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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Dec 28, 2015 at 15:08 comment added YCor Some assumptions are also needed to ensure that $\mathrm{Aut}(G)$ is a Lie group ($G$ virtually connected is enough), see the comments in math.stackexchange.com/questions/1589303/…
Dec 28, 2015 at 14:28 history edited Ali Taghavi
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S Feb 10, 2015 at 19:34 history bounty ended CommunityBot
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S Feb 2, 2015 at 18:28 history bounty started Ali Taghavi
S Feb 2, 2015 at 18:28 history notice added Ali Taghavi Authoritative reference needed
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S Jan 25, 2015 at 6:39 history bounty started Ali Taghavi
S Jan 25, 2015 at 6:39 history notice added Ali Taghavi Authoritative reference needed
Jan 20, 2015 at 9:53 comment added Dietrich Burde A reference is here; $G$ should be connected and simply connected to conclude that $Aut(\mathfrak{g})\simeq Aut(G)$.
Jan 20, 2015 at 9:05 comment added Ali Taghavi @DietrichBurde thank you very much for your comment. Could you please give a reference for your last part of your comment. Is it easy to proof?
Jan 19, 2015 at 21:51 comment added Dietrich Burde $Hol(G)$ is always a Lie group, since $G$ and $Aut(G)$ are Lie groups and the action is smooth. The Lie algebra is $\mathfrak{g}\rtimes Der(\mathfrak{g})$.
Jan 19, 2015 at 12:14 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 11:56 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 8:46 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 8:34 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 8:27 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 8:17 history edited Ali Taghavi CC BY-SA 3.0
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Jan 19, 2015 at 8:10 history asked Ali Taghavi CC BY-SA 3.0