# 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
    $[h_{i},h_{j}]=0,$
    $[e_{i},f_{i}]=h_{i},\,[e_{i},f_{j}]=0,i\neq j$,
    $[h_{i},e_{j}]=\langle \alpha _{i},\alpha _{j}\rangle e_{j},\,[h_{i},f_{j}]=-\langle \alpha _{i},\alpha _{j}\rangle f_{j}$,
    $\operatorname {ad} (e_{i})^{-\langle \alpha _{i},\alpha _{j}\rangle +1}(e_{j})=0,i\neq j$,
    $\operatorname {ad} (f_{i})^{-\langle \alpha _{i},\alpha _{j}\rangle +1}(f_{j})=0,i\neq j$.
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 $[\langle \alpha _{i},\alpha _{j}\rangle ]_{1\leq i,j\leq n}$ is called the Cartan matrix. Thus, with this notion, the theorem states that, given 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 (Serre 1966, Ch. VI, Appendix.) and (Kac 1990, Theorem 1.2.). 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:
* $[h_{i},h_{j}]=0$,
* $[e_{i},f_{i}]=h_{i}$, $[e_{i},f_{j}]=0,i\neq j$,
* $[h_{i},e_{j}]=a_{ij}e_{j},[h_{i},f_{j}]=-a_{ij}f_{j}$.
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=⨁$_{l=0}$$^∞$V$^{⊗l}${\textstyle 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 $a\in T,h\in {\mathfrak {h}},\lambda \in {\mathfrak {h}}^{*}$,
* $\pi (f_{i})a=v_{i}\otimes a,$
* $\pi (h)1=\langle \lambda ,\,h\rangle 1,\pi (h)(v_{j}\otimes a)=-\langle \alpha _{j},h\rangle v_{j}\otimes a+v_{j}\otimes \pi (h)a$, 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.
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).
* ${\widetilde {\mathfrak {n}}}_{+}$ (resp. ${\widetilde {\mathfrak {n}}}_{-}$) is a free Lie algebra generated by the $e_{i}$'s (resp. the $f_{i}$'s).
* As a vector space, ${\widetilde {\mathfrak {g}}}={\widetilde {\mathfrak {n}}}_{+}\bigoplus {\mathfrak {h}}\bigoplus {\widetilde {\mathfrak {n}}}_{-}$.
* ${\widetilde {\mathfrak {n}}}_{+}=\bigoplus _{0\neq \alpha \in Q_{+}}{\widetilde {\mathfrak {g}}}_{\alpha }$ where ${\widetilde {\mathfrak {g}}}_{\alpha }=\{x\in {\widetilde {\mathfrak {g}}}|[h,x]=\alpha (h)x,h\in {\mathfrak {h}}\}$ and, similarly, ${\widetilde {\mathfrak {n}}}_{-}=\bigoplus _{0\neq \alpha \in Q_{+}}{\widetilde {\mathfrak {g}}}_{-\alpha }$.
* (root space decomposition) ${\widetilde {\mathfrak {g}}}=\left(\bigoplus _{0\neq \alpha \in Q_{+}}{\widetilde {\mathfrak {g}}}_{-\alpha }\right)\bigoplus {\mathfrak {h}}\bigoplus \left(\bigoplus _{0\neq \alpha \in Q_{+}}{\widetilde {\mathfrak {g}}}_{\alpha }\right)$.
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) $[{\mathfrak {g}},{\mathfrak {g}}]={\mathfrak {g}}$.