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Let $X$ be a path-connected manifold nice enough such it's universal covering space $p:\widetilde{X} \to X$ exists, $k$ a field. Then there exist a wellknown correspondence

$$ \{\textit{linear}\text{ representations of }\pi_1(x,x)\} \leftrightarrow \{\text{local systems of }\textit{vector spaces}\text{ on }X\} $$

between $k$ linear finite dimensional representations of a fundamental group $\pi_1(X,x)$ and local systems of $k$ vector spaces.

The map in one direction is defined as follows: Take a $k$ linear rep $\rho: \pi_1(X,x) \longrightarrow \operatorname{GL}(V) $ where $V$ is a $k$ space and consider the associated $V$-bundle as quotient space $\widetilde{X}\times_{\rho} V :=(\widetilde{X}\times V)/\pi_1(X,x) $ where $\pi_1(X,x)$ acts on $\widetilde{X}\times V$ via

$$ g \cdot(x,v) := (g \cdot x, \rho(g)\cdot v ) $$

where $g$ acts at the left via monodromy on the covering space.

Obviously the projection to the first coordinate $p:\widetilde{X}\times_{\rho} V \to X$ has fiber $V$ and if we endow $V$ with the discrete topology we obtain a local system $\mathcal{F}_{\rho}$ on $X$ defined by sections

$$\mathcal{F}_{\rho}(U)= \{s:U \to p^{-1}(U) \ \vert p \cdot s =1_U \} $$

for open $U \subset X$. It's easy to check that if $U $ is contractible, then $p^{-1}(U)\cong U \times V$ and since $V$ has discrete topology, $\mathcal{F}_{\rho}(U) \cong V$, so it's a local system.

Question: Is there an explicit construction known to go in another direction? To start with an local system $\mathcal{F}$ with fibre $V$ and construct from it explicitly a representation $\rho_F: \pi_1(X,x) \longrightarrow \operatorname{GL}(V) $?

I know that it's rather easy to construct it abstractly: Let $g=[\gamma] \in \pi_1(X,x)$ be a class of a loop, then since $[0,1]$ is contractible, all local systems on $[0,1]$ are constant sheaves, therefore we have a chain of abstract isomorphisms

$$ \gamma^*\mathcal{F}_0 \cong \gamma^*\mathcal{F}([0,1])\cong \gamma^*\mathcal{F}_1 =V.$$

Can this isomorphism of $V$ be written down in explicit terms as an element of $\operatorname{GL}(V)$ if we pick a basis $e_1,\dotsc, e_n$ of $V \cong k^n$?

Motivation of the question: In Geordie Williamson's An illustrated guide to perverse sheaves in example 5.11 one considers for $X:= \mathbb{C}^*$ and $k:=\mathbb{C}$ the covering map $f:\mathbb{C}^* \to \mathbb{C}^*: z \mapsto z^m$. Let $\underline{k}$ be the constant sheaf on $\mathbb{C}^*$ with value $k=\mathbb{C}$ regarded as 1D vector space.

One considers the pushforward sheaf $f_*\underline{k} $ which has as stalk at $x=1$ the functions from the $m$-set $f^{-1}(x)$ to $k$, which is isomorphic to $k^m$.
And then it is claimed that $f_*\underline{k} $ is a local system determined by the action of the monodromy on the $m$-th roots of $1$.

And I was wondering how to check this claim explicitly, even though this sounds plausible. To come back to the question I posed above it suffices to check that $f_*\underline{k} $ induces the repr $\pi_1(\mathbb{C}^*,1) \cong \mathbb{Z} \to \operatorname{GL}_m(\mathbb{C})$ which maps the generator $1$ to $m$-cycle mapping for a fixed ordered basis $e_1,e_2,\dotsc, e_m$ of $k^m$ the basis vector $e_i$ to $e_{i+1}$.

