About minimal group actions? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Compact group actions and automatic propernessShrinking Group ActionsAction of discrete subgroups E(n) on $BbbR^n$The limit of minimal points is still minimal?about minimal point in non- autonomous discrete systemA question about how far a well known theorem of Sierpinski can be strengthenedCharacterisation of proper group actionsSet of recurrenceA group action on an infinite set with the cofinite topologyMinimal dynamical systems in $2^mathbb N$
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About minimal group actions?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Compact group actions and automatic propernessShrinking Group ActionsAction of discrete subgroups E(n) on $BbbR^n$The limit of minimal points is still minimal?about minimal point in non- autonomous discrete systemA question about how far a well known theorem of Sierpinski can be strengthenedCharacterisation of proper group actionsSet of recurrenceA group action on an infinite set with the cofinite topologyMinimal dynamical systems in $2^mathbb N$
$begingroup$
Let $G$ be infinite group and $G$ act on compact metric space $(X, d)$, $varphi:Gtimes Xrightarrow X$.
$varphi:Gtimes Xrightarrow X$ is called minimal action, whenever there is not proper closed set $Asubseteq X$ with $GAsubseteq A$. ($GA=varphi(g, a)$).
Question. Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$.
If $varphi:Gtimes Xrightarrow X$ is minimal action.
$varphi|H:Htimes Xrightarrow X$ is minimal action?
general-topology dynamical-systems
edited May 30 '16 at 10:48
Ali Barzanuni
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
$endgroup$
add a comment |
$begingroup$
Let $G$ be infinite group and $G$ act on compact metric space $(X, d)$, $varphi:Gtimes Xrightarrow X$.
$varphi:Gtimes Xrightarrow X$ is called minimal action, whenever there is not proper closed set $Asubseteq X$ with $GAsubseteq A$. ($GA=varphi(g, a)$).
Question. Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$.
If $varphi:Gtimes Xrightarrow X$ is minimal action.
$varphi|H:Htimes Xrightarrow X$ is minimal action?
general-topology dynamical-systems
edited May 30 '16 at 10:48
Ali Barzanuni
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
$endgroup$
1
$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13
add a comment |
$begingroup$
Let $G$ be infinite group and $G$ act on compact metric space $(X, d)$, $varphi:Gtimes Xrightarrow X$.
$varphi:Gtimes Xrightarrow X$ is called minimal action, whenever there is not proper closed set $Asubseteq X$ with $GAsubseteq A$. ($GA=varphi(g, a)$).
Question. Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$.
If $varphi:Gtimes Xrightarrow X$ is minimal action.
$varphi|H:Htimes Xrightarrow X$ is minimal action?
general-topology dynamical-systems
edited May 30 '16 at 10:48
Ali Barzanuni
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
$endgroup$
Let $G$ be infinite group and $G$ act on compact metric space $(X, d)$, $varphi:Gtimes Xrightarrow X$.
$varphi:Gtimes Xrightarrow X$ is called minimal action, whenever there is not proper closed set $Asubseteq X$ with $GAsubseteq A$. ($GA=varphi(g, a)$).
Question. Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$.
If $varphi:Gtimes Xrightarrow X$ is minimal action.
$varphi|H:Htimes Xrightarrow X$ is minimal action?
general-topology dynamical-systems
general-topology dynamical-systems
edited May 30 '16 at 10:48
Ali Barzanuni
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
edited May 30 '16 at 10:48
Ali Barzanuni
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
edited May 30 '16 at 10:48
Ali Barzanuni
31
edited May 30 '16 at 10:48
Ali Barzanuni
31
edited May 30 '16 at 10:48
Ali Barzanuni
31
31
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
asked May 30 '16 at 8:13
Ali BarzanouniAli Barzanouni
112
112
1
$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13
add a comment |
1
$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13
1
1
$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13
$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 sqcup S^1 = (theta,0),(theta,1)mid thetain S^1$ be the disjoint union of two circles and let $rhocolon mathbbZtimes S^1to S^1$ be rotation by some irrational angle $alpha$: $rho(theta)=[theta+alpha]$.
We extend this action to $X$ by defining the new action $tilderhocolon mathbbZtimes X to X$ by $$tilderho((theta, i)=(rho(theta),i+1)$$
with addition being mod $2$.
As $rho$ is minimal, so is $tilderho$ (prove this). The subgroup $H=2mathbbZ$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
$endgroup$
add a comment |
$begingroup$
Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $Gcurvearrowright X$, we have that $Hcurvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/Hcurvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $Gcurvearrowright X$ minimally".
Now we must have a true statement: $* Rightarrow Hcurvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $mathbbZ$ on $S^1$).
