Homology of subset of orbit space












2














Assume a finite group $G$ acts on a topological space $X$ and $Asubseteq X$. Denote by $q$ the quotient map from $X$ to the orbit space $X/G$ (we take the quotient topology). Moreover, let $H_n(A)=0$ for some $ngeq 1$.



Let us consider a subset $q(A)$ of $X/G$ with the induced subset topology.



When is it the case that $H_n(q(A))=0$? (I mean especially group properties of $G$) If the answer is not possible in general, it would be nice to point out specific situations too (e.g. $X$ - manifold, smooth manifold, Lie group, etc.)



The analogous problem for cohomology is also interesting to me.










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  • I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
    – Moishe Cohen
    Dec 30 '18 at 23:33










  • I am especially interested in the case when $G $ is perfect
    – piotrmizerka
    Dec 31 '18 at 10:11










  • Also, I would like to consider rather unproper inclusions $Asubset X$.
    – piotrmizerka
    Dec 31 '18 at 10:14










  • One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
    – Moishe Cohen
    Dec 31 '18 at 16:27
















2














Assume a finite group $G$ acts on a topological space $X$ and $Asubseteq X$. Denote by $q$ the quotient map from $X$ to the orbit space $X/G$ (we take the quotient topology). Moreover, let $H_n(A)=0$ for some $ngeq 1$.



Let us consider a subset $q(A)$ of $X/G$ with the induced subset topology.



When is it the case that $H_n(q(A))=0$? (I mean especially group properties of $G$) If the answer is not possible in general, it would be nice to point out specific situations too (e.g. $X$ - manifold, smooth manifold, Lie group, etc.)



The analogous problem for cohomology is also interesting to me.










share|cite|improve this question
























  • I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
    – Moishe Cohen
    Dec 30 '18 at 23:33










  • I am especially interested in the case when $G $ is perfect
    – piotrmizerka
    Dec 31 '18 at 10:11










  • Also, I would like to consider rather unproper inclusions $Asubset X$.
    – piotrmizerka
    Dec 31 '18 at 10:14










  • One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
    – Moishe Cohen
    Dec 31 '18 at 16:27














2












2








2


1





Assume a finite group $G$ acts on a topological space $X$ and $Asubseteq X$. Denote by $q$ the quotient map from $X$ to the orbit space $X/G$ (we take the quotient topology). Moreover, let $H_n(A)=0$ for some $ngeq 1$.



Let us consider a subset $q(A)$ of $X/G$ with the induced subset topology.



When is it the case that $H_n(q(A))=0$? (I mean especially group properties of $G$) If the answer is not possible in general, it would be nice to point out specific situations too (e.g. $X$ - manifold, smooth manifold, Lie group, etc.)



The analogous problem for cohomology is also interesting to me.










share|cite|improve this question















Assume a finite group $G$ acts on a topological space $X$ and $Asubseteq X$. Denote by $q$ the quotient map from $X$ to the orbit space $X/G$ (we take the quotient topology). Moreover, let $H_n(A)=0$ for some $ngeq 1$.



Let us consider a subset $q(A)$ of $X/G$ with the induced subset topology.



When is it the case that $H_n(q(A))=0$? (I mean especially group properties of $G$) If the answer is not possible in general, it would be nice to point out specific situations too (e.g. $X$ - manifold, smooth manifold, Lie group, etc.)



The analogous problem for cohomology is also interesting to me.







algebraic-topology finite-groups lie-groups homology-cohomology group-actions






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edited Dec 22 '18 at 19:25







piotrmizerka

















asked Dec 22 '18 at 18:45









piotrmizerkapiotrmizerka

334110




334110












  • I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
    – Moishe Cohen
    Dec 30 '18 at 23:33










  • I am especially interested in the case when $G $ is perfect
    – piotrmizerka
    Dec 31 '18 at 10:11










  • Also, I would like to consider rather unproper inclusions $Asubset X$.
    – piotrmizerka
    Dec 31 '18 at 10:14










  • One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
    – Moishe Cohen
    Dec 31 '18 at 16:27


















  • I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
    – Moishe Cohen
    Dec 30 '18 at 23:33










  • I am especially interested in the case when $G $ is perfect
    – piotrmizerka
    Dec 31 '18 at 10:11










