COMMUTING MAPS on SOME SUBSETS THAT ARE NOT CLOSED UNDER ADDITION a Dissertation Submitted to Kent State University in Partial F

COMMUTING MAPS ON SOME SUBSETS THAT ARE NOT CLOSED UNDER ADDITION

A dissertation submitted to Kent State University in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

Willian V. Franca

August 2013 Dissertation written by

Willian V. Franca

B.S., Universidade Federal do Rio de Janeiro Janeiro (UFRJ), Rio de Janeiro, Brazil, 2005

M.S., Universidade Federal do Rio de Janeiro Janeiro (UFRJ), Rio de Janeiro, Brazil, 2008

Ph.D., Kent State University, 2013

Approved by

—————————————— Chair, Doctoral Dissertation Committee Dr. Mikhail Chebotar

—————————————— Member, Doctoral Dissertation Committee Dr. Richard Aron

—————————————— Member, Doctoral Dissertation Committee Dr. Artem Zvavitch

—————————————— Member, Outside Discipline Dr. Sergey Anokhin

—————————————— Member, Graduate Faculty Representative Dr. Arden Ruttan

Accepted by

—————————————— Chair, Department of Mathematical Sciences Dr. Andrew Tonge

—————————————— Associate Dean, College of Arts and Sciences Dr. Raymond A. Craig

ii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... v

INTRODUCTION ...... 1

1 Preliminaries ...... 3

1.1 Additive Commuting Maps ...... 3

1.2 Commuting Traces of Multiadditive Maps ...... 4

2 Additive Commuting Maps on The Set of Invertible or Singular Matrices 6

2.1 Additive Commuting Maps on The Set of Singular Matrices ...... 6

2.2 Additive Commuting Maps on The Set of Invertible Matrices ...... 7

3 Commuting Traces of Multiadditive Maps on Invertible or Singular Ma-

trices ...... 11

3.1 Commuting Traces on The Set of Invertible Matrices ...... 11

3.2 Commuting Traces on The Set of Singular Matrices ...... 15

3.3 Applications ...... 16

3.4 Herstein’s Conjecture ...... 18

4 Commuting Traces of Multiadditive Maps on The Set of Rank-k Matrices 20

4.1 Additive Commuting Maps on The Set of Rank-k Matrices ...... 20

4.2 Commuting Traces on The Set of Rank-k Matrices ...... 23

5 Commuting Traces on Invertible or Singular Operators ...... 27

5.1 Commuting Traces on Invertible Operators ...... 27

iii 5.2 Commuting Traces on Singular Operators ...... 31

BIBLIOGRAPHY ...... 33

iv ACKNOWLEDGEMENTS

Thanks to my beloved wife for her love, understanding and support. Everything was just possible because she was by my side throughout my journey. She was there to brake my falls and to help me to get back on my feet after falling.

To my brother Rodolfo for all the support during my undergrad studies. To my mom for being so brave rasing me and my brother by herself for so many years. I will never forget it. To my former step father, Almir for his ﬁnancial support during my undergrad studies, without him I would not be able to get my bachelor degree. To my father, Brasilio who was passed away so long ago, wherever he is now I know that he is proud of the man that

I have become.

To Luiza, my Master degree advisor, for being so helpful during my masters. She helped in so many ways that it would take more than a thank note to explain. Thanks to her,

Artem, and Richard everything worked out and I had the opportunity of studying here at the Kent State University.

I want also to thank Misha for all his help with my research either suggesting interesting problems for me to work on or when he was just proofreading my papers, which I believe that was a very tedious work. I also want to thank him for answering my emails with all my questions and requests. I would terribly remiss not thanking him for his valuable ﬁshing lessons that he taught to me during last summer. Some of the math learned during my

Ph.D I will probably forget, but those lessons are unforgettable.

v INTRODUCTION

In 1961, during his AMS Hour talk, I. N. Herstein asked about the characterization of all additive maps from a unital simple ring to another one that preserve the multiplicative commutator. In particular, such maps preserve the set of invertible elements. In the context of matrix rings over ﬁelds and assuming that the map is linear we do have a Linear Preserver

Problem. Thus, in the matrix ring setting, we can connect Herstein’s conjecture and Linear

Preserver Problems.

Linear Preserver Problems is a very active area. This is true because the formulation of the problems is simple and natural, having quite often nice solutions. It is also common to reﬁne known results by weakening the assumptions and to extend existing results to other matrix spaces or algebras.

In [3], M. Breˇsarshowed how commuting maps also naturally appear when dealing with some Linear Preserver Problems, and there is no doubt that this is a relevant area of application for commuting maps. Our aim will be to use Breˇsar’stechnique to investigate

Herstein’s question in the case that our ring is the matrix ring over a ﬁeld.

In order to achieve our goal, we will start studying additive commuting maps on the set of invertible or singular elements. We will show that the only maps with such properties are the standard ones. Simultaneously, we will be demonstrating how results on functional identities [4] can be extended to sets that are not closed under addition. Later, we will describe multiadditive maps that are commuting on the set of invertible or singular elements.

Such description combined with Breˇsar’sidea will lead us to draw some conclusions about

Herstein’s conjecture when R is the matrix ring over a ﬁeld K (some assumptions about K are necessary).

Finally, we will deal with commuting maps on the set of rank-k matrices. This is not

1 2

only interesting because we will generalize some of the results obtained before but also because in the case that k = 1 we will see that there are commuting maps on the set of the rank-1 matrices that are not of the standard form. The ﬁfth and last chapter will be devoted to talk about commuting maps on the sets of invertible or singular elements when the ring R is the ring of all bounded operators from a real or complex separable Hilbert space H to itself. CHAPTER 1

Preliminaries

In this work n will denote a natural number greater or equal than 2, and Mn(K) will represent the ring of all n × n matrices over a ﬁeld K with center Z = K · I, where I is the identity matrix. Furthermore, it should be mentioned that the set of all nonzero elements

∗ of K will be denoted by K .

1.1 Additive Commuting Maps

Deﬁnition 1. Let R be a ring. An additive map f : R → R is called commuting if f(r)r = rf(r), equivalently, [f(r), r] = 0 for all r ∈ R.

There are some obvious examples of commuting maps, like additive maps µ : R → C having range in the center C of R and maps of the form f(r) = λr + µ(r), where λ ∈ C.

With those examples in mind, we can ask a natural question: Are those the only examples of commuting maps? The answer will be positive if some conditions are imposed, namely, assuming that R is either a unital simple ring or a prime ring (see [3] for details).

Deﬁnition 2. An additive map d : R → R is called a derivation if d(xy) = d(x)y + xd(y) for all x, y ∈ R.

In 1957 E. C. Posner [14] proved the ﬁrst important result regarding commuting maps.

He showed that the existence of a nonzero commuting derivation d : A → A on a prime ring

A implies that A is commutative. In 1955, I. M. Singer and J. Wermer [16] proved that every continuous derivation on a commutative Banach algebra A has its range in rad(A), where rad(A) denotes the Jacobson radical of A. Thirty years later, Thomas [17] proved

3 4

that the assumption of continuity is superﬂuous in the result proved by I. M. Singer and J.

Wermer. A natural conjecture that now appears is that Sinclair0s theorem [15] also holds without assuming continuity, that is, that every (possibly discontinuous) derivation on a

Banach algebra A leaves primitive ideals (annihilator of a simple left module) of A invariant.

This is usually called the noncommutative Singer-Wermer conjecture in the literature and it has a nice formulation in terms of commuting derivations (modulo Jacobson radical).

In [2] M. Breˇsarhas obtained the very ﬁrst results about commuting maps without any sort of assumption on the way that the map acts on the product of the elements (as in the case of derivations). One of the results that he proved was the following:

Theorem 3. Let A be a simple unital ring. Let f : A → A be a commuting map. Then f(a) = λa + µ(a), where λ is a central element of A and µ has its range in the center of A.

