Mixing
About closed subsets of the dual group $widehat G$
$begingroup$
Let $G$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:
$$widehat{G}:=operatorname{Hom}(G,S^1)$$
We endow $widehat{G}$ with the compact-open topology. Moreover For any subset $Ssubset G$, let's introduce the following notation:
$$S^circ:={fin widehat{G}: f(S)=1}subset widehat{G},.$$
My question is the following: Consider the family of subgroups of $widehat{G}$:
$$mathcal F:={C^circ:Csubset G text{ is closed and compact}}$$
Is it a local basis at $1$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
abstract-algebra general-topology group-theory topological-groups
asked Jan 7 at 17:02
manifoldmanifold
300213
$endgroup$
add a comment |
$begingroup$
Let $G$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:
$$widehat{G}:=operatorname{Hom}(G,S^1)$$
We endow $widehat{G}$ with the compact-open topology. Moreover For any subset $Ssubset G$, let's introduce the following notation:
$$S^circ:={fin widehat{G}: f(S)=1}subset widehat{G},.$$
My question is the following: Consider the family of subgroups of $widehat{G}$:
$$mathcal F:={C^circ:Csubset G text{ is closed and compact}}$$
Is it a local basis at $1$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
abstract-algebra general-topology group-theory topological-groups
asked Jan 7 at 17:02
manifoldmanifold
300213
$endgroup$
add a comment |
$begingroup$
Let $G$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:
$$widehat{G}:=operatorname{Hom}(G,S^1)$$
We endow $widehat{G}$ with the compact-open topology. Moreover For any subset $Ssubset G$, let's introduce the following notation:
$$S^circ:={fin widehat{G}: f(S)=1}subset widehat{G},.$$
My question is the following: Consider the family of subgroups of $widehat{G}$:
$$mathcal F:={C^circ:Csubset G text{ is closed and compact}}$$
Is it a local basis at $1$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
abstract-algebra general-topology group-theory topological-groups
asked Jan 7 at 17:02
manifoldmanifold
300213
$endgroup$
Let $G$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:
$$widehat{G}:=operatorname{Hom}(G,S^1)$$
We endow $widehat{G}$ with the compact-open topology. Moreover For any subset $Ssubset G$, let's introduce the following notation:
$$S^circ:={fin widehat{G}: f(S)=1}subset widehat{G},.$$
My question is the following: Consider the family of subgroups of $widehat{G}$:
$$mathcal F:={C^circ:Csubset G text{ is closed and compact}}$$
Is it a local basis at $1$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
abstract-algebra general-topology group-theory topological-groups
abstract-algebra general-topology group-theory topological-groups
asked Jan 7 at 17:02
manifoldmanifold
300213
asked Jan 7 at 17:02
manifoldmanifold
300213
asked Jan 7 at 17:02
manifoldmanifold
300213
asked Jan 7 at 17:02
manifoldmanifold
300213
asked Jan 7 at 17:02
manifoldmanifold
300213
300213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This answer is partial. I hope it will be useful for others.
I understood that $G$ is linearly topologized as there exists a local base $mathcal B$ at $1$ consisting of open subgroups of $G$. That is $bigcapmathcal B={1}$ and for each $H,H’inmathcal B$, $Hcap H’inmathcal B$. Next, by $operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1={|z|inBbb C:|z|=1}$ endowed with the topology and multiplication inherited from usual these of $Bbb C$.
Put $V_0={zin S^1:operatorname{Re} z>0}$. If $fin widehat{G}$ then there exist $Hinmathcal B$ such that $f(H)subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)={1}$. Conversely, any homomorphism $f:Gto S^1$ such that $f(H)={1}$ for some $Hinmathcal B$, is continuous. Thus for any $finwidehat G$ we have $operatorname{ker} finmathcal B$.
Then for any nonempty subset $S$ of $G$, $S^circ={fin widehat{G}: Ssubset operatorname{ker} f}={fin widehat{G}: langle Sranglesubset operatorname{ker} f}=langle Srangle^circ,$ where $langle Srangle$ is the subgroup of $G$ generated by $S$. Thus $mathcal F={H^circ: H$ is a compactly generated subgroup of $G}$.
I assume that the operation of $widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,ginwidehat G$ and $xin G$. Then the identity of $widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.
Thus $S^circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.
Is $mathcal F$ a local basis at $1$ of closed subgroups?
The local base $widehat{mathcal B}$ at $1$ of $widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $Vsubset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)in widehat{mathcal B}$ we have $C^circsubset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $Vsubset S^1$ of $1$ such that $(C’,V)subset C^circ$.
what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
Clearly, $S^circ$ is a subgroup of $widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^circ$ is closed. Indeed, suppose to the contrary that there exists $fin overline{S^circ}$ and $xin S$ such that $f(x)ne 1$. Then $({x},S^1setminus{1})$ is an open neighborhood of $f$, so there exists an element $gin ({x},S^1setminus{1})cap S^circ$. Since $gin ({x},S^1setminus{1})$, $g(x)ne 1$. On the other hand, since $gin S^circ $ and $xin S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $widehat G$ belongs to $mathcal F$.
