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

• (i) [x, x] = 0
• (ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0

(ii) is called the Jacobi identity.

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

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

Theorem Let ${\displaystyle V}$ be a finite-dimensional vector space.

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

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

Theorem ${\displaystyle {\mathfrak {g}}}$ is solvable if and only if ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$ is nilpotent.
Proof: Suppose ${\displaystyle {\mathfrak {g}}}$ is solvable. Then ${\displaystyle \operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]}$ is a subalgebra of ${\displaystyle {\mathfrak {b}}_{k}}$. Thus, ${\displaystyle \operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]\subset {\mathfrak {n}}_{k}}$. Hence, ${\displaystyle \operatorname {ad} [{\mathfrak {g}},{\mathfrak {g}}]}$ is nilpotent, and so ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$ is nilpotent. For the converse, note the exact sequence:

${\displaystyle 0\longrightarrow [{\mathfrak {g}},{\mathfrak {g}}]\longrightarrow {\mathfrak {g}}\longrightarrow {\mathfrak {g}}/{[{\mathfrak {g}},{\mathfrak {g}}]}\longrightarrow 0}$

Since both ${\displaystyle [{\mathfrak {g}},{\mathfrak {g}}]}$ and ${\displaystyle {\mathfrak {g}}/[{\mathfrak {g}},{\mathfrak {g}}]}$ are solvable, ${\displaystyle {\mathfrak {g}}}$ is solvable. ${\displaystyle \square }$

3 Therorem (Weyl's theorem) Every representation of a finite-dimensional semisimple Lie algebra:

${\displaystyle {\mathfrak {g}}\to \operatorname {End} (V)}$

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

References

• Algebraic Groups and Arithmetic Groups