Proof. See the original paper [1], or the survey paper [3, Corollary 3.3], or the book [4,

Corollary 5.28]).

It is also important to highlight that commuting maps give rise to the most basic and important example of functional identity. By replacing r by x1+x2, in the relation [f(r), r] =

0, we arrive at f(x1)x2 + f(x2)x1 − x2f(x1) − x1f(x2) = 0 for all x1, x2 ∈ R, which is a very

simple example of functional identity. For more information or examples see for instance [4].

1.2 Commuting Traces of Multiadditive Maps

Deﬁnition 4. Let R be a ring. A map G : Rm → R is m-additive if G is additive in each

component: G(x1, . . . , xi + yi, . . . , xm) = G(x1, . . . , xi, . . . , xm) + G(x1, . . . , yi, . . . , xm) for

all xi, yi ∈ R, and i ∈ {1, . . . , m}.

Deﬁnition 5. Let R be a ring and G : Rm → R an m-additive map. The map T : R → R deﬁned by T (x) = G(x, . . . , x) is known as the trace of G. We call T commuting if for each r ∈ R the equality G(r, . . . , r)r = rG(r, . . . , r) holds. Using the commutator form we can 5

rewrite the latter as [G(r, . . . , r), r] = G(r, . . . , r)r − rG(r, . . . , r) = 0. When m = 1 we have

G = T , that is, G is a commuting map.

Deﬁnition 6. Let R be a ring and G : Rm → R be an m-additive map. We say that the

trace of G is commuting on a subset S of R if G(s, . . . , s)s = sG(s, . . . , s) for all s ∈ S.

In [12, Theorem 3.1] P.-H. Lee, T.-L. Wong, J.-S. Lin, and R.-J. Wang proved the following:

Theorem 7. Let R be a prime ring. Let M : Rm → R an m-additive commuting map on

R. Assume that either char R = 0 or char R > m and that R is not algebraic of bounded degree less than or equal to m over its extended centroid C. Then, there exist µ0 ∈ C and maps µi : R → C, i ∈ {1, . . . , m}, such that each µi is the trace of an i-additive map and

m m−1 G(r, . . . , r) = µ0r + µ1(r)r + ... + µm−1(r)r + µm(r) for all r ∈ R.

Roughly speaking they showed that for each r ∈ R, the image M(r, . . . , r) has to be a polynomial in r of degree less than or equal to m with coeﬃcients in C.

Remark 8. Observe that we may assume that an m-additive map G : Rm → R, whose trace

commutes on some subset S of R, is symmetric when either char R = 0 or char R ≥ m + 1.

Indeed, since we can replace G with P G(s , . . . , s ) = G0 (s , . . . , s ), and it σ∈Sm σ(1) σ(m) 1 m is clear that G0 is symmetric and that [G0 (s, . . . , s), s] = 0 for each s ∈ S. Notice that

G0 (s, . . . , s) = m!G(s, . . . , s); Thus the trace of G0 is commuting if and only if the trace of

G is, since either char R = 0 or char R ≥ m + 1.

m We just want to recall that any m-additive map G : R → R is m-linear over Q (respectively Zq) when char R = 0 (respectively char R = q). This fact will be largely used. CHAPTER 2

Additive Commuting Maps on The Set of Invertible or Singular Matrices

The study of (linear) preserver problems on subsets of matrices remains an active research area for decades. In particular many important results in this area hold for subsets of matrices that are not closed under addition such as invertible matrices, singular matrices, matrices of rank one, etc (see the survey paper [13] for details). The theory of functional identities [4] is a relatively new subject and to the best of our knowledge all the results in this theory have been obtained so far only for the sets closed under addition. The purpose of this chapter is to show on the test example of a commuting map on invertible or singular matrices how results on functional identities can be extended to sets that are not closed under addition.

The results in this chapter have been published in [7].

2.1 Additive Commuting Maps on The Set of Singular Matrices

Theorem 9. Let K be any ﬁeld, and let G : Mn(K) → Mn(K) be an additive map such that

G(x)x = xG(x), for all singular x ∈ Mn(K). (2.1)

Then there exist an element λ ∈ Z and an additive map µ : Mn(K) → Z such that

G(x) = λx + µ(x), for all x ∈ Mn(K).

Proof. Let z, w ∈ Z be central elements. Observe that zeij is singular for all choices of i, j.

In addition, zeij + wekl is singular for all i, j, k, l ∈ {1, . . . , n} when n ≥ 3. Thus, using

6 7

(2.1) and additivity of G we see that:

G(zeij)zeij = zeijG(zeij), (2.2)

G(zeij)wekl + G(wekl)zeij = zeijG(wekl) + weklG(zeij), if n ≥ 3, (2.3) for all i, j ∈ {1, . . . , n}, and z, w ∈ Z.

Our goal is to show that (2.3) also holds when n = 2. First, observe that zeii + weip, and zeii + wepi are singular when p 6= i. Using (2.1), (2.2) and the additivity of G we have:

G(zeii)weip + G(weip)zeii = zeiiG(weip) + weipG(zeii), if p 6= i. (2.4)

The next step is to examine the entries of matrices G(ze11) and G(ze22). It follows from (2.2) that 0 = G(ze11)ze11 · e22 = ze11G(ze11) · e22, and e22 · G(ze11)ze11 = e22 · ze11G(ze11) = 0. Hence, e11G(ze11)e22 = e22G(ze11)e11 = 0 and so G(ze11) = e11G(ze11)e11 + e22G(ze11)e22. Similarly, G(ze22) = e11G(ze22)e11 + e22G(ze22)e22. This last relation tells us that G(e11) (respectively G(e22)) and ze22 (respectively ze11) commute.

It means that (2.3) holds for i = j = 1, and k = l = 2. More precisely, we obtained:

G(ze11)we22 + G(we22)ze11 = ze11G(we22) + we22G(ze11). (2.5)

Let A = ze11 +we12 +ze21 +we22. Since A is singular, we obtain applying (2.1), (2.2), (2.4) and (2.5) together that G(we12)ze21 + G(ze21)we12 = we12G(ze21) + ze21G(we12). Hence,

(2.3) holds when n = 2. Therefore, (2.3) holds for all n.

Writing any matrix x ∈ Mn(K) as the sum of its matrix entries xijeij we can conclude from (2.2) and (2.3) that G(x)x = xG(x) for all x ∈ Mn(K). The desired result follows now from Theorem 3.

2.2 Additive Commuting Maps on The Set of Invertible Matrices

Throughout this section K is any ﬁeld but Z2. 8

Proposition 10. Let G : Mn(K) → Mn(K) be an additive map such that

G(x)x = xG(x), for all invertible x ∈ Mn(K). (2.6)

Then G(z) ∈ Z for all z ∈ Z.

Proof. Let z ∈ Z be any nonzero central element and eij, i 6= j, be any non-diagonal matrix

unit. Note that I + eij is invertible. Since |K| > 2 there exists a nonzero element kij ∈ K

such that u = kij(I +eij) and v = z +u = z +kij(I +eij) are invertible. It follows from (2.6)

that G(z+u)(z+u) = (z+u)G(z+u), G(u)u = uG(u) and G(z)z = zG(z), so the additivity

of G yields G(z)u + G(u)z = uG(z) + zG(u). Since z is central, we get G(z)u = uG(z) and

taking into account that u = kij(I + eij) with invertible kij we obtain G(z)eij = eijG(z) for

all eij with i 6= j, or using the commutator form [G(z), eij] = G(z)eij − eijG(z) = 0. Note

that [G(z), eii] = [G(z), eijeji] = [G(z), eij]eji +eij[G(z), eji] = 0. Therefore G(z) commutes

with all matrix units and must be a central element.

