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For $g$ an element of a group and $\chi$ an irreducible character, there are two easy bounds for the character value $\chi(g)$: First, the bound $|\chi(g)|\leq \chi(1)$ by the dimension of the representation and, second, the bound $|\chi(g) |\ \leq \sqrt{ |Z(g)|}$ by the centralizer arising from the formula $\sum_{\chi} |\chi(g)|^2= |Z(g)|$.

I am interested on upper bounds for characters of the symmetric group that improve slightly on the second bound.

More specifically, fix $\delta> 0$ small. I want to know for which $g\in S_n$ there exists an irreducible character $\chi$ with $|\chi(g)| > \sqrt{ |Z(g)|} e^{- \delta \sqrt{n}}$.

Recall that for a permutation $g\in S_n$ with $m_k$ cycles of size $k$ for all $k$, so that $\sum_k m_k k =n$, we have $|Z(g)| = \prod_k (k^{m_k} m_k!)$.

For example, known upper bounds for the dimension of representations of the symmetric group imply there does not exist such a character for $g$ the identity. On the other hand, such a character does exist if $g$ is an $n$-cycle, since the right side is less than $1$. I suspect and hope that this can only happen for $g$ containing some relatively large cycles.

I found works in the literature giving bounds for $\chi(g)$ that improve on the $|\chi(g)|\leq \chi(1)$ by a multiplicative factor (Asymptotics of characters of symmetric groups related to Stanley character formula by Féray and Śniady) or a power (Characters of symmetric groups: sharp bounds and applications by Michael Larsen and Aner Shalev) but it's not obvious if it's possible to transform them into a bound of the form I need.

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  • $\begingroup$ Some various combinatorial formulas for Sn-characters can be found here: symmetricfunctions.com/murnaghanNakayama.htm $\endgroup$ Commented Feb 20, 2023 at 16:01
  • $\begingroup$ @PerAlexandersson Thanks! It seems to me that one of these formulas will be helpful if there exists a relatively simple combinatorial proof of the bound $\chi(g)^2 \leq |Z(g)|$, because then one can try to study the proof more carefully to see when it can be sharp. So far I don't see how to do that for any of them, but it could be possible... $\endgroup$
    – Will Sawin
    Commented Feb 20, 2023 at 17:21
  • $\begingroup$ My first instinct is to try Murnaghan-Nakayama, but I don't know if that will work. Best-case scenario would be that the number of border-strip tableaux is already $\le \sqrt{|Z(g)|}$ (so that you don't even have to worry about sign-cancellation), but maybe this is false? $\endgroup$ Commented Feb 20, 2023 at 18:44
  • $\begingroup$ @TimothyChow For $n=4$ if $g$ is a $3$-cycle and a $1$-cycle and the character corresponds to the partition $3+1$ then $\sqrt{ | Z(g)|} =\sqrt{3}<2$ and the number of border-strip tableaux is either $2$ or $0$ depending on the order of the cycles. So maybe the truth or falsity depends on the order. Another evidence is that this bound is true is if all the cycles are the same size so that there is no choice of order, by Corollary 9 of the paper A bijection proving orthogonality of the characters of $S_n$ by Dennis E White mentioned in a deleted answer. $\endgroup$
    – Will Sawin
    Commented Feb 20, 2023 at 20:46
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    $\begingroup$ Of course this does not in any way preclude bounds of the form you are interested in, but it might also be worth pointing out on the negative side of things that there have been recent dramatic advances which show that symmetric group character values are computationally hard: arxiv.org/abs/2207.05423 $\endgroup$ Commented Feb 20, 2023 at 22:43

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