So by a logic argument, we deduce that your question is false.
answered Apr 2 at 7:33
ShakibaShakiba
806
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 sqcup S^1 = (theta,0),(theta,1)mid thetain S^1$ be the disjoint union of two circles and let $rhocolon mathbbZtimes S^1to S^1$ be rotation by some irrational angle $alpha$: $rho(theta)=[theta+alpha]$.
We extend this action to $X$ by defining the new action $tilderhocolon mathbbZtimes X to X$ by $$tilderho((theta, i)=(rho(theta),i+1)$$
with addition being mod $2$.
As $rho$ is minimal, so is $tilderho$ (prove this). The subgroup $H=2mathbbZ$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
$endgroup$
add a comment |
$begingroup$
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 sqcup S^1 = (theta,0),(theta,1)mid thetain S^1$ be the disjoint union of two circles and let $rhocolon mathbbZtimes S^1to S^1$ be rotation by some irrational angle $alpha$: $rho(theta)=[theta+alpha]$.
We extend this action to $X$ by defining the new action $tilderhocolon mathbbZtimes X to X$ by $$tilderho((theta, i)=(rho(theta),i+1)$$
with addition being mod $2$.
As $rho$ is minimal, so is $tilderho$ (prove this). The subgroup $H=2mathbbZ$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
$endgroup$
add a comment |
$begingroup$
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 sqcup S^1 = (theta,0),(theta,1)mid thetain S^1$ be the disjoint union of two circles and let $rhocolon mathbbZtimes S^1to S^1$ be rotation by some irrational angle $alpha$: $rho(theta)=[theta+alpha]$.
We extend this action to $X$ by defining the new action $tilderhocolon mathbbZtimes X to X$ by $$tilderho((theta, i)=(rho(theta),i+1)$$
with addition being mod $2$.
As $rho$ is minimal, so is $tilderho$ (prove this). The subgroup $H=2mathbbZ$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
$endgroup$
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 sqcup S^1 = (theta,0),(theta,1)mid thetain S^1$ be the disjoint union of two circles and let $rhocolon mathbbZtimes S^1to S^1$ be rotation by some irrational angle $alpha$: $rho(theta)=[theta+alpha]$.
We extend this action to $X$ by defining the new action $tilderhocolon mathbbZtimes X to X$ by $$tilderho((theta, i)=(rho(theta),i+1)$$
with addition being mod $2$.
As $rho$ is minimal, so is $tilderho$ (prove this). The subgroup $H=2mathbbZ$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
answered May 30 '16 at 12:02
Dan RustDan Rust
23.1k115084
23.1k115084
add a comment |
add a comment |
$begingroup$
Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $Gcurvearrowright X$, we have that $Hcurvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/Hcurvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $Gcurvearrowright X$ minimally".
Now we must have a true statement: $* Rightarrow Hcurvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $mathbbZ$ on $S^1$).
So by a logic argument, we deduce that your question is false.
answered Apr 2 at 7:33
ShakibaShakiba
806
$endgroup$
add a comment |
$begingroup$
Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $Gcurvearrowright X$, we have that $Hcurvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/Hcurvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $Gcurvearrowright X$ minimally".
Now we must have a true statement: $* Rightarrow Hcurvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $mathbbZ$ on $S^1$).
So by a logic argument, we deduce that your question is false.
answered Apr 2 at 7:33
ShakibaShakiba
806
$endgroup$
add a comment |
$begingroup$
Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $Gcurvearrowright X$, we have that $Hcurvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/Hcurvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $Gcurvearrowright X$ minimally".
Now we must have a true statement: $* Rightarrow Hcurvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $mathbbZ$ on $S^1$).
So by a logic argument, we deduce that your question is false.
answered Apr 2 at 7:33
ShakibaShakiba
806
$endgroup$
Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $Gcurvearrowright X$, we have that $Hcurvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/Hcurvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $Gcurvearrowright X$ minimally".
Now we must have a true statement: $* Rightarrow Hcurvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $mathbbZ$ on $S^1$).
So by a logic argument, we deduce that your question is false.
answered Apr 2 at 7:33
ShakibaShakiba
806
edited Apr 14 at 15:37
answered Apr 2 at 7:33
ShakibaShakiba
806
answered Apr 2 at 7:33
ShakibaShakiba
806
answered Apr 2 at 7:33
ShakibaShakiba
806
806
add a comment |
add a comment |
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$begingroup$
Let $G$ be a finite group acting minimally on $X$ and let $xin X$ and $H=operatorname Stab x$. Here the action of $H$ is not minimal.
$endgroup$
– R_D
May 30 '16 at 9:13