Help identify this generalized sign of real representations

Question

Let $V$ be a real representation of a finite group $G$.

Define $\mathbb Z[I]_{I\leq G}$ to be the ring over the integers generated by subgroups of $G$ with multiplication corresponding to intersection of subgroups.

Associate an element $\sigma_V\in \mathbb Z[I]_{I\leq G}$ so that $$\sigma_V= \sum_{I\leq G} n_II$$ and for all subgroups $H$ of $G$, the following equality holds: $$\sum_{I\geq H}n_I=(-1)^{dim V^H}$$

Question: Is $\sigma_V$ some standard representation theoretic thingy? (My knowledge of representation theory doesn't go much further than Wikipedia...)

I am using $\sigma_V$ as a kind of generalized sign of the determinant of complex conjugation on $V\oplus iV$. I think that it has the following nice properties: $$\sigma_{V\oplus W}=\sigma_V\sigma_W$$ $$\sigma_V^2=1G$$

If $I\leq G$ is normal, then $n_I$ is an integer divisible by the index of $I\leq G$. A second question: is $n_{\{1\}}$ always $0$ or $\pm \lvert G\rvert$? This is the case in the few examples I have computed, but I have no general reason to expect it to be true.

For example: if $G=S_n$, and $\rho$ is the representation which permutes the coordinates of $\mathbb R^n$, then $n_I=(-1)^k k!$ if $I$ is a subgroup of $S_n$ corresponding to partitioning $\{1,\dotsc,n\}$ into $k$ nonempty subsets, and $n_I=0$ otherwise.

Answer 1

I think the answer to your second question is "no". Note first that by the M\"obius inversion formula (see for example Chapter 3 of Richard Stanley's ``Enumerative Combinatorics, Volume 1"), for any real $G$-module $V$, $$ n_{\{1\}}=\sum_{H \leq G}\mu(1,H)(-1)^{\dim V^H}, $$ where $\mu$ is the M\"obius function on the subgroup lattice of $G$. (As an aside, it follows from this that if all the dimensions of the fixed point subspaces have the same parity, then $n_{\{1\}}=0$.)

Now let $V$ be the regular module for $G$. Then, for each $H \leq G$, $$ \dim V^H=[G:H]. $$ Let ${\mathcal S}_2(G)$ be the set of all subgroups of $G$ that contain at least one Sylow $2$-subgroup of $G$ (that is, have odd index in $G$). Since the sum of $\mu(1,H)$ over all subgroups $H$ of $G$ is zero, $$ n_{\{1\}}=-2\sum_{H \in {\mathcal S}_2(G)}\mu(1,H). $$ Now one can take $G=PSL_2(31)$. A Sylow $2$-subgroup of $G$ is dihedral of order $32$ and is maximal in $G$. By results of Philip Hall (the paper is called "The Eulerian functions of a group", and can be found in Hall's collected works), $\mu(1,G)=2|G|$ and, for any Sylow $2$-subgroup $D$ of $G$, $\mu(1,D)=0$. So, $$ n_{\{1\}}=-4|G|. $$ A similar but slightly more complicated argument argument should yield the same answer if $31$ is replaced by any prime $p$ satisfying $p \equiv \pm 1 \bmod 16$ and $p \equiv \pm 1 \bmod 5$. (One has to handle dihedral overgroups of Sylow $2$-subgroups, but these will satisfy $\mu(1,D)=0$.)