Exercise 3, § 4

Let $(\mathfrak{g},\mathfrak{h})$ be a split semi-simple Lie algebra, R the corresponding root system, and B a basis of R. For any $\alpha \in B$ (resp. for any pair $(\alpha , \beta )\in B^2$), let $R(\alpha )$ (resp. $R(\alpha , \beta )$) be the closed subset of R formed by the $\pm \alpha$ (resp. the smallest closed subset of R containing $\pm \alpha$ and $\pm \beta$ ). Let $\mathfrak{g}(\alpha )$ (resp. $\mathfrak{g}(\alpha , \beta )$) be the derived algebra of the algebra $\mathfrak{h}+\mathfrak{g}^{R(\alpha)}$ (resp. $\mathfrak{h}+\mathfrak{g}^{R(\alpha ,\beta)}$), cf. §3.

a) Show that $\mathfrak{g}(\alpha ) =kH_{\alpha}\oplus \mathfrak{g}^{\alpha}\oplus \mathfrak{g}^{-\alpha}$; it is isomorphic to $\mathfrak{s}\mathfrak{l}(2, k)$.

b) Show that $\mathfrak{g}(\alpha , \beta )$ is semi-simple, and that it is generated by $\mathfrak{g}(\alpha )$ and $\mathfrak{g}(\beta )$. Its root system can be identified with $R(\alpha , \beta )$.

c) Let $\mathfrak{s}$ be a Lie algebra (not necessarily finite dimensional). For all $\alpha \in B$, let $f_{\alpha}$ be a homomorphism from $\mathfrak{g}(\alpha )$ to $\mathfrak{s}$. Assume that, for any pair $(\alpha , \beta )$, there exists a homomorphism $f_{\alpha \beta}:\mathfrak{g}(\alpha , \beta )\rightarrow \mathfrak{s}$ that extends both $f_{\alpha}$ and $f_{\beta}$. Show that, in that case, there exists a unique homomorphism $f:\mathfrak{g}\rightarrow \mathfrak{s}$ that extends the $f_{\alpha}$. (Use Prop. 4 (i).)