Timeline for Is every connected semisimple linear Lie group the connected component of (the real points of) an algebraic group?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2019 at 20:12 | vote | accept | Jerry | ||
| Aug 2, 2019 at 19:41 | comment | added | Jim Humphreys | This kind of question points up the need for an article (historically correct) spelling out the precise relationship between (real) Lie groups and linear algebraic groups in the connected semisimple case. There are many fragments in the literature, of course, including Chevalley's proof that compact semisimple groups are algebraic. | |
| Aug 2, 2019 at 9:10 | answer | added | Francois Ziegler | timeline score: 7 | |
| Jul 28, 2019 at 15:46 | comment | added | Victor Petrov | One can show this using Tannaka duality: consider the category of the finite dimensional representations and show that it is Tannakian. | |
| Jul 28, 2019 at 7:35 | comment | added | Venkataramana | The fact that a perfect real lie algebra is algebraic may be deeper. But the Lie algebra $\mathfrak g$ of the linear semi-simple lie group $G \subset GL_n({\mathbb R})$ can be complexified; the latter ${\mathfrak g}_{\mathbb C}$ has a compact real form, which by Weyl's theorem, has an algebraic compact subgroup inside $GL_n({\mathbb C})$, whence its Zariski closure $G({\mathbb C})$ has the same Lie algebra as ${\mathfrak g}_{\mathbb C}$. Consequently, $G({\mathbb R})=G({\mathbb C})\cap GL_n({\mathbb R})$ has the same Lie algebra $\mathfrak g$ and contains $G$. | |
| Jul 28, 2019 at 7:27 | comment | added | Ben McKay | I am pretty sure there is an easy proof in Onishchik and Vinberg, Lie Groups and Algebraic Groups. | |
| Jul 28, 2019 at 6:00 | history | edited | Francois Ziegler | CC BY-SA 4.0 |
Top-level tag added
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| Jul 27, 2019 at 20:18 | history | edited | Jim Humphreys |
edited tags
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| Jul 27, 2019 at 14:24 | history | edited | YCor | CC BY-SA 4.0 |
moved question to text
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| Jul 27, 2019 at 10:52 | comment | added | YCor | Well, look at the notion of "algebraic Lie subalgebra". To be perfect is a sufficient (not necessary) condition for a Lie subalgebra (of the Lie algebra of an algebraic group in char. zero) to be algebraic. | |
| Jul 27, 2019 at 1:43 | comment | added | Jerry | @YCor Thanks. I am still confused: as perfectness is a property about structures inside $\mathfrak g$ itself and does not involves relations between $g\in G$ and $g'\notin G$, how does this give rise to the definition of $H$ inside $GL_n$? | |
| Jul 26, 2019 at 22:52 | comment | added | YCor | Yes, call it $G$ and view it as continuously embedded into $GL_n$. Then its Lie algebra being perfect, it is the Lie algebra of an $\mathbf{R}$-defined subgroup $H$ of $GL_n$. Then $G$ is an open subgroup of $H(\mathbf{R})$. | |
| Jul 26, 2019 at 20:58 | history | asked | Jerry | CC BY-SA 4.0 |