Connected Lie group for which every connected Lie subgroup is simply connected












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    @user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
    $endgroup$
    – zzy
    Dec 18 '18 at 6:19
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    @user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
    $endgroup$
    – zzy
    Dec 18 '18 at 6:19














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 6:13









zzyzzy

2,6111420




2,6111420












  • $begingroup$
    @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




$begingroup$
@user10354138 For simply connected nilpotent Lie group, there is no torus inside it.
$endgroup$
– zzy
Dec 18 '18 at 6:19










1 Answer
1






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oldest

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0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 9:34

























answered Dec 18 '18 at 9:26









Tsemo AristideTsemo Aristide

58.7k11445




58.7k11445












  • $begingroup$
    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












  • $begingroup$
    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


















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