The Unapologetic Mathematician

Projecting Onto Invariants

Given a $G$-module $V$, we can find the $G$-submodule $V^G$ of $G$-invariant vectors. It’s not just a submodule, but it’s a direct summand. Thus not only does it come with an inclusion mapping $V^G\to V$, but there must be a projection $V\to V^G$. That is, there’s a linear map that takes a vector and returns a $G$-invariant vector, and further if the vector is already $G$-invariant it is left alone.

Well, we know that it exists, but it turns out that we can describe it rather explicitly. The projection from vectors to $G$-invariant vectors is exactly the “averaging” procedure we ran into (with a slight variation) when proving Maschke’s theorem. We’ll describe it in general, and then come back to see how it applies in that case.

Given a vector $v\in V$, we define

$\displaystyle\bar{v}=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv$

This is clearly a linear operation. I say that $\bar{v}$ is invariant under the action of $G$. Indeed, given $g'\in G$ we calculate

\displaystyle\begin{aligned}g'\bar{v}&=g'\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv\\&=\frac{1}{\lvert G\rVert}\sum\limits_{g\in G}(g'g)v\\&=\bar{v}\end{aligned}

since as $g$ ranges over $G$, so does $g'g$, albeit in a different order. Further, if $v$ is already $G$-invariant, then we find

\displaystyle\begin{aligned}\bar{v}&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}gv\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}v\\&=v\end{aligned}

so this is indeed the projection we’re looking for.

Now, how does this apply to Maschke’s theorem? Well, given a $G$-module $V$, the collection of sesquilinear forms on the underlying space $V$ forms a vector space itself. Indeed, such forms correspond to correspond to Hermitian matrices, which form a vector space. Anyway, rather than write the usual angle-brackets, we will write one of these forms as a bilinear function $B:V\times V\to\mathbb{C}$.

Now I say that the space of forms carries an action from the right by $G$. Indeed, we can define

$\displaystyle\left[Bg\right](v_1,v_2)=B(gv_1,gv_2)$

It’s straightforward to verify that this is a right action by $G$. So, how do we “average” the form to get a $G$-invariant form? We define

$\displaystyle\bar{B}(v,w)=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}B(gv,gw)$

which — other than the factor of $\frac{1}{\lvert G\rvert}$ — is exactly how we came up with a $G$-invariant form in the proof of Maschke’s theorem!

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November 13, 2010 -

1 Comment »

1. [...] can get a more explicit description to verify this equivalence by projecting onto the invariants. Given a tensor , we consider it instead as a tensor in . Now, this is far from unique, since many [...]

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