Commutativity preserving linear maps on central simple algebras

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Journal of Algebra 284 (2005) 102–110

www.elsevier.com/locate/jalgebr

Commutativity preserving linear mapson central simple algebras

Matej Brešara,1, Peter Šemrlb,∗,1

a Department of Mathematics, University of Maribor, PEF, Koroška 160, SI-2000 Maribor, Sloveniab Department of Mathematics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia

Received 4 May 2004

Available online 26 November 2004

Communicated by Gordon James

Abstract

Let F be a field with char(F) = 0, and letA be a finite dimensional central simple algebra oveF

with dimF A �= 4. If a linear mapφ :A → A satisfiesφ(x2)φ(x) = φ(x)φ(x2) for all x ∈ A (inparticular, ifφ preserves commutativity), thenφ is either a standard commutativity preserving mor its range is commutative. 2004 Elsevier Inc. All rights reserved.

Keywords:Linear commutativity preserver; Finite dimensional central simple algebra

1. Introduction

Let A be an algebra over a fieldF, and letφ : A → A be a linear map. We say thatφ

preserves commutativityif

φ(x)φ(y) = φ(y)φ(x) wheneverxy = yx. (1)

* Corresponding author.E-mail addresses:[email protected] (M. Brešar), [email protected] (P. Šemrl).

1 Partially supported by a grant from MŠZŠ.

0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2004.09.014

M. Brešar, P. Šemrl / Journal of Algebra 284 (2005) 102–110 103

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A simple example is a map of the form

φ(x) = λψ(x) + µ(x)1 for all x ∈ A,

whereλ ∈ F, ψ is an automorphism or an antiautomorphism ofA, µ is a linear functionaon A, and1 is the unity ofA. We will say that such map is astandard commutativitypreserving map.

The problem of describing commutativity preserving linear maps is one of thestudied linear preserver problems. The usual solution is that they are standard. Thlem was first considered in the case whereA = Mn(F) is the algebra ofn × n matricesoverF [6], and has been afterwards studied in various much more general algebras (5] for references). A rather general theorem,and also the first ring-theoretic result in tharea, was obtained in [3] by the first author. This theorem treats the case whereA is a primealgebra satisfying certain technical conditions, and instead of the commutativity presassumption it only requires a milder assumption that

φ(x2)φ(x) = φ(x)φ

(x2) for all x ∈A. (2)

The paper [3], particularly the idea to consider the condition (2), initiated a systestudy of different topics in ring theory and functional analysis; the reader is referred4] for historical accounts and further references. In the present paper we shall use thin a somewhat different way.

A standard assumption when treating a commutativity preserving mapφ (or a mapsatisfying (2)) is thatφ is bijective (or at least surjective). An exception is the recent resuby Omladic, Radjavi, and the second author:

Theorem 1.1 [5]. Let F be an algebraically closed field withchar(F) = 0, and letA =Mn(F) with n �= 2. If a linear mapφ :A → A preserves commutativity, thenφ is either astandard commutativity preserving map or its range is commutative.

Can the assumption thatF is algebraically closed be omitted? This is the questioninitiated the present paper. The answer is certainly not clear from the proof of Theorewhich combines linear algebraic and analytic tools together with the transfer princimodel theoretic algebra, which enables to pass from the case whereF = C to the casewhereF is an arbitrary algebraically closed field with char(F) = 0. Anyhow, we shall seethat this assumption can indeed be omitted, and in fact a considerably more generacan be proved. Our main result, Theorem 3.1, in particular shows that Theorem 1.1if A is a finite dimensional central simple algebra over any fieldF with char(F) = 0. Thebasic idea of our approach is to deal with the condition (2) instead of with (1), which mit possible for us to use the scalar extension argument, and thereby to reduce thecase to the one where Theorem 1.1 can be used.

In order to establish that the conditions (1) and (2) are equivalent, we will, in Sectfirst consider a more general problem concerning certain bilinear maps. Section 3voted to the proof of the main result.

