Is the direct sum in Maschke's Theorem an orthogonal decomposition?

Question

I am reading a paper on coding theory, and it uses a statement, which was claimed to be a reformulation of Maschke's Theorem. But I felt that was false...

Let's say $\mathcal(V):=\mathcal{F}_2^n$ is the big ambient space. $\sigma \in S_n$ is a permutation of odd prime order $p$. $\sigma$ acts on our ambient space by permuting the order of the coordinates.

Let $\mathcal{V}(\sigma)$ be the space in $\mathcal{V}$ that is fixed by the action of $\sigma$. (So those are the ones that have the same value on the orbit of $\sigma$.). Below is what I think what the paper meant by it's a reformulation of Maschke's theorem.

Let $\pi$ be a projection from $\mathcal{V}$ onto $\mathcal{V}(\sigma)$, and let $H$ be the group generated by $\sigma$. (So just $1, \sigma, \sigma^2, \dots$.) Define

\begin{align*} \phi: \mathcal{V} &\rightarrow \mathcal{V}(\sigma) \\ x &\mapsto \sum_{g \in H} g \pi(g^{-1}x) \end{align*}

Then it's easy to check $\phi$ is still a projection, and $\phi$ composed with the inclusion map is identity on $\mathcal{V}(\sigma)$. By splitting lemma,

This uses the fact that $p$ is odd. Since if $p$ is even, the sum will always be 0.

What confuses me is that the paper claimed that $$ \mathcal{V} = \mathcal{V}(\sigma) \oplus \mathcal{V}(\sigma)^\bot$$ where $\mathcal{V}(\sigma)^\bot$ is the dual space of $\mathcal{V}(\sigma)$ using the usual inner product defined on $\mathbb{F}_2^n$.

My feeling is that those two decompositions are not necessarily the same one. Or are they? If so, is there a way to show that?

Answer 1

The decompositions are the same because actually there is only one $H$-invariant complement to $\mathcal{V}(\sigma)$ in $V$. Consider $e = \sum_{g\in H} g$ as operator on $V$. Then $v\in \mathcal{V}(\sigma) = \operatorname{Fix}_V(H)$ iff $ev=v$, as is easy to show. So if $W$ is any complement to $\mathcal{V}(\sigma)$, and if $W$ is $H$-invariant, then $eW \subseteq W\cap \mathcal{V}(\sigma) = \{0\}$ and thus $W = (1-e)V$.
This means also that no matter what projection $\pi$ you start with, you will get $\phi = e$.

In your concrete situation, $\mathcal{V}(\sigma)^{\perp}$ is $H$-invariant because $\sigma$ acts "orthogonally" with respect to the inner product.

This can be done in a more general context: Suppose some finite group $H$ acts on some $F$-vector space $V$ and the characteristic of $F$ does not divide the group order $|H|$: Then $e = (1/|H|)\sum_{g\in H} g$ is the only $H$-invariant projection onto the fixed space of $H$ on $V$. Moreover, if $f\colon V\times V \to F$ is an $H$-invariant bilinear form, then $f(ev,w) = f(v,ew)$ and so $f(eV, (1-e)V) = 0$. Thus if $f$ is non-degenerate on $V$, then the restriction of $f$ to the fixed space is also non-degenerate.
Another remark is that the uniqueness above is a special case of the uniqueness of the decomposition of a semisimple module into its homogeneous components.