The purpose of this post is to show that the category of finite-dimensional representations of a semismple Lie algebra is  a semisimple category; there is thus an analogy with Maschke’s theorem, except in this case the proofs are more complicated.  They can be simplified somewhat if one uses the cohomology of Lie algebras (i.e., appropriate Ext groups), which I may talk more about, but most likely only later.  Here we will give the proofs based on linear algebra.

The first step is to construct certain central elements in the enveloping algebra.

Casimir elements

Let ${B}$ be a nondegenerate invariant bilinear form on the Lie algebra ${\mathfrak{g}}$. (E.g. ${\mathfrak{g}}$ could be semisimple and ${B}$ the Killing form.) Given a basis ${e_i \in \mathfrak{g}}$, we can consider the dual basis ${f_j}$ with respect to it, i.e. such that ${B(e_i, f_j) = \delta_{ij}}$. Consider the Casimir element

$\displaystyle C := \sum e_i f_i \in U \mathfrak{g}.$

I claim that ${C}$ is independent of the choice of ${e_i}$ and is in the center of the enveloping algebra. First off, consider the isomorphisms of ${\mathfrak{g}}$-modules,

$\displaystyle \hom_k( \mathfrak{g}, \mathfrak{g}) \simeq \mathfrak{g} \otimes \mathfrak{g}^{\vee} \simeq \mathfrak{g} \otimes \mathfrak{g} .$

The last one is given by the form ${B}$.

Now the identity, an invariant element of ${\hom_k( \mathfrak{g}, \mathfrak{g})}$, is sent to ${\sum e_i \otimes f_i \in \mathfrak{g} \otimes \mathfrak{g}}$. Since there is a well-defined homomorphism of vector spaces,

$\displaystyle f: \mathfrak{g} \otimes \mathfrak{g} \rightarrow U \mathfrak{g}, \ a \otimes b \rightarrow ab$

we see that ${C}$ is unique. Moreover, we can make ${U\mathfrak{g}}$ into a ${\mathfrak{g}}$-module by the adjoint—or, equivalently, commutator—mapping, i.e. ${x \cdot a := xa - ax}$ for ${a \in U\mathfrak{g}, x \in \mathfrak{g}}$. Then ${f}$ becomes a ${\mathfrak{g}}$-homomorphism, because

$\displaystyle x \cdot f(a \otimes b) = xab - abx$

while

$\displaystyle f( [x,a] \otimes b + a \otimes [x.b]) = xab - axb + axb - abx.$

So ${C}$ is then an invariant element under this action of ${\mathfrak{g}}$ on ${U\mathfrak{g}}$, which means that ${C}$ is in the center of ${U\mathfrak{g}}$.

Complete reducibility

Lemma 1 Let ${\mathfrak{g} \subset gl(V)}$ be a semisimple subalgebra. Let ${B_V}$ be the bilinear form via ${x,y \mapsto \mathrm{Tr}(xy)}$. Then ${B_V}$ is nondegenerate on ${\mathfrak{g}}$.

Now ${B_V}$ is the form associated to the representation ${V}$ so is invariant. The kernel of the form is thus an ideal ${\mathfrak{z}}$, and by the second version of Cartan’s solvability criterion, ${\mathfrak{z}}$ is solvable. This proves the lemma. One half the proof of Cartan’s semisimplicity criterion can be generalized, as it shows.

Fix a semisimple Lie algebra ${\mathfrak{g}}$ over a field ${k}$ of characteristic zero. Let ${M}$ be a simple ${\mathfrak{g}}$-representation, i.e. containing no proper subrepresentations besides zero.

Lemma 2 (Raising of invariants) Consider an exact sequence of ${\mathfrak{g}}$-modules$\displaystyle 0 \rightarrow V \rightarrow W \rightarrow k \rightarrow 0$

where ${k}$ is acted upon trivially. Then it splits.

We prove this by induction on ${\dim V}$. If ${V}$ is not simple, then we can take a smaller submodule ${X}$ and consider the sequence

$\displaystyle 0 \rightarrow V/X \rightarrow W/X \rightarrow k \rightarrow 0.$

We get by the inductive hypothesis a ${\mathfrak{g}}$-section ${k \rightarrow W/X}$, whose image is a one-dimensional submodule ${\tilde{X}/X \subset W/X}$. Then ${\tilde{X} + V = W}$.

