If a group $G$ acts on a topological space $M$, and a representation of $G$ on a vector space $V$, why $M \times_G V$ is a local constant sheaf over $M/G$?
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$\begingroup$ What structure maps are you pulling back? $\endgroup$– Harry GindiDec 6, 2010 at 15:22
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4$\begingroup$ @Harry: the notation is not for a fiber product. It means the quotient of $M×V$ by the diagonal $G$-action (perhaps with an inverse thrown in somewhere). $\endgroup$– SheikraisinrollbankDec 6, 2010 at 15:41
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$\begingroup$ Maybe I'm mistaken, but if $G={1}$ is the trivial group then what you wrote is not a constant sheaf. But maybe you should explain better what sheaf you are actually talking about. G invariant sections of the bundle? $\endgroup$– Michael BächtoldDec 6, 2010 at 15:48
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$\begingroup$ Presumably what the OP means is why (or under what circumstances, I'm slightly uncomfortable with the paucity of hypotheses) is $M \times_G V$ locally trivial, rather than locally constant. $\endgroup$– SheikraisinrollbankDec 6, 2010 at 15:55
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2 Answers
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Dear HYYY,
You should assume that the group $G$ acts freely and properly discontinuously on $M$. Also, you should equip $V$ with the discrete topology. (This way you will get a local system rather than a vector bundle, which was a point of confusion in the comments above.)
If $G$ acts freely and properly discontinuously, then if $m$ is any point of $M$, it has a neighbourhood $U$ such that $Ug$ is disjoint from $U$ for all non-trivial $g \in G$. (Here I am writing the $G$-action on $M$ on the right, as is implicit in the question.)
Thus $U$ maps injectively into $M/G$, and $U \times V$ maps injectively into $M\times_G V$. Hence $M\times_G V$ is locally constant (because, as we have just shown, its pull-back over the open subset $U$ of $M/G$ is constant).
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$\begingroup$ @Emerton, thanks!but how can we see the action of $G$ on $V$ from this local system? If we choose a point $x$ in $M$,do we get a section $c\in V$ of this local system over the point $[x]$ in $M/G$?And then if we choose another point $y$ which in the same orbit of $x$,assume $yg=x$,then we get another section $c'\in V$ over $[y]=[x]$,is that $gc=c'$? $\endgroup$– HYYYDec 7, 2010 at 7:30
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$\begingroup$ Are you asking how to go backwards, i.e. how to recover $V$ from the associated local system? $\endgroup$– EmertonDec 8, 2010 at 4:25
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For a bit of intuition, consider $V = \mathbb{C}^2$ and $M = \mathbb{C}$ with $G = \mathbb{Z}/2$ acting diagonally as $\pm 1$ in both cases.
Then for any open set $U$ not containing the origin in the base, the sheaf is just the constant sheaf $\mathbb{C}^2$. If, however, $U$ does contain the origin, then it is the constant sheaf $\mathbb{C}^2/\pm 1$, and so this sheaf is locally constant.
In general, the fibre over a point $[x]$ in the base should be a copy of $V/stab_G(x)$. You should be able to study these to obtain your result.