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
add a comment |
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$begingroup$
This answer is partial. I hope it will be useful for others.
I understood that $G$ is linearly topologized as there exists a local base $mathcal B$ at $1$ consisting of open subgroups of $G$. That is $bigcapmathcal B={1}$ and for each $H,H’inmathcal B$, $Hcap H’inmathcal B$. Next, by $operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1={|z|inBbb C:|z|=1}$ endowed with the topology and multiplication inherited from usual these of $Bbb C$.
Put $V_0={zin S^1:operatorname{Re} z>0}$. If $fin widehat{G}$ then there exist $Hinmathcal B$ such that $f(H)subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)={1}$. Conversely, any homomorphism $f:Gto S^1$ such that $f(H)={1}$ for some $Hinmathcal B$, is continuous. Thus for any $finwidehat G$ we have $operatorname{ker} finmathcal B$.
Then for any nonempty subset $S$ of $G$, $S^circ={fin widehat{G}: Ssubset operatorname{ker} f}={fin widehat{G}: langle Sranglesubset operatorname{ker} f}=langle Srangle^circ,$ where $langle Srangle$ is the subgroup of $G$ generated by $S$. Thus $mathcal F={H^circ: H$ is a compactly generated subgroup of $G}$.
I assume that the operation of $widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,ginwidehat G$ and $xin G$. Then the identity of $widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.
Thus $S^circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.
Is $mathcal F$ a local basis at $1$ of closed subgroups?
The local base $widehat{mathcal B}$ at $1$ of $widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $Vsubset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)in widehat{mathcal B}$ we have $C^circsubset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $Vsubset S^1$ of $1$ such that $(C’,V)subset C^circ$.
what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
Clearly, $S^circ$ is a subgroup of $widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^circ$ is closed. Indeed, suppose to the contrary that there exists $fin overline{S^circ}$ and $xin S$ such that $f(x)ne 1$. Then $({x},S^1setminus{1})$ is an open neighborhood of $f$, so there exists an element $gin ({x},S^1setminus{1})cap S^circ$. Since $gin ({x},S^1setminus{1})$, $g(x)ne 1$. On the other hand, since $gin S^circ $ and $xin S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $widehat G$ belongs to $mathcal F$.
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
add a comment |
$begingroup$
This answer is partial. I hope it will be useful for others.
I understood that $G$ is linearly topologized as there exists a local base $mathcal B$ at $1$ consisting of open subgroups of $G$. That is $bigcapmathcal B={1}$ and for each $H,H’inmathcal B$, $Hcap H’inmathcal B$. Next, by $operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1={|z|inBbb C:|z|=1}$ endowed with the topology and multiplication inherited from usual these of $Bbb C$.
Put $V_0={zin S^1:operatorname{Re} z>0}$. If $fin widehat{G}$ then there exist $Hinmathcal B$ such that $f(H)subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)={1}$. Conversely, any homomorphism $f:Gto S^1$ such that $f(H)={1}$ for some $Hinmathcal B$, is continuous. Thus for any $finwidehat G$ we have $operatorname{ker} finmathcal B$.
Then for any nonempty subset $S$ of $G$, $S^circ={fin widehat{G}: Ssubset operatorname{ker} f}={fin widehat{G}: langle Sranglesubset operatorname{ker} f}=langle Srangle^circ,$ where $langle Srangle$ is the subgroup of $G$ generated by $S$. Thus $mathcal F={H^circ: H$ is a compactly generated subgroup of $G}$.
I assume that the operation of $widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,ginwidehat G$ and $xin G$. Then the identity of $widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.
Thus $S^circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.
Is $mathcal F$ a local basis at $1$ of closed subgroups?
The local base $widehat{mathcal B}$ at $1$ of $widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $Vsubset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)in widehat{mathcal B}$ we have $C^circsubset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $Vsubset S^1$ of $1$ such that $(C’,V)subset C^circ$.
what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
Clearly, $S^circ$ is a subgroup of $widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^circ$ is closed. Indeed, suppose to the contrary that there exists $fin overline{S^circ}$ and $xin S$ such that $f(x)ne 1$. Then $({x},S^1setminus{1})$ is an open neighborhood of $f$, so there exists an element $gin ({x},S^1setminus{1})cap S^circ$. Since $gin ({x},S^1setminus{1})$, $g(x)ne 1$. On the other hand, since $gin S^circ $ and $xin S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $widehat G$ belongs to $mathcal F$.
