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The standard polynomial in $r$ non-commuting indeterminates $x_1,\ldots,x_r$ is defined by $${\mathcal S}_r(x_1,\ldots,x_r):=\sum_{\sigma\in S_r}\epsilon(\sigma)x_{\sigma(1)}x_{\sigma(2)}\cdots x_{\sigma(r)}\,$$ where $S_r$ is the symmetric group in $r$ letters and $\epsilon$ is the signature. Each monomial is a word in the letters $x_j$, affected of a sign $\pm1$.

When $r=2$, ${\mathcal S}_2(x_1,x_2)=x_1x_2-x_2x_1$ is the commutator.

Let us apply $\mathcal S_r$ to $n\times n$ matrices with entries in some field $k$ (say $\mathbb C$). We know the following

  • if $r$ is even, then ${\rm Tr}{\mathcal S}_r(A_1,\ldots,A_r)\equiv0$.
  • if $r$ is even, then ${\mathcal S}_r(A_1+\alpha_1 I_n,\ldots,A_r+\alpha_rI_n)={\mathcal S}_r(A_1,\ldots,A_r)$.
  • if $r=2n$, then ${\mathcal S}_{2n}(A_1,\ldots,A_{2n})\equiv0_n$ (theorem of Amitsur and Levitski).

What is the image $S_{n,p}$ of $M_n(k)\times\cdots\times M_n(k)$ under $\mathcal S_r$ when $r=2p$ is even ? Is it an algebraic variety (i.e defined by polynomial equations) ? Is it smaller as $p$ increases ? The latter question has two interpretations: either $\dim S_{n,p+1}\leq\dim S_{n,p}$ (?), or $S_{n,p+1}\subset\dim S_{n,p}$ (?).

Edit. I proved recently that for real $2\times2$ matrices, equipped with the Schur-Frobenius norm, the image of the unit ball under $S_2$ is the ball of radius $\sqrt2$, and the image under $S_3$ is an ellipsoid. In particular, the image of $M_2({\mathbb R})$ is $M_2({\mathbb R})$ itself, in both cases.

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How do you get a map from $M_2(\mathbb{R})$ to itself via $S_2$? Do you fix one of the factors? – auniket Nov 19 '15 at 12:56
@auniket. That was a shortcoming. I meant the image of $M_2({\mathbb R})^r$, when $r=2$ or $3$. – Denis Serre Nov 19 '15 at 15:14

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