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?

  • $\begingroup$ mathoverflow.net/questions/17786/… may help. $\endgroup$ – KReiser May 18 at 6:18
  • $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – abdul May 22 at 15:18
  • $\begingroup$ @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 $U$. $\endgroup$ – Balarka Sen May 22 at 18:36

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