Fix $n$ such that $n > 2$. Let $\chi_{(n−1,1)}$ denote the character of the Specht module $S_{(n−1,1)}$
How do I prove that the value of $\chi_{(n−1,1)}$ on a permutation $\sigma\in S_n$ is one less than the number of fixed points (or cycles of length 1) in $\sigma$?
From my understanding a character of a representation is the trace of the corresponding matrix, however, in this case there is no matrix or representation so I don't know how to go about this or what a "character of a permutation" even is.
Best Answer
Let's start with the permutation representation of $S_n$ on $n$-dimensional space $\mathbf{C}^n$. The representing matrix of a permutation $w \in S_n$ is just the permutation matrix with a $1$ in position $(i,j)$ if $w(j)=i$ and $0$ otherwise.
The trace of this matrix is the number of $1$'s on the diagonal, or in other words the number of fixed points of $w$. Now the representation $\mathbf{C}^n$ decomposes as the direct sum of the trivial representation of $S_n$ on the space of vectors with all entries equal and the Specht module $S_{(n-1,1)}$, which may be realized as the space of vectors in $\mathbf{C}^n$ whose coordinates sum to $0$. Thus the character of the Specht module is the difference of the character of $\mathbf{C}^n$ and the trivial character, which gives the result you asked for.