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Probability of commutation in a compact group
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It is well known that if $G$ is a finite group, then the probability that two elements commutte is either $1$ (if $G$ is abelian) or less than or equal to $frac58$.
If instead $K$ is a compact group, there exists a unique probability over $K$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $K$ may have several connected components, being a (semi-)direct product of its neutral component $K_0$ with a finite group. If $K_0$ is abelian (a torus), then we are led back to the finite-case result.
So let me assume that $K=K_0$, that is $K$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $0<p<1$ ?
pr.probability lie-groups
asked 1 hour ago
Denis Serre
28.5k791191
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It is well known that if $G$ is a finite group, then the probability that two elements commutte is either $1$ (if $G$ is abelian) or less than or equal to $frac58$.
If instead $K$ is a compact group, there exists a unique probability over $K$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $K$ may have several connected components, being a (semi-)direct product of its neutral component $K_0$ with a finite group. If $K_0$ is abelian (a torus), then we are led back to the finite-case result.
So let me assume that $K=K_0$, that is $K$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $0<p<1$ ?
pr.probability lie-groups
asked 1 hour ago
Denis Serre
28.5k791191
About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago
add a comment |Â
up vote
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up vote
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It is well known that if $G$ is a finite group, then the probability that two elements commutte is either $1$ (if $G$ is abelian) or less than or equal to $frac58$.
If instead $K$ is a compact group, there exists a unique probability over $K$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $K$ may have several connected components, being a (semi-)direct product of its neutral component $K_0$ with a finite group. If $K_0$ is abelian (a torus), then we are led back to the finite-case result.
So let me assume that $K=K_0$, that is $K$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $0<p<1$ ?
pr.probability lie-groups
asked 1 hour ago
Denis Serre
28.5k791191
It is well known that if $G$ is a finite group, then the probability that two elements commutte is either $1$ (if $G$ is abelian) or less than or equal to $frac58$.
If instead $K$ is a compact group, there exists a unique probability over $K$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $K$ may have several connected components, being a (semi-)direct product of its neutral component $K_0$ with a finite group. If $K_0$ is abelian (a torus), then we are led back to the finite-case result.
So let me assume that $K=K_0$, that is $K$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $0<p<1$ ?
pr.probability lie-groups
pr.probability lie-groups
asked 1 hour ago
Denis Serre
28.5k791191
asked 1 hour ago
Denis Serre
28.5k791191
asked 1 hour ago
Denis Serre
28.5k791191
asked 1 hour ago
Denis Serre
28.5k791191
asked 1 hour ago
Denis Serre
28.5k791191
28.5k791191
About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago
add a comment |Â
About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago
About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago
About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago
add a comment |Â
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No: if $K$ is a compact Lie group, the commuting probability is positive iff $K_0$ is abelian. As you noticed, $Leftarrow$ is trivial.
Conversely, assume that $K_0$ is not abelian. We can view $K$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $2dim K$), then it contains a product of two cosets: $aK_0times bK_0$ for some $a,bin K$. So $agbh=bhag$ for all $g,hin K_0$.
Putting $g,h=1$, we get $ab=ba$. Putting $h=1$, we get $agb=bag=1$ for all $h$ which using $ab=ba$ yields $gb=bg$ for all $gin K_0$. Putting $g=1$, we get $abh=bha$ for all $h$, which using $ab=ba$ yields $ah=ha$ for all $hin K_0$. Thus $a,b$ commute and centralize $K_0$. So the formula simplifies as $hg=gh$ for all $g,hin K_0$.
answered 48 mins ago
YCor
25.7k277121
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No: if $K$ is a compact Lie group, the commuting probability is positive iff $K_0$ is abelian. As you noticed, $Leftarrow$ is trivial.
Conversely, assume that $K_0$ is not abelian. We can view $K$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $2dim K$), then it contains a product of two cosets: $aK_0times bK_0$ for some $a,bin K$. So $agbh=bhag$ for all $g,hin K_0$.
Putting $g,h=1$, we get $ab=ba$. Putting $h=1$, we get $agb=bag=1$ for all $h$ which using $ab=ba$ yields $gb=bg$ for all $gin K_0$. Putting $g=1$, we get $abh=bha$ for all $h$, which using $ab=ba$ yields $ah=ha$ for all $hin K_0$. Thus $a,b$ commute and centralize $K_0$. So the formula simplifies as $hg=gh$ for all $g,hin K_0$.
answered 48 mins ago
YCor
25.7k277121
add a comment |Â
up vote
4
down vote
No: if $K$ is a compact Lie group, the commuting probability is positive iff $K_0$ is abelian. As you noticed, $Leftarrow$ is trivial.
Conversely, assume that $K_0$ is not abelian. We can view $K$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $2dim K$), then it contains a product of two cosets: $aK_0times bK_0$ for some $a,bin K$. So $agbh=bhag$ for all $g,hin K_0$.
Putting $g,h=1$, we get $ab=ba$. Putting $h=1$, we get $agb=bag=1$ for all $h$ which using $ab=ba$ yields $gb=bg$ for all $gin K_0$. Putting $g=1$, we get $abh=bha$ for all $h$, which using $ab=ba$ yields $ah=ha$ for all $hin K_0$. Thus $a,b$ commute and centralize $K_0$. So the formula simplifies as $hg=gh$ for all $g,hin K_0$.
answered 48 mins ago
YCor
25.7k277121
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No: if $K$ is a compact Lie group, the commuting probability is positive iff $K_0$ is abelian. As you noticed, $Leftarrow$ is trivial.
Conversely, assume that $K_0$ is not abelian. We can view $K$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $2dim K$), then it contains a product of two cosets: $aK_0times bK_0$ for some $a,bin K$. So $agbh=bhag$ for all $g,hin K_0$.
Putting $g,h=1$, we get $ab=ba$. Putting $h=1$, we get $agb=bag=1$ for all $h$ which using $ab=ba$ yields $gb=bg$ for all $gin K_0$. Putting $g=1$, we get $abh=bha$ for all $h$, which using $ab=ba$ yields $ah=ha$ for all $hin K_0$. Thus $a,b$ commute and centralize $K_0$. So the formula simplifies as $hg=gh$ for all $g,hin K_0$.
answered 48 mins ago
YCor
25.7k277121
No: if $K$ is a compact Lie group, the commuting probability is positive iff $K_0$ is abelian. As you noticed, $Leftarrow$ is trivial.
Conversely, assume that $K_0$ is not abelian. We can view $K$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $2dim K$), then it contains a product of two cosets: $aK_0times bK_0$ for some $a,bin K$. So $agbh=bhag$ for all $g,hin K_0$.
Putting $g,h=1$, we get $ab=ba$. Putting $h=1$, we get $agb=bag=1$ for all $h$ which using $ab=ba$ yields $gb=bg$ for all $gin K_0$. Putting $g=1$, we get $abh=bha$ for all $h$, which using $ab=ba$ yields $ah=ha$ for all $hin K_0$. Thus $a,b$ commute and centralize $K_0$. So the formula simplifies as $hg=gh$ for all $g,hin K_0$.
answered 48 mins ago
YCor
25.7k277121
answered 48 mins ago
YCor
25.7k277121
answered 48 mins ago
YCor
25.7k277121
answered 48 mins ago
YCor
25.7k277121
25.7k277121
add a comment |Â
add a comment |Â
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About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
1 hour ago