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In the study of representation theory of $S_n$, we know that the irreducible characters of $\chi_\lambda$ of $S_n$ are indexed by partitions $\lambda \vdash n$. There are several methods in determining the dimension of each $S^\lambda$, $f^\lambda$. One of the method is by considering

$ f^\lambda = \frac{n!}{\prod \text{hook length}} $

Let $\lambda' \vdash n$ be a partition obtained by taking 'transpose' in Ferrer's diagram of $\lambda$. For example, if $\lambda = (5,4,1)$, then $\lambda' = (3,2,2,2,1)$. Using the formula of $f^\lambda$ above, we thus have $f^\lambda = f^{\lambda'}$.

After some observations, I found out that when $n \geq 8$, then the dimension of $S^\lambda$ is unique up to transpose. In other words,

"Given any $\lambda \vdash n$, then there exists no other $\alpha \vdash n$ such that $f^\lambda = f^{\alpha}$ except when $\alpha = \lambda$. "

Is the above result well-known or established by anyone?

Thanks for the help!

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  • $\begingroup$ Out of curiosity, why is this tagged with spectral-graph-theory? $\endgroup$
    – Steven Sam
    Commented Mar 6, 2013 at 3:01
  • $\begingroup$ Oops, because this question arises when I'm working on Cayley graph on $S_n$. Write $U_\lambda$ for the sum of all copies of $S^\lambda$ in $\mathbb{C} S_n$, then $\mathbb{C} S_n = \bigoplus_{\lambda \vdash n} U_\lambda$ and each $U_\lambda$ is an eigenspace of Cayley graph on $S_n$ with some generating set $X$. The corresponding eigenvalue will be $\eta_\lambda = \frac{1}{f^\lambda} \sum_{x \in S} \chi_\lambda (x)$ That is the motivation behind. $\endgroup$
    – terrylsc
    Commented Mar 6, 2013 at 3:08
  • $\begingroup$ So I guess maybe some experts in spectral graph theory may also be interested in this though XD $\endgroup$
    – terrylsc
    Commented Mar 6, 2013 at 3:11

2 Answers 2

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The opposite is true. It is a result of D. Craven, settling a conjecture of A. Moreto, that given any $k$, for all large enough $n$, there are at least $k$ distinct irreducible representations of $S_n$ all of the same dimension.

http://arxiv.org/pdf/0709.0897.pdf

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  • $\begingroup$ Ah, thanks! I wonder there is a general properties/characterization on the dimension of $S^\lambda$ besides its formula =) $\endgroup$
    – terrylsc
    Commented Mar 6, 2013 at 4:34
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An even stronger counter-example is presented in Example 8 in my paper. It shows that $\lambda = (8,5,4)$ and $\mu=(7,7,2,1)$ both have the same multi-set of hook values. This of course implies $f^\lambda=f^\mu$.

What is interesting though is that the corresponding Ehrhart functions for the posets defined by $\lambda$ and $\mu$ are different.

Perhaps you can salvage your conjecture, and ask if for $|\lambda|>8$, the Ehrhart polynomial is unique (up to transposition). The value of $f^\lambda$ can be recovered from the leading coefficient of the Ehrhart polynomial.

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    $\begingroup$ For the record, earlier counterexamples to the uniqueness of hook multisets were written down in Joan E. Herman, Fan R.K. Chung, Some results on hook lengths, Discrete Math. 20 (1) (1977/78) 33–40. They say the question was raised by contemporaneous work of Logan--Shepp and treated it as an open problem.They don't quite state it this way, but they note that the ordered pair $(A, B)$ where $A$ is the multiset of hook lengths of $\lambda$ and $B$ is the multiset of hook lengths of the "box complement" of $\lambda$ does uniquely determine $\lambda$ up to transposition. $\endgroup$
    – Joshua P. Swanson
    Commented Jul 3, 2020 at 2:09

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