Let $\chi_\lambda(\pi)$ be the irreducible characters of the permutation group $S_n$, where $\pi\in S_n$ and $\lambda\vdash n$. We know that the average value over $S_n$ vanishes unless $\lambda$ corresponds to the trivial representation,
$$\sum_{\pi\in S_n}\chi_\lambda(\pi)=n!\delta_{\lambda,(n)}.$$
My question is: do we know how to compute the average over a smaller permutation group,
$$\sum_{\pi\in S_k\subset S_n}\chi_\lambda(\pi), \quad k<n?$$
(in this sum $S_k$ is seen as the group of permutations of the first $k$ symbols, with $n-k$ fixed points).
The sum is trivial for $\lambda=(n)$, of course, but what about general $\lambda$?
Best Answer
This is $k!$ times the dimension of the trivial subspace of the restriction of the Specht module $V_{\lambda}$ to $S_k$. We can figure out what this restriction is using the branching rule for the symmetric groups, which says that the restriction of $V_{\lambda}$ to $S_{n-1}$ breaks up as a direct sum $\bigoplus V_{\lambda'}$ where $\lambda'$ runs over all ways to remove a box from $\lambda$ (regarded as a Young diagram).
By induction it follows that the restriction to $S_k$ breaks up as a direct sum $\bigoplus V_{\lambda'}$ where $\lambda'$ runs over all ways to remove $n-k$ boxes from $\lambda$. So the dimension of the trivial subspace is the number of times we get the Young diagram $(k)$ consisting of a single row of length $k$. I don't know if there's a nice way of describing this number in general but for any particular Young diagram it shouldn't be too hard to work out.
Edit: The special case that $\lambda$ itself has longest row of length $k$ is pretty straightforward. Then $\lambda$ is $(k)$ stacked on top of (or below, depending on your conventions for drawing Young diagrams) another Young diagram $\mu$, and the $n-k$ boxes we remove must be all the boxes of $\mu$, in some order. So the dimension must be $f^{\mu}$ which we can compute using the hook length formula.
If the longest row of $\lambda$ has length less than $k$ then the dimension is zero. The hard case (I think) is that the longest row of $\lambda$ has length greater than $k$. Then we are removing boxes both from the longest row and from $\mu$ but there are constraints on what order we can do this in so that every intermediate step remains a Young diagram.