Examples of non-abelian simply connected nilpotent Lie groups.
0
$begingroup$
I am searching for examples of connected locally compact group $G = N rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. Please enlighten me.
P.S. I added the ergodic theory tag because I believe such groups are seen there.
group-theory lie-groups ergodic-theory
asked Dec 13 '18 at 0:16
MamboMambo
335113
$endgroup$
add a comment |
0
$begingroup$
I am searching for examples of connected locally compact group $G = N rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. Please enlighten me.
P.S. I added the ergodic theory tag because I believe such groups are seen there.
group-theory lie-groups ergodic-theory
asked Dec 13 '18 at 0:16
MamboMambo
335113
$endgroup$
1
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17
add a comment |
0
0
0
$begingroup$
I am searching for examples of connected locally compact group $G = N rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. Please enlighten me.
P.S. I added the ergodic theory tag because I believe such groups are seen there.
group-theory lie-groups ergodic-theory
asked Dec 13 '18 at 0:16
MamboMambo
335113
$endgroup$
I am searching for examples of connected locally compact group $G = N rtimes H$, where $N$ is a simply connected nilpotent non-abelian Lie group, $H$ is linear reductive and $H$ operates on $N$ without non-trivial fixed points. Please enlighten me.
P.S. I added the ergodic theory tag because I believe such groups are seen there.
group-theory lie-groups ergodic-theory
group-theory lie-groups ergodic-theory
asked Dec 13 '18 at 0:16
MamboMambo
335113
asked Dec 13 '18 at 0:16
MamboMambo
335113
asked Dec 13 '18 at 0:16
MamboMambo
335113
asked Dec 13 '18 at 0:16
MamboMambo
335113
asked Dec 13 '18 at 0:16
MamboMambo
335113
335113
1
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17
add a comment |
1
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17
1
1
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17
add a comment |
1 Answer
1
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The (real or complex) upper triangular group in size $ge 3$ is a trivial example (with $H$ the diagonal group).
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
$endgroup$
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
|
show 1 more comment
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1 Answer
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1 Answer
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1
$begingroup$
The (real or complex) upper triangular group in size $ge 3$ is a trivial example (with $H$ the diagonal group).
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
$endgroup$
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
|
show 1 more comment
1
$begingroup$
The (real or complex) upper triangular group in size $ge 3$ is a trivial example (with $H$ the diagonal group).
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
$endgroup$
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
|
show 1 more comment
1
1
1
$begingroup$
The (real or complex) upper triangular group in size $ge 3$ is a trivial example (with $H$ the diagonal group).
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
$endgroup$
The (real or complex) upper triangular group in size $ge 3$ is a trivial example (with $H$ the diagonal group).
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
answered Dec 13 '18 at 19:34
YCorYCor
7,778929
7,778929
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
|
show 1 more comment
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
I believe upper uni-triangular group is nilpotent. but not this.
$endgroup$
– Mambo
Dec 13 '18 at 20:39
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Yes: $G$ is the upper triangular group, $N$ is the subgroup of $G$ with diagonal $1$.
$endgroup$
– YCor
Dec 13 '18 at 20:55
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
$begingroup$
Sorry. I don't understand the action of $H$ on $N$.
$endgroup$
– Mambo
Dec 13 '18 at 21:11
1
1
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
$begingroup$
The action of $H$ on $N$ is by conjugation. You have a result of 1st year undergraduate, which tells you that if a group $G$ has two subgroups $H,N$, with $N$ normal, $Hcap N={1}$ and $HN=G$, then $G$ is semidirect product $Nrtimes H$ with $H$ acting on $N$ by conjugation.
$endgroup$
– YCor
Dec 13 '18 at 21:15
1
1
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
$begingroup$
It stabilizes $N$, which does not means it fixes $N$ (which means fix each point of $N$). In this precise case, no element $neq 1$ of $N$ is fixed by $H$, because $H$ equals its own centralizer in the group of invertible matrices.
$endgroup$
– YCor
Dec 13 '18 at 21:41
|
show 1 more comment
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1
$begingroup$
MO crosspost: mathoverflow.net/questions/317573
$endgroup$
– YCor
Dec 13 '18 at 21:17