Mixing
Killing cohomology of surfaces in finite covers
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Let $S$ be a closed orientable surface and $R$ a commutative ring. Given a nonzero element $alpha in H^1(S;R)$ is there a finite cover $p : tilde{S} to S$ such that $p^*(alpha) = 0 in H^1(tilde{S}; R)$?
algebraic-topology surfaces geometric-topology
asked Feb 1 at 23:18
user101010user101010
1,972516
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Let $S$ be a closed orientable surface and $R$ a commutative ring. Given a nonzero element $alpha in H^1(S;R)$ is there a finite cover $p : tilde{S} to S$ such that $p^*(alpha) = 0 in H^1(tilde{S}; R)$?
algebraic-topology surfaces geometric-topology
asked Feb 1 at 23:18
user101010user101010
1,972516
$endgroup$
add a comment |
$begingroup$
Let $S$ be a closed orientable surface and $R$ a commutative ring. Given a nonzero element $alpha in H^1(S;R)$ is there a finite cover $p : tilde{S} to S$ such that $p^*(alpha) = 0 in H^1(tilde{S}; R)$?
algebraic-topology surfaces geometric-topology
asked Feb 1 at 23:18
user101010user101010
1,972516
$endgroup$
Let $S$ be a closed orientable surface and $R$ a commutative ring. Given a nonzero element $alpha in H^1(S;R)$ is there a finite cover $p : tilde{S} to S$ such that $p^*(alpha) = 0 in H^1(tilde{S}; R)$?
algebraic-topology surfaces geometric-topology
algebraic-topology surfaces geometric-topology
asked Feb 1 at 23:18
user101010user101010
1,972516
asked Feb 1 at 23:18
user101010user101010
1,972516
asked Feb 1 at 23:18
user101010user101010
1,972516
asked Feb 1 at 23:18
user101010user101010
1,972516
asked Feb 1 at 23:18
user101010user101010
1,972516
1,972516
add a comment |
add a comment |
2 Answers
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No; in fact, if $R$ is torsion-free, there never is such a finite cover. If $p:tilde{S}to S$ is a finite cover, then $p_*pi_1(tilde{S})$ has finite index in $pi_1(S)$. We can naturally identify $H^1(S;R)$ with $operatorname{Hom}(pi_1(S),R)$ and similarly for $tilde{S}$. If $p^*(alpha)=0$, that means that $alpha$ vanishes on $p_*pi_1(tilde{S})$. But that implies $alpha$ vanishes on all of $pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free). Thus $p^*(alpha)=0$ implies $alpha=0$.
Alternatively, you could use Poincaré duality: if $alphaneq 0$, there exists $betain H^1(S;R)$ such that $alphacupbetaneq 0$, and then $p^*(alphacupbeta)neq 0$ since $p^*:H^2(S;R)to H^2(tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
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No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) cong text{Hom}(pi_1 S, G)$ for any $S, G$. Given any element $phi in text{Hom}(pi_1 S, G)$, if $f: S' to S$ is a covering space, the correspinding map $phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $text{ker}(phi) subset pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $text{ker}(phi)$ has infinite index, then the cover corresponding to $text{ker}(phi)$ is an infinite cover. For instance, if $G$ is infinite and $phi$ is surjective (or at least its image has finite index), then you cannot kill $phi$ in a finite cover.
Because every subgroup of $Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;Bbb Z)$ in a finite cover.
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2 Answers
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2 Answers
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$begingroup$
No; in fact, if $R$ is torsion-free, there never is such a finite cover. If $p:tilde{S}to S$ is a finite cover, then $p_*pi_1(tilde{S})$ has finite index in $pi_1(S)$. We can naturally identify $H^1(S;R)$ with $operatorname{Hom}(pi_1(S),R)$ and similarly for $tilde{S}$. If $p^*(alpha)=0$, that means that $alpha$ vanishes on $p_*pi_1(tilde{S})$. But that implies $alpha$ vanishes on all of $pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free). Thus $p^*(alpha)=0$ implies $alpha=0$.
Alternatively, you could use Poincaré duality: if $alphaneq 0$, there exists $betain H^1(S;R)$ such that $alphacupbetaneq 0$, and then $p^*(alphacupbeta)neq 0$ since $p^*:H^2(S;R)to H^2(tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
$endgroup$
add a comment |
$begingroup$
No; in fact, if $R$ is torsion-free, there never is such a finite cover. If $p:tilde{S}to S$ is a finite cover, then $p_*pi_1(tilde{S})$ has finite index in $pi_1(S)$. We can naturally identify $H^1(S;R)$ with $operatorname{Hom}(pi_1(S),R)$ and similarly for $tilde{S}$. If $p^*(alpha)=0$, that means that $alpha$ vanishes on $p_*pi_1(tilde{S})$. But that implies $alpha$ vanishes on all of $pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free). Thus $p^*(alpha)=0$ implies $alpha=0$.
