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1

$begingroup$

This question arised from something in Ergodic theory, however this is not necessary to state or answer the question.

Suppose that $G$ is a compact abelian group and $gin G$.

Are the following two equivalent:

1. For every continuous character $chi:Grightarrow S^1$, $chi(g)$ generates a dense subgroup of $chi(G)$.




2. $g$ generates a dense subgroup of $G$.

Clearly $2$ implies $1$, but I can't figure out whether the opposite holds.

For those who are interested in the connection with ergodic theory. Property $2$ means that $(G,R_g)$ is an ergodic system, and I ask whether this is equivalent to property 1 which means that all the factors $(chi(G),R_chi(g))$ are ergodic.

topological-groups ergodic-theory

share|cite|improve this question

asked Mar 23 at 17:25

YankoYanko

8,3492830

$endgroup$

add a comment | 

1

$begingroup$

This question arised from something in Ergodic theory, however this is not necessary to state or answer the question.

Suppose that $G$ is a compact abelian group and $gin G$.

Are the following two equivalent:

1. For every continuous character $chi:Grightarrow S^1$, $chi(g)$ generates a dense subgroup of $chi(G)$.




2. $g$ generates a dense subgroup of $G$.

Clearly $2$ implies $1$, but I can't figure out whether the opposite holds.

For those who are interested in the connection with ergodic theory. Property $2$ means that $(G,R_g)$ is an ergodic system, and I ask whether this is equivalent to property 1 which means that all the factors $(chi(G),R_chi(g))$ are ergodic.

topological-groups ergodic-theory

share|cite|improve this question

asked Mar 23 at 17:25

YankoYanko

8,3492830

$endgroup$

add a comment | 

$begingroup$

This question arised from something in Ergodic theory, however this is not necessary to state or answer the question.

Suppose that $G$ is a compact abelian group and $gin G$.

Are the following two equivalent:

1. For every continuous character $chi:Grightarrow S^1$, $chi(g)$ generates a dense subgroup of $chi(G)$.




2. $g$ generates a dense subgroup of $G$.

Clearly $2$ implies $1$, but I can't figure out whether the opposite holds.

For those who are interested in the connection with ergodic theory. Property $2$ means that $(G,R_g)$ is an ergodic system, and I ask whether this is equivalent to property 1 which means that all the factors $(chi(G),R_chi(g))$ are ergodic.

topological-groups ergodic-theory

share|cite|improve this question

asked Mar 23 at 17:25

YankoYanko

8,3492830

$endgroup$

share|cite|improve this question

asked Mar 23 at 17:25

YankoYanko

8,3492830

share|cite|improve this question

asked Mar 23 at 17:25

YankoYanko

8,3492830

asked Mar 23 at 17:25

YankoYanko

8,3492830

asked Mar 23 at 17:25

YankoYanko

8,3492830

2 Answers
2

active

oldest

votes

1

$begingroup$

Short proof: if 2 fails, the group $G/overlinelangle grangle$ is a nontrivial compact abelian group and hence has a nontrivial continuous character $chi$. Then $chi(g)=1$ generates the trivial subgroup, which is not dense in $chi(G)$, so 1 fails.

share|cite|improve this answer

edited Mar 30 at 13:54

answered Mar 24 at 20:09

YCorYCor

8,5171129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is no different than the answer above is it?
  $endgroup$
  – Yanko
  Mar 25 at 9:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Yanko yes you're right. I didn't notice that there were two answers in your post, and thought a briefer one would be useful, but actually it's the same argument with slightly different wording. (PS beginning of your answer "is non-trivial" should be "is a proper subgroup".)
  $endgroup$
  – YCor
  Mar 30 at 13:54
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The answer is Yes!.

Suppose by contradiction that $1$ holds but $2$ doesn't. Let $H=overlineleft<gright>$, then by assumption $Hleq G$ is a proper subgroup.

Look at $G/H$. This is a non-trivial group and so admits a non-trivial character $chi:G/Hrightarrow S^1$. We may compose with $Grightarrow G/H$ to obtain a character $tildechi:Grightarrow S^1$ which is trivial on $H$. In particular $tildechi(g)$ is trivial, however $tildechi(G) = chi(G/H)$ is a non-trivial subgroup of $S^1$ - contradiction.

This completes the proof.

Edit: I just figured even another answer for that!

Let $sin G$, for all $chiinhat G$ there exists a net $n_alpha$ such that $chi(g^n_alpha)rightarrow chi(s)$.

