Serre's theorem on a semisimple Lie algebra

In abstract algebra, specifically the theory of Lie algebras, Serre's theorem states: given a (finite reduced) root system $$\Phi$$, there exists a finite-dimensional semisimple Lie algebra whose root system is the given $$\Phi$$.

Statement
The theorem states that: given a root system $$\Phi$$ in a Euclidean space with an inner product $$$$, $$\langle \beta, \alpha \rangle = 2(\alpha, \beta)/(\alpha, \alpha), \beta, \alpha \in E$$ and a base $$\{ \alpha_1, \dots, \alpha_n \}$$ of $$\Phi$$, the Lie algebra $$\mathfrak g$$ defined by (1) $$3n$$ generators $$e_i, f_i, h_i$$ and (2) the relations
 * $$[h_i, h_j] = 0,$$
 * $$[e_i, f_i] = h_i, \, [e_i, f_j] = 0, i \ne j$$,
 * $$[h_i, e_j] = \langle \alpha_i, \alpha_j \rangle e_j, \, [h_i, f_j] = -\langle \alpha_i, \alpha_j \rangle f_j$$,
 * $$\operatorname{ad}(e_i)^{-\langle \alpha_i, \alpha_j \rangle+1}(e_j) = 0, i \ne j$$,
 * $$\operatorname{ad}(f_i)^{-\langle \alpha_i, \alpha_j \rangle+1}(f_j) = 0, i \ne j$$.

is a finite-dimensional semisimple Lie algebra with the Cartan subalgebra generated by $$h_i$$'s and with the root system $$\Phi$$.

The square matrix $$[\langle \alpha_i, \alpha_j \rangle]_{1 \le i, j \le n}$$ is called the Cartan matrix. Thus, with this notion, the theorem states that, give a Cartan matrix A, there exists a unique (up to an isomorphism) finite-dimensional semisimple Lie algebra $$\mathfrak g(A)$$ associated to $$A$$. The construction of a semisimple Lie algebra from a Cartan matrix can be generalized by weakening the definition of a Cartan matrix. The (generally infinite-dimensional) Lie algebra associated to a generalized Cartan matrix is called a Kac–Moody algebra.

Sketch of proof
The proof here is taken from and. Let $$a_{ij} = \langle \alpha_i, \alpha_j \rangle$$ and then let $$\widetilde{\mathfrak g}$$ be the Lie algebra generated by (1) the generators $$e_i, f_i, h_i$$ and (2) the relations:
 * $$[h_i, h_j] = 0$$,
 * $$[e_i, f_i] = h_i$$, $$[e_i, f_j] = 0, i \ne j$$,
 * $$[h_i, e_j] = a_{ij} e_j, [h_i, f_j] = -a_{ij} f_j$$.

Let $$\mathfrak{h}$$ be the free vector space spanned by $$h_i$$, V the free vector space with a basis $$v_1, \dots, v_n$$ and $T = \bigoplus_{l=0}^{\infty} V^{\otimes l}$ the tensor algebra over it. Consider the following representation of a Lie algebra:
 * $$\pi : \widetilde{\mathfrak g} \to \mathfrak{gl}(T)$$

given by: for $$a \in T, h \in \mathfrak{h}, \lambda \in \mathfrak{h}^*$$, It is not trivial that this is indeed a well-defined representation and that has to be checked by hand. From this representation, one deduces the following properties: let $$\widetilde{\mathfrak{n}}_+$$ (resp. $$\widetilde{\mathfrak{n}}_-$$) the subalgebras of $$\widetilde{\mathfrak g}$$ generated by the $$e_i$$'s (resp. the $$f_i$$'s).
 * $$\pi(f_i)a = v_i \otimes a,$$
 * $$\pi(h)1 = \langle \lambda, \, h \rangle 1, \pi(h)(v_j \otimes a) = -\langle \alpha_j, h \rangle v_j \otimes a + v_j \otimes \pi(h)a$$, inductively,
 * $$\pi(e_i)1 = 0, \, \pi(e_i)(v_j \otimes a) = \delta_{ij} \alpha_i(a) + v_j \otimes \pi(e_i)a$$, inductively.
 * $$\widetilde{\mathfrak{n}}_+$$ (resp. $$\widetilde{\mathfrak{n}}_-$$) is a free Lie algebra generated by the $$e_i$$'s (resp. the $$f_i$$'s).
 * As a vector space, $$\widetilde{\mathfrak g} = \widetilde{\mathfrak{n}}_+ \bigoplus \mathfrak{h} \bigoplus \widetilde{\mathfrak{n}}_-$$.
 * $$\widetilde{\mathfrak{n}}_+ = \bigoplus_{0 \ne \alpha \in Q_+} \widetilde{\mathfrak g}_{\alpha}$$ where $$\widetilde{\mathfrak g}_{\alpha} = \{ x \in \widetilde{\mathfrak g}|[h, x] = \alpha(h) x, h \in \mathfrak h \}$$ and, similarly, $$\widetilde{\mathfrak{n}}_- = \bigoplus_{0 \ne \alpha \in Q_+} \widetilde{\mathfrak g}_{-\alpha}$$.
 * (root space decomposition) $$\widetilde{\mathfrak g} = \left( \bigoplus_{0 \ne \alpha \in Q_+} \widetilde{\mathfrak g}_{-\alpha} \right) \bigoplus \mathfrak h \bigoplus \left( \bigoplus_{0 \ne \alpha \in Q_+} \widetilde{\mathfrak g}_{\alpha} \right)$$.

For each ideal $$\mathfrak i$$ of $$\widetilde{\mathfrak g}$$, one can easily show that $$\mathfrak i$$ is homogeneous with respect to the grading given by the root space decomposition; i.e., $$\mathfrak i = \bigoplus_{\alpha} (\widetilde{\mathfrak g}_{\alpha} \cap \mathfrak i)$$. It follows that the sum of ideals intersecting $$\mathfrak h$$ trivially, it itself intersects $$\mathfrak h$$ trivially. Let $$\mathfrak r$$ be the sum of all ideals intersecting $$\mathfrak h$$ trivially. Then there is a vector space decomposition: $$\mathfrak r = (\mathfrak r \cap \widetilde{\mathfrak n}_-) \oplus (\mathfrak r \cap \widetilde{\mathfrak n}_+)$$. In fact, it is a $$\widetilde{\mathfrak g}$$-module decomposition. Let
 * $$\mathfrak g = \widetilde{\mathfrak g}/\mathfrak r$$.

Then it contains a copy of $$\mathfrak h$$, which is identified with $$\mathfrak h$$ and
 * $$\mathfrak g = \mathfrak{n}_+ \bigoplus \mathfrak{h} \bigoplus \mathfrak{n}_-$$

where $$\mathfrak{n}_+$$ (resp. $$\mathfrak{n}_-$$) are the subalgebras generated by the images of $$e_i$$'s (resp. the images of $$f_i$$'s).

One then shows: (1) the derived algebra $$[\mathfrak g, \mathfrak g]$$ here is the same as $$\mathfrak g$$ in the lead, (2) it is finite-dimensional and semisimple and (3) $$[\mathfrak g, \mathfrak g] = \mathfrak g$$.