The Unapologetic Mathematician

Mathematics for the interested outsider

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 - Posted by | Algebra, Group theory, Representation Theory

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 [...]

    Pingback by Tensors Over the Group Algebra are Invariants « The Unapologetic Mathematician | November 15, 2010 | Reply


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