104 M. Brešar, P. Šemrl / Journal of Algebra 284 (2005) 102–110

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2. A related problem on bilinear maps

If a linear mapφ from an algebraA into itself satisfies (2), does thenφ satisfy (1), i.e.,does it preserve commutativity? The result from [3] mentioned above in particular sthat this is true under rather mild assumptions concerningA, provided, however, thatφis bijective. But in the present paper we are interested in arbitrary linear commutativipreserving maps, so this result can be viewed only as an indication that our questireasonable one. Let us first show that the answer is not always affirmative.

Example 2.1. LetA be a noncommutative algebra such thatx2 = 0 for allx ∈A (for exam-ple, the nonunital Grassmann algebra in two generators). Then every linear mapφ :A→Atrivially satisfies (2). Pickx, y ∈ A such thatxy �= 0. Thenxy andy are linearly independent, and so we can chooseφ so thatφ(xy) = x andφ(y) = y. Henceφ does not preservcommutativity. Incidentally, we can also adjoin a unity toA and defineφ(1) = 1; this givesan example of a unital algebra satisfying (2) but not (1).

In order to obtain an affirmative answer for certain algebras, we shall in fact consmore general question which is perhaps of some independent interest. Before stating itfirst fix the notation for this section. Throughout,F will be a field with char(F) = 0 andAwill be a unital algebra overF. Further,V will be a vector space overF andf :A×A → Vwill be a bilinear map.

Our question is: does the condition

f (x, y) = 0 wheneverxy = yx (3)

follow from the condition

f (x, x) = f(x2, x

) = 0 for all x ∈ A. (4)

The connection with the aforementioned question is simple. Namely, if a linearφ : A → A satisfies (2), then a mapf defined byf (x, y) = φ(x)φ(y) − φ(y)φ(x) sat-isfies (4). Therefore, the affirmative answer to the latter question implies the affirmanswer to the former question.

For x, y ∈ A we shall write[x, y] = xy − yx andx ◦ y = xy + yx. It will be useful tohave the linearized form of (4), i.e.,

f (x1, x2) + f (x2, x1) = 0, (5)

f (x2 ◦ x3, x1) + f (x3 ◦ x1, x2) + f (x1 ◦ x2, x3) = 0. (6)

We begin with

Lemma 2.2. If f satisfies(4), thenf (p(x), q(x)) = 0 for all x ∈ A and all polynomialsp,q ∈ F[X].


case

-

at-

.2,h. This

s

Proof. Fix x ∈A. It suffices to show thatf (xr, xs) = 0 for all r, s � 0. First of all, settingx1 = x andx2 = x3 = 1 in (6) and using (5), we see that this is true ifr = 0 or s = 0.Now fix an integern � 3, and setvs = f (xn−s , xs), s = 1, . . . , n − 1. By (5) we havevs = −vn−s . Accordingly, settingx1 = xi , x2 = xj , andx3 = xn−i−j , it follows from (6)thatvi +vj = vi+j wheneveri � 1,j � 1, andi +j < n. This clearly implies thatvs = sv1

for everys = 1, . . . , n − 1. However, sincevs = −vn−s , it follows thatnv1 = 0. Thereforev1 = 0, and hence alsovs = 0 for everys = 1, . . . , n − 1. This proves thatf (xr , xs) = 0wheneverr > 0 ands > 0. �Remark 2.3. Using Bergman’s centralizer theorem [2] it follows immediately fromLemma 2.2 that the answer to our question is affirmative, i.e., (4) implies (3), in thewhereA = F〈X〉 is the free algebra on a setX.

For anyx ∈ A we setC(x) = {y ∈ A | [x, y] = 0}. Consider the following two conditions:

(a) If n ∈ A is such thatn2 = 0, then every element inC(n) is of the formn ◦ u + eve

whereu,v, e ∈A, e = e2, anden = ne = 0.(b) If x ∈ A is arbitrary, then every element inC(x) is of the form

∑p

i=1 ni +∑q

j=1 ejpj (x) whereni, ej ∈ C(x), n2i = 0, ej = e2

j , andpj ∈ F[X].

These conditions might appear somewhat artificial, but we shall see that they arise nurally in the matrix algebra.

Lemma 2.4. If A satisfies the conditions(a)and(b), then(4) implies(3).

