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Transitive action of a discrete group on a compact space

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8

1

$begingroup$

Let $G$ be a discrete countable group acting on a compact, Hausdorff space $X$. Assume that the action is transitive. Namely, $Gcdot x=X$, for all $xin X$.

Does it follow that $X$ is finite?

I thought that the answer should be yes, as $Gcdot x$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.

general-topology group-theory compactness group-actions

share|cite|improve this question

edited 8 hours ago

Servaes

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asked 8 hours ago

KikoKiko

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Notice that under these conditions $G/G_xcong Gcdot x=X$ homeomorphism (see theorem 6.2 in the book "Crossed Products" by Williams). Thus $G/G_x$ is discrete and compact, so finite, and so is $X$.
  $endgroup$
  – Shirly Geffen
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ShirlyGeffen $G/G_x$ need not be homeomorphic to $Gx$. The standard map is a continuous bijection only. There are counterexamples when it is not a homeomorphism. It is true however when $G$ is compact for example (not our case). The theorem most likely assumes something more about $G$ or the action.
  $endgroup$
  – freakish
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @freakish It is a homeomorphism if $Gcdot x$ is locally closed in $X$, which is the case here.
  $endgroup$
  – Shirly Geffen
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ShirlyGeffen the theorem also assumes that $X$ is second countable. Which isn't clear for me why that holds. And with that assumption this can be easily proved using Baire category theorem instead.
  $endgroup$
  – freakish
  6 hours ago
  
  

  
  
  
  

add a comment | 

8

1

$begingroup$

Let $G$ be a discrete countable group acting on a compact, Hausdorff space $X$. Assume that the action is transitive. Namely, $Gcdot x=X$, for all $xin X$.

Does it follow that $X$ is finite?

I thought that the answer should be yes, as $Gcdot x$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.

general-topology group-theory compactness group-actions

share|cite|improve this question

edited 8 hours ago

Servaes

34.1k44 gold badges4444 silver badges103103 bronze badges

asked 8 hours ago

KikoKiko

55122 silver badges1111 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Notice that under these conditions $G/G_xcong Gcdot x=X$ homeomorphism (see theorem 6.2 in the book "Crossed Products" by Williams). Thus $G/G_x$ is discrete and compact, so finite, and so is $X$.
  $endgroup$
  – Shirly Geffen
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ShirlyGeffen $G/G_x$ need not be homeomorphic to $Gx$. The standard map is a continuous bijection only. There are counterexamples when it is not a homeomorphism. It is true however when $G$ is compact for example (not our case). The theorem most likely assumes something more about $G$ or the action.
  $endgroup$
  – freakish
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @freakish It is a homeomorphism if $Gcdot x$ is locally closed in $X$, which is the case here.
  $endgroup$
  – Shirly Geffen
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ShirlyGeffen the theorem also assumes that $X$ is second countable. Which isn't clear for me why that holds. And with that assumption this can be easily proved using Baire category theorem instead.
  $endgroup$
  – freakish
  6 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $G$ be a discrete countable group acting on a compact, Hausdorff space $X$. Assume that the action is transitive. Namely, $Gcdot x=X$, for all $xin X$.

Does it follow that $X$ is finite?

I thought that the answer should be yes, as $Gcdot x$ is then a discrete compact space, and so must be finite. However, I am not sure anymore that I can claim that it is a discrete space. I'd appreciate any help.

general-topology group-theory compactness group-actions

share|cite|improve this question

edited 8 hours ago

Servaes

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asked 8 hours ago

KikoKiko

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$endgroup$

share|cite|improve this question

edited 8 hours ago

Servaes

34.1k44 gold badges4444 silver badges103103 bronze badges

asked 8 hours ago

KikoKiko

55122 silver badges1111 bronze badges

share|cite|improve this question

edited 8 hours ago

Servaes

34.1k44 gold badges4444 silver badges103103 bronze badges

asked 8 hours ago

KikoKiko

55122 silver badges1111 bronze badges

edited 8 hours ago

Servaes

34.1k44 gold badges4444 silver badges103103 bronze badges

edited 8 hours ago

Servaes

34.1k44 gold badges4444 silver badges103103 bronze badges

asked 8 hours ago

KikoKiko

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asked 8 hours ago

KikoKiko

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2 Answers
2

active

oldest

votes

3

$begingroup$

So generally an image of a discrete space can be any topological space. But in this scenario it works. First of all note that $X$ is countable as an image of a countable set.

Assume that $X$ is not discrete. Then there is a point $x_0in X$ which is not isolated. But since $G$ acts on $X$ transitively and $xmapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable since it is a perfect set (in fact it contains a copy of the Cantor set). For the proof see here. Contradiction.

