Dimension of Specht Modules $S^\lambda$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:01:45Z http://mathoverflow.net/feeds/question/123690 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123690/dimension-of-specht-modules-s-lambda Dimension of Specht Modules $S^\lambda$ terrylsc 2013-03-06T01:30:22Z 2013-03-06T04:04:07Z <p>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</p> <p>$f^\lambda = \frac{n!}{\prod \text{hook length}}$</p> <p>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 $f^\lambda = f^{\lambda'}$.</p> <p>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,</p> <p>"Given any $\lambda \vdash n$, then there exists no other $\alpha \vdash n$ such that $f^\lambda = f^{\alpha}$ except when $\alpha = \lambda$. "</p> <p>Is the above result well-known or established by anyone?</p> <p>Thanks for the help!</p> http://mathoverflow.net/questions/123690/dimension-of-specht-modules-s-lambda/123701#123701 Answer by John Shareshian for Dimension of Specht Modules $S^\lambda$ John Shareshian 2013-03-06T04:04:07Z 2013-03-06T04:04:07Z <p>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.</p> <p><a href="http://arxiv.org/pdf/0709.0897.pdf" rel="nofollow">http://arxiv.org/pdf/0709.0897.pdf</a></p>