Theorem 11. Let G : Mn(K) → Mn(K) be an additive map such that

G(x)x = xG(x), for all invertible x ∈ Mn(K). (2.7)

Then there exist an element λ ∈ Z and an additive map µ : Mn(K) → Z such that

G(x) = λx + µ(x), for all x ∈ Mn(K). (2.8)

Proof. Let z ∈ Z. Since, |K| > 2 there exists a nonzero element a = a(z) ∈ K such that

zeij +aI is invertible. It follows from (2.7) that G(zeij +aI)(zeij +aI) = (zeij +aI)G(zeij +

aI), so the additivity of G and Proposition 10 tells us that

G(zeij)zeij = zeijG(zeij), for all i, j ∈ {1, . . . , n}, and z ∈ Z. (2.9)

Let i, j ∈ {1, . . . , n}. First, let’s assume that i 6= j. Once again, since |K| > 2 there

exists a nonzero element a = a(z, w) ∈ K such that (zeij +wekl)+aI is invertible. It follows 9

from (2.7) that G(zeij +wekl +aI)(zeij +wekl +aI) = (zeij +wekl +aI)G(zeij +wekl +aI).

Using (2.9), the additivity of G and Proposition 10, we see that:

G(zeij)wekl + G(wekl)zeij = zeijG(wekl) + weklG(zeij), (2.10)

for all k, l ∈ {1, . . . , n}, and for all z, w ∈ Z. Now, we want to show that the above equation

holds when i = j. Note that there is nothing to be proved in the case that k 6= l. So, we

can assume without loss of generality that k = l, and k 6= i. By the same argument that we

have used before, we see that G(zeii + wekk)(zeii + wekk) = (zeii + wekk)G(zeii + wekk) if

|K| ≥ 4. From (2.9), and additivity of G, we see that (2.10) holds when i = j, and |K| ≥ 4.

It remains to cover the case when |K| = 3. Observe that G(eii + ekk)(eii + ekk) =

(eii + ekk)G(eii + ekk). Hence,

G(eii)ekk + G(ekk)eii = eiiG(ekk) + ekkG(eii). (2.11)

On the other hand, G(zx) = zG(x) for all x ∈ Mn(K), and z ∈ Z, when K = Z3. Thus, if we

multiply both sides of the above equation by zw, where z, w ∈ Z, we get that G(zeii)wekk +

G(wekk)zeii = zeiiG(wekk) + wekkG(zeii). Then, ﬁnally we can conclude that

G(zeij)wekl + G(wekl)zeij = zeijG(wekl) + weklG(zeij), (2.12)

for all i, j, k, l ∈ {1, . . . , n}, and for all z, w ∈ Z. Therefore, G(x)x = xG(x) for all x ∈

Mn(K), since G is additive and Mn(K) is additively generated by zeij. The result follows now from Theorem 3.

It is interesting to note that the technical condition on the cardinality of K is necessary in the case of the Theorem 11 and it can be seen in the following example.

Example 1. The set of all invertible matrices of M2(Z2) consists of six elements: 10

          1 0 1 1 1 0 0 1 1 1           A1 =  , A2 =  , A3 =  , A4 =  , A5 =  , 0 1 0 1 1 1 1 0 1 0   0 1   and A6 =  . 1 1

Note that A1, A2, A3 and A5 additively generate M2(Z2). We deﬁne G(A1) = G(A5) =

A5 and G(A2) = G(A3) = 0, so we get G(A4) = G(A2 +A3) = 0 and G(A6) = G(A1 +A5) =

0. Clearly, G is commuting on invertible elements of M2(Z2), so it satisﬁes (2.7), but it is not of the form (2.8) as it maps the identity element A1 to a noncentral element A5.

Let us conclude with an easy corollary, just to illustrate how Theorem 11 can be applied to some Linear Algebra questions.

Corollary 12. Let G : Mn(K) → Mn(K) be an additive map such that for every inner

automorphism ϕ : Mn(K) → Mn(K), the maps G and ϕ commute, that is G(ϕ(x)) =

ϕ(G(x)) for all x ∈ Mn(K). Then there exist an element λ ∈ Z and an additive map

µ : Mn(K) → Z such that

G(x) = λx + µ(x), for all x ∈ Mn(K).

Proof. Note that for every invertible matrix y we have G(y−1xy) = y−1G(x)y for all x ∈

−1 −1 Mn(K). In particular, taking x = y, we obtain G(y yy) = G(y) = y G(y)y, or yG(y) =

G(y)y for all invertible y ∈ Mn(K). Now Theorem 11 completes the proof. CHAPTER 3

Commuting Traces of Multiadditive Maps on Invertible or Singular Matrices

In the second chapter we have obtained results related to additive commuting maps on the set of invertible or singular matrices. Now, motivated by the result proved by P.-H.

Lee, T.-L. Wong, J.-S. Lin, and R.-J. Wang (Theorem 7) we will investigate the case when

R = Mn(K) to see if the conclusion holds when the trace of an m-multiadditive map is commuting either in the set of invertible or singular elements when either char K = 0 or char K ≥ m + 1. The results in this chapter will appear in [9].

3.1 Commuting Traces on The Set of Invertible Matrices

m Proposition 13. Let n > m > 1 be natural numbers. Let G : Mn(K) → Mn(K) be a symmetric m-additive map such that

[G(x, . . . , x), x] = 0 for all invertible x ∈ Mn(K). (3.1)

Assume that char K = 0 or char K ≥ m + 1. Assume also that K contains at least m + 3 elements. Then G(z, . . . , z) ∈ Z for all z ∈ Z.

Proof. Let s be the smallest even number greater or equal than m, that is, s = m if m is

even, and s = m + 1 if m is odd. Fix z ∈ Z, and let eij, i 6= j, be any non-diagonal matrix

s unit. For each a ∈ {±1,..., ± 2 }, let ya = az+u, where u = k(I +eij), and k ∈ K. Note that

ya = az + u is invertible if and only if −az is not an eigenvalue of u. So, since |K| ≥ m + 3,

s we can ﬁnd a nonzero element k ∈ K such that ya is invertible for all a ∈ {±1,..., ± 2 }.

11 12

It follows from (3.1) that 0 = [G(u, . . . , u), u] = [G(z, . . . , z), z] = [G(ya, . . . , ya), ya], and this last bracket can be written as [G(ya, . . . , ya), u] = 0, since ya = az + u. In particular,

s [G(ya, . . . , ya) + G(y−a, . . . , y−a), u] = 0 for all a ∈ {1,..., 2 }. Now, since G is symmetric, s m-additive, and ya = az + u we can obtain for each a ∈ {1,..., 2 } that: m X m−ζ m G(ya, . . . , ya) = a G(z, . . . , z, u, . . . , u), (3.2) ζ | {z } ζ=0 ζ and m X m−ζ m−ζ m G(y−a, . . . , y−a) = (−1) a G(z, . . . , z, u, . . . , u). (3.3) ζ | {z } ζ=0 ζ By keeping in mind the equations (3.2), (3.3), and the relation [G(u, . . . , u), u] = 0 we

see that [G(ya, . . . , ya) + G(y−a, . . . , y−a), u] = 0 becomes:

s−2 X2 m am−2ζ [G(z, . . . , z, u, . . . , u), u] = 0, when m is even, (3.4) 2ζ | {z } ζ=0 2ζ and s−4 X2 m am−(2ζ+1) [G(z, . . . , z, u, . . . , u), u] = 0, when m is odd. (3.5) 2ζ + 1 | {z } ζ=0 2ζ + 1

