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 $x,y\in {𝐑}^3$, 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 $\mathrm{Der}(V)=\{D\in \mathrm{Ext}(V):D(xy)=(Dx)y+xDy\}$. A member of $\mathrm{Der}(V)$ is called a derivation. Define $[x,y]=xy-yx$. Then $[x,y]\in \mathrm{Der}(V)$.

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

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

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

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

$0⟶[𝔤,𝔤]⟶𝔤⟶𝔤/[𝔤,𝔤]⟶0$

Since both $[𝔤,𝔤]$ and $𝔤/[𝔤,𝔤]$ are solvable, $𝔤$ is solvable. $◻$

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

$𝔤\to \mathrm{End}(V)$

is completely reducible.
Proof: It suffices to prove that every $𝔤$-submodule has a $𝔤$-submodule complement. Furthermore, the proof reduces to the case when $W$ is simple (as a module) and has codimension one. Indeed, given a $𝔤$-submodule $W$, let $E\subset \mathrm{Hom}(V,W)$ 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 $𝔤$-submodule of codimension 1. Let $c$ be a Casimir element of $𝔤\to \mathrm{End}(V)$. It follows that $V$ is the direct sum of $W$ and the kernel of $c$. $◻$ (TODO: obviously, the proof is very sketchy; we need more details.)

References

• Algebraic Groups and Arithmetic Groups