Are the fixed point sets homeomorphic?
$begingroup$
Assume $f:Grightarrow H$ is a continuous epimorphism of topological groups and $Kleq H$ is a subgroup of $H$. One can consider the actions of subgroups $K$ and $f^{-1}(K)$ on $H/K$ and $G/f^{-1}(K)$ respectively (here $H/K$ and $G/f^{-1}(K)$ denote right coset spaces - with quotient topology and the actions are given by left multiplications).
Consider the fixed point sets of these actions, $Fix(G/f^{-1}(K),f^{-1}(K))$ and $Fix(H/K,K)$ with the induced subset topologies.
Is it true that $Fix(G/f^{-1}(K),f^{-1}(K))$ is homeomorphic to $Fix(H/K,K)$? If not in egneral, then maybe at least in the case $K$ and $f^{-1}(K)$ are finite?
general-topology algebraic-topology group-actions fixedpoints
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
$endgroup$
add a comment |
$begingroup$
Assume $f:Grightarrow H$ is a continuous epimorphism of topological groups and $Kleq H$ is a subgroup of $H$. One can consider the actions of subgroups $K$ and $f^{-1}(K)$ on $H/K$ and $G/f^{-1}(K)$ respectively (here $H/K$ and $G/f^{-1}(K)$ denote right coset spaces - with quotient topology and the actions are given by left multiplications).
Consider the fixed point sets of these actions, $Fix(G/f^{-1}(K),f^{-1}(K))$ and $Fix(H/K,K)$ with the induced subset topologies.
Is it true that $Fix(G/f^{-1}(K),f^{-1}(K))$ is homeomorphic to $Fix(H/K,K)$? If not in egneral, then maybe at least in the case $K$ and $f^{-1}(K)$ are finite?
general-topology algebraic-topology group-actions fixedpoints
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
$endgroup$
1
$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20
add a comment |
$begingroup$
Assume $f:Grightarrow H$ is a continuous epimorphism of topological groups and $Kleq H$ is a subgroup of $H$. One can consider the actions of subgroups $K$ and $f^{-1}(K)$ on $H/K$ and $G/f^{-1}(K)$ respectively (here $H/K$ and $G/f^{-1}(K)$ denote right coset spaces - with quotient topology and the actions are given by left multiplications).
Consider the fixed point sets of these actions, $Fix(G/f^{-1}(K),f^{-1}(K))$ and $Fix(H/K,K)$ with the induced subset topologies.
Is it true that $Fix(G/f^{-1}(K),f^{-1}(K))$ is homeomorphic to $Fix(H/K,K)$? If not in egneral, then maybe at least in the case $K$ and $f^{-1}(K)$ are finite?
general-topology algebraic-topology group-actions fixedpoints
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
$endgroup$
Assume $f:Grightarrow H$ is a continuous epimorphism of topological groups and $Kleq H$ is a subgroup of $H$. One can consider the actions of subgroups $K$ and $f^{-1}(K)$ on $H/K$ and $G/f^{-1}(K)$ respectively (here $H/K$ and $G/f^{-1}(K)$ denote right coset spaces - with quotient topology and the actions are given by left multiplications).
Consider the fixed point sets of these actions, $Fix(G/f^{-1}(K),f^{-1}(K))$ and $Fix(H/K,K)$ with the induced subset topologies.
Is it true that $Fix(G/f^{-1}(K),f^{-1}(K))$ is homeomorphic to $Fix(H/K,K)$? If not in egneral, then maybe at least in the case $K$ and $f^{-1}(K)$ are finite?
general-topology algebraic-topology group-actions fixedpoints
general-topology algebraic-topology group-actions fixedpoints
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
edited Dec 21 '18 at 10:52
piotrmizerka
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
asked Dec 20 '18 at 15:33
piotrmizerkapiotrmizerka
339110
339110
1
$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20
add a comment |
1
$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20
1
1
$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20
add a comment |
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$begingroup$
Is $f$ a homomorphism? Then the fixed point sets are $N_H(K)/K$ and $N_G(K)N/KN$, where $N=ker(f)$., And these are isomorphic groups (via $f$).
$endgroup$
– Hempelicious
Dec 21 '18 at 2:11
$begingroup$
Thinking about this more, $f$ always induces a continuous bijection between the fixed point sets. If $f$ is a quotient map, then you definitely get a homeomorphism. But it's not always true: let $G$ be $mathbb{R}$ with the discrete topology, $H$ be $mathbb{R}$ with the usual topology, $f$ the identity map and $K$ the rationals.
$endgroup$
– Hempelicious
Dec 21 '18 at 17:20