAx
=
=
=
=
?

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4 BIN MENG
Now we introduce some notations. Lx:= {θx,g | g ∈ HA}, Rf := {θx,f | x ∈
HA}, LCI
x
:= {θx,g| g ∈ CI}, RCI
f
:= {θy,f| y ∈ CI}.
The following lemma play important roles in this paper.
Lemma 10. Let Φ be a additive rank one preserving map. For every x ∈ HA
there exists either a y ∈ HA such that Φ(LCI
Φ(LCI
x) ⊆ Rf.
x) ⊆ LCI
y
or an f ∈ CI such that
Proof. Assume that there exists a x0∈ HAsuch that Φ(LCI
for all x ∈ HA,f ∈ CI. Then there are f1,f2∈ CI such that Φ(θx0,f1) = θx1,g1,
Φ(θx0,f2) = θx1,g2, where x1?= α1x, x2?= α2x, for all x ∈ HA,α1,α2∈ A and g1?=
β1g2, g2?= β2g1, for all α,β ∈ A. Since Φ is rank one preserving, Φ(θx0,f1+f2) =
θx3,g3for some x3 ∈ HA,g3 ∈ CI. On the other hand, from Lemma 5, we get
θx1,g1+ θx2,g2?= θx3,g3. Thus we reach a contradiction.
x0⊆ LCI
x, Φ(LCI
x0) ⊆ Rf,
?
Lemma 11. At least one of the following is true
(1)for all x ∈ HA, there exits y ∈ HAsuch that Φ(LCI
(2)for all x ∈ HA, there exits f ∈ CI such that Φ(LCI
x) ⊆ LCI
⊆ Rf
y;
x
Proof. If there are DimAΦ(LCI
θy,g2implying g1= αg2.) then both (1) and (2) hold.
Now we consider DimAΦ(LCI
reach a contradicion, that Φ(LCI
HA, g1∈ CI. Since DimAΦ(LCI
θy0,g, g ∈ CI with g ?= αg1, ∀α ∈ A. Since DimAΦ(Rf) ≥ 2, we know there exist
k ∈ HA, such that Φ(θx1,k) = θz,g1with z ?= βx, y0 ?= βx, ∀x ∈ HA, β ∈ A.
Letting m ∈ CI such that Φ(θx0,k) = θy0,m, then Φ(θx0+x1,k) = θz,g1+θy0,m. Since
z ?= αx,y0?= βx, ∀x ∈ HA,α,β ∈ A, there exists a λ ∈ A such that m = λg1and
Φ(θx0,k) = θy0,m= θy0,λg1. On the other hand, Φ(θx0,h) = θy0,g, Φ(θx1,h) = θl,g1,
where h,g,g1∈ CI. From g1?= αg, for all α ∈ A and Φ(θx0+x1,h) = θy0,g+ θl,g1,
we infer that there are α0,β0which are invertible such that y0= α0y′
Therefore, Φ(θx1,g) = θβ0y′
x) = 1 (∀f1,f2∈ CI, Φ(θx,f1) = θy,g1, Φ(θx,f2) =
x) ≥ 2,DimAΦ(Rf) ≥ 2. Otherwise suppose, to
x0) ⊆ LCI
x) ≥ 2, there exists a h ∈ CI, such that Φ(θx0,h) =
y0, Φ(LCI
x1) ⊆ Rg1, where x0,x1∈ HA, y0∈
0, l = β0y′
0.
0,g1, Φ(θx0,h) = θα1y′
0,gand
Φ(θx0+x1,h+k) = Φ(θx0,h) + Φ(θx0,k) + Φ(θx1,h) + Φ(θx1,k)
θα0y′
θα0y′
0+β0y′
Since g ?= αg1,z ?= βx,y0?= γx, for all α,β,γ ∈ A, we know Φ(θx0+x1,h+k) is not
rank one which contradicts to the assumption of Φ.
=
0,g+ θα0y′
0,λg1+ θβ0y′
0,g1+ θz,g1
=
0,g+ θ(λ∗α0y′
0+z),g1
?
Corollary 12. Then one of the following is true
(i)for all f ∈ HAthere exists a f ∈ HAwith Φ(RCI
(ii)for all f ∈ HAthere exists y ∈ CI with Φ(RCI
f) ⊆ RCI
f) ⊆ Ly
g;
Lemma 13. Suppose ImΦ is neither contained in any Lyor contained in any Rg
and DimAΦ(LCI
x) ≥ 2. Then
(i)If for all x ∈ HAthere exists a y ∈ HAsuch that Φ(LCI
(ii)If for all x ∈ HAthere exists a y ∈ CI such that Φ(LCI
f ∈ HAthere exists a z ∈ CI such that Φ(RCI
x, then Φ(RCI
x) ⊆ Ry. Then for all
f) ⊆ RCI
g;
f) ⊆ Lz.
