Timeline for When is the Ad (Adjoint Representation) Morphism a Closed Map
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 5, 2014 at 1:10 | vote | accept | Selene Routley | ||
| Oct 4, 2014 at 12:58 | answer | added | YCor | timeline score: 5 | |
| Oct 1, 2014 at 12:41 | comment | added | user27920 | Yes, sorry, my edit to fix my oversight about passing to the universal cover crossed with your comment. | |
| Oct 1, 2014 at 12:40 | comment | added | YCor | Yes I had in mind connected Lie groups and the case of virtually connected Lie groups is analogous (although the assumption would then be that $\mathrm{Hom}(G^\circ,\mathbf{R})=0$). | |
| Oct 1, 2014 at 12:37 | comment | added | user27920 | @YCor: Probably you intend to assume $G$ has finite component group (or else infinite torsion abelian groups provide counterexamples), and then you may as well take $G$ to be connected. If $G$ is simply connected then your hypothesis says exactly that $\mathfrak{g}$ is perfect, so then 7.9 in Borel's textbook applied to any homomorphic image of $\mathfrak{g}$ in $\mathfrak{gl}_n(\mathbf{R})$ or of $\mathfrak{g}_{\mathbf{C}}$ in $\mathfrak{gl}_n(\mathbf{C})$ does the job (by finiteness of $\pi_0(H(\mathbf{R}))$ for linear algebraic $H$ over $\mathbf{R}$ for $\mathbf{R}$-representations). | |
| Oct 1, 2014 at 8:30 | comment | added | YCor | I'd first like to find a proof of my "guess" before posting an answer. | |
| Oct 1, 2014 at 8:15 | comment | added | Selene Routley | @user52824 Please see my new edit and my comment to Yves above. | |
| Oct 1, 2014 at 8:14 | comment | added | Selene Routley | @YCor Yves, please see my edit at the bottom of the question. If you want to expand your comment into an answer I think that will be helpful. Otherwise I'll be gathering all the helpful comments into a Community Wiki answer. | |
| Oct 1, 2014 at 8:13 | history | edited | Selene Routley | CC BY-SA 3.0 |
added 351 characters in body
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| Oct 1, 2014 at 8:03 | comment | added | YCor | I guess that if a Lie group $G$ satisfies $\mathrm{Hom}(G,\mathbf{R})=0$, then every linear representation of $G$ has a closed image (this would be optimal since otherwise $G$ admits a 2-dimensional complex rep with non-closed image). | |
| Oct 1, 2014 at 1:41 | comment | added | user27920 | @WetSavannaAnimalakaRodVance: Yeah, I have no idea how to prove closedness directly, and in the context of differential geometry it seems rather delicate issue how to tell that a connected Lie subgroup is closed just from the data of the corresponding Lie subalgebra. The approach through algebraic geometry provides a genuinely extra technique which I have no idea how to replace from a purely $C^{\infty}$ point of view. The relation between Lie groups and matrix groups is quite essential to the theory beyond the compact case. | |
| Oct 1, 2014 at 0:38 | comment | added | Selene Routley | @user52824 So it seems that it's probably not just my overlooking something trivial. I had a gut feeling one had to assume semisimplicity (which of course follows by Cartan's criterion from the nondegeneracy of the Killing form) or something along those lines, but no idea how to prove it. | |
| Sep 30, 2014 at 23:56 | comment | added | user27920 | Two things to add to YCor's comment: (1) since that example has trivial center, it shows the author's claim #1 is false, (2) in practice, a very natural criterion which ensures closedness is that $\mathfrak{g}$ is semisimple, but the only proof I am aware of for closedness in that case is via non-trivial theorems from the theory of linear algebraic groups. | |
| Sep 30, 2014 at 18:31 | comment | added | YCor | Here's an example in which it's not closed. Let $q$ be irrational, and consider the 5-dimensional Lie group $G=\mathbf{C}^2\rtimes\mathbf{R}$ where $t\in\mathbf{R}$ acts by $t\cdot (z,x)=(e^{it}z,e^{iqt}w)$. Then the adjoint map has a non-closed image. To see this, embed the latter as a dense normal Lie subgroup in $G'=(\mathbf{C}\rtimes U(1))^2$ and restrict the adjoint representation of $G'$ to a faithful rep of $G'$ on the Lie algebra of $G$. | |
| Sep 30, 2014 at 14:22 | history | asked | Selene Routley | CC BY-SA 3.0 |