Sheaf cohomology and singular cohomology












4














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







share|cite|improve this question






















  • 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
















4














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







share|cite|improve this question






















  • 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














4












4








4


1





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






active

oldest

votes


















0














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.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011663%2fsheaf-cohomology-and-singular-cohomology%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 11:12









        Nicolas Hemelsoet

        5,7452417




        5,7452417






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011663%2fsheaf-cohomology-and-singular-cohomology%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
            Read more

            “minimization” problem in Euclidean space related to orthonormal basis

            Image
            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
            Read more