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
add a comment |
$begingroup$
This answer is partial. I hope it will be useful for others.
I understood that $G$ is linearly topologized as there exists a local base $mathcal B$ at $1$ consisting of open subgroups of $G$. That is $bigcapmathcal B={1}$ and for each $H,H’inmathcal B$, $Hcap H’inmathcal B$. Next, by $operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1={|z|inBbb C:|z|=1}$ endowed with the topology and multiplication inherited from usual these of $Bbb C$.
Put $V_0={zin S^1:operatorname{Re} z>0}$. If $fin widehat{G}$ then there exist $Hinmathcal B$ such that $f(H)subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)={1}$. Conversely, any homomorphism $f:Gto S^1$ such that $f(H)={1}$ for some $Hinmathcal B$, is continuous. Thus for any $finwidehat G$ we have $operatorname{ker} finmathcal B$.
Then for any nonempty subset $S$ of $G$, $S^circ={fin widehat{G}: Ssubset operatorname{ker} f}={fin widehat{G}: langle Sranglesubset operatorname{ker} f}=langle Srangle^circ,$ where $langle Srangle$ is the subgroup of $G$ generated by $S$. Thus $mathcal F={H^circ: H$ is a compactly generated subgroup of $G}$.
I assume that the operation of $widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,ginwidehat G$ and $xin G$. Then the identity of $widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.
Thus $S^circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.
Is $mathcal F$ a local basis at $1$ of closed subgroups?
The local base $widehat{mathcal B}$ at $1$ of $widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $Vsubset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)in widehat{mathcal B}$ we have $C^circsubset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $Vsubset S^1$ of $1$ such that $(C’,V)subset C^circ$.
what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
Clearly, $S^circ$ is a subgroup of $widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^circ$ is closed. Indeed, suppose to the contrary that there exists $fin overline{S^circ}$ and $xin S$ such that $f(x)ne 1$. Then $({x},S^1setminus{1})$ is an open neighborhood of $f$, so there exists an element $gin ({x},S^1setminus{1})cap S^circ$. Since $gin ({x},S^1setminus{1})$, $g(x)ne 1$. On the other hand, since $gin S^circ $ and $xin S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $widehat G$ belongs to $mathcal F$.
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
$endgroup$
This answer is partial. I hope it will be useful for others.
I understood that $G$ is linearly topologized as there exists a local base $mathcal B$ at $1$ consisting of open subgroups of $G$. That is $bigcapmathcal B={1}$ and for each $H,H’inmathcal B$, $Hcap H’inmathcal B$. Next, by $operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1={|z|inBbb C:|z|=1}$ endowed with the topology and multiplication inherited from usual these of $Bbb C$.
Put $V_0={zin S^1:operatorname{Re} z>0}$. If $fin widehat{G}$ then there exist $Hinmathcal B$ such that $f(H)subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)={1}$. Conversely, any homomorphism $f:Gto S^1$ such that $f(H)={1}$ for some $Hinmathcal B$, is continuous. Thus for any $finwidehat G$ we have $operatorname{ker} finmathcal B$.
Then for any nonempty subset $S$ of $G$, $S^circ={fin widehat{G}: Ssubset operatorname{ker} f}={fin widehat{G}: langle Sranglesubset operatorname{ker} f}=langle Srangle^circ,$ where $langle Srangle$ is the subgroup of $G$ generated by $S$. Thus $mathcal F={H^circ: H$ is a compactly generated subgroup of $G}$.
I assume that the operation of $widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,ginwidehat G$ and $xin G$. Then the identity of $widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.
Thus $S^circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.
Is $mathcal F$ a local basis at $1$ of closed subgroups?
The local base $widehat{mathcal B}$ at $1$ of $widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $Vsubset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)in widehat{mathcal B}$ we have $C^circsubset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $Vsubset S^1$ of $1$ such that $(C’,V)subset C^circ$.
what is the relationship between closed subsets (or subgroups) of $widehat{G}$ and $mathcal F$?
Clearly, $S^circ$ is a subgroup of $widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^circ$ is closed. Indeed, suppose to the contrary that there exists $fin overline{S^circ}$ and $xin S$ such that $f(x)ne 1$. Then $({x},S^1setminus{1})$ is an open neighborhood of $f$, so there exists an element $gin ({x},S^1setminus{1})cap S^circ$. Since $gin ({x},S^1setminus{1})$, $g(x)ne 1$. On the other hand, since $gin S^circ $ and $xin S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $widehat G$ belongs to $mathcal F$.
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
answered Jan 10 at 18:10
Alex RavskyAlex Ravsky
40.4k32282
40.4k32282
add a comment |
add a comment |
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