Using a diagonal argument we can choose the same sub-net $n_beta$ for all characters, also by moving into a sub-sub-net if necessary we may assume that $g^n_betarightarrow hin G$. Since all the characters are continuous we have that $chi(h)=chi(s)$ for all $chi$. Using the fact that characters separates points we have that $h=s$, which means that $g^n_beta$ converges to $s$. In particular $sin overlineleft<gright>$

share|cite|improve this answer

edited Mar 30 at 14:01

answered Mar 23 at 18:56

YankoYanko

8,3492830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should be talking about nets rather than sequences
  $endgroup$
  – mathworker21
  Mar 29 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 I don't understand the comment. I mean I know what's a net but I don't see why it is relevant here.
  $endgroup$
  – Yanko
  Mar 29 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know "there exists a sequence $n_k$ such that $chi(g^n_k) to chi(s)$"?
  $endgroup$
  – mathworker21
  Mar 29 at 11:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 $chi(s)$ lies in the closure of $chi(g^n):ninmathbbN$. You're right that $n_k$ is not necessarily increasing (we can assume it is increasing but not monotonically increasing). Is that your point?
  $endgroup$
  – Yanko
  Mar 29 at 11:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. My point is that in topological spaces, a point $x$ lies in the closure of a set $E$ iff there is a net $x_alpha$ of points in $E$ converging to $x$. I wonder how you could think my object is to the $n_k$'s being increasing, when I brought up nets.
  $endgroup$
  – mathworker21
  Mar 29 at 11:09
  
  

  
  
  
  

 | 
show 2 more comments

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1

$begingroup$

Short proof: if 2 fails, the group $G/overlinelangle grangle$ is a nontrivial compact abelian group and hence has a nontrivial continuous character $chi$. Then $chi(g)=1$ generates the trivial subgroup, which is not dense in $chi(G)$, so 1 fails.

share|cite|improve this answer

edited Mar 30 at 13:54

answered Mar 24 at 20:09

YCorYCor

8,5171129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is no different than the answer above is it?
  $endgroup$
  – Yanko
  Mar 25 at 9:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Yanko yes you're right. I didn't notice that there were two answers in your post, and thought a briefer one would be useful, but actually it's the same argument with slightly different wording. (PS beginning of your answer "is non-trivial" should be "is a proper subgroup".)
  $endgroup$
  – YCor
  Mar 30 at 13:54
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Short proof: if 2 fails, the group $G/overlinelangle grangle$ is a nontrivial compact abelian group and hence has a nontrivial continuous character $chi$. Then $chi(g)=1$ generates the trivial subgroup, which is not dense in $chi(G)$, so 1 fails.

share|cite|improve this answer

edited Mar 30 at 13:54

answered Mar 24 at 20:09

YCorYCor

8,5171129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is no different than the answer above is it?
  $endgroup$
  – Yanko
  Mar 25 at 9:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Yanko yes you're right. I didn't notice that there were two answers in your post, and thought a briefer one would be useful, but actually it's the same argument with slightly different wording. (PS beginning of your answer "is non-trivial" should be "is a proper subgroup".)
  $endgroup$
  – YCor
  Mar 30 at 13:54
  
  

  
  
  
  

add a comment | 

$begingroup$

Short proof: if 2 fails, the group $G/overlinelangle grangle$ is a nontrivial compact abelian group and hence has a nontrivial continuous character $chi$. Then $chi(g)=1$ generates the trivial subgroup, which is not dense in $chi(G)$, so 1 fails.

share|cite|improve this answer

edited Mar 30 at 13:54

answered Mar 24 at 20:09

YCorYCor

8,5171129

$endgroup$

share|cite|improve this answer

edited Mar 30 at 13:54

answered Mar 24 at 20:09

YCorYCor

8,5171129

answered Mar 24 at 20:09

YCorYCor

8,5171129

answered Mar 24 at 20:09

YCorYCor

8,5171129

1

$begingroup$

The answer is Yes!.

Suppose by contradiction that $1$ holds but $2$ doesn't. Let $H=overlineleft<gright>$, then by assumption $Hleq G$ is a proper subgroup.

Look at $G/H$. This is a non-trivial group and so admits a non-trivial character $chi:G/Hrightarrow S^1$. We may compose with $Grightarrow G/H$ to obtain a character $tildechi:Grightarrow S^1$ which is trivial on $H$. In particular $tildechi(g)$ is trivial, however $tildechi(G) = chi(G/H)$ is a non-trivial subgroup of $S^1$ - contradiction.

This completes the proof.

Edit: I just figured even another answer for that!

Let $sin G$, for all $chiinhat G$ there exists a net $n_alpha$ such that $chi(g^n_alpha)rightarrow chi(s)$.

