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Could someone please explain to me what the map, $g$, is doing intuitively?

$f:V\to V$ is a complex linear map and $f^n=\operatorname{id}$, for some $n>1$

$A$ is a subspace of $V$ and $f(A)\subset A$

$p:V\to A$ with $p|_A=\operatorname{id}_A$

$g:V\to A$ with $g(v)={1\over n}\sum_{k=0}^{n-1} f^k pf^{-k}(v)$


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Is it possible that you meant $f(A)\subset A$ rather than $f(A)\subset V$? As written, that condition is redundant. – Michael Hardy May 22 '12 at 19:48
@MichaelHardy: I am very sorry!! You are very right. Edited. – mambo May 22 '12 at 19:49
up vote 1 down vote accepted

First, $f(A)\subset A$ and $f^n=\operatorname{id}$ imply $f(A)=A$. We can consider $k\mapsto f^k$ as a representation of the cyclic group $\mathbb Z_n$ on $V$. Then $f(A)=A$ means that $A$ is an invariant subspace.

If we further assume that $p$ is linear, then $p$ is a projection; in this case $g$ is the projection used in the proof of Maschke's theorem; its kernel is an invariant complementary subspace to $A$.

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