With (3.5), the identity [G(ya, . . . , ya), u] = 0 becomes:

s−2 X2 m am−2ζ [G(z, . . . , z, u, . . . , u), u] = 0, when m is odd. 2ζ | {z } ζ=0 2ζ

s Therefore, for each a ∈ {1,..., 2 } we have obtained an equation of the form (3.4) when s s m is either even or odd. It means that we have 2 equations in 2 unknowns, namely m 2ζ [G(z, . . . , z, u, . . . , u), u], where u appears exactly in 2ζ components of G, and ζ ∈ 13

s−2 {1,..., 2 }. Using matrix notation we can rewrite these systems in the following way:     1 1 1 ... 1 m[G(z, . . . , z), u]    0     .   m m−2 m−4 m−(s−2)   .   2 2 2 ... 2   .           3m 3m−2 3m−4 ... 3m−(s−2)   m[G(z, . . . , z, u, . . . , u), u]  = 0.    2ζ       . . . . .   .   . . . . .   .          s m s m−2 s m−4 s m−(s−2) m ( 2 ) ( 2 ) ( 2 ) ... ( 2 ) s−2 [G(z, u, . . . , u), u] Because the determinant of the Vandermonde matrix formed by the coeﬃcients of the system is not zero, we get that [G(z, . . . , z), u] = 0. As u = k(I + eij) and k is a nonzero

element of K, we conclude that [G(z, . . . , z), eij] = 0. Finally, we see that

[G(z, . . . , z), eii] = [G(z, . . . , z), eijeji] = [G(z, . . . , z), eij]eji + eij[G(z, . . . , z), eji] = 0.

It means that G(z, . . . , z) commutes with all matrix units, so G(z, . . . , z) must be a central element for all z ∈ Z.

m Theorem 14. Let m, n be natural numbers, where n > m > 1. Let G : Mn(K) → Mn(K) be an m-additive map such that

[G(x, . . . , x), x] = 0, for all invertible x ∈ Mn(K). (3.6)

2 Assume that char K = 0 or char K ≥ m+1. Assume also that K contains at least m +2m+3 elements. Then, there exist µ0 ∈ Z and maps µi : Mn(K) → Z, i ∈ {1, . . . , m}, such that

m m−1 each µi is the trace of an i-additive map and G(x, . . . , x) = µ0x + µ1(x)x + ... +

µm−1(x)x + µm(x) for all x ∈ Mn(K).

Proof. Without loss of generality, we may assume that G is symmetric. Once again, let

s be the smallest even number greater or equal than m, that is, s = m if m is even, and

s = m + 1 if m is odd. 14

Pm+1 For each r ∈ {1, . . . , m+1}, let zr ∈ K and let ir, jr ∈ {1, . . . , n}. Put b = r=1 zreirjr . The rank of b is at most m + 1, and so b has at most m + 1 nonzero eigenvalues. For any

s Pm+1 a ∈ {±1,..., ± 2 } and nonzero k ∈ K the matrix ya = akI + b = akI + r=1 zreirjr is

2 invertible as long as −ak is not one of the nonzero eigenvalues of b. Since |K| ≥ m +2m+3,

s we can choose a k such that ya is invertible for all a ∈ {±1,..., ± 2 }. It follows from (3.6) s and ya = akI + b that [G(ya, . . . , ya), b] = [G(ya, . . . , ya), ya] = 0 for all a ∈ {±1,..., ± 2 }. Consequently,

s [G(y , . . . , y ) + G(y , . . . , y ), b] = 0 for all a ∈ {1,..., }. (3.7) a a −a −a 2

Now, since G is symmetric, m-additive and ya = akI + b, we conclude that m X r m G(ya, . . . , ya) = a G(kI, . . . , kI, b, . . . , b), (3.8) r | {z } r=0 r s for each a ∈ {±1,..., ± 2 }. Thus, if we take into the account that G(kI, . . . , kI) ∈ Z (Proposition 13) and the equation (3.8) we can derive from (3.7) that:

[G(b, b), b] = 0 if m = 2, and s−2 X2 m a2r [G(kI, . . . , kI, b, . . . , b), b] = 0 if m ≥ 3. 2r | {z } r=0 2r As in the proof of the Proposition 13, for each m ≥ 3 we have:     1 1 1 ... 1   [G(b, . . . , b), b]      2 4 s−2   m   1 2 2 ... 2   [G(kI, kI, b, . . . , b), b]     2       1 32 34 ... 3s−2   .  = 0.    .       . . . . .     . . . . .   m     [G(kI, . . . , kI, b, . . . , b), b]   s − 2   s 2 s 4 s s−2  | {z } 1 ( 2 ) ( 2 ) ... ( 2 ) m−(s−2) Therefore, " m+1 m+1 m+1 # X X X G( zreirjr ,..., zreirjr ), zreirjr = 0, (3.9) r=1 r=1 r=1 15

Pm+1 because [G(b, . . . , b), b] = 0, and b = r=1 zreirjr for all m ≥ 2. Pm+1 Notice that the previous argument works if the element b = r=1 zreirjr is replaced P with cA = r∈A zreirjr , where A ⊆ {1, . . . , m+1}, and ir, jr, zr are exactly as before. That is, in such cases, we have " # X X X G( zreirjr ,..., zreirjr ), zreirjr = [G(cA, . . . , cA), cA] = 0. (3.10) r∈A r∈A r∈A

With (3.10) for A ( {1, . . . , m + 1}, we can linearize the identity [G(b, . . . , b), b] = 0. Namely, expanding (3.9) using the fact that G is m-additive, and the identities (3.10) for

A ( {1, . . . , m + 1}, we will arrive at

X [G(ziσ(1) eiσ(1)jσ(1) , . . . , ziσ(m) eiσ(m)jσ(m) ), ziσ(m+1) eiσ(m+1)jσ(m+1) ] = 0. (3.11) σ∈Sm+1

Hence, writing any matrix x ∈ Mn(K) as the sum of its entries xijeij we can conclude from (3.11) that [G(x, . . . , x), x] = 0 for all x ∈ Mn(K), since G is m-additive. With n > m > 1, the desired result now follows from Theorem 7.

3.2 Commuting Traces on The Set of Singular Matrices

Theorem 15. Let m, n be natural numbers, where n > m + 1, and m > 1. Let G :

m Mn(K) → Mn(K) be a symmetric m-additive map such that

[G(x, . . . , x), x] = 0, for all singular x ∈ Mn(K). (3.12)

Assume that char K = 0 or char K ≥ m + 1. Then, there exist µ0 ∈ Z and maps µi :

Mn(K) → Z, i ∈ {1, . . . , m}, such that each µi is the trace of an i-additive map and

m m−1 G(x, . . . , x) = µ0x + µ1(x)x + ... + µm−1(x)x + µm(x) for all x ∈ Mn(K).

Proof. For each r ∈ {1, . . . , m + 1}, let zr be a central element and let ir, jr ∈ {1, . . . , n}. P It is clear that cA = r∈A zreirjr is singular for all A ⊆ {1, . . . , m + 1}, since n ≥ m + 2. It 16

follows from (3.12) that [G(cA, . . . , cA), cA] = 0. Thus, we have: " # X X X G( zreirjr ,..., zreirjr ), zreirjr = 0, for all A ⊆ {1, . . . , m + 1}. (3.13) r∈A r∈A r∈A

In particular, for A = {1, . . . , m + 1} we get that

" m+1 m+1 m+1 # X X X G( zreirjr ,..., zreirjr ), zreirjr = 0. r=1 r=1 r=1

Thus, (as we did in the proof of the Theorem 14) if we apply the m-additivity of G

together with the identity (3.13) (for each A ( {1, . . . , m}), we can obtain from the last equation that:

X [G(ziσ(1) eiσ(1)jσ(1) , . . . , ziσ(m) eiσ(m)jσ(m) ), ziσ(m+1) eiσ(m+1)jσ(m+1) ] = 0. (3.14) σ∈Sm+1

Therefore, [G(x, . . . , x), x] = 0 for all x ∈ Mn(K) since G is m-additive and Mn(K) is

additively generated by zijeij, where zij ∈ Z. The result follows now from Theorem 7.

