Connected Lie group for which every connected Lie subgroup is simply connected
$begingroup$
Let $G$ be a simply connected Lie group. If $G$ is nilpotent, we know every connected Lie subgroup of $G$ is simply connected by Baker-Campbell-Hausdorff formula. What about the converse? If every connected Lie subgroup $G$ is simply connected, then is $G$ nilpotent? If not, can we classify them?
algebraic-topology lie-groups nilpotent-groups
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
$endgroup$
add a comment |
$begingroup$
Let $G$ be a simply connected Lie group. If $G$ is nilpotent, we know every connected Lie subgroup of $G$ is simply connected by Baker-Campbell-Hausdorff formula. What about the converse? If every connected Lie subgroup $G$ is simply connected, then is $G$ nilpotent? If not, can we classify them?
algebraic-topology lie-groups nilpotent-groups
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
$endgroup$
$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19
add a comment |
$begingroup$
Let $G$ be a simply connected Lie group. If $G$ is nilpotent, we know every connected Lie subgroup of $G$ is simply connected by Baker-Campbell-Hausdorff formula. What about the converse? If every connected Lie subgroup $G$ is simply connected, then is $G$ nilpotent? If not, can we classify them?
algebraic-topology lie-groups nilpotent-groups
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
$endgroup$
Let $G$ be a simply connected Lie group. If $G$ is nilpotent, we know every connected Lie subgroup of $G$ is simply connected by Baker-Campbell-Hausdorff formula. What about the converse? If every connected Lie subgroup $G$ is simply connected, then is $G$ nilpotent? If not, can we classify them?
algebraic-topology lie-groups nilpotent-groups
algebraic-topology lie-groups nilpotent-groups
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
asked Dec 18 '18 at 6:13
zzyzzy
2,6171420
2,6171420
$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19
add a comment |
$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19
$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19
$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19
add a comment |
1 Answer
1
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$begingroup$
Consider the group of matrices $pmatrix{a &bcr 0&c}, a,c>0$ every connected subgroup of this group is simply connected but this group is solvable and not nilpotent.
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
Consider the group of matrices $pmatrix{a &bcr 0&c}, a,c>0$ every connected subgroup of this group is simply connected but this group is solvable and not nilpotent.
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
add a comment |
$begingroup$
Consider the group of matrices $pmatrix{a &bcr 0&c}, a,c>0$ every connected subgroup of this group is simply connected but this group is solvable and not nilpotent.
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
add a comment |
$begingroup$
Consider the group of matrices $pmatrix{a &bcr 0&c}, a,c>0$ every connected subgroup of this group is simply connected but this group is solvable and not nilpotent.
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
$endgroup$
Consider the group of matrices $pmatrix{a &bcr 0&c}, a,c>0$ every connected subgroup of this group is simply connected but this group is solvable and not nilpotent.
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
edited Dec 18 '18 at 9:34
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
answered Dec 18 '18 at 9:26
Tsemo AristideTsemo Aristide
58.4k11445
58.4k11445
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
add a comment |
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
Thank you! Is there any classification? Must it be solvable? And you use that $mathbb R_{>0}$ is simply connected which does not appear over $mathbb C$.
$endgroup$
– zzy
Dec 18 '18 at 16:15
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
I believe that all connected subgroups of the universal cover of $Sl(2,mathbb{R})$ are simply connected.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 18:31
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
$begingroup$
How do you see that?
$endgroup$
– zzy
Dec 19 '18 at 23:35
add a comment |
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$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19