Homology of subset of orbit space
$begingroup$
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
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
$endgroup$
add a comment |
$begingroup$
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
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
$endgroup$
$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27
add a comment |
$begingroup$
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
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
$endgroup$
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
algebraic-topology finite-groups lie-groups homology-cohomology group-actions
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
edited Dec 22 '18 at 19:25
piotrmizerka
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
asked Dec 22 '18 at 18:45
piotrmizerkapiotrmizerka
339110
339110
$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27
add a comment |
$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27
$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $X$.
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
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 $X$.
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
$endgroup$
add a comment |
$begingroup$
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 $X$.
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
$endgroup$
add a comment |
$begingroup$
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 $X$.
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
$endgroup$
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 $X$.
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
answered Jan 4 at 21:57
Moishe KohanMoishe Kohan
48.2k344110
48.2k344110
add a comment |
add a comment |
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$begingroup$
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
$endgroup$
– Moishe Kohan
Dec 30 '18 at 23:33
$begingroup$
I am especially interested in the case when $G $ is perfect
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:11
$begingroup$
Also, I would like to consider rather unproper inclusions $Asubset X$.
$endgroup$
– piotrmizerka
Dec 31 '18 at 10:14
$begingroup$
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.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 16:27