MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In representation theory of $S_n$, we know that for $n \geq 9$, the only Specht modules $S^\alpha$ of dimension $f^\alpha < {n-1 \choose 2} - 1$ are:

$S^{(n)}$ and $S^{(1^n)}$ with dimension $1$, $S^{(n-1,1)}$ and $S^{(2,1^{n-2})}$ with dimension $n-1$.

By simple calculations, we know that the next smallest is of dimension ${n-1 \choose 2} - 1$ (it is calculated by considering $\alpha = (n-2,2)$ and $(2^2,1^{n-4})$). The question is:

Are $ (n-2,2)$ and $(2^2,1^{n-4})$ the only Specht modules with dimension ${n-1 \choose 2} - 1$?

Further Question: The next smallest dimension after ${n-1 \choose 2} - 1$ is ${n-1 \choose 2} $. Are $(n-2,1^2)$ and $(3,1^{n-3})$ the only Specht modules with dimension ${n-1 \choose 2}$?

Thanks for any help!

share|cite|improve this question
I guess your other question, which is solved, is stronger than this one. – Harry Huang Mar 6 '13 at 15:58
but that doesn't answer to this, isn't it? Correct me if I'm wrong :) – terrylsc Mar 6 '13 at 23:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.