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Let ${\mathcal A}$ be the algebra spanned by the words in two letters $x$ and $y$. Its (infinite) basis is $1,x,y,x^2,xy,yx,y^2,...$

Let ${\mathcal A}_0$ be the sub-space (warning: not the sub-algebra) spanned by all the commutators $[w,z]$. Finally, let $E$ be the subspace of words $t$ such that $[x,y]t\in{\mathcal A}_0$.

Problem 1: identify $E$.

Motivation: If $A$ and $B$ are $n\times n$ matrices, and if $t\in E$, then the trace of $[A,B]t(A,B)$ is zero. Therefore the whole space $E(A,B)$ of matrices of the form $t(A,B)$ with $t\in E$ is in the orthogonal of $[A,B]^T$, for the standard scalar product $\langle M,N\rangle:={\rm Tr}(M^TN)$. The space $E(A,B)^T$ of transpose matrices is thus contained in $[A,B]^\bot$.

Problem 2: for which pairs $(A,B)$ do we have the equality $E(A,B)^T=[A,B]^\bot$ ?

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  • $\begingroup$ What is $(A,B)$? The anticommutator? $\endgroup$
    – Mark Meckes
    Commented Sep 29, 2010 at 14:55
  • $\begingroup$ You mean in $t(A,B)$ ? Since $t$ is a word in $x$ and $y$, it is the matrix obtained by substituing $A$ to $x$ and $B$ to $y$. $\endgroup$
    – Denis Serre
    Commented Sep 29, 2010 at 15:01

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