# Is every connected semisimple linear Lie group the connected component of (the real points of) an algebraic group?

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.

• 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 $H(\mathbf{R})$. – YCor Jul 26 at 22:52
• @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 $GL_n$? – Jerry Jul 27 at 1:43
• 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. – YCor Jul 27 at 10:52
• I am pretty sure there is an easy proof in Onishchik and Vinberg, Lie Groups and Algebraic Groups. – Ben McKay Jul 28 at 7:27
• 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 $G$. – Venkataramana Jul 28 at 7:35