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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 $\pm1$. Finally, let us normalize $$T_r:={1\over r!}{\mathcal S}_r.$$

When $r=2$, ${\mathcal S}_2(x_1,x_2)=x_1x_2-x_2x_1$ is the commutator. When using the Frobenius norm $\|M\|=\sqrt{Tr(M^*M)}$ , we know from [1] that $\|[A,B]\|\le\sqrt2\|A\|\cdot\|B\|$, and $\sqrt2$ is the least constant in this inequality, whatever the dimension $n\ge2$.

If $r\ge2$, what is the least constant $c(r,n)$ in the inequality $$\|T_r(A_1,\ldots,A_r)\|\le c(r;n)\prod_{j=1}^r\|A_j\|,\quad\forall A_1,\ldots,A_r\in M_n(\mathbb{C})?$$

Because $n\mapsto c(r,n)$ is non-decreasing, we set $\tau(r)=\lim_{n\rightarrow+\infty}=\sup_n c(r,n)$.

By Amitsur--Levitski Theorem (see [2] for a short proof, or Section 4.4 of the second edition of my book on Matrices), we have $c(r,n)=0$ when $r\ge2n$. From above, we have $c(2,n)=\sqrt2/2$. I can also prove the inequalities $$c(r+s,n)\le c(r,n)\cdot c(s,n),\qquad \tau(r+s)\le\tau(r)\tau(s).$$ In particular, the limit $$\rho:=\lim_{r\rightarrow+\infty}\tau(r)^{1/r}$$ exists and is also the infimum of the sequence. What is its value ?

Finally, is there a quantitative counterpart of Amitsur--Levitski, in the sense that $r\mapsto c(r,n)$ would drop to a small value at some threshold value, less than $2n$ ?

[1] Seak-Weng Vong, Xiao-Qing Jin, Proof of Böttcher and Wenzel’s conjecture. Oper. Matrices, 2 (2008), pp 435--442.

[2] S. Rosset. A new proof of the Amitsur-Levitski identity. Israel J. Math., 23 (1976), pp 187--188.

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Is it really $1/n!$ and not $1/r!$ in the definition of $T_r$? –  Darsh Ranjan Sep 14 '10 at 16:23
    
Of course, it is $1/r!$. Sorry for the typo. I shall correct it. –  Denis Serre Sep 14 '10 at 19:15
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Isn't $\tau(r)$ equal to the supremum of the $c(r,n)$ as $n$ tends to infinity? –  Andreas Thom Oct 5 '10 at 20:22
    
Is $\tau(3)$ known? –  Andreas Thom Oct 5 '10 at 20:23
    
@Andreas. Yes, $\tau$ is $\sup c$. I correct it. –  Denis Serre Oct 6 '10 at 5:35

1 Answer 1

P.G. Dixon has studied topologically nilpotent Banach algebras, those Banach algebras $A$ for which $$ \sup\{||x_1x_2... x_n||^{1/n}: x_i \in A,\; ||x_i|| < 1 \;(1 < i < n)\}\to 0 $$ as $n \to \infty$. Possibly his papers: "Topologically nilpotent Banach algebras and factorization", Proc. Roy. Soc. Edin. A, 119 (1991), 329-341 and (with V. Müller) , "A note on topologically nilpotent Banach algebras" Studia Math., 102 (1992), 269-275 would be of interest. The latter can be downloaded here

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Interesting, although I do not think that Dixon's paper has something to do with my question. Strange yet that Dixon did not cite (did not know ?) the paper A note on the joint spectral radius by G.-C. Rota and G. Strang (1960). –  Denis Serre Oct 6 '10 at 13:13

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