# What local system really is

I know a local system is a locally constant sheaf. But why does a local system on the topological space $$X$$ correspond to $$\tilde{X}\times_G V$$, where $$G$$ is the fundamental group of $$X$$, $$\tilde{X}$$ is the universal covering space of $$X$$, and $$V$$ is a $$G$$-module? How do you recover the locally free sheaf from $$\tilde{X} \times_G V$$?

The group $G$ acts properly discontinuously on $\tilde{X}$, and so if $x$ is any point of $\tilde{X}$, it admits a neighbourhood $U$ s.t. that $U g$ is disjoint from $U$ if $g \in G$ is non-trivial. Thus the natural map from $U$ to $\tilde{X}/G = X$ is an embedding.

Thus the natural map from $U \times V$ to $\tilde{X}\times_G V$ is also an embedding, and so $\tilde{X}\times_G V$ is locally constant (i.e. locally a product).

More detailed remarks:

• We should equip $V$ with its discrete topology

• The object $\tilde{X}\times_G V$ is not itself actually a sheaf, but is rather the espace etale of a sheaf. To get the actual sheaf we consider the natural projection $\tilde{X}\times_G V \to \tilde{X}/G = X$, and form the associated sheaf of sections. Over the open set $U \hookrightarrow X,$ this restricts to the sheaf of sections to the projection $U\times V \to U$, whose sections are precisely the constant sheaf on $U$ attached to the vector space $V$. (Here is where we see that it is important to equip $V$ with the discrete topology.) Thus our original sheaf of sections is locally constant, as claimed.

• so does all the local system arise in this way? Does this way more natural than the original way(the usual definition of sheaf)?
– abc
Dec 7, 2010 at 7:35
• @abc: Yes, all local systems arise in this way, and this a very natural way to think about them. Very often people will speak of the local system corresponding to a $G$-module $V$, and this is what they mean. Dec 7, 2010 at 8:58
• Can I ask how one defines $\tilde{X}\times_G V$ ? The notation makes it seem like a pullback but I don't see any natural maps $\tilde{X}\to G$ or $V\to G$ ? Is it just the product modulo $(x,gv) = (xg,v)$ for $x\in\tilde{X},v\in V,g\in G$ ? Jan 25, 2019 at 17:14
• For the future reader, the answer to Maxime's comment is yes. Jul 24, 2021 at 20:31