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In the proof of Maschke's theorem, we magically transform linear transformations $T$ into maps that preserve the group ring structure. The method is to define

$ \bar T(v) = \frac{1}{|G|} \sum\limits_{g \in G} g^{-1} T(g(v))$

Note for this to make sense, we have to be able to divide by $ |G|$. In other words, the characteristic of $ F$ does not divide $ |G|$. When $ T$ is already an $ FG$-module homomorphism:

$ \bar T(v) = \frac{1}{|G|} \sum\limits_{g \in G} g^{-1} T(g(v)) = \frac{1}{|G|} \sum\limits_{g \in G} g^{-1} gT(v) = \frac{1}{|G|} \sum\limits_{g \in G} T(v) = T(v)$

In essence, we would like $ g^{-1}T(g(v))$ to just be $T(v)$ for any linear transformation $ T$, but it's not. So we form a function by taking averages and trying to water down the deviation. And it works! We end up with an honest $ FG$-module morphism.

I saw another example of this in Guillemin/Pollack's Differential topology. Here they start with a n-tensor $T$ and form an alternating tensor:

$Alt(T) = \frac{1}{n!} \sum\limits_{g \in S_n} sgn(g) T^g$

My question is: Where does this kind of "averaging to pick up structure" work? Can we turn set maps into equivariant maps in the G-set setting? Or something different like averaging continuous maps to get smooth ones. The obvious obstruction is being able to divide, so is there a way around this? So a second question is: when you can't do this averaging, what kind of workarounds exist?

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    The technique is also useful for constructing G-invariant metrics on Lie groups and homogeneous spaces, and to construct the Haar measure on compact groups.2011-08-17

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In both cases you mention, the "averaging" construction is projection onto the $G$-invariant subspace.

Namely, suppose you are working with the category of (usually finite-dimensional) representations of some algebraic object: an algebra, a group, a Lie algebra, etc. Usually there is a trivial representation of some sort, which is one-dimensional and has an uninteresting representation structure. For instance, given a group, then there is the representation on the one-dimensional space with the trivial action. Given a Lie algebra, the analog is the one-dimensional representation where everything acts by zero.

Let $A$ be some algebraic object that can act on $k$-vector spaces, so we can form the category of $A$-representations, and suppose $A$ has a trivial representation $\mathbf{1}$. Suppose all the representations of interest are finite-dimensional vector spaces. Let $V$ be an object of our category (of $A$-representations); then $V$ contains a largest subobject V' \subset V on which the $A$-action is trivial: that is, V' is a sum of copies of $\mathbf{1}$. In the case of a group, the largest trivial subobject is the fixed point space of the entire group. In the case of a Lie algebra, the largest trivial subobject consists of vectors that are annihilated by everything in the Lie algebra.

Now the question is whether V' admits a complement in $V$: that is, does the inclusion A' \to A split? If it does, then there is an $A$-equivariant projection V \to V', which you should think of the generalization of averaging. In the case of finite groups (where the characteristic of $k$ is prime to the order), there is such a projection: this follows by Maschke's theorem (semisimplicity of the group algebra), and the idea is the averaging argument. Given a $k$-linear projection V \to V' (which always exists for vector spaces), one "averages" it to get an $A$-linear projection. One case where the analogy with averaging is less obvious is where $A$ is a semisimple Lie algebra, where it is also true that such projection operators exist. (Admittedly one can associate to $A$ a compact Lie group and then for that an averaging argument does work, with $A$ replaced by said Lie group.)

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    @Matt: Dear Matt, you're right.2011-08-17