Proof. We begin with a general observation. Suppose thats ∈ A ande = e2 ∈ A are suchthates = se = 0. Then takingx1 = s, x2 = e, andx3 = eye in (6), we obtainf (eye, s) = 0for everyy ∈A.

Let n ∈ A be such thatn2 = 0. We claim thatf (w,n) = 0 for everyw ∈ C(n). In viewof (a) and the above observation it suffices to show thatf (n ◦ u,n) = 0 for everyu ∈ A.But this follows immediately from (6) by takingx1 = x2 = n andx3 = u.

Now let x ∈ A be arbitrary. We have to show thatf (x, y) = 0 for everyy ∈ C(x). Ifn ∈ C(x) andn2 = 0, then we already know thatf (x,n) = 0. Therefore, in view of (b) itremains to prove thatf (x, ep(x)) = 0 wheree = e2 ∈ C(x) andp ∈ F[X]. Note that

f(x, ep(x)

) = f(ex,p(ex)

) + f((1 − e)x, ep(x)

) − λ0f (ex,1 − e),

whereλ0 is the constant term ofp. The first term on the right-hand side is 0 by Lemma 2and the second and the third term are 0 by the observation from the first paragrapproves the lemma. �Lemma 2.5. If F is algebraically closed, thenA = Mn(F), n � 1, satisfies the condition(a)and(b).


rank

oughat

that

Proof. Let us first prove (a). It is a standard fact that every square zero matrix ofk � 1 is similar to a matrix of the form

N =[0 I 0

0 0 00 0 0

],

where the diagonal matrices are of the sizek × k, k × k, and(n − 2k) × (n − 2k), respec-tively; if k = n/2, then the last row and the last column are absent. Therefore, it is ento consider only such square zero matricesN . A straightforward computation shows thW ∈ C(N) if and only if

W =[

A B C

0 A 00 D T

],

whereA,B,C,D,T are any matrices of the appropriate size. Note thatW = N ◦U +EV E

where

U =[

B 0 0A 0 C

D 0 0

], E =

[0 0 00 0 00 0 I

], V =

[0 0 00 0 00 0 T

].

Moreover,E = E2 andEN = NE = 0. This proves (a).Now let X ∈ A be an arbitrary matrix. Without loss of generality we may assume

X is in Jordan canonical form, that is

X =

J1 0 . . . 00 J2 . . . 0...

.... . .

...

0 0 . . . Jk

,

whereJ1, . . . , Jk are Jordan blocks. It is easy to check thatY ∈ C(X) if and only if

Y =

Y11 Y12 . . . Y1k

Y21 Y22 . . . Y2k...

.... . .

...

Yk1 Yk2 . . . Ykk

,

whereYij , 1� i, j � k, are matrices of the appropriate size satisfyingJiYij = Yij Jj . Ac-cordingly, each of the matrices

N12 =

0 Y12 . . . 00 0 . . . 0...

.... . .

...

, N13 =

0 0 Y13 . . . 00 0 0 . . . 0...

......

. . ....

, etc.,

0 0 . . . 0 0 0 0 . . . 0


pre-arefore,

l

n

e

lies inC(X) and satisfiesN2ij = 0. Further, it is well known (and easy to verify) thatJiYii =

YiiJi implies thatYii = pi(Ji) for somepi ∈ F[X]. Therefore,

Y11 0 . . . 00 0 . . . 0...

.... . .

...

0 0 . . . 0

= E1p1(X),

0 0 . . . 00 Y22 . . . 0...

.... . .

...

0 0 . . . 0

= E2p2(X), etc.,

where

E1 =

I 0 . . . 00 0 . . . 0...

.... . .

...

0 0 . . . 0

, E2 =

0 0 . . . 00 I . . . 0...

.... . .

...

0 0 . . . 0

, etc.,

are idempotents commuting withX. Thus, (b) holds true. �Lemmas 2.4 and 2.5 together yield

Theorem 2.6. Let F be an algebraically closed field withchar(F) = 0, let A = Mn(F),n � 1, and letV be a vector space overF. If a bilinear mapf : A ×A → V satisfies(4),thenf satisfies(3).