3.3 Applications

Let us illustrate how the results obtained in the sections 3.1 and 3.2 can be used to answer questions that are related to linear preserver problems. In [6] J. Dieudonn´eproved that an invertible linear operator θ : Mn(K) → Mn(K) mapping the set of singular matrices into

t itself is of the form θ(x) = P xQ or θ(x) = P x Q for some P,Q ∈ Mn(K) with det(PQ) 6= 0,

where K is any ﬁeld. Later, C. Cao, X. Zhang have characterized linear maps θ from Mn(K)

to itself that preserve the set of invertible matrices when char K is neither 2 nor 3 (see [5] for details).

Using Breˇsar’stechnique (see [3, Theorem 5.5] for details) as an application of the

Theorem 14 we give a simple proof of the result proved by C. Cao, X. Zhang in [5, Theorem

3.2]. 17

Corollary 16. Let K be a ﬁeld with char K 6= 2, 3 and let n > 2. Assume also that

K contains at least 11 elements. Let θ : Mn(K) → Mn(K) be a linear map such that

θ : GLn(K) → GLn(K) is a bijection, where GLn(K) is the set of all invertible matrices.

2 Assume also that θ satisﬁes the condition [θ(x ), θ(x)] = 0 for all x ∈ GLn(K). Then

θ(x) = λφ(x), where λ ∈ K, λ 6= 0, and φ is either an isomorphism or antiisomorphism

from Mn(K) to itself.

2 Proof. The idea of this proof consists of ﬁnding a biadditive map G : Mn(K) → Mn(K)

−1 −1 such that [G(x, x), x] = 0 for all x ∈ GLn(K). Set G(x, y) = θ(θ (x)θ (y)). Clearly,

G is bilinear. Fix x ∈ GLn(K). It follows from the hypothesis that there exists a unique

u ∈ GLn(K) such that x = θ(u). By noticing that u is invertible we get that G(x, x)x = G(θ(u), θ(u))θ(u) = θ(u2)θ(u) = θ(u)θ(u2) = xG(x, x). It means that [G(x, x), x] = 0 for

all x ∈ GLn(K). Now, we are under the conditions of the Theorem 14. Hence, there are

µ0 ∈ K, µ1, µ2 : Mn(K) → Z, where µ1 is additive and µ2 is biadditive such that

−1 2 2 G(x, x) = θ(θ (x) ) = µ0x + µ1(x)x + µ2(x, x) for all x ∈ Mn(K). (3.15)

On the other hand, θ is bijective in Mn(K) (because θ is linear and bijective in GLn(K)). Thus, replacing θ−1(x) by y in the equation (3.15) we can conclude that [θ(y2), θ(y)] = 0

for all y ∈ Mn(K). Then, [3, Theorem 5.5] tells us that θ(x) = λφ(x) + µ(x)I, where λ ∈ K, λ 6= 0, µ is a linear functional, and φ is either an isomorphism or antiisomorphism from

Mn(K) to itself. It remains to show that µ is identically zero. Notice that we only need to

prove that µ(eij) = 0 for all i, j ∈ {1, . . . , n}, since µ is linear and {eij : i, j ∈ {1, . . . , n}}

spans Mn(K) as a vector space. Let A be a nilpotent matrix of degree 2, that is, A2 = 0. Since, φ is either an isomorphism

or an antiisomorphism we see that φ(A) is also nilpotent of degree 2. Notice that θ(A) is

singular, because A is singular, and θ : GLn(K) → GLn(K) is a bijection. From the relation det(λφ(A)+µ(A)I) = det(θ(A)) = 0, we realize that −µ(A) is an eigenvalue of the nilpotent 18

element λφ(A), so µ(A) = 0. In particular, µ(eij) = 0, if i, j ∈ {1, . . . , n}, and i 6= j.

Now, let A be a singular idempotent matrix, that is, A2 = A, and A 6= I. So, φ(A)

is also idempotent (because φ is an (anti)isomorphism), and θ(A) is singular (because θ :

GLn(K) → GLn(K) is a bijection). Then, as before, the equality det(λφ(A) + µ(A)I) = det(θ(A)) = 0 tells us that −µ(A) is an eigenvalue of λφ(A), therefore µ(A) is either −λ or

0. In particular, µ(eii) is either −λ or 0, for all i ∈ {1, . . . , n}.

Our aim is to show that µ(eii) = 0 for all i ∈ {1, . . . , n}. Suppose there exists k ∈

{1, . . . , n} such that µ(ekk) = −λ. It is clear that Aj = ekk + ejj is idempotent for all j ∈ {1, . . . , n}\{k}. Thus, we see that µ(ejj) = 0, otherwise µ(Aj) = µ(ekk)+µ(ejj) = −2λ, which is not possible, since n > 2 and, Aj is idempotent for all j ∈ {1, . . . , n}\{k}. On the other hand, φ(I) = I since φ is an (anti)isomorphism. Thus, we see that θ(I) =

Pn λφ(I) + µ(I)I = λI + i=1 µ(eii)I = 0, contradicting the fact that θ(I) ∈ GLn(K). So, the proof is completed.

Again using Breˇsar’stechnique as an application of the Theorem 15 we can easily obtain the following result that was ﬁrst proved by J. Dieudonn´e(see [6] for details).

Corollary 17. Let K be a ﬁeld with char K 6= 2, 3, and let n > 3 be a natural number.

Let θ : Mn(K) → Mn(K) be a linear map that is bijective on singular matrices. Assume

2 also that θ satisﬁes the condition [θ(x ), θ(x)] = 0 for all singular matrix x ∈ Mn(K). Then

θ(x) = λφ(x), where λ ∈ K, λ 6= 0, and φ is either an isomorphism or antiisomorphism

from Mn(K) to itself.

3.4 Herstein’s Conjecture

I. N. Herstein, at the end of his AMS Hour talk [11], suggested some possible areas of

research. One of such problems was exactly the following: 19

The study of multiplicative analogs and variants of the Lie question for the group of invertible elements in simple rings with unit elements. That is, characterize all additive mapping of a simple ring R into a simple ring R0 such that f(xyx−1y1) = f(x)f(y)f(x)−1f(y)−1

for all invertible x, y ∈ R.

Now, let us show how using Corollary 16, Herstein’s question can be answered for the

known case of matrices over ﬁelds.

Theorem 18. Let K be a ﬁeld with char K 6= 2, 3 and let n > 2. Assume also that K

contains at least 11 elements. Let θ : Mn(K) → Mn(K) be a bijective linear map such that

−1 1 −1 −1 θ(xyx y ) = θ(x)θ(y)θ(x) θ(y) for all invertible x, y ∈ Mn(K). Then θ(x) = λφ(x),

where λ ∈ K, λ 6= 0, and φ is either an isomorphism or anti-isomorphism from Mn(K) to itself.

