Sheaf cohomology and singular cohomology
Let $X$ be a manifold, $mathcal{F}$ be an abelian sheaf on $X$. We can consider the etale space $F$ of $mathcal{F}$, which is the disjoint union of stalks, whose topology is generated by local sections.
Is there some the relation between sheaf cohomology of $mathcal{F}$ and cohomology of topological space $F$?
(Say, is there a relation between $H^i(X,mathcal{F})$ and singular cohomology of $H^i(F,mathbb{Z})$? Or in a nicer situation, when $mathcal{F}$ is a locally constant $Lambda=mathbb{Z}/lmathbb{Z}$ sheaf, is there relation between $H^i(X,mathcal{F})$ and $H^i(F,Lambda)$? )
algebraic-topology sheaf-theory sheaf-cohomology
asked Nov 24 at 15:08
Qixiao
2,7971628
add a comment |
Let $X$ be a manifold, $mathcal{F}$ be an abelian sheaf on $X$. We can consider the etale space $F$ of $mathcal{F}$, which is the disjoint union of stalks, whose topology is generated by local sections.
Is there some the relation between sheaf cohomology of $mathcal{F}$ and cohomology of topological space $F$?
(Say, is there a relation between $H^i(X,mathcal{F})$ and singular cohomology of $H^i(F,mathbb{Z})$? Or in a nicer situation, when $mathcal{F}$ is a locally constant $Lambda=mathbb{Z}/lmathbb{Z}$ sheaf, is there relation between $H^i(X,mathcal{F})$ and $H^i(F,Lambda)$? )
algebraic-topology sheaf-theory sheaf-cohomology
asked Nov 24 at 15:08
Qixiao
2,7971628
The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22
add a comment |
Let $X$ be a manifold, $mathcal{F}$ be an abelian sheaf on $X$. We can consider the etale space $F$ of $mathcal{F}$, which is the disjoint union of stalks, whose topology is generated by local sections.
Is there some the relation between sheaf cohomology of $mathcal{F}$ and cohomology of topological space $F$?
(Say, is there a relation between $H^i(X,mathcal{F})$ and singular cohomology of $H^i(F,mathbb{Z})$? Or in a nicer situation, when $mathcal{F}$ is a locally constant $Lambda=mathbb{Z}/lmathbb{Z}$ sheaf, is there relation between $H^i(X,mathcal{F})$ and $H^i(F,Lambda)$? )
algebraic-topology sheaf-theory sheaf-cohomology
asked Nov 24 at 15:08
Qixiao
2,7971628
Let $X$ be a manifold, $mathcal{F}$ be an abelian sheaf on $X$. We can consider the etale space $F$ of $mathcal{F}$, which is the disjoint union of stalks, whose topology is generated by local sections.
Is there some the relation between sheaf cohomology of $mathcal{F}$ and cohomology of topological space $F$?
(Say, is there a relation between $H^i(X,mathcal{F})$ and singular cohomology of $H^i(F,mathbb{Z})$? Or in a nicer situation, when $mathcal{F}$ is a locally constant $Lambda=mathbb{Z}/lmathbb{Z}$ sheaf, is there relation between $H^i(X,mathcal{F})$ and $H^i(F,Lambda)$? )
algebraic-topology sheaf-theory sheaf-cohomology
algebraic-topology sheaf-theory sheaf-cohomology
asked Nov 24 at 15:08
Qixiao
2,7971628
asked Nov 24 at 15:08
Qixiao
2,7971628
asked Nov 24 at 15:08
Qixiao
2,7971628
asked Nov 24 at 15:08
Qixiao
2,7971628
asked Nov 24 at 15:08
Qixiao
2,7971628
2,7971628
The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22
add a comment |
The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22
The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22
The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22
add a comment |
1 Answer
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If $mathcal F$ is a locally constant sheaf of stalk $Bbb Z/ell Bbb Z$, the étalé space $F$ is a degree $ell$ cover, and the monodromy of this cover $f : F to X$ is given by $mathcal F$. What you can say is basically that for any $j$, there is an isomorphism
$$H^j(F, underline{Bbb Z}) cong H^j(X,f_*underline{Bbb Z}) (*)$$
This is more or less the desired relation, since $f_*underline{Bbb Z}$ is "morally" $mathcal F$. (**)
We will do two examples with $X = S^1$ and $ell = 2$. We always write $f : F to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 sqcup S^1$, so the cohomology of $F$ is $Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*Bbb Z = Bbb Z oplus Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z mapsto z^2$. The left hand side is $Bbb Z$ in degree $0$ and $Bbb Z$ in degree $1$.
