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I am reading Geordie Williamson's guide to perverse sheaves and stuck on Example 5.11.

Consider the map $f:\mathbb{C}^* \to \mathbb{C}^*: z \mapsto z^m$. Let $\underline{k}$ be the constant sheaf on $\mathbb{C}^*$ with values in $k:=\mathbb{C}$ (regarded as 1D $\mathbb{C}$-space) and we want to understand the pushforward sheaf $f_*\underline{k}$.

Obviously the stalks of $f_*\underline{k}$ are $\mathbb{C}^m$ because $f$ is a covering. In detail, let $U \subset \mathbb{C}^*$ contractible and small enough such that $f$ trivializes over $U$ to disjoint union $f^{-1}(U)= \coprod_{i=1}^m V_i$, then since $\underline{k}(U)=k$ can be identified with continuous functions $U \to k=\mathbb{C}$ where $k$ is endowed with discrete topology, then $f_*\underline{k}(U)$ is naturally identifid with set of $k$-valued continuous functions $V_1 \coprod ... \coprod V_m \to k$, which can be identified with $k^m$.

Then the part I don't understand: It is claimed that $f_*\underline{k}$ is a local system determined by the action of the monodromy on the $m$-th roots of $1$ (an $m$-cycle).

Why is that the case and how can it be verified? Secondly, the converse question is how can I construct from the local system $f_*\underline{k}$ the monodromy representation $\pi_1(\mathbb{C}^*,1) \to GL_m(k)$ in explicit terms?

I know how to construct explicitly from a given monodromy representation $\pi_1(\mathbb{C}^*,1) \to GL_n(k)$ a local system (the construction is presented e.g. here ). But if not provides an explicit construction in another direction.

JackYo
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Given a local system $\mathcal{L}$ on a path connected space $X$ and a point $x$, you can define the corresponding monodromy on $\mathcal{L}_x$ as follows : take a path $\gamma : [0,1]\to X$ such that $\gamma(0)=\gamma(1)=x$. Then take the pullback $\gamma^{-1}\mathcal{L}$. Since $[0,1]$ is a contractible space, $\gamma^{-1}\mathcal{L}$ is trivial and thus it induces for any $t \in [0,1]$ the map $\Gamma([0,1],\gamma^{-1}\mathcal{L})\to (\gamma^{-1}\mathcal{L})_t$ is an isomorphism.

The monodromy is the map $$\mathcal{L}_x = (\gamma^{-1}\mathcal{L})_0 \overset{\sim}\longleftarrow \Gamma([0,1],\gamma^{-1}\mathcal{L}) \overset{\sim}\longrightarrow (\gamma^{-1}\mathcal{L})_1 = \mathcal{L}_x$$ You can show that it only depends on the homotopy class of $\gamma$ in $\pi_1(X,x)$ and that it is indeed a well defined action of $\pi_1(X,x)$.

Concretely, let's have a look at the local system $f_*\underline{k}$ where $f:\mathbb{C}^*\to\mathbb{C}^*$ is the $m$-fold covering $z\mapsto z^m$. Let $\gamma:[0,1]\to \mathbb{C}^*$ be the map $t\mapsto e^{2i\pi t}$. The path $\gamma$ has $m$ liftings $\gamma_0,\gamma_1,...,\gamma_{m-1}$ where $\gamma_n:[0,1]\to\mathbb{C}^*$ is the map $t\mapsto e^{2i\pi(t+n)/m}$. The main ingredient is the fact that $\gamma_n(0)=\gamma_{n-1}(1)$ for all $n\in\mathbb{Z}/m\mathbb{Z}$.

Before a more precise look, let's introduce some notations. We will write $\eta = e^{2i\pi/m}$. So we have $\eta^{n}=\gamma_n(0)=\gamma_{n-1}(1)$. And to avoid confusion between the point $1\in[0,1]$ and $1\in\mathbb{C}^*$, I will now on write the latter $x=\gamma(0)=\gamma(1)$.