Proof. We only prove (i). Assume, to reach a contradiction, that we have simula-
tously Φ(LCI
y
and Φ(RCI
x) ⊆ LCI
f) ⊆ Lz, for some z ∈ CI. Since DimAΦ(LCI
x) ≥ 2

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ADDITIVE PRESERVING RANK ONE MAPS ON HILBERT C∗-MODULES5
there are k′,k′′,g′,g′′∈ CI with g′?= αg′′such that Φ(θx,k′) = θy,g′, Φ(θx,k′′) =
θy,g′′.
Since ImΦ is not contained in any Lg, there are x1 ∈ HA, k1 ∈ CI with
Φ(θx1,k1) = θy1,g1such that y1?= αh, y ?= βh, for all α,β ∈ A,h ∈ HA. Conse-
quently Φ(LCI
y1, Φ(θx1,k′) = θy1,u. Therefore, Φ(θx+x1,k′) = θy,g′+θy1,uand
then u = λg′for some invertible λ ∈ A i.e. Φ(θx1,k′) = θλ∗y1,g′ and Φ(RCI
It follows Φ(RCI
tion.
x1) ⊆ LCI
k′ ) ⊆ Rg′.
k′ ) ? Lz which implying Φ(Rf) ⊆ Rg. So we reach a contradic-
?
Lemma 14. Suppose ImΦ is not contained in any Lynor Rfand DimAΦ(LCI
2. If for all x ∈ CI, there exists a y′∈ HA such that Φ(LCI
exists a y ∈ CI such that Φ(LCI
y.
x) ≥
x) ⊆ LCI
y′ , then there
x) ⊆ LCI
Proof. It follows from Φ(LCI
the conclusion of the Lemma were wrong, we will have Φ(LCI
y ∈ CI. Thus there exist g1?= αg2,y1?= αy,y2?= βy,y1,y2∈ CI, for all y ∈ CI
with Φ(θx,f1) = θy1,g1, Φ(θx,f2) = θy2,g2. We claim that y1?= αx,y2?= βx, for all
x ∈ HA. If not, there exists a y′
that α0,β0 are invertible and y′
Φ(LCI
y, for all y ∈ CI. So y1?= αx,y2?= βx but this this contradicting to
that Φ(θx,f1+f2) is rank one.
x) ⊆ LCI
y′
that Φ(LCI
x) ? Rf, for all f ∈ CI.
x) ? LCI
If
y, for all
0∈ HAsuch that y1α0y′
0is coordinately invertible which contradicting to
0,y2= β0y′
0. Then it follows
x) ? LCI
?
Lemma 15. If Φ(LCI
x) ⊆ LCI
y, then Φ(Lx) ⊆ Ly.
Proof. For all f ∈ HA, f =?
?
Similarly we can prove
i
αifi, fi ∈ CI, we have Φ(θx,f) = Φ(θx,P
i
αifi) =
i
α∗
iΦ(θx,fi) =?
i
α∗
iθy,gi= θy,P
i
gi. Then Φ(Lx) ⊆ Ly.
?
Lemma 16. If Φ(LCI
x) ⊆ Rf, for some f ∈ CI, then Φ(Lx) ⊆ Rf.
Lemma 17. For all x ∈ CI,f ∈ HA, there is a map τx,f such that Φ(λθx,f) =
τx,f(λ)Φ(θx,f).
Proof. For all x ∈ HA,f ∈ CI, Φ(θx,f) = θy,g, f,g ∈ CI. Therefore Φ(λθx,f) =
Φ(θλx,f) = θr(λx),g= θy,r′(λf)with g ∈ CI. For a e ∈ ε, ?e,g?r(λx) = ?e,r′(λf)?y.
It follows from g ∈ CI that r(λx) = αy, with α = ?e,g?−1?e,r′(λf)?. Denote by
τx,f(λ) = α. Then Φ(λθx,f) = τx,fΦ(θx,f).
?
3. Main results
Theorem 18. Φ : F(HA) → F(HA) is a sujective rank one preserving additive
map. ImΦ is neither contained in any Lynor contained in any RgThen one of the
following is true: (1) For all x,f ∈ HA, Φ(θx,f) = θAx,Cf where A,C are injective
quasi-modular maps on HA;
(2)For all x,f ∈ HA, Φ(θx,f) = θCf,Axwhere A,c are injective conjugate quasi-
modular maps on HA.
Proof. For any x ∈ HA, there exists a y ∈ HA such that Φ(Lx) ⊆ Ly. For all
f ∈ HAwith Φ(θx,f) = θy,Cxf.
When x in CI, and y can be in CI, we have following claims.