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 $\$ of $\Phi$, the Lie algebra $\mathfrak g$ defined by (1) $3n$ generators $e_i, f_i, h_i$ and (2) the relations
: _i, h_j= 0,
: _i, f_i= h_i, \, _i, f_j= 0, i \ne j,
: _i, e_j= \langle \alpha_i, \alpha_j \rangle e_j, \, _i, f_j= -\langle \alpha_i, \alpha_j \rangle f_j,
:$\operatorname(e_i)^(e_j) = 0, i \ne j$,
:$\operatorname(f_i)^(f_j) = 0, i \ne j$.
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 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 $a_ = \langle \alpha_i, \alpha_j \rangle$ and then let $\widetilde$ be the Lie algebra generated by (1) the generators $e_i, f_i, h_i$ and (2) the relations:
* _i, h_j= 0,
* _i, f_i= h_i, _i, f_j= 0, i \ne j,
* _i, e_j= a_ e_j, _i, f_j= -a_ f_j.

Let $\mathfrak$ be the free vector space spanned by $h_i$, ''V'' the free vector space with a basis $v_1, \dots, v_n$ and $T = \bigoplus_^ V^$ the tensor algebra over it. Consider the following representation of a Lie algebra:
:$\pi : \widetilde \to \mathfrak(T)$
given by: for $a \in T, h \in \mathfrak, \lambda \in \mathfrak^*$,
*$\pi(f_i)a = v_i \otimes a,$
*$\pi(h)1 = \langle \lambda, \, h \rangle 1, \pi(h)(v_j \otimes a) = -\langle \alpha_j, h \rangle v_j \otimes a + v_j \otimes \pi(h)a$, inductively,
*$\pi(e_i)1 = 0, \, \pi(e_i)(v_j \otimes a) = \delta_ \alpha_i(a) + v_j \otimes \pi(e_i)a$, 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 $\widetilde_+$ (resp. $\widetilde_-$) the subalgebras of $\widetilde$ generated by the $e_i$'s (resp. the $f_i$'s).
*$\widetilde_+$ (resp. $\widetilde_-$) is a free Lie algebra generated by the $e_i$'s (resp. the $f_i$'s).
*As a vector space, $\widetilde = \widetilde_+ \bigoplus \mathfrak \bigoplus \widetilde_-$.
*$\widetilde_+ = \bigoplus_ \widetilde_$ where $\widetilde_ = \$ and, similarly, $\widetilde_- = \bigoplus_ \widetilde_$.
*(root space decomposition) $\widetilde = \left( \bigoplus_ \widetilde_ \right) \bigoplus \mathfrak h \bigoplus \left( \bigoplus_ \widetilde_ \right)$.

For each ideal $\mathfrak i$ of $\widetilde$, one can easily show that $\mathfrak i$ is homogeneous with respect to the grading given by the root space decomposition; i.e., $\mathfrak i = \bigoplus_ (\widetilde_ \cap \mathfrak i)$. 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: $\mathfrak r = (\mathfrak r \cap \widetilde_-) \oplus (\mathfrak r \cap \widetilde_+)$. In fact, it is a $\widetilde$-module decomposition. Let
:$\mathfrak g = \widetilde/\mathfrak r$.
Then it contains a copy of $\mathfrak h$, which is identified with $\mathfrak h$ and
:$\mathfrak g = \mathfrak_+ \bigoplus \mathfrak \bigoplus \mathfrak_-$
where $\mathfrak_+$ (resp. $\mathfrak_-$) 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/math> 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.

References

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Theorems about algebras
Lie algebras