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Let $j : \mathbb C^* \to \mathbb C$ be the inclusion. Write $U = \mathbb C^*$ and $X = \mathbb C$. I want to calculate the cohomology groups $H^*(X,j_* L)$ where $L$ is a local system on $U$.

It is clear that $H^0 (X,j_* L)$ = $H^0(U,L)$ by definition. However I'm less sure about $H^1(X,j_*L)$.

Using the sequence $0 \to j_! L \to j_* L \to (j_*L)_0 \to 0 $ we see that $H^0(X,j_!L) = 0$ and deduce $H^1(X,j_!L) \cong H^1(X,j_*L)$. However it doesn't seem to help much. Also I am not sure about a good cover to compute the cohomology. I feel the answer should be $H^1(U,L)$ but I can't see it unfortunately.

Edit : if I'm not mistaken, looking at the Leray spectral sequence shows that (as vector spaces) $H^1(U,L) \cong H^1(X,j_*L) \oplus H^0(X,R^1 j_* L)$. However an easy calculation gives $H^0(X,R^1 j_*L) \cong H^1(U,L)$ hence we must have $H^1(X,j_*L) = 0$ which really surprises me. I would love to see if someone agrees with my argument, and also have a more elementary argument to obtain the same result.

curious
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  • Can you expand on your easy calculation ? – Roland Mar 21 '24 at 12:06
  • @Roland : sure! The stalk of $R^1 j_L$ is zero if $x \neq 0$. The stalk at zero is $(R^1 j_L)0 = H^1(D^,L) = H^1(U,L)$ where $D^$ was a small disk minus zero. So $ H^0( R^1 j*L) = H^1(U,L)$. – curious Mar 21 '24 at 12:58
  • I think your result is valid, you can even prove that for any sheaf $\mathcal{F}$ on $\mathbb{C}$ such that $\mathcal{F}|_{\mathbb{C}^}$ is locally constant, then $H^1(\mathbb{C},\mathcal{F})=0$. For these sheaves and any small disk around $0$, the restriction map $H^(\mathbb{C},\mathcal{F})\to H^*(D,\mathcal{F}|_D)$ is an isomorphism. Hence $H^i(\mathbb{C},\mathcal{F})\simeq\varinjlim H^i(D,\mathcal{F}|_D)$ but this last group is zero for any $i>0$ (this is true for any sheaf in any space) – Roland Mar 25 '24 at 10:07
  • @Roland : that's great! so if I could commute the limit with $H^i$, I obtain $H^i(pt, G)$ for some sheaf $G$ on a point, which is clearly zero if $i>0$. Do you know a reference how to commutes these limits (I think H^i can already be seen as a limit over all covering). Or maybe the argument is simpler? (if you write this an answer I would be super happy to accept! If you don't have time I can type it myself). Thanks a lot for the answer!! – curious Mar 26 '24 at 12:29
  • Well, I didn't said cohomology commutes with this direct limit (though this is true in that case, but this is exceptionnal). The claim is that the presheaf $U\mapsto H^i(U,\mathcal{F}|U)$ has zero stalks if $i>0$. This is true for any sheaf on any space. There are several ways to see it. For example, this presheaf is the derived functor of the inclusion $Sh(X)\to PSh(X)$, so its sheafification is the derived functor of the identity which is zero. Another way : $U\mapsto H^i(f^{-1}(U),\mathcal{F})$ is a presheaf whose associated sheaf is $R^if*\mathcal{F}$.... – Roland Mar 26 '24 at 14:21
  • ... so the associated sheaf to $U\mapsto H^i(U,\mathcal{F})$ is $R^i\operatorname{id}*\mathcal{F}$ which is obviously $0$ as $\operatorname{id}*$ is exact. Anyway, in this argument, there is no Cech cohomology involved, this is just basic homological algebra and composition of functors. – Roland Mar 26 '24 at 14:22
  • By “local system” do you just mean sheaf or do you mean a specific kind of sheaf? – FShrike Mar 31 '24 at 13:26
  • I should add that $A\cong A\oplus B$ doesn’t always force $B=0$ – FShrike Mar 31 '24 at 17:08
  • @Roland : sorry for the delay! This is a great answer, thank you. – curious Apr 03 '24 at 15:47
  • @FShrike I mean a local system : https://en.wikipedia.org/wiki/Local_system – curious Apr 03 '24 at 15:48

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