By definition, $f_*\underline{k}$ is the sheaf such that section on $U$ are locally constant function on $f^{-1}(U)$ with value in your coefficient ring $k=\mathbb{C}$. As you said, the stalk at $x$ is $(f_*\underline{k})_x = k ^m$ where on the $n$-th coordinate (starting at 0) is the value of the locally constant function at $\eta^n=e^{2i\pi n/m}$. So maybe, it is best to write $$(f_*\underline{k})_x = \operatorname{Maps}(f^{-1}(x),k) = \operatorname{Maps}(\{\eta^0, \eta^1,...,\eta^{m-1}\}, k)\simeq k^m$$

Now, the pullback $\gamma^{-1}f_*\underline{k}$ is the sheaf of tuples of $m$ functions $(y_0, y_1, ..., y_{m-1})$ where each $y_n$ is defined on $[0,1]$ and takes a locally constant value in $k$. But we need to think of each $y_n$ as a locally constant function defined on the image of the path $\gamma_n$.

The stalk at $t\in[0,1]$ is their values, namely $(y_0, y_1, ..., y_{m-1})_t = (y_0(t),...,y_{m-1}(t))$. But now, to make the identification $$(\gamma^{-1}f_*\underline{k})_t = (f_*\underline{k})_{\gamma(t)}$$ we need to remember that each $y_n$ is correspond to a locally constant function on the image of $\gamma_n$. Thus, given a $(y_0,...,y_{m-1})_t \in (\gamma^{-1}f_*\underline{k})_t$, its image in $(f_*\underline{k})_{\gamma(t)}$ is the map $f^{-1}(\gamma(t))\to k$ such that $\gamma_n(t)\mapsto y_n(t)$.

It follows that the two maps of interest are : $$\begin{array}{rcccl} \Gamma([0,1],\gamma^{-1}f_*\underline{k})&\longrightarrow&(\gamma^{-1}f_*\underline{k})_0&\longrightarrow&(f_*\underline{k})_x\\ (y_0,...,y_n)&\longmapsto&(y_0(0),...,y_{m-1}(0))&\longmapsto&(\eta^n=\gamma_n(0)\mapsto y_n(0))\end{array} $$ and $$\begin{array}{rcccl} \Gamma([0,1],\gamma^{-1}f_*\underline{k})&\longrightarrow&(\gamma^{-1}f_*\underline{k})_1&\longrightarrow&(f_*\underline{k})_x\\ (y_0,...,y_n)&\longmapsto&(y_0(1),...,y_{m-1}(1))&\longmapsto&(\eta^n=\gamma_{n-1}(1)\mapsto y_{n-1}(1))\end{array} $$ Of course, each function $y_n$ is actually constant so $y_n(0)=y_n(1)$.

Hence, the monodromy $$(f_x\underline{k})_x = (\gamma^{-1}f_x\underline{k})_0 \overset{\sim}\longleftarrow \Gamma([0,1],\gamma^{-1}f_x\underline{k}) \overset{\sim}\longrightarrow (\gamma^{-1}f_x\underline{k})_1 = (f_x\underline{k})_x$$ is a cycle. If we write $(f_*\underline{k})_x\simeq k^m$ as a set of tuples where on the $n$-th coordinate (starting at 0) is the image of $\eta^n$, then the monodromy is $(y_0,y_1,...,y_{m-1})\mapsto (y_{m-1},y_0,y_1,...y_{m-2})$.

