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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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Out of curiosity, why is this tagged with spectral-graph-theory? –  Steven Sam Mar 6 '13 at 3:01
    
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. –  terrylsc Mar 6 '13 at 3:08
    
So I guess maybe some experts in spectral graph theory may also be interested in this though XD –  terrylsc Mar 6 '13 at 3:11
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1 Answer 1

up vote 11 down vote accepted

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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Ah, thanks! I wonder there is a general properties/characterization on the dimension of $S^\lambda$ besides its formula =) –  terrylsc Mar 6 '13 at 4:34
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