TITLE: Irreducible representations of the symmetric group on homology of simplicial complex
QUESTION [6 upvotes]: I am following Wall's paper A note on symmetry of singularities and I have some questions regarding representation theory and the homology of some objects:
Consider an action of $\Sigma_k$ on a finite simplicial complex $X$ of dimension $n$, such that the action is simplicial and a simplex is fixed iff it is point-wise fixed. Consider $\sigma\in\Sigma_k$, then
$$\chi_{Top}(X^\sigma)=\chi_{\Sigma_k}(X)(\sigma):= \sum_i(-1)^i \text{trace } \sigma^*:H_i(X)\rightarrow H_i(X), $$
where $X^\sigma$ is the fixed subcomplex fixed by $\sigma$ and $\chi_{Top}$ the Euler-Poincaré characteristic.
My questions are:
1.- Is the sum of characters $\chi_{\Sigma_k}(X)$ called the equivariant Euler characteristic? Because I believe the equivariant Euler characteristic is the Euler characteristic of $H_*(X)^G$ in other texts.
2.- Can one say anything about the character on a single homology group instead of the alternated sum of all the groups? For example, compare the trace of $\sigma$ on one $H_i$ and $H_i(X^\sigma)$. I have a lot of troubles trying to isolate one homology group from this equality (with the extra structure that I have in my problem).
3.- Can one say something about the isotypes corresponding to a single irreducible representation of $\Sigma_k$? For example, compare the number of copies of the $sign$ representations and $H_*(X^\sigma)$ for some $\sigma$.
If it is useful, in my problem I know that every $X^\sigma$ has lower dimension than $X$, if $\sigma\neq 1_{\Sigma_k}$.
Edit: I forgot to say that I am only interested on the free part of the homology groups, so everything makes sense if we take complex coefficients.

REPLY [2 votes]: Given a friendly self map $\sigma: X \rightarrow X$ of a finite complex, the alternating sum of the traces you write down defines $L(\sigma)$, the Lefschetz number of $\sigma$.  It is standard that this number is 0 if $\sigma$ has no fixed points, and the proof of this involves all the homology groups together.  It is also clear that if $\sigma$ is homotopic to the identity, then $L(\sigma)$ is just the Euler characteristic.  Finally, if $A \subset X$ is a subspace taken to itself by $\sigma$, then
$$ L(\sigma:X \rightarrow X) = L(\sigma: A \rightarrow A) + L(\sigma: (X,A) \rightarrow (X,A)).$$
Letting $A=X^{\sigma}$, one learns that $L(\sigma: X \rightarrow X) = \chi(X^{\sigma})$, which is the displayed equality in your post.
The fact that your $\sigma$ is part of a group action doesn't really seem that relevant (except that the fixed point space will still be a finite complex).