Roland
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  • the part which still not got is why in the pullback $\gamma^{-1}f_*\underline{k}$ the $m$-tuples $(y_0,y_1,...,y_{m-1})$ each $y_n$ can be identified with loc const fct $\gamma_n([0,1]) \to k$? Is this identification "heuristical" or can be turned in a formal one? The most natural way for me to attempt establishing this identification is by precomposing loc const maps $\gamma_n([0,1]) \to k$ with $\gamma_n$, which would give back the the $n$-th component $y_n$. – JackYo Sep 25 '23 at 17:50
  • Can this argument be formalized to something like: Assume $\mathcal{F}$ is local system, $g: X \to Y$ a map, then the "global sections" of $g^{-1}\mathcal{F}$ can be identified with $\mathcal{F}(U)$ where $g(X) \subset U$ and $U \subset Y$ is a kind of "narrowest neighborhood of $g(X)$. But does such $U$ always exist? If $\mathcal{F}$ is general sheaf, I would say no, but for $\mathcal{F}$ local system possibly there is more flexibility to find such $U$? – JackYo Sep 25 '23 at 17:51
  • a general remark: so far I know the stalks of inverse image sheaf $g^{-1}\mathcal{F}$ can be identified naturally $(g^{-1}\mathcal{F})x = \mathcal{F}{g(x)} $, but for sections I don't know any correspodnence between $g^{-1}\mathcal{F}$ and $\mathcal{F}$ for general sheaves. But what about the case when $\mathcal{F}$ is a local system? But maybe in this particular situation the sections of them can be somehow related to each other? – JackYo Sep 25 '23 at 17:55
  • Well, your second comment is actually almost the definition of the pullback presheaf. You need to sheafify in order to get the sections of the sheaf pullback. And yes, you can think of these sections that way. As for your third comment, you always have a natural map (coming from the $(f^{-1},f_*)$ adjunction) $\Gamma(U,\mathcal{F})\to\Gamma(f^{-1}(U),f^{-1}\mathcal{F})$ which "looks" like and should be thought as $s\mapsto s\circ f$. – Roland Sep 25 '23 at 21:07
  • so we have this natural comparison map, but the problem with certain degree of control over the sections of the pullback sheaf $f^{1} \mathcal{F}$ in terms of the sections of $ \mathcal{F}$ using this natural map from adjunction seems to lie in the sheafification process, since this can in general change the sections of the presheaf in rather severe way. That's why I'm not sure how "precise" is this identification in your answer between the global sections of $\gamma^{-1}f_*\underline{k}$ given as these $m$-tuples $(y_0,..., y_{m-1})$ with $m$-tuples $(\hat{y_0},..., \hat{y_{m-1}})$ – JackYo Sep 26 '23 at 00:09
  • appearing as sections of $f_k(U)$ for appropriate $U$ containing $\gamma$. Is this an exact identification in the sense that the natural map $\Gamma(U,f_k)\to\Gamma([0,1],\gamma^{-1}(f_*k))$ is indeed a isomorphism or only a "heuristic" way to regard them as "more or less nearly the same"?, If the former is indeed a honest isom, how to see it? – JackYo Sep 26 '23 at 00:20
  • There are several way to make it more precise. When I was writing the answer, I thought it was clear. The map $\Gamma(U,f_k)\to\Gamma([0,1], \gamma^{-1}f_k))$ is never an isomorphism (there is no $U$ containing $\gamma([0,1])$ making it an iso). For such a $U$, we indeed have $\Gamma(U, f_*k)=k$ while the right hand side is $k^m$. The sheafification process is essential here. But locally on $[0,1]$, we indeed have such isomorphisms which is enough to make the identification. – Roland Sep 26 '23 at 15:40
  • Maybe a way to see it, is that what I wrote is indeed an inverse image : let $\mathcal{F}$ be the sheaf on $[0,1]$ of $m$-tuples of locally constant functions in $k$ and define the map $\Psi:\gamma^{-1}f_k\to\mathcal{F}$ (use the adjunction $f_k\to\gamma_*\mathcal{F}$). Finally, prove that $\Psi$ is an isomorphism (this is easy using stalks). You will see that $\Psi$ is the reason you can make the identification as in my answer. – Roland Sep 26 '23 at 15:57