Proof. Let x be an invertible matrix. Taking x = y, we can deduce from the condition

θ(xyx−1y1) = θ(x)θ(y)θ(x)−1θ(y)−1 that θ(I) = I. Using y = x2 and θ(I) = I we arrive at

2 2 θ(x)θ(x ) = θ(x )θ(x), which is true for all invertible x ∈ Mn(K). Now, the conclusion can be derived from Corollary 16. CHAPTER 4

Commuting Traces of Multiadditive Maps on The Set of Rank-k Matrices

We have proved that if G : Mn(K) → Mn(K) is an additive map that commutes on the set of invertible matrices then G has the form

G(x) = λx + µ(x), (4.1) where λ ∈ Z, µ : Mn(K) → Z is an additive map, and K is any ﬁeld but Z2. Thus, writing this result in terms of rank, we could say that: If G : Mn(K) → Mn(K) is an additive map such that [G(x), x] = 0 for each rank-n matrix x ∈ Mn(K) then G has the form (4.1). So, it gave rise to a natural question: For a ﬁxed natural number k satisfying 1 ≤ k ≤ n − 1 does an additive map G from Mn(K) to itself with the property that [G(x), x] = 0 for every rank-k matrix x ∈ Mn(K) have the description (4.1)? Under the assumption that K is a

ﬁeld such that either char K = 0 or char K > 3 the answer will be yes when 1 < k ≤ n − 1. In the case that k = 1 we have a negative answer for n ≥ 3. Also, we will discuss the m-additive case.

The results in the ﬁrst section were published in [8].

4.1 Additive Commuting Maps on The Set of Rank-k Matrices

Lemma 19. Let n ≥ 3 be a natural number. Fix a natural number k satisfying 1 < k ≤ n, and numbers i, j ∈ {1, . . . n}. Then there exists a matrix B = B(i, j) ∈ Mn(K) such that zeij + tB has rank k for all nonzero z, t ∈ K.

Proof. Fix i, j, k as described in the statement. For each v ∈ {2, . . . , k} choose iv, jv ∈

20 21

{1, . . . , n} such that i 6= iu, j 6= ju, iu 6= iv, and ju 6= jv when u 6= v (u, v ∈ {2, . . . , k}). Set

Pk ∗ B = v=2 eivjv . Clearly zeij + tB has rank k for all z, t ∈ K .

In the same way we can prove the following:

Lemma 20. Let n ≥ 3 be a natural number. Fix a natural number k satisfying 1 < k ≤ n, and numbers i, j, k, l ∈ {1, . . . n}, where (i, j) 6= (k, l). Then there exists a matrix B =

B(i, j, k, l) ∈ Mn(K) such that zeij + wekl + tB has rank k for all nonzero z, w, t ∈ K.

Theorem 21. Let n ≥ 3 be a natural number. Fix a natural number k satisfying 1 < k ≤

n − 1. Assume that char K = 0 or char K > 3. Let G : Mn(K) → Mn(K) be an additive map such that

[G(x), x] = 0 for every rank-k matrix x ∈ Mn(K). (4.2)

Then there exist an element λ ∈ Z and an additive map µ : Mn(K) → Z such that

G(x) = λx + µ(x) for each x ∈ Mn(K). (4.3)

Proof. Let z, w ∈ K, and i, j, k, l ∈ {1, . . . , n}, where (i, j) 6= (k, l). All the identities that will be obtained during this proof can be easily checked when either z or w are zero. So, we

may assume that z, w are nonzero elements of K. Using Lemma 19, we can ﬁnd a matrix

B ∈ Mn(K) such that zeij + tB has rank k for all nonzero t ∈ K. Then equation (4.2) tells

∗ us that [G(zeij + tB), zeij + tB] = 0 for all t ∈ K . Replacing t by 1, −1 and 2 we arrive at

(because char K > 3):

[G(zeij), zeij] = 0 for all z ∈ K, and i, j ∈ {1, . . . , n}. (4.4)

On the other hand, Lemma 20 guarantees the existence of an element D ∈ Mn(K) such

that zeij +wekl+tD = c+tD has rank k. Again, equation (4.2) provides [G(c+tD), c+tD] =

0. So, it follows from the same argument that we used in the proof of equation (4.4) that

[G(c), c] = 0, which becomes

[G(zeij), wekl] + [G(wekl), zeij] = 0, (4.5) 22

since c = zeij + wekl (note that equation (4.4) has been used).

Hence, [G(zeij), wekl] + [G(wekl), zeij] = 0, for all i, j, k , l ∈ {1, . . . n}, and z, w ∈ K.

Thus, [G(x), x] = 0 for each x ∈ Mn(K), because G is additive and Mn(K) is additively

generated by zeij. The desired result now is obtained from Theorem 3.

Now, we will study commuting maps on the set of rank-1 matrices.

Theorem 22. Let K be any ﬁeld, and n = 2. Let G : M2(K) → M2(K) be an additive map such that

[G(x), x] = 0 for every rank-1 matrix x ∈ M2(K). (4.6)

Then there exist an element λ ∈ Z and an additive map µ : M2(K) → Z such that

G(x) = λx + µ(x) for each x ∈ M2(K). (4.7)

Proof. The result follows directly from Theorem 9 (take n = 2).

Surprisingly, Theorem 21 fails when k = 1 and n ≥ 3.

Example 2. Let K be any ﬁeld. Fix a natural number n ≥ 3. Let us deﬁne a linear map from

Mn(K) to itself in the following way: G(ze11) = −zen2, G(ze1n) = ze12, G(ze21) = zen1, Pn G(ze2n) = j=2 zejj, and G(zeij) = 0 otherwise, where z ∈ K. The map G is commuting on the set of rank-1 matrices. Indeed, let x be a rank-1 matrix. Thus, x can be written in the following way: n n X X x = βj xieij , j=1 i=1 23

where xv, βv ∈ K for all v ∈ {1, . . . , n}, and βl = 1 for some l ∈ {1, . . . , n}. Therefore, for the above x, we have:

 n  X G(x) = −β1x1en2 + βnx1e12 + β1x2en1 + βnx2ejj. j=2

By standard computations, we see that:

 n n  X X G(x)x = βnx2  βj xieij = xG(x). j=1 i=1

Hence, [G(x), x] = 0 for all rank-1 matrix x ∈ Mn(K), but G is not of the form (4.1) as it

maps the identity element I to a noncentral element −en2.

4.2 Commuting Traces on The Set of Rank-k Matrices

We will state the following result that is a mere generalization of the Lemma 20.

Lemma 23. Let K be a ﬁeld, and let m, n be natural numbers, where n ≥ m + 1, and

m > 1. Let Mn(K) be the ring of all n × n matrices over K. Fix a natural number k satisfying (m + 1) ≤ k ≤ n. Let v be a natural number such that 1 ≤ v ≤ m + 1. For each

r ∈ {1, . . . , v}, let ir, jr ∈ {1, . . . n}, where (iu, ju) 6= (iv, jv) when u 6= v. Then there exists Pv a matrix B = B((ir, jr)(r ∈ {1, . . . , v}) ∈ Mn(K) such that r=1 zreirjr + tB has rank k

for all nonzero z1, . . . , zv, t ∈ K.

The argument needed in the proof of the next theorem is similar to the one used in the

proof of the Theorem 14, but for completeness sake we will provide all the details.

Theorem 24. Let K be a ﬁeld, and let m, n be natural numbers, where n ≥ m + 1, and

m > 1. Let Mn(K) be the ring of all n × n matrices over K with center Z = K · I, where I is the identity matrix. Fix a natural number k satisfying (m + 1) ≤ k ≤ n. Let

m G : Mn(K) → Mn(K) be a symmetric m-additive map such that

[G(x, . . . , x), x] = 0 for every rank-k matrix x ∈ Mn(K). (4.8) 24

Assume that char K = 0 or char K > m + 1. Assume also that K contains at least m + 4

elements. Then, there exist µ0 ∈ Z and maps µi : Mn(K) → Z, i ∈ {1, . . . , m}, such that

m m−1 each µi is the trace of an i-additive map and G(x, . . . , x) = µ0x + µ1(x)x + ... +

µm−1(x)x + µm(x) for each x ∈ Mn(K).