Alternatively, you could use Poincaré duality: if $alphaneq 0$, there exists $betain H^1(S;R)$ such that $alphacupbetaneq 0$, and then $p^*(alphacupbeta)neq 0$ since $p^*:H^2(S;R)to H^2(tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
$endgroup$
add a comment |
$begingroup$
No; in fact, if $R$ is torsion-free, there never is such a finite cover. If $p:tilde{S}to S$ is a finite cover, then $p_*pi_1(tilde{S})$ has finite index in $pi_1(S)$. We can naturally identify $H^1(S;R)$ with $operatorname{Hom}(pi_1(S),R)$ and similarly for $tilde{S}$. If $p^*(alpha)=0$, that means that $alpha$ vanishes on $p_*pi_1(tilde{S})$. But that implies $alpha$ vanishes on all of $pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free). Thus $p^*(alpha)=0$ implies $alpha=0$.
Alternatively, you could use Poincaré duality: if $alphaneq 0$, there exists $betain H^1(S;R)$ such that $alphacupbetaneq 0$, and then $p^*(alphacupbeta)neq 0$ since $p^*:H^2(S;R)to H^2(tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
$endgroup$
No; in fact, if $R$ is torsion-free, there never is such a finite cover. If $p:tilde{S}to S$ is a finite cover, then $p_*pi_1(tilde{S})$ has finite index in $pi_1(S)$. We can naturally identify $H^1(S;R)$ with $operatorname{Hom}(pi_1(S),R)$ and similarly for $tilde{S}$. If $p^*(alpha)=0$, that means that $alpha$ vanishes on $p_*pi_1(tilde{S})$. But that implies $alpha$ vanishes on all of $pi_1(S)$ (a homomorphism from a group to $R$ which vanishes on a finite index subgroup must vanish on the whole group, since $R$ is torsion-free). Thus $p^*(alpha)=0$ implies $alpha=0$.
Alternatively, you could use Poincaré duality: if $alphaneq 0$, there exists $betain H^1(S;R)$ such that $alphacupbetaneq 0$, and then $p^*(alphacupbeta)neq 0$ since $p^*:H^2(S;R)to H^2(tilde{S};R)$ is multiplication by the degree of $p$ which is injective if $R$ is torsion-free.
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
answered Feb 1 at 23:49
Eric WofseyEric Wofsey
193k14220352
193k14220352
add a comment |
add a comment |
$begingroup$
No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) cong text{Hom}(pi_1 S, G)$ for any $S, G$. Given any element $phi in text{Hom}(pi_1 S, G)$, if $f: S' to S$ is a covering space, the correspinding map $phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $text{ker}(phi) subset pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $text{ker}(phi)$ has infinite index, then the cover corresponding to $text{ker}(phi)$ is an infinite cover. For instance, if $G$ is infinite and $phi$ is surjective (or at least its image has finite index), then you cannot kill $phi$ in a finite cover.
Because every subgroup of $Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;Bbb Z)$ in a finite cover.
$endgroup$
add a comment |
$begingroup$
No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) cong text{Hom}(pi_1 S, G)$ for any $S, G$. Given any element $phi in text{Hom}(pi_1 S, G)$, if $f: S' to S$ is a covering space, the correspinding map $phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $text{ker}(phi) subset pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $text{ker}(phi)$ has infinite index, then the cover corresponding to $text{ker}(phi)$ is an infinite cover. For instance, if $G$ is infinite and $phi$ is surjective (or at least its image has finite index), then you cannot kill $phi$ in a finite cover.
Because every subgroup of $Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;Bbb Z)$ in a finite cover.
$endgroup$
add a comment |
$begingroup$
No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) cong text{Hom}(pi_1 S, G)$ for any $S, G$. Given any element $phi in text{Hom}(pi_1 S, G)$, if $f: S' to S$ is a covering space, the correspinding map $phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $text{ker}(phi) subset pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $text{ker}(phi)$ has infinite index, then the cover corresponding to $text{ker}(phi)$ is an infinite cover. For instance, if $G$ is infinite and $phi$ is surjective (or at least its image has finite index), then you cannot kill $phi$ in a finite cover.
Because every subgroup of $Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;Bbb Z)$ in a finite cover.
$endgroup$
No. This has nothing to do with $S$ being a surface. In what follows all we assume is that $S$ can run the classification theory of covers (path-connected, locally path-connected, semilocally simply connected). It holds for any CW complex.
There is a natural isomorphism $H^1(S;G) cong text{Hom}(pi_1 S, G)$ for any $S, G$. Given any element $phi in text{Hom}(pi_1 S, G)$, if $f: S' to S$ is a covering space, the correspinding map $phi f_* = 0$ if and only if $f$ factors through the cover corresponding to $text{ker}(phi) subset pi_1 S$. (We use here the classification of covering spaces over reasonable spaces.)
If $text{ker}(phi)$ has infinite index, then the cover corresponding to $text{ker}(phi)$ is an infinite cover. For instance, if $G$ is infinite and $phi$ is surjective (or at least its image has finite index), then you cannot kill $phi$ in a finite cover.
Because every subgroup of $Bbb Z$ is zero or infinite, in particular, you cannot kill any nontrivial class of $H^1(S;Bbb Z)$ in a finite cover.
answered Feb 1 at 23:46
user98602
add a comment |
add a comment |
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