Timeline for Are complex semisimple Lie groups matrix groups?
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| Feb 26, 2010 at 11:24 | comment | added | BCnrd | Once you have the equivalence over C in the connected case, it comes for free in the case of finite component groups, since the "algebraicity" is obtained by translation from the identity. So everything I said above holds with connnectedness relaxed to "finite component group". And Hoshchild's book proves in any Lie group with finite component group that maximal compacts meet every connected component, every compact is contained in one, and they're all conjugate. Thus, you get the compact real form from the connected case as well. | |
| Feb 26, 2010 at 10:53 | comment | added | Faisal | Thanks for the answer, Brian. Unfortunately it's not quite what I'm looking for: I was mainly wondering about what happens for disconnected groups. I'm also curious about the availability of a ``nice'' compact real form for such groups. | |
| Feb 26, 2010 at 10:02 | history | answered | BCnrd | CC BY-SA 2.5 |