There is another exact sequence

$\displaystyle 0 \rightarrow X \rightarrow \tilde{X} \rightarrow k \rightarrow 0$

from which we can take a section ${k \rightarrow \tilde{X}}$, whose image ${Y}$ satisfies ${Y+X = \tilde{X}}$. In particular, ${Y+V = Y+ V+X = \tilde{X} + V =W}$.

So we may assume ${V}$ simple, and work with our exact sequence ${0 \rightarrow V \rightarrow W \rightarrow k \rightarrow 0}$ as before.

We may make one further reduction, namely to assume that the action of ${\mathfrak{g}}$ on ${V}$ is faithful. If ${\mathfrak{n}}$ is the kernel of this action, it is an ideal—and, as a direct summand in ${\mathfrak{g}}$, semisimple itself (as a Lie algebra). I claim that this ideal acts trivially on ${W}$. Now, ${\mathfrak{g}}$ sends ${ W}$ into ${V}$ because it acts trivially on ${k}$. Since ${[\mathfrak{n}, \mathfrak{n}]=\mathfrak{n}}$, it follows that ${\mathfrak{n}}$ must act trivially on ${W}$. So we can treat our sequence as a sequence of ${\mathfrak{g}/\mathfrak{n}}$-modules, where this quotient is a semisimple Lie algebra (as $\mathfrak{n}$ was a direct factor).

We have now made the reduction of the previous lemma to the following claim:

Lemma 3 (Special case) Let ${V}$ be a faithful, simple representation of the semisimple Lie algebra ${\mathfrak{g}}$ and consider an exact sequence$\displaystyle 0 \rightarrow V \rightarrow W \rightarrow k \rightarrow 0 ;$

this splits.

We have the nondegenerate bilinear form ${B_V}$ and the corresponding Casimir element ${C}$. Then ${C}$ acts trivially on ${k}$; so ${CW \subset V}$. Moreover, I claim ${CV = V}$. Indeed, ${C}$ is a ${\mathfrak{g}}$-endomorphism of ${V}$, and once we prove it is nonzero, it will follow that it is an isomorphism. But

$\displaystyle \mathrm{Tr}_V(C) = \sum \mathrm{Tr}_V( e_i f_i) = \dim V \neq 0$

since we are in characteristic zero.

As a result, we take a nonzero vector ${v}$ annihilated by ${C}$ as the splitting. Then ${\mathfrak{g}v = C \mathfrak{g} v = \mathfrak{g} Cv = 0}$, so ${v}$ is invariant.

Finally, we can do the general case:

Theorem 4 (Weyl) Let ${W \subset V}$ be a ${\mathfrak{g}}$-submodule of the finite-dimensional representation ${V}$ of the semisimple Lie algebra ${\mathfrak{g}}$. Then ${W}$ has a ${\mathfrak{g}}$-complement.

(I.e. the category ${Rep(\mathfrak{g})}$ is semisimple.)

We can assume that ${W}$ is simple (i.e. contains no proper subrepresentations). It then follows by induction that every finite-dimensional ${U\mathfrak{g}}$-module is a semisimple module, which implies the theorem.

So, there is an exact sequence of ${\mathfrak{g}}$-modules

$\displaystyle \hom_{k}(V,W) \rightarrow \hom_k(W,W) \rightarrow 0.$

We consider the submodule ${\hom_{\mathfrak{g}}(W,W)}$ of ${\hom_k(W,W)}$ consisting of multiples of the identity and the inverse image ${A \subset \hom_k(V,W)}$; this is then a ${\mathfrak{g}}$-submodule as well. There is an exact sequence

$\displaystyle A \rightarrow \hom_{\mathfrak{g}}(W,W) \rightarrow 0.$

By the previous result on raising invariants, we can find a 1-dimensional ${\mathfrak{g}}$-submodule of ${A}$ which maps isomorphically onto ${\hom_{\mathfrak{g}}(W,W)}$. Note that any 1-dimensional ${\mathfrak{g}}$-module is the trivial (action by zero) one because ${\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]}$. This 1-dimensional module is generated by some ${\mathfrak{g}}$-invariant ${f: V \rightarrow W}$ which restricts to a nonzero multiple of the identity on ${W}$. An appropriate multiple gives a ${\mathfrak{g}}$-invariant projection ${V \rightarrow W}$ and thus a complement.

This result is hugely important, as we will see in the future.