In the sequel we shall need the following special case of this theorem.

Corollary 2.7. LetF be an algebraically closed field withchar(F) = 0, and letA = Mn(F),n � 1. If a linear mapφ :A →A satisfies(2), thenφ satisfies(1).

3. The main result

The 2× 2 matrix algebra is quite exceptional with regard to the commutativityserver problem. Every linear mapφ :M2(F) → M2(F) that sends the identity into a scalmultiple of the identity preserves commutativity (see, e.g., [5, Theorem 1.1]). Theralgebras of dimension 4 must be excluded in general results.

Theorem 3.1. Let F be a field withchar(F) = 0, and letA be a finite dimensional centrasimple algebra overF with dimF A �= 4. If a linear mapφ :A → A satisfies(2), thenφ iseither a standard commutativity preserving map or its range is commutative.

Proof. Let F be the algebraic closure ofF, and letA = F ⊗F A be the scalar extensioof A. As it is well known,A ∼= Mn(F) wheren = √

dimF A. Thusn �= 2 in view of ourassumption. The theorem trivially holds true in the case whenn = 1, and so we may assumthatn � 3. Define anF-linear mapφ : A → A by φ(λ ⊗ x) = λ ⊗ φ(x). Considering the


-dlatterre

her

linearized form of (2), that is (6) withf (x, y) = [φ(x),φ(y)], we see thatφ satisfies thesame condition, i.e.,

[φ(x2 ◦ x3),φ(x1)

] + [φ(x3 ◦ x1),φ(x2)

] + [φ(x1 ◦ x2),φ(x3)

] = 0

for all x1, x2, x3 ∈ A. Settingx1 = x2 = x3 = x, we thus see thatφ satisfies[φ(x2),

φ(x)] = 0 for everyx ∈ A. Therefore Corollary 2.7 tells us thatφ preserves commutativity. Now we are in a position to apply Theorem 1.1. Accordingly,φ is either a standarcommutativity preserving map or its range is commutative. It is immediate that in thecase the range ofφ is commutative as well. Therefore, itsuffices to consider the case whethere exist an automorphism or an antiautomorphismψ of A, a linear functionalµ onA,and a nonzero scalarλ ∈ F such that

φ(x) = λψ(x) + µ(x)1, (7)

where1 = 1 ⊗ 1 is the identity element ofA. We shall treat only the case whereψ is anautomorphism. Then the case whereψ is an antiautomorphism requires only some ratobvious modifications.

Givenx, y ∈ A we have

λ2ψ(1⊗ xy) = λ2ψ((1⊗ x)(1⊗ y)

) = λψ(1⊗ x) · λψ(1⊗ y).

In view of (7) this can be rewritten as

λφ(1⊗ xy) − λµ(1⊗ xy)1 = (φ(1⊗ x) − µ(1⊗ x)1

) · (φ(1⊗ y) − µ(1⊗ y)1),

that is,

λ ⊗ φ(xy) − λµ(1⊗ xy) ⊗ 1

= (1⊗ φ(x) − µ(1⊗ x) ⊗ 1

) · (1⊗ φ(y) − µ(1⊗ y) ⊗ 1).

Setµ(x) = µ(1 ⊗ x) for everyx ∈ A. Clearly,µ is anF-linear map fromA into F. Notethat the last identity can be written as

λ ⊗ φ(xy) − 1⊗ φ(x)φ(y)

= −µ(x) ⊗ φ(y) − µ(y) ⊗ φ(x) + (λµ(xy) + µ(x)µ(y)

) ⊗ 1. (8)

Suppose thatλ /∈ F. Thenλ and 1 are linearly independent overF and so it followsfrom (8), by using a standard tensor product argument, that for every pairx, y ∈ A, bothelementsφ(xy) andφ(x)φ(y) lie in the linear span ofφ(x), φ(y) and1. In particular,there areαx,y, βx,y, γx,y ∈ F such that

φ(x)φ(y) = αx,yφ(x) + βx,yφ(y) + γx,y1.