∗ Proof. For each r ∈ {1, . . . , m + 1}, let zr ∈ K and let ir, jr ∈ {1, . . . , n}, where (iu, ju) 6= Pm+1 (iv, jv) when u 6= v. By Lemma 23 there exists a matrix B ∈ Mn(K) such that r=1 zreirjr +

tB = c + tB has rank k for all nonzero t ∈ K. It follows from (4.8) that [G(c + tB, . . . , c +

∗ tB), c + tB] = 0 for all t ∈ K . Hence, [G(c + tB, . . . , c + tB), c + tB] + [G(c + sB, . . . , c +

∗ sB), c + sB] = 0 for all s, t ∈ K . Letting 0 6= s = −t, and using the symmetricity of G, we get that:

ζ X m X m [G(tB, . . . , tB, c, . . . , c), c] + [G(tB, . . . , tB, c, . . . , c), tB] = 0, (4.9) 2h | {z } 2h + 1 | {z } h=0 2h h=0 2h+1

m m−1 where ζ = 2 , and = ζ − 1 when m is even and ζ = = 2 when m is odd. For convenience let us set:

m α(h) = [G(B,...,B, c, . . . , c), c], where h ∈ {0, . . . , ζ}. 2h | {z } 2h

m γ(h) = [G(B,...,B, c, . . . , c),B], where h ∈ {0, . . . , }. 2h + 1 | {z } 2h+1 Observe that for each t ∈ {1, . . . , + 2} we have obtained an equation of the form (4.9).

It means that we got ( + 2) distinct equations of such form (this is true because char 25

K > m + 1, and |K| ≥ m + 4). Thus, using matrix notation we have the following:     α(0)   1 1 1 ... 1          α(1) + γ(0)   2 4 a     1 2 2 ... 2           α(2) + γ(1)   1 32 34 ... 3a    = 0,        α(3) + γ(2)   . . . . .     . . . . .       .     .  2 4 a   1 ( + 2) ( + 2) ... ( + 2)   y where a = m, (respectively a = m + 1) and y = α(ζ) + γ() (respectively y = γ()) when m is even (respectively m is odd).

Because the determinant of the Vandermonde matrix formed by the coeﬃcients of the system is not zero, we get that α(0) = [G(c, . . . , c), c] = 0, and this implies that:

" m+1 m+1 m+1 # X X X G( zreirjr ,..., zreirjr ), zreirjr = 0, (4.10) r=1 r=1 r=1 Pm+1 since c = r=1 zreirjr . Now, ﬁx v ∈ {1, . . . , m}. Notice that the previous argument works if the element

Pm+1 Pv ∗ c = r=1 zreirjr is replaced by cv = r=1 zreirjr , where zr ∈ K , ir, jr ∈ {1, . . . , n} for each r ∈ {1, . . . , v} with (iu, ju) 6= (iv, jv) when u 6= v. Thus, in the same fashion we can prove that [G(cv, . . . , cv), cv] = 0, which yields:

" v v v # X X X G( zreirjr ,..., zreirjr ), zreirjr = 0, for all v ∈ {1, . . . , m}. (4.11) r=1 r=1 r=1 So, after using that G is m-additive and the identity (4.11) (for each v ∈ {1, . . . , m}), the equality (4.10) becomes:

X [G(ziσ(1) eiσ(1)jσ(1) , . . . , ziσ(m) eiσ(m)jσ(m) ), ziσ(m+1) eiσ(m+1)jσ(m+1) ] = 0. (4.12) σ∈Sm+1

Note that the above equation holds trivially when zr = 0 for some r ∈ {1, . . . , m + 1}.

Hence, writing any matrix x ∈ Mn(K) as the sum of its entries xijeij we can conclude from 26

(4.12) that [G(x, . . . , x), x] = 0 for each x ∈ Mn(K), since G is m-additive. The desired result now follows from Theorem 7.

Observe that Theorems 14 and 15 can be derived from the previous theorem by taking k = n and k = 2 respectively. CHAPTER 5

Commuting Traces on Invertible or Singular Operators

From now on K will be either the field of real or complex numbers. Denote by H an infinite dimensional separable Hilbert space over K and by {φ1, φ2,...} a fixed orthonor- P∞ mal basis for H, that is, x = i=1 < x, φi > φi for each x ∈ H, where < . > denotes the inner product in H. As usual B(H) stands for the Banach space of all bounded operators from H to itself. We have described all commuting traces of an m-additive

n m n map M : B(K ) → B(K ) such that [M(x, . . . , x), x] = 0 for all invertible or singular

n x ∈ B(K ) = Mn(K)(m ≥ 1). Since B(H) is a prime ring, we will investigate to see if the results obtained in the second and third chapters also hold in the inﬁnite dimensional case.

Let I ∈ B(H) be the identity operator. The spectrum σ(T ) of T ∈ B(H) is deﬁned by

σ(T ) = {λ ∈ K | λI − T is not invertible}.

The resolvent set ν(T ) is deﬁned by ν(T ) = K \ σ(T ). It is well known that the spectrum

σ(T ) of T is a compact set in K bounded by ||T ||. In particular, the resolvent ν(T ) is an unbounded open set that contains { ∈ K | > ||T ||}. The results of this chapter have been submitted for publication in [10].

5.1 Commuting Traces on Invertible Operators

Proposition 25. Let m ≥ 1 be a natural number. Let G : B(H)m → B(H) be a symmetric m-additive map such that

[G(T,...,T ),T ] = 0 for all invertible T ∈ B(H). (5.1)

27 28

Then G(kI, . . . , kI) ∈ Z for all k ∈ K, where Z = K · I.

∗ Proof. First of all observe that the result holds trivially when k = 0. Now, ﬁx k ∈ K , and

let eij, i 6= j be the operator eij(x) =< x, φi > φj ∈ B(H). Let s be the smallest even number greater or equal than m, that is, s = m if m is even, and s = m + 1 if m is odd.

We will show that [G(kI, . . . , kI), eij] = 0.

∗ For each a ∈ K , let ya = akI+u, where u = (I+eij). Note that ya = akI+u is invertible

1 if and only if −ak ∈ ν(u) = K \{1}. Therefore ya is invertible if a 6= − k . So, we can ﬁnd s a nonzero rational number b such that ya is invertible for all a ∈ {±b, ±2b, . . . , ± 2 b} (take 1 b satisfying |b| > k ). It follows from (5.1) that 0 = [G(u, . . . , u), u] = [G(kI, . . . , kI), kI] =

[G(ya, . . . , ya), ya], and this last bracket can be written as

[G(ya, . . . , ya), u] = 0, (5.2)

since ya = akI + u. For m = 1 we conclude from (5.2) and [G(u), u] = 0 that [G(kI), u] =

[G(kI), eij] = 0 because u = (I + eij). It remains to prove that [G(kI, . . . , kI), eij] = 0 for m ≥ 2. Using (5.2) one more time, we see that [G(ya, . . . , ya) + G(y−a, . . . , y−a), u] = 0 for

s all a ∈ {b, 2b, . . . , 2 b}. Now, since G is symmetric, m-additive, and ya = akI + u we can s obtain for each a ∈ {b, 2b, . . . , 2 b} that: m X m−ζ m G(ya, . . . , ya) = a G(kI, . . . , kI, u, . . . , u), (5.3) ζ | {z } ζ=0 ζ and m X m−ζ m−ζ m G(y−a, . . . , y−a) = (−1) a G(kI, . . . , kI, u, . . . , u). (5.4) ζ | {z } ζ=0 ζ By keeping in mind the equations (5.3), (5.4), and the relation [G(u, . . . , u), u] = 0 we

see that [G(ya, . . . , ya) + G(y−a, . . . , y−a), u] = 0 becomes:

s−2 X2 m am−2ζ [G(kI, . . . , kI, u, . . . , u), u] = 0, when m is even, (5.5) 2ζ | {z } ζ=0 2ζ 29

and

s−4 X2 m am−(2ζ+1) [G(kI, . . . , kI, u, . . . , u), u] = 0, when m is odd. (5.6) 2ζ + 1 | {z } ζ=0 2ζ + 1

