Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Any traceless $n\times n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?

When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:

  1. multiples of the identity may have trace $=0$.
  2. a matrix may have a spectrum equal to the whole field.

Are all traceless matrices commutators? If not, for which $n\in\mathbb{N}\setminus\lbrace 0 \rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.

In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]\cong k.$$

Now, the trace is a non-zero linear map $\mathrm{tr}:M_n(k)\to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.

N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.

Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168

share|improve this answer
    
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer. –  Olivier Bégassat Mar 27 '12 at 21:57
    
Yes. That's what I wrote :) –  Mariano Suárez-Alvarez Mar 27 '12 at 22:01
    
But are traceless matrices always a particular commutator? –  Olivier Bégassat Mar 27 '12 at 22:08
    
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case! –  Olivier Bégassat Apr 29 '12 at 10:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.