Is every connected semisimple linear Lie group the identity connected component of (the real points of) an algebraic group?
I was told some fact along this line is true but could not find any reference after searching for a while.
$\begingroup$
$\endgroup$
7
-
4$\begingroup$ 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 $H(\mathbf{R})$. $\endgroup$– YCorCommented Jul 26, 2019 at 22:52
-
4$\begingroup$ 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. $\endgroup$– YCorCommented Jul 27, 2019 at 10:52
-
1$\begingroup$ I am pretty sure there is an easy proof in Onishchik and Vinberg, Lie Groups and Algebraic Groups. $\endgroup$– Ben McKayCommented Jul 28, 2019 at 7:27
-
5$\begingroup$ 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 $G$. $\endgroup$– VenkataramanaCommented Jul 28, 2019 at 7:35
-
1$\begingroup$ 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. $\endgroup$– Jim HumphreysCommented Aug 2, 2019 at 19:41
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
(Comments converted into an answer:)
Mostow (to whom I think this is often attributed?) gives a detailed proof in (1949, Lemmas 2.2, 2.3).
Borel (2001, pp. 152, 114) notes that algebraicity of perfect (e.g. semisimple) linear Lie algebras was claimed by Cartan in (1897, p. 547), and spells out what “it may not too far fetched to believe” would have been his simple proof.
answered Aug 2, 2019 at 9:10