  • Also, I would like to consider rather unproper inclusions $Asubset X$.
    – piotrmizerka
    Dec 31 '18 at 10:14










  • One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
    – Moishe Cohen
    Dec 31 '18 at 16:27
















I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
– Moishe Cohen
Dec 30 '18 at 23:33




I am not sure what kind of an answer your are looking for. Consider for instance the action of $Z_2$ on $X=A=S^k$, $nne k$, such that the generator of $Z_2$ acts as the antipodal map. topospaces.subwiki.org/wiki/Homology_of_real_projective_space
– Moishe Cohen
Dec 30 '18 at 23:33












I am especially interested in the case when $G $ is perfect
– piotrmizerka
Dec 31 '18 at 10:11




I am especially interested in the case when $G $ is perfect
– piotrmizerka
Dec 31 '18 at 10:11












Also, I would like to consider rather unproper inclusions $Asubset X$.
– piotrmizerka
Dec 31 '18 at 10:14




Also, I would like to consider rather unproper inclusions $Asubset X$.
– piotrmizerka
Dec 31 '18 at 10:14












One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
– Moishe Cohen
Dec 31 '18 at 16:27




One can easily modify my examples to satisfy your two conditions. In any case, you should edit your question to reflect what you are actually interested in.
– Moishe Cohen
Dec 31 '18 at 16:27










1 Answer
1






active

oldest

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2





+50









Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)cong {mathbb Z}_2ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on
$$
X= S^5times S^7,
$$

see Theorem 1.1 of



Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu,
Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.



It follows that $H_2(X/G)cong H_2(G)ne 0$, while $H_2(X)=0$.



If you want $A$ a proper subset of $X$, take $X$ to be the product
$$
S^5times S^7 times S^{2019}
$$

and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product
$$
S^5times S^7 times{p}subset X.
$$



On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $Gtimes Xto X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.






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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2





    +50









    Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)cong {mathbb Z}_2ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on
    $$
    X= S^5times S^7,
    $$

    see Theorem 1.1 of



    Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu,
    Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.



    It follows that $H_2(X/G)cong H_2(G)ne 0$, while $H_2(X)=0$.



    If you want $A$ a proper subset of $X$, take $X$ to be the product
    $$
    S^5times S^7 times S^{2019}
    $$

    and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product
    $$
    S^5times S^7 times{p}subset X.
    $$



    On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $Gtimes Xto X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.






    share|cite|improve this answer


























      2





      +50









      Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)cong {mathbb Z}_2ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on
      $$
      X= S^5times S^7,
      $$

      see Theorem 1.1 of



      Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu,
      Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.



      It follows that $H_2(X/G)cong H_2(G)ne 0$, while $H_2(X)=0$.



      If you want $A$ a proper subset of $X$, take $X$ to be the product
      $$
      S^5times S^7 times S^{2019}
      $$

      and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product
      $$
      S^5times S^7 times{p}subset X.
      $$



      On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $Gtimes Xto X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.






      share|cite|improve this answer
























        2





        +50







        2





        +50



        2




        +50




        Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)cong {mathbb Z}_2ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on
        $$
        X= S^5times S^7,
        $$

        see Theorem 1.1 of



        Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu,
        Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.



        It follows that $H_2(X/G)cong H_2(G)ne 0$, while $H_2(X)=0$.



        If you want $A$ a proper subset of $X$, take $X$ to be the product
        $$
        S^5times S^7 times S^{2019}
        $$

        and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product
        $$
        S^5times S^7 times{p}subset X.
        $$



        On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $Gtimes Xto X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.






        share|cite|improve this answer












        Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)cong {mathbb Z}_2ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on
        $$
        X= S^5times S^7,
        $$

        see Theorem 1.1 of



        Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu,
        Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.



        It follows that $H_2(X/G)cong H_2(G)ne 0$, while $H_2(X)=0$.



        If you want $A$ a proper subset of $X$, take $X$ to be the product
        $$
        S^5times S^7 times S^{2019}
        $$

        and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product
        $$
        S^5times S^7 times{p}subset X.
        $$



        On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $Gtimes Xto X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 21:57









        Moishe CohenMoishe Cohen

        46.1k342104




        46.1k342104






























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