Timeline for Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic?
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| Jul 20, 2011 at 18:39 | comment | added | Pierre | oh OK, complexifying preserves Lie algebra isomorphisms so it preserves covers, nice -- that double covers are turned into double covers is a little more interesting (though your argument is fine), indeed the change in $\pi_1$ is interesting. If by "complexifying" you mean going from compact to complex reductive (say from $U_n$ to $GL_n( \mathbb{C}$) then the fundamental group (indeed the homotopy type) is unchanged; but if by "complexifying" you include the passage from $SL_2(\mathbb{R})$ to $SL_2(\mathbb{C})$, then things happen, as you point out. I had never thought of this. | |
| Jul 20, 2011 at 18:01 | history | edited | Allen Knutson | CC BY-SA 3.0 |
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| Jul 20, 2011 at 17:35 | comment | added | André Henriques | I depends how you define "complexification". At the Lie algebra level, it's pretty easy to complexify. At the Lie group level, it's not so clear what you mean in general... and indeed, some real Lie groups cannot be complxified. If the group is algebraic, then of course, you can complexify without problem, and covers (aka Lie algebra isomorphisms) complexify to covers. | |
| Jul 20, 2011 at 16:25 | comment | added | Pierre | oop, is a cover of the complexification). | |
| Jul 20, 2011 at 16:25 | comment | added | Pierre | Is it clear that the "complexification" functor preserves covering maps? (so that the complexification of a cover is | |
| Jul 8, 2011 at 7:34 | vote | accept | Marc Palm | ||
| Jul 7, 2011 at 21:04 | history | answered | André Henriques | CC BY-SA 3.0 |