  • There are more fancy ways to make the identifications (proper base change,...) but in the end the result is the same. I guess, these kind of sheaf theoretic operations just require practice. – Roland Sep 26 '23 at 16:09
  • That over $[0,1]$ $\gamma^{-1}f_*k $ isomorphic to the sheaf of $m$-tuples of locally constant functions in $k$ you denoted by $\mathcal{F}$ I understand. Abstractly, this follows also from the fact that $[0,1]$ is contractible und therefore all local systems are isomorphic, that's clear to me. – JackYo Sep 26 '23 at 23:11
  • The part which is still not clear to me is the statement you made that for any such $m$-tuple $(y_0,..., y_{m-1})$ beeing a global section of $\gamma^{-1}f_*k$ due to our established identification with $\mathcal{F}$, that we "need to think of each $y_n$ from it as a locally constant function defined on the image of the path $\gamma_n$". This identification is exactly that one I still not understand. How is this identification formally established? – JackYo Sep 26 '23 at 23:11
  • As I explained in first comment it seems plausible that this identification can be realized "by hand" by precomposing $\gamma_n([0,1]) \to k$ with $\gamma_n$. But I'm wondering if this identification can be established conceptionally it terms of a honest morphism of sheaves like you did e.g. for $\mathcal{F} \to f_f^{-1}\mathcal{F}$ and $\Psi:\gamma^{-1}f_k\to\mathcal{F}$, whose existence are derived more less by pure formal reasons, adjunctions, etc. – JackYo Sep 26 '23 at 23:12
  • I think you are asking the wrong question. You can very well say that $\gamma^{-1}f_k$ is the sheaves of tuple $(y_0,...,y_{m-1})$ without talking about each of them being defined on the image of $\gamma_0,...,\gamma_{m-1}$. The point is what is the isomorphism $(\gamma^{-1}f_k)t\simeq (f*k)_{\gamma(t)}$ ? If you don't think that $y_n$ is defined on the image, this isomorphism may not seem obvious whereas it is supposed to be canonical. If I think as each $y_n$ has having a canonical factorization through $\gamma_n$, the isomorphisms at the stalks will appears more clearly. – Roland Sep 27 '23 at 20:58
  • When I read old books on sheaf theory (in french), a sheaf was called "a sheaf of stalks". For instance, the sheaf of stalks of holomorphic functions, the sheaf of stalks of sections of $p:X\to S$... I suppose I think of sheaves a bit that way : a sheaf contain the stalks and how they glue. Stalks behave very well with respect to pullback so this way of thinking the sheaves is helpful when I want to take a pullback. Here $f_k$ is the sheaves whose stalks are maps $f^{-1}(x)\to k$, so $\gamma^{-1}f_k$ is the sheaf on $[0,1]$ whose stalks are maps $f^{-1}(\gamma(t))\to k$. – Roland Sep 27 '23 at 21:06
  • "Gluing all this", I see that $\gamma^{-1}f_*k$ is a sheaf of constant functions on each $\gamma_n([0,1])$. But they also are function on $[0,1]$ so we may say it is the sheaf of $m$ locally constant functions on $[0,1]$ which factors through $\gamma_0,...,\gamma_{m-1}$. Hence this sentence, maybe a bit sloppy and misleading, but which is more a helpful way to think of the stalks than a real construction. – Roland Sep 27 '23 at 21:18
  • possibly I understand your point now. I think my whole confusion arose from where we can consider something up to isomorphism and where not. If we invoke the definition of inverse sheaf, then $(\gamma^{-1}f_k)t \cong (fk){\gamma(t)}$ is not only an identification via certain isomorphism, but more less by definition a honest identity, on the other hand $\Psi:\gamma^{-1}f*k \cong \mathcal{F}$ is certain isomorphism. Therefore keeping this identification in mind locally the sections of $\mathcal{F}$ - the tuples $(y_0,..., y_{m-1})$ can be regarded locally (=on level of stalks) as – JackYo Sep 28 '23 at 16:13
  • locally constant functions $\gamma_n([0,1])$ via this identification $\Psi$ and identity $(\gamma^{-1}f_k)t = (fk)_{\gamma(t)}$, where in the latter the sections locally have exactly that shape. Thank you a lot for your explanations, the intuitive picture you exposed unveils a very insightful viewpoint on this issue! – JackYo Sep 28 '23 at 16:27