-

ions,of

Commuting this relation withφ(x) it follows that

φ(x)[φ(y),φ(x)

] = βx,y

[φ(y),φ(x)

],

and so in particularφ(x)[φ(y),φ(x)] commutes with[φ(y),φ(x)], that is,

[φ(x),

[φ(y),φ(x)

]][φ(y),φ(x)

] = 0

for all x, y ∈ A. A complete linearization of this identity gives

∑π∈S3

∑σ∈S2

[φ(xπ(1)),

[φ(yσ(1)), φ(xπ(2))

]][φ(yσ(2)), φ(xπ(3))

] = 0

for all x1, x2, x3, y1, y2 ∈ A, whereSn denotes the symmetric group of degreen. Thisclearly yields

∑π∈S3

∑σ∈S2

[φ(xπ(1)),

[φ(yσ(1)), φ(xπ(2))

]][φ(yσ(2)), φ(xπ(3))

] = 0

for all x1, x2, x3, y1, y2 ∈ A. Settingx1 = x2 = x3 = x andy1 = y2 = y it follows that

[φ(x),

[φ(y),φ(x)

]][φ(y),φ(x)

] = 0

for all x, y ∈ A. According to (7) this can be written as

λ5[ψ(x),[ψ(y),ψ(x)

]][ψ(y),ψ(x)

] = 0.

Sinceλ �= 0 andψ is an automorphism, this gives

[x, [y, x]][y, x] = 0

for all x, y ∈ A. However,Mn(F) ∼= A does not satisfy this identity (recall thatn �= 1).Indeed, just takex = E11 andy = E12 − E21 (here,Eij denotes a matrix unit). This contradiction shows thatλ ∈ F.

Now writeλ for λ, and rewrite (8) as

1⊗ (λφ(xy) − φ(x)φ(y)

) + µ(x) ⊗ φ(y)

= −µ(y) ⊗ φ(x) + (λµ(xy) + µ(x)µ(y)

)⊗ 1. (9)

Suppose there isx ∈ A such thatµ(x) /∈ F, i.e.,µ(x) and 1 are linearly independent overF.Then (9) implies that for everyy ∈ A, φ(y) lies in the linear span ofφ(x) and 1. Inparticular, the range ofφ is commutative. This is of course one of the desired conclusbut in fact in the present setting it cannot occur (namely, the commutativity of the range


f

-y

he in-androach

44.map-

28.5.

φ yields the commutativity of the range ofφ, which in turn implies the commutativity oA - a contradiction). Thereforeµ(x) ∈ F for everyx ∈A. Accordingly, (9) shows that

λφ(xy) − φ(x)φ(y) + µ(x)φ(y) = −µ(y)φ(x) + (λµ(xy) + µ(x)µ(y)

)1

for all x, y ∈ A. This implies thatψ :A →A, defined byψ(x) = λ−1φ(x)−λ−1µ(x)1, isan algebra homomorphism. Clearly,ψ �= 0. But then, sinceA is simple and finite dimensional,ψ is an automorphism. That is,φ(x) = λψ(x)+µ(x)1 is a standard commutativitpreserving map. �

The statement of Theorem 3.1 is purely algebraic. However, as mentioned in ttroduction, its proof indirectly involves various tools, including those from analysismodel theoretic algebra. It would be interesting to find a more direct algebraic appwhich would give a new insight into the matter.

References

[1] P. Ara, M. Mathieu, Local Multipliers ofC∗-Algebras, Springer-Verlag, Berlin, 2003.[2] G.M. Bergman, Centralizers in free associative algebras, Trans. Amer. Math. Soc. 137 (1969) 327–3[3] M. Brešar, Commuting traces of biadditive mappings, commutativity-preserving mappings and Lie

pings, Trans. Amer. Math. Soc. 335 (1993) 525–546.[4] M. Brešar, Commuting maps: A survey, Taiwanese J. Math., in press.[5] M. Omladic, H. Radjavi, P. Šemrl, Preserving commutativity, J. Pure Appl. Algebra 156 (2001) 309–3[6] W. Watkins, Linear maps that preserve commuting pairs of matrices, Linear Algebra Appl. 14 (1976) 29–3

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Commutativity preserving linear maps on central simple algebras