A note on the commutativity of C∗-algebras

2016-11-11JIANGRunliang

华东师范大学学报（自然科学版） 2016年2期

关键词：上海师范大学博士后同构

JIANG Run-liang

（Department of Mathematics，Shanghai Normal University，Shanghai200234，China）

A note on the commutativity of C∗-algebras

JIANG Run-liang

（Department of Mathematics，Shanghai Normal University，Shanghai200234，China）

There are many characterizations for commutative C∗-algebras.In this note，we prove that a C∗-algebra A is not commutative if and only if there is a C∗-subalgebra B in A′′（the enveloping Von Neumann algebra of A）such that B is∗-isomorphic to M2（C）. In terms of this result，we can recover some characterizations for the commutativity of C∗-algebras appeared before.

commutative C∗-algebras；enveloping Von Neumann algebra

Article ID：1000-5641（2016）02-0030-05

0　Introduction

There are many characterizations for commutative C∗-algebras.For examples，Kaplansky showed that a C∗-algebra A is commutative if and only if the only nilpotent element in A is 0（cf.［1，2.12.21］）；Nakamoto in［2］gave a spectral characterization for the commutativity of a C∗-algebra；Ogasawara，Sherman，Wu，Ji and Tomiyama gave different characterizations of commutative C∗-algebras by means of the order structure，see［3-6］for details.In［7］，Jeang and Ko proved that for non-constant continuous functions f and g defined on a closed interval I1and I2respectively，if f（x）g（y）=g（y）f（x）for all self-adjoint elements x and y in A with σ（x）⊆I1and σ（y）⊆I2，then A is commutative.

In this note，we give the following lemma that a C∗-algebra A is not commutative if and only if there is a C∗-subalgebra B in A′′（the enveloping Von Neumann algebra of A）such thatB is∗-isomorphic to M2（C）.In terms of this result，we can recover some characterizations for the commutativity of C∗-algebras appeared in［3-7］.

1　Basic Theories

Throughout the paper，C and R are the complex number field，the real number field respectively and A is always a C∗-algebra.If A has a unit e，we set σ（a）to be the spectrum of an element a∶σ（a）=｛λ∈C|λe-a is not invertible in A｝.If A is non-unital，then we set σ（a）=｛λ∈C|λe-a is not invertible in｝，whereis the unitization of A.We put Asa=｛a∈A|a∗=a｝.An element a of Asais positive if σ（a）⊆R+.We write a≥0 to mean a is positive，and denoted by A+the set of all positive elements of A.For a,b∈Asa，we write a≥b to mean a-b∈A+.Please see［8］for more details.

For convenience，we assume that A is a C∗-subalgebra of B（H）for certain complex Hilbert space H such that｛a（ξ）|a∈A,ξ∈H｝is dense in H，where B（H）is the C∗-algebra consisting of all bounded linear operators from H to H（cf.［8 Theorem 3.4.1］）.Then A is strongly dense in A′′（cf.［8，Theorem 4.1.4］）.For T∈B（H），let Ran（T）（resp.Ker T）denote the range（resp.null space）of T.

Recall that a continuous function f∶R→ C is called to be strongly continuous if for every Hilbert space K and each net｛Tλ｝λ∈Λ⊆B（K）saconverging strongly to an operator T∈B（K）sa，we have｛f（Tλ）｝λ∈Λconverges strongly to f（T）.

The following lemma comes from［8，Theorem 4.3.2］.

Lemma 1If f∶R→C is a continuous bounded function，then f is strongly continuous.

Recall from［5，Definition 1］that a continuous real-valued function f defined on a subset S of R is called to be operator monotonic increasing（resp.decreasing）on A associated with S if f（x）≤f（y）（resp.f（x）≥f（y））inwhenever x,y∈Asawith x≤y and σ（x）∪σ（y）⊆S.

Proposition 2Let f be a continuous real-valued function defined on R（or R+）.If f is operator monotonic on A associated with R（or R+），then f becomes operator monotonic on A′′associated with R（or R+）.

ProofWe just suppose that f is monotonic increasing on A associated with R.The proof of the remains is similar.

Let a,b∈（A′′）sawith a≤b.Set c=b-a≥0.By Kaplansky's density theorem（cf.［8，Theorem 4.3.3］），there are a net｛aλ｝λ∈Λ⊆Asaand a net｛cλ｝λ∈Λ⊆A+with‖aλ‖≤‖a‖and‖cλ‖≤‖c‖，∀λ∈Λ，such that aλconverges strongly to a and cλconverges strongly to c.Set bλ=aλ+cλ,∀λ∈Λ.Then‖bλ‖≤‖a‖+‖c‖，aλ≤bλ，∀λ∈Λ，and bλconverges strongly to b.

Put k=‖a‖+‖c‖and define a continuous real-valued function f0on R by

Since σ（a），σ（b），σ（aλ），σ（bλ）⊆［-k,k］，∀λ∈Λ，we have

Noting that f0is continuous bounded function defined on R，by Lemma 1，f0（aλ）converges strongly to f0（a）and f0（bλ）converges strongly to f0（b），hence f（aλ）converges strongly to f（a）and f（bλ）converges strongly to f（b）.Since f is operator monotonic increasing on A and aλ≤bλ，we have f（aλ）≤f（bλ），∀λ∈Λ，and consequently，f（a）≤f（b）.