Using a diagonal argument we can choose the same sub-net $n_beta$ for all characters, also by moving into a sub-sub-net if necessary we may assume that $g^n_betarightarrow hin G$. Since all the characters are continuous we have that $chi(h)=chi(s)$ for all $chi$. Using the fact that characters separates points we have that $h=s$, which means that $g^n_beta$ converges to $s$. In particular $sin overlineleft<gright>$

share|cite|improve this answer

edited Mar 30 at 14:01

answered Mar 23 at 18:56

YankoYanko

8,3492830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should be talking about nets rather than sequences
  $endgroup$
  – mathworker21
  Mar 29 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 I don't understand the comment. I mean I know what's a net but I don't see why it is relevant here.
  $endgroup$
  – Yanko
  Mar 29 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know "there exists a sequence $n_k$ such that $chi(g^n_k) to chi(s)$"?
  $endgroup$
  – mathworker21
  Mar 29 at 11:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 $chi(s)$ lies in the closure of $chi(g^n):ninmathbbN$. You're right that $n_k$ is not necessarily increasing (we can assume it is increasing but not monotonically increasing). Is that your point?
  $endgroup$
  – Yanko
  Mar 29 at 11:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. My point is that in topological spaces, a point $x$ lies in the closure of a set $E$ iff there is a net $x_alpha$ of points in $E$ converging to $x$. I wonder how you could think my object is to the $n_k$'s being increasing, when I brought up nets.
  $endgroup$
  – mathworker21
  Mar 29 at 11:09
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

The answer is Yes!.

Suppose by contradiction that $1$ holds but $2$ doesn't. Let $H=overlineleft<gright>$, then by assumption $Hleq G$ is a proper subgroup.

Look at $G/H$. This is a non-trivial group and so admits a non-trivial character $chi:G/Hrightarrow S^1$. We may compose with $Grightarrow G/H$ to obtain a character $tildechi:Grightarrow S^1$ which is trivial on $H$. In particular $tildechi(g)$ is trivial, however $tildechi(G) = chi(G/H)$ is a non-trivial subgroup of $S^1$ - contradiction.

This completes the proof.

Edit: I just figured even another answer for that!

Let $sin G$, for all $chiinhat G$ there exists a net $n_alpha$ such that $chi(g^n_alpha)rightarrow chi(s)$.

Using a diagonal argument we can choose the same sub-net $n_beta$ for all characters, also by moving into a sub-sub-net if necessary we may assume that $g^n_betarightarrow hin G$. Since all the characters are continuous we have that $chi(h)=chi(s)$ for all $chi$. Using the fact that characters separates points we have that $h=s$, which means that $g^n_beta$ converges to $s$. In particular $sin overlineleft<gright>$

share|cite|improve this answer

edited Mar 30 at 14:01

answered Mar 23 at 18:56

YankoYanko

8,3492830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should be talking about nets rather than sequences
  $endgroup$
  – mathworker21
  Mar 29 at 3:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 I don't understand the comment. I mean I know what's a net but I don't see why it is relevant here.
  $endgroup$
  – Yanko
  Mar 29 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know "there exists a sequence $n_k$ such that $chi(g^n_k) to chi(s)$"?
  $endgroup$
  – mathworker21
  Mar 29 at 11:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathworker21 $chi(s)$ lies in the closure of $chi(g^n):ninmathbbN$. You're right that $n_k$ is not necessarily increasing (we can assume it is increasing but not monotonically increasing). Is that your point?
  $endgroup$
  – Yanko
  Mar 29 at 11:08
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. My point is that in topological spaces, a point $x$ lies in the closure of a set $E$ iff there is a net $x_alpha$ of points in $E$ converging to $x$. I wonder how you could think my object is to the $n_k$'s being increasing, when I brought up nets.
  $endgroup$
  – mathworker21
  Mar 29 at 11:09
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

The answer is Yes!.

Suppose by contradiction that $1$ holds but $2$ doesn't. Let $H=overlineleft<gright>$, then by assumption $Hleq G$ is a proper subgroup.

Look at $G/H$. This is a non-trivial group and so admits a non-trivial character $chi:G/Hrightarrow S^1$. We may compose with $Grightarrow G/H$ to obtain a character $tildechi:Grightarrow S^1$ which is trivial on $H$. In particular $tildechi(g)$ is trivial, however $tildechi(G) = chi(G/H)$ is a non-trivial subgroup of $S^1$ - contradiction.

This completes the proof.

Edit: I just figured even another answer for that!

Let $sin G$, for all $chiinhat G$ there exists a net $n_alpha$ such that $chi(g^n_alpha)rightarrow chi(s)$.

Using a diagonal argument we can choose the same sub-net $n_beta$ for all characters, also by moving into a sub-sub-net if necessary we may assume that $g^n_betarightarrow hin G$. Since all the characters are continuous we have that $chi(h)=chi(s)$ for all $chi$. Using the fact that characters separates points we have that $h=s$, which means that $g^n_beta$ converges to $s$. In particular $sin overlineleft<gright>$

share|cite|improve this answer

edited Mar 30 at 14:01

answered Mar 23 at 18:56

YankoYanko

8,3492830

$endgroup$

share|cite|improve this answer

edited Mar 30 at 14:01

answered Mar 23 at 18:56

YankoYanko

8,3492830

answered Mar 23 at 18:56

YankoYanko

8,3492830

answered Mar 23 at 18:56

YankoYanko

8,3492830