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Nov 28, 2023 at 21:03 comment added R. van Dobben de Bruyn @Mathstudent as I defined $\exp \colon I \to S^1$ to be a generator of $\pi_1(S^1)$, this means I'm taking the convention that $\exp(x) = e^{2\pi i x}$ (or its multiplicative inverse, I suppose). I think this notation is not uncommon; for instance it is the Lie theoretic exponential on $S^1$.
Nov 28, 2023 at 8:12 comment added Mathstudent Typo: p should be $exp(2\pi i/m(x+a))$
Sep 25, 2023 at 0:14 comment added JackYo isomorphism instead of isom classes, which alone -as I tried to understand above - not suffice to reconstruct the monodromy map?
Sep 25, 2023 at 0:12 comment added JackYo So this would mean that the reason that the latter would be the "wrong image candidate" for the image of $[\gamma]$ under the monodromy map reflects the error above to naively assume that the image of $[\gamma]$ is already determined by the isom class of $\gamma^* \mathcal{F}$ alone and ignoring the neccessarity to specify as additional data a concrete isom $i: \gamma^* \mathcal{F} \to \underline k^r$. Is this "the philosophy behind" the ism sheaf which captures the neccessarity here to keep track for concrete
Sep 25, 2023 at 0:04 comment added JackYo distinguishes the $g_n \cdot g_{n-1} \cdot ... \cdot g_1 \in \operatorname{GL}_r(k)$ constructed in comments above from the cocycle $(g_i)_i^n$ comming from the concrete trivialization of $\gamma^* \mathcal{F}$ on $I$ which in turn "pulls back" from trivialization $(\bigcup U_i)_i$, which looks like a "promising candidate" for the image of $[\gamma]$ under monodromy map, in contrast to "wrong image candidate" which one might try naively to contruct from $\gamma^* \mathcal{F}$ over $I$ itself as trivialization in $I$, which would give an identity map, that would be look fishy.
Sep 25, 2023 at 0:02 comment added JackYo And here so far I understand your point correctly with the meaning of the hom sheaf and it's intrinsic nature for keeping track of choosen isomorphisms, that in contrast the isom class of the local system $\gamma^* \mathcal{F}$ on $I$ alone not suffice to determine the image of $[\gamma]$ under to monodromy map, but $\gamma^* \mathcal{F}$ together with a concrete choice of an isomorphism $i: \gamma^* \to \underline k^r$. And this additional data piece is exactly what
Sep 24, 2023 at 23:59 comment added JackYo Now going a step back, say we only want to know where the class of a concrete picked loop $[\gamma]$ going to be mapped by the induced monodromy map (so we are interested really only in where $[\gamma]$ is mapped, not the whole map), and pose the question which "datum" is already sufficient to know it? Surely, in terms from above the restriction of $\mathcal{F}$ to the trivializing family $(U_i)_i \subset X$ containing $\gamma$ suffice, but the question is if we can reconstruct to image of $[\gamma]$ by a datum purely on the side of $I$?
Sep 24, 2023 at 23:57 comment added JackYo sidenote: in last sentence I of course meant $\mathcal{F} \vert _{\bigcup U_i}$. But there is another issue I not completely understand. Maybe this is precisely what you mean by using $\mathscr Isom(\mathscr F,\underline k^r)$ with $\operatorname{GL}_r(\underline k)$-torsor structure as object that captures the whole "information" simultaneously where each class of path $\gamma$ going to be mapped, ie it knows completely the assoc map $\pi_1(X,x), \to \operatorname{GL}_r(k)$.
Sep 24, 2023 at 21:00 comment added R. van Dobben de Bruyn I suppose this is the hands-on way to work this out. The more high-brow way to say the same thing is that you consider the espace étalé of the $\operatorname{GL}_r(\underline k)$-torsor and use path lifting. This torsor can be directly obtained from $\mathscr F$ as the sheaf hom $\mathscr Isom(\mathscr F,\underline k^r)$. Path lifting is exactly this thing with the intervals that you mention. (The advantage of this methods over the one in the answer is that you can do it for all paths simultaneously, instead of coming up with a trivialisation of $\gamma^*\mathscr F$ for each path $\gamma$.)
Sep 24, 2023 at 20:13 comment added JackYo Remark: If what I have wrote in the comments above is correct, then it seems that using this cocycle approach the key to obtain the "right" cocycle which defines the monodrmy map lies in the aspect that we trivialize $\gamma^* \mathcal{F}$ in $I$ over a cover which comes from a "cover" of image $\gamma$. Othrwise we could for example triavialize $\gamma^* \mathcal{F}$ over single chart cover $I$, but that would not give to as the monodromy image of class $[\gamma]$. It seems that what we are really pursueing is the cocycle of the restriction $\mathcal{F} \vert _{\bigcap U_i}$. But I'm not sure
Sep 24, 2023 at 19:55 comment added JackYo 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?
Sep 24, 2023 at 19:55 comment added JackYo 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}$.
Sep 24, 2023 at 19:54 comment added JackYo 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).
Sep 24, 2023 at 19:54 comment added JackYo 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.
Sep 24, 2023 at 19:54 comment added JackYo 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 $X$.
Sep 24, 2023 at 18:53 vote accept JackYo
Sep 24, 2023 at 1:09 history answered R. van Dobben de Bruyn CC BY-SA 4.0