The next result is easy but useful.

Lemma 3Let A be a C∗-algebra.Then the following statements are equivalent∶

1.A is not commutative；

2.A′′is not commutative；

3.There exists a C∗-subalgebra B in A′′such that B is∗-isomorphic to M2（C）.

ProofThe implications（1）⇔（2）and（3）⇒（2）are obvious.We now prove（2）⇒（3）.

Since A′′is not commutative，it follows from［1，2.12.21］that A′′contains a non-zero nilpotent element a（i.e.，a2=0 and a≠0）.Thus，Ran（a）⊆Kera.

where λi∈C,i=1,2,3,4.It is easy to check that ψ is a∗-isomorphism.

2　Main Results

In this section，we will recover some characterizations for the commutativity of C∗-algebras appeared in［3-7］by using Lemma 3.

Definition 4We call a continuous monotone function f is matrix monotone of order n if f induces a monotone function on the matrix algebra Mn（C）.

Combining Lemma 3 with Proposition 2，we have the following corollary proved in［6］.

Corollary 5Let A be a C∗-algebra.If there exists a continuous monotone function f on R+（or R）which is not matrix monotone of order 2 but operator monotonic on A，then A must be commutative.

ProofIf A is not commutative，then A′′is not commutative either.Thus A′′has a C∗-subalgebra B which is*-isomorphic to M2（C）.f is operator monotonic on A，so f must be operator monotonic on A′′，and f is also operator monotonic on B.Hence，f becomes a monotone function on M2（C），a contradiction.So A must be commutative.

Corollary 6Let A be a C∗-algebra.If A satisfies one of the following conditions，then A is commutative.

1.exp（x）≤exp（y）for all x,y∈A+with x≤y（cf.［5］）；

2.x2≤y2for all x,y∈A+with x≤y（cf.［3］）；

3.Let f+（t）=max｛t,0｝and f-（t）=-min｛t,0｝，∀t∈R.For a∈Asa，put a+=f+（a）and a-=f-（a）.If a≤b for a,b∈Asa，then a+≤b+（resp.b-≤a-）（cf.［4］）.

Proof（1）Since exp（t）is not matrix monotone of order 2 by［6，Proposition 1］，we obtain that A is commutative by Corollary 5.

For a,b∈Asawith a≤b，we have-b≤-a and hence（-a）-≤（-b）-by the assumption. Noting that（-a）-=a+and（-b）-=b+，we get the assertion by above argument.

Corollary 7Let f and g be non-constant continuous functions defined on closed intervals I1and I2respectively.Suppose that there exist non-zero elements xi∈Asawith σ（xi）⊆Ii,i=1,2.If f（x）g（y）=g（y）f（x）for all x,y∈Asawith σ（x）⊆I1and σ（y）⊆I2，then A is commutative.

ProofAssume that I1=［µ1,µ2］and I2=［µ3,µ4］.We first show that for x,y∈（A′′）sa. with σ（x）⊆I1and σ（y）⊆I2，f（x）g（y）=g（y）f（x）.

If A is unital，by Kaplansky's density theorem，there exist two nets｛xλ｝λ∈Λand｛yλ｝λ∈Λin A+such that‖xλ‖≤‖x-µ1‖，‖yλ‖≤‖y-µ3‖，∀λ∈Λ，and xλ（resp.yλ）converges strongly to x-µ1（resp.y-µ3）.Thus，σ（xλ+µ1）⊆I1，σ（yλ+µ3）⊆I2，∀λ∈Λ，and xλ+µ1（resp.yλ+µ3）converges strongly to x（resp.y）.Extend f and g to the continuous bounded functions on R by

by the hypothesis，we get that f（x）g（y）=g（y）f（x）by Lemma 1.

If A is non-unital，then I1and I2contain 0.Setwhenµ1=0 orµ2=0 and=［-s,s］when s=min｛-µ1,µ2｝＞0.Then⊆I1and for each x∈（A′′）sawith σ（x）⊆there is a net｛xλ｝λ∈Λ⊆Asawith σ（xλ）⊆∀λ∈Λ，such that xλconverges strongly to x by Kaplansky's density theorem.Similarly，there exists a closed intervalin I2such that for each y∈（A′′）sawith σ（y）⊆we can find a net｛yλ｝λ∈Λin（A′′）sawith σ（yλ）⊆，∀λ∈Λ， such that yλconverges strongly to y.Therefore，f（x）g（y）=g（y）f（x）for any x,y∈（A′′）sawith σ（x）⊆and σ（y）⊆by Lemma 1.

Now suppose that A is not commutative.Then there is a C∗-subalgebra B in A′′such that B is∗-isomorphic to M2（C）.Thus，we have f（x）g（y）=g（y）f（x）for any x,y∈M2（C）sawith σ（x）⊆and σ（y）⊆LetandSet