Local system associated to monodromy representation

How can I associate a local system to a representation $$\rho: \pi_1(X) \to \mathbb C^*$$?

I have seen some construction, but it doesn't click for me. I know that the idea is to use a diagonal action of $$\pi_1(X)$$ on $$\tilde X \times \mathbb C$$, where $$P: \tilde X \to X$$ is the universal cover. But why does one obtain a local system in this way?

For example, I have seen the construction (with $$L_{\tilde X}$$ the trivial sheaf on $$\tilde X$$): $$\Gamma(U,\mathcal L):= \{s \in \Gamma(P^{-1}(U,L_{\tilde X})):~ \forall u \in P^{-1}(U), \forall \gamma \in \pi_1(X,v),~ s(\gamma.u)=\rho(\gamma).s(u) \}.$$

I guess the sheaf axioms are inherited from $$L_{\tilde X}$$. But why is this locally constant?

My approach: $$P$$ is a covering map, so we automatically get nice neighbourhoods to work with. But then how do I proceed?

• mathoverflow.net/questions/17786/… may help. – KReiser May 18 at 6:18
• Thank you for the link. Unfortunately, in the answers, there is only "This sheaf can be checked to be locally constant". I will have to look at the provided references probably – abdul May 22 at 9:43

For any representation $$\rho : \pi_1 X \to \text{GL}(\mathbf{V})$$ of the fundamental group in a complex vector space $$\mathbf{V}$$, denote $$L = \tilde{X} \times_\rho \mathbf{V}$$ obtained from quotienting $$\tilde{X} \times \mathbf{V}$$ by the diagonal $$\pi_1(X)$$-action $$g\cdot(y, v) = (gy, \rho(g)v)$$ as you described. The topology on $$\tilde{X} \times \mathbf{V}$$ is given so that it is the espace etale of a constant sheaf on $$\tilde{X}$$ (i.e., $$\mathbf{V}$$ is discrete) which descends down to a quotient topology on $$L$$.

To be completely explicit, denote $$q : \tilde{X} \times \mathbf{V} \to L$$ to be the quotient map, $$p : \tilde{X} \to X$$ to be the universal covering projection and $$\pi : L \to X$$ to be the projection obtained from the $$\pi_1(X)$$-equivariant coordinate projection $$\text{proj} : \tilde{X} \times \mathbf{V} \to \tilde{X}$$ after quotienting by $$\pi_1(X)$$ on domain and range. The diagram commutes:

$$\require{AMScd} \begin{CD} \tilde{X} \times \mathbf{V} @>{q}>> L\\ @V{\text{proj}}VV @VV{\pi}V \\ \tilde{X} @>{p}>> X \end{CD}$$

For any $$x \in X$$ one can choose a neighborhood $$U$$ evenly covered by $$p : \tilde{X} \to X$$, so that $$p^{-1}(U) = \bigsqcup V_i$$ where $$V_i$$ are the slices over $$U$$, $$p|V_i : V_i \to U$$ are homeomorphisms. Then $$(p \circ \text{proj})^{-1}(U) = \bigsqcup V_i \times \mathbf{V}$$ and since $$\pi_1(X)$$ acts freely by homeomorphisms on the collection of slices $$\{V_i\}$$ in $$\tilde{X}$$ and by linear homeomorphisms on $$\mathbf{V}$$ there's a homeomorphism $$q\left((p \circ \text{proj})^{-1}(U)\right) \to U \times \mathbf{V}$$ which respects the individual projections to $$U$$. By commutativity of the square above, $$q \left((p \circ \text{proj})^{-1}(U)\right) = \pi^{-1}(U)$$. To encapsulate, then, we have a homeomorphism $$\pi^{-1}(U) \to U \times \mathbf{V}$$ making the following square commute:

$$\require{AMScd} \begin{CD} \pi^{-1}(U) @>{\cong}>> U \times \mathbf{V}\\ @V{\pi}VV @VV{\text{proj}}V \\ U @>{\text{id}}>> U \end{CD}$$

Now that the topology is worked out, note that the sheaf $$\mathscr{F}$$ associated to $$\rho$$ is nothing other than the one obtained from thinking of $$L$$ as the espace etale. Indeed, for any $$U \subset X$$ associate the complex vector space $$\mathscr{F}(U) = \{s : U \to X | s \pi = \text{id}_U\}$$. It's fairly straightforward to prove this is a sheaf; just check the identity and the gluing axioms by hand.

The local triviality of $$\mathscr{F}$$ is a corollary of the topological triviality of $$L$$ that we obtained above. For any $$x \in X$$, we can find a neighborhood $$U$$ such that $$L$$ is topologically trivial over $$U$$, i.e., the last commutative diagram holds. This would imply for every open subset $$O \subset U$$, $$\mathscr{F}(O) \cong \{s : O \to U \times \mathbf{V} : \text{proj}\circ s = \text{id}_O\} \cong C^0(O, \mathbf{V})$$. But remember $$\mathbf{V}$$ has the discrete topology, so this is just isomorphic to $$\mathbf{V}$$. Under this isomorphism, the restriction maps are all identity. In other words, $$\mathscr{F}|_U$$ is trivial.

• Thank you, but right at the end, isn't it needed to argument slighlty different? Is it really enough to show $\mathscr F(U)= V$ in order to show $\mathscr F|_U$ is trivial? Wouldn't this mean that for example the sheaf of holomorphic functions on a compact complex manifold is trivial? – abdul May 22 at 15:18
• @abdul Yes, of course. What I mean is, for any open subset $O \subset U$, $\mathscr{F}(O) \cong \mathbf{V}$, which follows from the topological triviality of $L|_U$ again. That's what the constant sheaf means; the leaf space pulls back to $U \times \mathbf{V}$ above $U$. – Balarka Sen May 22 at 18:36