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  • $\begingroup$ Did you really mean $\operatorname{GL}_m(\mathcal C)$ at the end, and note $\operatorname{GL}_m(\mathbb C)$? \\ TeX note: \mathrm is not for "\rm in a math-mode environment", but for "math in \rm". For the latter, you want something like \textrm, although actually it's just \text. Note $\text{local systems}$ \text{local systems} vs. $\mathrm{local systems}$ \mathrm{local systems}. Similarly for \mathit, which in your case should have been \textit. I edited accordingly. $\endgroup$
    – LSpice
    Commented Sep 23, 2023 at 23:49
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    $\begingroup$ Starting with a local system, can you define the map as follows? Take your fiber $V$ over a basepoint $x \in X$, pull it back via a local isomorphism $p$ to a point $\tilde{x} \in \tilde{X}$ over $x$, apply a deck transformation $g$, and then define the representation to act on $V$ via $\rho(g) = \pi_* \circ g \circ (p_*)^{-1}$. If you have a reasonably explicit description of your covering map I think you can make this pretty explicit. Perhaps you run into a similar problem by trying to explicitly describe the pullback of your local system to a local system on $\tilde{X}$ though $\endgroup$
    – Vik78
    Commented Sep 24, 2023 at 0:44
  • $\begingroup$ The word "explicit" is ambiguous. "Algorithmic" is better, but then the input data has to be given in a suitable form. One way to define a flat bundle in a computable form is via a 1-cocycle for a sufficiently good finite open cover. The next issue is the fundamental group: one can describe a cover by its nerve. The 1-skeleton of the nerve will determine the group generators (once a maximal tree is chosen). Given all this, there is a relatively straightforward algorithm for computation of images of generators. $\endgroup$ Commented Sep 24, 2023 at 2:27
  • $\begingroup$ @MoisheKohan: Could you explain in a bit more details how to describe a cover from the nerve data - ie the simplicial model of classifying space - of the fundamental group as you suggest? $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 18:54
  • $\begingroup$ Later. But the direction is the opposite one: a cover determines its nerve, a cocycle defined via the cover defines a group homomorphism. $\endgroup$ Commented Sep 24, 2023 at 19:08

2 Answers 2

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Here is a way to fill in the details. For simplicity, write $I = [0,1]$ for the interval, and $\exp \colon I \to S^1$ for the function $x \mapsto e^{2\pi i x}$. So let $\gamma = \exp \colon I \to S^1$ be a generator of $\pi_1(S^1,1)$. To get the map $\gamma_* \colon f_*\underline k \to f_*\underline k$, we need to find a trivialisation of $\gamma^*f_* \underline k$. Luckily, this is not too hard. First, consider the pullback square $$\begin{array}{ccc}X & \stackrel p \to & S^1 \\ \!\!\!\!{\small q}\downarrow & & \downarrow{\small f}\!\!\! \\ I & \stackrel\gamma\to & S^1.\!\end{array}$$ Then $X$ is a disjoint union $\coprod_{a \in \mathbf Z/m\mathbf Z} I$ of $m$ copies of $I$ that each map isomorphically to $I$ under $q$, and $p \colon X \to S^1$ is given on the $a$-th copy by $x \mapsto \exp((x+a)/m)$. In other words, $X$ is obtained from the top copy of $S^1$ by breaking it up into $m$ pieces by cutting at the roots of unity.

Now we claim that $\gamma^*f_*\underline k$ is naturally isomorphic to $q_*p^*\underline k$. There is always a map $$\gamma^*f_*\underline k \to \gamma^*f_*p_*p^* \underline k = \gamma^*\gamma_*q_*p^*\underline k \to q_*p^*\underline k$$ coming from the unit $1 \to p_*p^*$ and counit $\gamma^*\gamma_* \to 1$ of the adjunctions $p^* \dashv p_*$ and $\gamma^* \dashv \gamma_*$ respectively. In this case, the map $\gamma^*f_* \underline k \to q_*p^*\underline k$ is an isomorphism, either by checking directly at stalks, or by invoking (an easy case of) the proper base change theorem.

Of course $p^*\underline k$ is just the constant sheaf $\underline k$ (this holds for pullback of any constant sheaf along any continuous map), and $q_*\underline k$ is isomorphic to $\bigoplus_{a \in \mathbf Z/m\mathbf Z}\underline k$, giving the required trivialisation. Under this isomorphism, the fibre at $0 \in I$ is $\bigoplus_{a \in \mathbf Z/m\mathbf Z} \underline k_{\exp(a/m)}$, whereas the fibre at $1 \in I$ is $\bigoplus_{a \in \mathbf Z/m\mathbf Z} \underline k_{\exp((a+1)/m)}$ (where we express everything in terms of the stalks of $\underline k$ in the upper copy of $S^1$). Both of these are isomorphic to $(f_*\underline k)_1$, but in ways that differ by a cyclic shift. $\square$

Summary. Don't just compute the isomorphism types of various vector spaces, but remember the isomorphisms. All the data is there, but only once you choose a trivialisation of $\gamma^* \mathscr F$.

