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Let $n\in\mathbb N$ be a natural number and $\lambda=(a,b)\vdash n$ a partition of $n$ into two parts, i.e. $a\ge b$ and $a+b=n$. In this special case, is there a simple description of the character $\chi_\lambda$ of the irreducible $S_n$-representation corresponding to $\lambda$? I have tried to deduce something from the Frobenius character formula and also using the Murnaghan-Nakayama recursion, but so far I couldn't really come up with a simple description. I would really appreciate any references/theorems in that direction.

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I think this is what you want, but it's just a link, so I just leave it as a comment: ma.rhul.ac.uk/~uvah099/Maths/labels.pdf –  Aaron Mar 5 '13 at 17:05
    
Unfortunately, I don't see how that helps me =/. –  user38451 Mar 5 '13 at 17:30
    
Sorry, it's my bad, he only have a formula relating characters for permutation module and Specht module of tow rows partition there...I think it's worth to try to ask this question on Mathoverflow instead if you don't get good answer within a week time. –  Aaron Mar 5 '13 at 20:44

1 Answer 1

This is probably way too little too late, but someone just linked to this question from elsewhere on the site and seeing as it was never answered I thought I'd give an answer.

Consider the natural action of $S_n$ on ($\mathbb{C}$-linear combinations of) $b$-element subsets of $\{1,2,\dots,n\}$. This is not quite the representation you want but its character values are easy to compute. For a given permutation the fixed subsets are exactly those which are unions of cycles of the permutation. So for a permutation of cycle type $m_1, m_2, \dots$, (that is, it has $m_1$ $1$-cycles, $m_2$ $2$-cycles, etc.) the character value is the number of ways to write $b$ as a sum of at most $m_1$ ones, at most $m_2$ twos, etc. This is just the coefficient of $x^b$ in $\Pi(1+x^i)^{m_i}$.

Now as I said, this isn't quite what you wanted, but it's close. Rather than being $\chi (n-b,b)$ this is $\chi (n-b,b) + \chi (n-b+1,b-1)+ \chi (n-b+2,b-2) + \dots \chi (n)$ (by say the Pieri rule). In particular though you can take the answer you got above for $b$ and subtract the answer for $b-1$, and this will give you the irreducible character you are looking for.

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