16
$\begingroup$

What is the length of the shortest word $w\in F_2$ such that $w(x,y)$ is trivial for every $x,y\in S_n$?

There is a simple argument showing that we must have $\ell(w)\geq n$. See here for instance. It seems likely however that $\ell(w)$ must be super-polynomial in $n$.

What is the state of the art for lower bounds?

(As for upper bounds, the paper "Identical relations in symmetric groups and separating words with reversible automata" by Gimadeev and Vyalyi finds an upper bound of $\exp(\sqrt{n}\log n)$.)

$\endgroup$
3
  • $\begingroup$ Would it be possible to be a little more explicit? What is $S_n$? The symmetric group? What is $F_2$, and what is $w$ or $\mathcal{l}(w)$? $\endgroup$ Apr 10, 2013 at 11:23
  • 3
    $\begingroup$ $F_2$ is a free group in $x$ and $y$. Sean is asking about the shortest identity $w(x,y)$ such that $w(a, b)=1$ for any $a,b\in S_n$. $\endgroup$ Apr 10, 2013 at 11:37
  • $\begingroup$ Apparently, the upper bound has been significantly improved by Kozma and Thom to $\exp(C \log^4n \log \log n)$. See arxiv.org/abs/1403.2324 . $\endgroup$ Aug 25, 2020 at 19:05

1 Answer 1

15
$\begingroup$

As far as I know, the state of the art gives some improvements to those bounds, but they are not huge. For a better lower bound: given a nontrivial element $w \in F_2$ of word length $\ell$, using a result of Buskin (Economical separability in free groups, Sib. Math. J., 50 (2009), 603-608) there exists a subgroup, $H$, of index $\ell/2+2$ that does not contain $w$. By looking at the action of $F_2$ on $F_2/ H$ we get a representation of $F_2$ into $S_{\ell/2+2}$, that does not kill $w$. Therefore, in order for $w$ to be trivial in any representation of $F_2$ into $S_n$ we must have that $n \leq \ell /2 + 2$, or $2(n-2) \leq \ell$.

There are also better upper bounds known (see, for instance, Asymptotic growth and least common multiples in groups (me and Ben McReynolds), Bulletin of the LMS (2011)).

Your question is equivalent to quantifying residual finiteness of free groups (the non-normal case), for which the precise answer is still unknown (the best known bounds are from the papers above).

$\endgroup$
2
  • $\begingroup$ Thank you, this is a great answer. I didn't know about either of these papers. $\endgroup$ Apr 10, 2013 at 14:39
  • 2
    $\begingroup$ It's amazing that the lower bound $10n$ is open! $\endgroup$ Apr 10, 2013 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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