We now compute the right hand side. The sheaf $mathcal f_*Bbb Z$ has stalks $Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V to V$ ($V$ is the stalk at $1 in S^1$ of $L$) the following complex computes the monodromy : $0 to V overset{T - id}{to} V to 0$. In our case, this is the complex $$0 to Bbb Z^2 overset{pmatrix{-1 & 1 \ 1 & -1}}{to} Bbb Z^2 to 0$$
which has cohomology $Bbb Z$ in degree $0$ and $1$, as expected.
(*) More precisely, they have different coefficients but the same monodromy at a point $x in X$ : we have $(f_*underline{Bbb{Z}})_x cong Bbb Z oplus dots oplus Bbb Z$ ($ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $pi_1(X;x)$ acts by permuting these copy like $mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*underline{mathbb Z})_x$ and $mathcal F_x$ is $pi_1(X,x)$-equivariant.
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
add a comment |
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If $mathcal F$ is a locally constant sheaf of stalk $Bbb Z/ell Bbb Z$, the étalé space $F$ is a degree $ell$ cover, and the monodromy of this cover $f : F to X$ is given by $mathcal F$. What you can say is basically that for any $j$, there is an isomorphism
$$H^j(F, underline{Bbb Z}) cong H^j(X,f_*underline{Bbb Z}) (*)$$
This is more or less the desired relation, since $f_*underline{Bbb Z}$ is "morally" $mathcal F$. (**)
We will do two examples with $X = S^1$ and $ell = 2$. We always write $f : F to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 sqcup S^1$, so the cohomology of $F$ is $Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*Bbb Z = Bbb Z oplus Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z mapsto z^2$. The left hand side is $Bbb Z$ in degree $0$ and $Bbb Z$ in degree $1$.
We now compute the right hand side. The sheaf $mathcal f_*Bbb Z$ has stalks $Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V to V$ ($V$ is the stalk at $1 in S^1$ of $L$) the following complex computes the monodromy : $0 to V overset{T - id}{to} V to 0$. In our case, this is the complex $$0 to Bbb Z^2 overset{pmatrix{-1 & 1 \ 1 & -1}}{to} Bbb Z^2 to 0$$
which has cohomology $Bbb Z$ in degree $0$ and $1$, as expected.
(*) More precisely, they have different coefficients but the same monodromy at a point $x in X$ : we have $(f_*underline{Bbb{Z}})_x cong Bbb Z oplus dots oplus Bbb Z$ ($ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $pi_1(X;x)$ acts by permuting these copy like $mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*underline{mathbb Z})_x$ and $mathcal F_x$ is $pi_1(X,x)$-equivariant.
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
add a comment |
If $mathcal F$ is a locally constant sheaf of stalk $Bbb Z/ell Bbb Z$, the étalé space $F$ is a degree $ell$ cover, and the monodromy of this cover $f : F to X$ is given by $mathcal F$. What you can say is basically that for any $j$, there is an isomorphism
$$H^j(F, underline{Bbb Z}) cong H^j(X,f_*underline{Bbb Z}) (*)$$
This is more or less the desired relation, since $f_*underline{Bbb Z}$ is "morally" $mathcal F$. (**)
We will do two examples with $X = S^1$ and $ell = 2$. We always write $f : F to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 sqcup S^1$, so the cohomology of $F$ is $Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*Bbb Z = Bbb Z oplus Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z mapsto z^2$. The left hand side is $Bbb Z$ in degree $0$ and $Bbb Z$ in degree $1$.
We now compute the right hand side. The sheaf $mathcal f_*Bbb Z$ has stalks $Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V to V$ ($V$ is the stalk at $1 in S^1$ of $L$) the following complex computes the monodromy : $0 to V overset{T - id}{to} V to 0$. In our case, this is the complex $$0 to Bbb Z^2 overset{pmatrix{-1 & 1 \ 1 & -1}}{to} Bbb Z^2 to 0$$
which has cohomology $Bbb Z$ in degree $0$ and $1$, as expected.
