# Lie Algebra of a connected simple linear algebraic group

Let $G$ be a linear algebraic group and $A=K[G]$ (K is a field of characterstic 0) be the coordinate ring of $G$. In Humphreys, the Lie algebra of $G$ is defined as the space of left invariant derivations of $A$ , i.e, $\mathfrak{L}(G) = \{\delta \in Der \hspace{1mm} A$ | $\delta \lambda_x = \lambda_x \delta$ for all $x \in G$ }

Note that $\lambda_x : K[G] \to K[G]$ sending $\hspace{1mm} f(y) \mapsto f(x^{-1}y)$.

1. Now, if we are given a connected simple algebraic group over $C$ (i.e , the defining polynomial equations of the underlying variety are over $C$), how do we see that its Lie algebra is simple ?

2. Further, given a simple Lie algebra how do we see that there exists a simply connected algebraic group $G$ with Lie algebra isomorphic to this given simple Lie algebra?

Please give reference to specific material which I can read to understand these claims. Any hints to the above claims would be highly appreciated. Thank you !

• Note that part 2 uses in an essential way what the base field is, as there are simple Lie algebras in positive characteristic which do not come from any algebraic groups. – Tobias Kildetoft Jan 20 '16 at 19:00
• Part 1 is also false in characteristic $p$, e.g., for $G=\mathrm{PGL}_p$. – YCor Jan 22 '16 at 17:54
• @TobiasKildetoft Is the simply connected algebraic group $G$ claimed to exist as in point 2 of the question defined over $C$ ? – Jagdeep Singh Feb 18 '16 at 8:50
• @YCor Is the simply connected algebraic group $G$ claimed to exist as in point 2 of the question defined over $C$ ? – Jagdeep Singh Feb 18 '16 at 8:50
• As far as I recall, this holds in general for reductive groups, and the part about being simply connected is not important for this particular claim (i.e. being defined over $\mathbb{Z}$). – Tobias Kildetoft Feb 18 '16 at 9:12

## 1 Answer

Among the possible references are Milne's lectures notes. In Proposition $4.1$ it is proved that a connected algebraic group $G$ is semisimple if and only if its Lie algebra is semisimple. Then 1. follows together with Theorem $4.5$. Theorem $4.22$ shows the claim 2, for fields of characteristic zero.