With (5.6), the identity [G(ya, . . . , ya), u] = 0 becomes:

s−2 X2 m am−2ζ [G(kI, . . . , kI, u, . . . , u), u] = 0, when m is odd. 2ζ | {z } ζ=0 2ζ

s Therefore, for each a ∈ {b, 2b . . . , 2 b} we have obtained an equation of the form (5.5) s s when m is either even or odd. It means that we have 2 equations in 2 unknowns, namely m 2ζ [G(kI, . . . , kI, u, . . . , u), u], where u appears exactly in 2ζ components of G, and ζ ∈ s−2 {0, 1,..., 2 }. Using matrix notation we can rewrite these systems in the following way:     bm bm−2 bm−4 . . . bm−(s−2) m[G(kI, . . . , kI), u]    0     .   m m−2 m−4 m−(s−2)   .   (2b) (2b) (2b) ... (2b)   .           (3b)m (3b)m−2 (3b)m−4 ... (3b)m−(s−2)   m[G(kI, . . . , kI, u, . . . , u), u]  = 0.    2ζ       . . . . .   .   . . . . .   .          s m s m−2 s m−4 s m−(s−2) m ( 2 b) ( 2 b) ( 2 b) ... ( 2 b) s−2 [G(kI, u, . . . , u), u]

Because the determinant of the Vandermonde matrix formed by the coeﬃcients of the system is not zero, we get that [G(kI, . . . , kI), u] = 0. As u = (I + eij), we conclude that

[G(kI, . . . , kI), eij] = 0, when i 6= j. Finally, we see that

[G(kI, . . . , kI), eii] = [G(kI, . . . , kI), eijeji] =

= [G(kI, . . . , kI), eij]eji + eij[G(z, . . . , z), eji] = 0.

It means that G(kI, . . . , kI) commutes with all ﬁnite rank operators of the form eij =<

x, φi > φj. Therefore, G(kI, . . . , kI) ∈ Z for all k ∈ K. 30

Theorem 26. Let m ≥ 1 be a natural number. Let G : B(H)m → B(H) be an m-additive

map such that

[G(T,...,T ),T ] = 0 for all invertible T ∈ B(H). (5.7)

Then, there exist µ0 ∈ Z and maps µi : B(H) → Z, i ∈ {1, . . . , m}, such that each µi is the

m m−1 trace of an i-additive map and G(T,...,T ) = µ0T +µ1(T )T +...+µm−1(T )T +µm(T )

for all T ∈ B(H), where Z = K · I.

Proof. Without loss of generality, we may assume that G is symmetric. Once again, let

s be the smallest even number greater or equal than m, that is, s = m if m is even, and

s = m + 1 if m is odd. Our goal is to show that [G(T,...,T ),T ] = 0 for all T ∈ B(H). Fix

T ∈ B(H). Since { ∈ K | > ||T ||} ⊂ ν(T ), we can ﬁnd a nonzero number λ ∈ K such that

s ya = T + aλI is invertible for all a ∈ {±1,..., ± 2 }. For m = 1 we obtain after employing the Proposition 25 in the identity [G(T + λI),T ] = [G(T + λI),T + λI] = 0 (equation

(5.7)) that [G(T ),T ] = 0. From now on, we may take m ≥ 2. It follows from (5.7) and

s ya = T + aλI that [G(ya, . . . , ya),T ] = [G(ya, . . . , ya), ya] = 0 for all a ∈ {±1,..., ± 2 }. Consequently,

s [G(y , . . . , y ) + G(y , . . . , y ),T ] = 0 for all a ∈ {1,..., }. (5.8) a a −a −a 2

Now, since G is symmetric, m-additive and ya = T + aλI, we conclude that m X r m G(ya, . . . , ya) = a G(λI, . . . , λI,T,...,T ), (5.9) r | {z } r=0 r s for each a ∈ {±1,..., ± 2 }. Thus, if we take into the account that G(λI, . . . , λI) ∈ Z (Proposition 25) and the equation (5.9) we can derive from (5.8) that:

[G(T,T ),T ] = 0 if m = 2, and s−2 X2 m a2r [G(λI, . . . , λI,T,...,T ),T ] = 0 if m ≥ 3. 2r | {z } r=0 2r 31

As in the proof of the Proposition 25, for each m ≥ 3 we have:

    1 1 1 ... 1   [G(T,...,T ),T ]      2 4 s−2   m   1 2 2 ... 2   [G(λI, λI, T, . . . , T ),T ]     2       1 32 34 ... 3s−2   .  = 0.    .       . . . . .     . . . . .   m     [G(λI, . . . , λI, T, . . . , T ),T ]   s − 2   s 2 s 4 s s−2  | {z } 1 ( 2 ) ( 2 ) ... ( 2 ) m−(s−2)

Therefore, [G(T,...,T ),T ] = 0 for all T ∈ B(H). With this in hand, the desired result now follows from Theorem 7.

5.2 Commuting Traces on Singular Operators

Theorem 27. Let m ≥ 1 be a natural number. Let G : B(H)m → B(H) be a symmetric

m-additive map such that

[G(T,...,T ),T ] = 0 for all singular T ∈ B(H). (5.10)

Then, there exist µ0 ∈ Z and maps µi : B(H) → Z, i ∈ {1, . . . , m}, such that each µi is the

m m−1 trace of an i-additive map and G(T,...,T ) = µ0T +µ1(T )T +...+µm−1(T )T +µm(T )

for all T ∈ B(H), where Z = K · I.

Proof. We shall proceed as we did in the proof of the Theorem 26; That is, we will show

that [G(T,...,T ),T ] = 0 for all T ∈ B(H). Fix T ∈ B(H). Let us deﬁne the ﬁnite rank

operator S ∈ B(H) as the following:

m+2 X 1 S = − < x, φ > T (φ ). (5.11) n n n n=1

By construction, we see that T + jS is singular for all j ∈ {1, . . . , m + 2}, because

(T + jS)(φj) = 0. Thus, [G(T + jS, . . . , T + jS),T + jS] = 0 (equation 5.10) for all j ∈ {1, . . . , m + 2}. Using the symmetricity and the m-additivity of G, we arrive at 32

m X m jh [G(S, . . . , S,T,...,T ),T ] + h | {z } h=0 h m X m jh+1 [G(S, . . . , S,T,...,T ),S] = 0. (5.12) h | {z } h=0 h For convenience let us set:

m α(h) = [G(S, . . . , S,T,...,T ),T ], where h ∈ {0, . . . , m}, h | {z } h and m γ(h) = [G(S, . . . , S,T,...,T ),S], where h ∈ {0, . . . , m}. h | {z } h Observe that for each j ∈ {1, . . . , m + 2} we have obtained an equation of the form

(5.12). Thus, using matrix notation we have the following:     α(0)   1 1 1 ... 1          α(1) + γ(0)   1 2 m+1     1 2 2 ... 2           α(2) + γ(1)   1 31 32 ... 3m+1    = 0.    .     .   . . . . .     . . . . .           α(m) + γ(m − 1)  1 2 m+1   1 (m + 2) (m + 2) ... (m + 2)   γ(m)

Therefore, α(0) = [G(T,...,T ),T ] = 0, and this is true for all T ∈ B(H) since T was arbitrary. The result follows now from Theorem 7. BIBLIOGRAPHY

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