(*) More precisely, they have different coefficients but the same monodromy at a point $x in X$ : we have $(f_*underline{Bbb{Z}})_x cong Bbb Z oplus dots oplus Bbb Z$ ($ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $pi_1(X;x)$ acts by permuting these copy like $mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*underline{mathbb Z})_x$ and $mathcal F_x$ is $pi_1(X,x)$-equivariant.
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
add a comment |
If $mathcal F$ is a locally constant sheaf of stalk $Bbb Z/ell Bbb Z$, the étalé space $F$ is a degree $ell$ cover, and the monodromy of this cover $f : F to X$ is given by $mathcal F$. What you can say is basically that for any $j$, there is an isomorphism
$$H^j(F, underline{Bbb Z}) cong H^j(X,f_*underline{Bbb Z}) (*)$$
This is more or less the desired relation, since $f_*underline{Bbb Z}$ is "morally" $mathcal F$. (**)
We will do two examples with $X = S^1$ and $ell = 2$. We always write $f : F to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 sqcup S^1$, so the cohomology of $F$ is $Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*Bbb Z = Bbb Z oplus Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z mapsto z^2$. The left hand side is $Bbb Z$ in degree $0$ and $Bbb Z$ in degree $1$.
We now compute the right hand side. The sheaf $mathcal f_*Bbb Z$ has stalks $Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V to V$ ($V$ is the stalk at $1 in S^1$ of $L$) the following complex computes the monodromy : $0 to V overset{T - id}{to} V to 0$. In our case, this is the complex $$0 to Bbb Z^2 overset{pmatrix{-1 & 1 \ 1 & -1}}{to} Bbb Z^2 to 0$$
which has cohomology $Bbb Z$ in degree $0$ and $1$, as expected.
(*) More precisely, they have different coefficients but the same monodromy at a point $x in X$ : we have $(f_*underline{Bbb{Z}})_x cong Bbb Z oplus dots oplus Bbb Z$ ($ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $pi_1(X;x)$ acts by permuting these copy like $mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*underline{mathbb Z})_x$ and $mathcal F_x$ is $pi_1(X,x)$-equivariant.
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
If $mathcal F$ is a locally constant sheaf of stalk $Bbb Z/ell Bbb Z$, the étalé space $F$ is a degree $ell$ cover, and the monodromy of this cover $f : F to X$ is given by $mathcal F$. What you can say is basically that for any $j$, there is an isomorphism
$$H^j(F, underline{Bbb Z}) cong H^j(X,f_*underline{Bbb Z}) (*)$$
This is more or less the desired relation, since $f_*underline{Bbb Z}$ is "morally" $mathcal F$. (**)
We will do two examples with $X = S^1$ and $ell = 2$. We always write $f : F to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 sqcup S^1$, so the cohomology of $F$ is $Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*Bbb Z = Bbb Z oplus Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z mapsto z^2$. The left hand side is $Bbb Z$ in degree $0$ and $Bbb Z$ in degree $1$.
We now compute the right hand side. The sheaf $mathcal f_*Bbb Z$ has stalks $Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V to V$ ($V$ is the stalk at $1 in S^1$ of $L$) the following complex computes the monodromy : $0 to V overset{T - id}{to} V to 0$. In our case, this is the complex $$0 to Bbb Z^2 overset{pmatrix{-1 & 1 \ 1 & -1}}{to} Bbb Z^2 to 0$$
which has cohomology $Bbb Z$ in degree $0$ and $1$, as expected.
(*) More precisely, they have different coefficients but the same monodromy at a point $x in X$ : we have $(f_*underline{Bbb{Z}})_x cong Bbb Z oplus dots oplus Bbb Z$ ($ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $pi_1(X;x)$ acts by permuting these copy like $mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*underline{mathbb Z})_x$ and $mathcal F_x$ is $pi_1(X,x)$-equivariant.
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
answered Nov 28 at 11:12
Nicolas Hemelsoet
5,7452417
5,7452417
add a comment |
add a comment |
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The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf.
– Pedro Tamaroff♦
Nov 24 at 15:22