Remark. There are alternative viewpoints that are also fruitful. For instance, locally trivialising $\mathscr F$ and comparing the trivialisations on the intersections gives a cocycle in $H^1(X,\operatorname{GL}_r(\underline k))$ (cohomology in the constant sheaf of non-abelian groups $\operatorname{GL}_r(\underline k)$). This set also classifies $\operatorname{GL}_r(k)$-torsors over $X$, which is the same thing as $\operatorname{GL}_r(k)$-covering spaces. These are again in correspondence with $\operatorname{Hom}(\pi_1(X,x),\operatorname{GL}_r(k))$. But we have gained a point of view using local trivialisations of $\mathscr F$ on $X$, which is often easier to work with than global trivialisations of $\gamma^* \mathscr F$ on $I$.

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    $\begingroup$ concerning your remark on alternative viewpoint of the monodromy action associated to local system $\mathcal{F}$ in terms of a cocycle in $H^1(X,\operatorname{GL}_r(\underline k))$. do I understand it correctly that you pursue following stategy: Let $\gamma \subset X$ be a loop in "nice enough" space $X$. $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:54
  • $\begingroup$ Let $(U_i)_i^n \subset X$ be a family of open subsets of $X$ with following properties: (1) $\gamma \subset \bigcup_i U_i $ (2) $U_i$ and $U_i \cap U_{i+1}$ non empty & contractible, while $U_i \cap U_j$ for $\vert i-j \vert >1$ empty, except for $i=1, j=n$ (3)$\mathcal{F}$ trivializes over every $U_i$ (4) $U_i \cap \gamma $ contractible (5) $\gamma^{-1}(U_i)= [k_{i,1}, k_{i,2}] \subset I $ is an interval. $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:54
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    $\begingroup$ Then by construction of $U_i$ we have $[k_{i,1}, k_{i,2}] \cap [k_{i+1,1}, k_{i+1,2}]= [k_{i+1,1}, k_{i,2}]$ and $[k_{i,1}, k_{i,2}] \cap [k_{j,1}, k_{j,2}]$ for $\vert i-j \vert >1$ empty. Then $\gamma^* \mathcal{F}$ trivializes over each $[k_{i,1}, k_{i,2}]$. Fix now for each $i$ an explicit isom $f_j: k^n \cong \gamma^* \mathcal{F}([i_j,i_{j+1}])$ (I think here exactly appears the point you emphasised in your summary that we need an explicit choice). $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:54
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    $\begingroup$ We remember the datum $(g_i)_i^n$ comming from isoms $g_i: k^n \cong \gamma^* \mathcal{F}([k_{i,1}, k_{i,2}]) \cong \gamma^* \mathcal{F}([k_{i+1,1}, k_{i,2}]) \cong \mathcal{F}([k_{i+1,1}, k_{i+1,2}]) \cong k^n$ given as composition of $f_j$, can restriction and $ f_{j+1}^{-1}$. $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:55
  • $\begingroup$ Then the associated monodromy rep should be given by mapping the class $[\gamma]$ to product $g_n \cdot g_{n-1} \cdot ... \cdot g_1$. By the the way $(g_i)_i^n$ would be the cocyce data of $\gamma^* \mathcal{F}$. Is the argument correct? $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:55
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In general, I would say that there is no way around the fact that the answer uses the fact that a sheaf on $[0,1]$ is constant.

Does this help? Instead of the pushforward of the constant sheaf $k$ of one-dimensional vector spaces, first think about the pushforward of the constant sheaf $1$ of one-element sets. Its stalk at $x$ is $f^{-1}(x)$. Think about the monodromy. In general, if $f:E\to X$ is any covering space, think about the monodromy of $f_\ast 1$ where $1$ is the constant singleton sheaf.

Here is another piece of advice: since you know what representation $(V,\rho)$ you're supposed to be getting, why not start with an explicit description of a universal covering space of $X=\mathbb C^\ast$ and try to see a bijection between $\tilde X\times_\rho V$ and the union of the stalks of $f_\ast k$?

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  • $\begingroup$ Below R. van Dobben de Bruyn's answer I gave in the comments a sketch how I would like to try to constuct the monodromy repr from cycle data of $\gamma^* \mathcal{F}$. But if I'm not missing something the same construction should go through with $\mathcal{F}:=f_*1$ since the only difference would be that in your example $\mathcal{F}:=f_*1$ would be a set valued local system, so the $g_i$ would be live in $\text{Aut}(f^{-1}(x)) $, so the product $g_n \cdot g_{n-1} \cdot ...\cdot g_1$ still makes sense. $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:02
  • $\begingroup$ Is this the kind of approach you suggested to apply to $f_*1$ to extract the monodromy in second paragraph of your answer or did you had there another approach in mind? $\endgroup$
    – JackYo
    Commented Sep 24, 2023 at 19:03

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