# Topics in Abstract Algebra/Lie algebras

Let $V$ be a vector space. $(V,[,])$ is called a Lie algebra if it is equipped with the bilinear operator $V\times V\to V$ , denoted by $[,]$ , subject to the properties: for every $x,y,z\in V$ • (i) [x, x] = 0
• (ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0

(ii) is called the Jacobi identity.

Example: For $x,y\in \mathbf {R} ^{3}$ , define $[x,y]=x\times y$ , the cross product of $x$ and $y$ . The known properties of the cross products show that $(R^{3},[,])$ is a Lie algebra.

Example: Let $\operatorname {Der} (V)=\{D\in \operatorname {Ext} (V):D(xy)=(Dx)y+xDy\}$ . A member of $\operatorname {Der} (V)$ is called a derivation. Define $[x,y]=xy-yx$ . Then $[x,y]\in \operatorname {Der} (V)$ .

Theorem Let $V$ be a finite-dimensional vector space.

• (i) If ${\mathfrak {g}}\subset {\mathfrak {gl}}_{k}(V)$ is a Lie algebra consisting of nilpotent elements, then there exists $v\in V$ such that $x(v)=0$ for every $x\in {\mathfrak {g}}$ .
• (ii) If ${\mathfrak {g}}$ is solvable, then there exists a common eigenvalue $v\in V$ .

Theorem (Engel) ${\mathfrak {g}}$ is nilpotent if and only if $\operatorname {ad} (x)$ is nilpotent for every $x\in {\mathfrak {g}}$ .
Proof: The direct part is clear. For the converse, note that from the preceding theorem that $\operatorname {ad} ({\mathfrak {g}})$ is a subalgebra of ${\mathfrak {n}}_{k}$ . Thus, $\operatorname {ad} ({\mathfrak {g}})$ is nilpotent and so is ${\mathfrak {g}}$ . $\square$ Theorem ${\mathfrak {g}}$ is solvable if and only if $[{\mathfrak {g}},{\mathfrak {g}}]$ is nilpotent.
Proof: Suppose ${\mathfrak {g}}$ is solvable. Then $\operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]$ is a subalgebra of ${\mathfrak {b}}_{k}$ . Thus, $\operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]\subset {\mathfrak {n}}_{k}$ . Hence, $\operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]$ is nilpotent, and so $[{\mathfrak {g}},{\mathfrak {g}}]$ is nilpotent. For the converse, note the exact sequence:

$0\longrightarrow [{\mathfrak {g}},{\mathfrak {g}}]\longrightarrow {\mathfrak {g}}\longrightarrow {\mathfrak {g}}/{[{\mathfrak {g}},{\mathfrak {g}}]}\longrightarrow 0$ Since both $[{\mathfrak {g}},{\mathfrak {g}}]$ and ${\mathfrak {g}}/[{\mathfrak {g}},{\mathfrak {g}}]$ are solvable, ${\mathfrak {g}}$ is solvable. $\square$ 3 Therorem (Weyl's theorem) Every representation of a finite-dimensional semisimple Lie algebra:

${\mathfrak {g}}\to \operatorname {End} (V)$ is completely reducible.
Proof: It suffices to prove that every ${\mathfrak {g}}$ -submodule has a ${\mathfrak {g}}$ -submodule complement. Furthermore, the proof reduces to the case when $W$ is simple (as a module) and has codimension one. Indeed, given a ${\mathfrak {g}}$ -submodule $W$ , let $E\subset \operatorname {Hom} (V,W)$ be the subspace consisting of elements $f$ such that $f|_{W}$ is a scalar multiplication. Since any commutator of elements $f\in E$ is zero (that is, multiplication by zero), it is clear that $E/[E,E]$ has dimension 1. $E$ may not be simple, but by induction on the dimension of $E$ , we can assume that. Hence, $E$ has complement of dimension 1, which is spanned by, say, $f$ . It follows that $V$ is the direct sum of $W$ and the kernel of $f$ . Now, to complete the proof, let $W$ be a simple ${\mathfrak {g}}$ -submodule of codimension 1. Let $c$ be a Casimir element of ${\mathfrak {g}}\to \operatorname {End} (V)$ . It follows that $V$ is the direct sum of $W$ and the kernel of $c$ . $\square$ (TODO: obviously, the proof is very sketchy; we need more details.)