Since $X$ is discrete and compact then it has to be finite.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

freakishfreakish

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$endgroup$

add a comment | 

0

$begingroup$

Since $G$ is countable and $Gcdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $omega^alphacdot n+1$, where $alpha+1$ is the Cantor-Bendixon rank of $X$, and $ngeq 1$ is the cardinality of $X^(alpha)$.

Now we need to rule out $alphageq 1$. For such cases, note that $G$ can only map $omega^alphacdot m$, $m>0$ to another point $omega^alphacdot m'$, $m'>0$, because every neighbourhood of $omega^alphacdot m$ contains a homeomorphic copy of $omega^alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $alpha=0$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

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add a comment | 

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3

$begingroup$

So generally an image of a discrete space can be any topological space. But in this scenario it works. First of all note that $X$ is countable as an image of a countable set.

Assume that $X$ is not discrete. Then there is a point $x_0in X$ which is not isolated. But since $G$ acts on $X$ transitively and $xmapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable since it is a perfect set (in fact it contains a copy of the Cantor set). For the proof see here. Contradiction.

Since $X$ is discrete and compact then it has to be finite.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

freakishfreakish

14.2k11 gold badge88 silver badges3333 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

So generally an image of a discrete space can be any topological space. But in this scenario it works. First of all note that $X$ is countable as an image of a countable set.

Assume that $X$ is not discrete. Then there is a point $x_0in X$ which is not isolated. But since $G$ acts on $X$ transitively and $xmapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable since it is a perfect set (in fact it contains a copy of the Cantor set). For the proof see here. Contradiction.

Since $X$ is discrete and compact then it has to be finite.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

freakishfreakish

14.2k11 gold badge88 silver badges3333 bronze badges

$endgroup$

add a comment | 

$begingroup$

So generally an image of a discrete space can be any topological space. But in this scenario it works. First of all note that $X$ is countable as an image of a countable set.

Assume that $X$ is not discrete. Then there is a point $x_0in X$ which is not isolated. But since $G$ acts on $X$ transitively and $xmapsto gx$ is a homeomorphism then this shows that no point in $X$ is isolated. But a compact Hausdorff space without isolated points has to be uncountable since it is a perfect set (in fact it contains a copy of the Cantor set). For the proof see here. Contradiction.

Since $X$ is discrete and compact then it has to be finite.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

freakishfreakish

14.2k11 gold badge88 silver badges3333 bronze badges

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

freakishfreakish

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answered 7 hours ago

freakishfreakish

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answered 7 hours ago

freakishfreakish

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0

$begingroup$

Since $G$ is countable and $Gcdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $omega^alphacdot n+1$, where $alpha+1$ is the Cantor-Bendixon rank of $X$, and $ngeq 1$ is the cardinality of $X^(alpha)$.

Now we need to rule out $alphageq 1$. For such cases, note that $G$ can only map $omega^alphacdot m$, $m>0$ to another point $omega^alphacdot m'$, $m'>0$, because every neighbourhood of $omega^alphacdot m$ contains a homeomorphic copy of $omega^alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $alpha=0$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

18.2k22 gold badges1212 silver badges3232 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Since $G$ is countable and $Gcdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $omega^alphacdot n+1$, where $alpha+1$ is the Cantor-Bendixon rank of $X$, and $ngeq 1$ is the cardinality of $X^(alpha)$.

Now we need to rule out $alphageq 1$. For such cases, note that $G$ can only map $omega^alphacdot m$, $m>0$ to another point $omega^alphacdot m'$, $m'>0$, because every neighbourhood of $omega^alphacdot m$ contains a homeomorphic copy of $omega^alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $alpha=0$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

18.2k22 gold badges1212 silver badges3232 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since $G$ is countable and $Gcdot x=X$, we have $X$ is countable, compact, Hausdorff. So by Sierpinski-Mazurkiewicz theorem (this seems like an overkill, but I can't think of any simplifications at the moment), $X$ is homeomorphic to $omega^alphacdot n+1$, where $alpha+1$ is the Cantor-Bendixon rank of $X$, and $ngeq 1$ is the cardinality of $X^(alpha)$.

Now we need to rule out $alphageq 1$. For such cases, note that $G$ can only map $omega^alphacdot m$, $m>0$ to another point $omega^alphacdot m'$, $m'>0$, because every neighbourhood of $omega^alphacdot m$ contains a homeomorphic copy of $omega^alpha+1$. So the action of the homeomorphism group (hence $G$) is not transitive unless $alpha=0$.

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

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$endgroup$

share|cite|improve this answer

answered 7 hours ago

user10354138user10354138

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answered 7 hours ago

user10354138user10354138

18.2k22 gold badges1212 silver badges3232 bronze badges

answered 7 hours ago

user10354138user10354138

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