Weyl's complete reducibility theorem for semisimple Lie algebra: homological algebra proof
In this article, we reproduce the proof of Weyl’s complete reducibility theorem:
Every finite-dimensional representation of a semisimple Lie algebra is completely reducible.
that uses homological algebra, which can be found here. I recently learned this from one of the exercises comes from the Koszul duality in the local Langlands program, and I liked it because it is very clean and easy to remember.
Universal enveloping algebra and Casimir element
For a Lie algebra $\mathfrak{g}$, universal enveloping algebra $\mathcal{U}(\mathfrak{g})$ is defined as
\[\mathcal{U}(\mathfrak{g}) := T(\mathfrak{g}) / \langle x \otimes y - y \otimes x - [x, y] : x, y \in \mathfrak{g} \rangle\]where $T(\mathfrak{g})$ is the tensor algebra
\[T(\mathfrak{g}) := \bigoplus_{k \geq 0} \mathfrak{g}^{\otimes k} = \mathbb{C} \oplus \mathfrak{g} \oplus (\mathfrak{g} \otimes \mathfrak{g}) \oplus \cdots.\]It has the following universal property: for any associative unital algebra $A$ and a Lie algebra homomorphism $\mathfrak{g} \to A$ (where the Lie bracket on $A$ is defined via commutator $[a, b] := ab - ba$) lifts to $\mathcal{U}(\mathfrak{g}) \to A$.
It is often important to understand the center $\mathcal{Z} = \mathcal{Z}(\mathcal{U}(\mathfrak{g}))$ of the univesal enveloping algebra for many reasons - the one reason is it help us to classify irreducible representations of $\mathfrak{g}$. More precisely, the action of center commutes with other elements’ action, and by Schur’s lemma they should act as scalars, i.e. a $\mathfrak{g}$-representation admits a central character $\chi : \mathcal{Z}(\mathcal{U}(\mathfrak{g})) \to \mathbb{C}^{\times}$. One may try to classify representation by fixing a central character first.
For a given representation $\rho : \mathfrak{g} \to \mathrm{End}(V)$, we have a special elemtn $C_V \in \mathcal{Z}$ that satisfy the following lemma:
Lemma. Let $(\rho, V)$ be a nontrivial irreducible representation of a semisimple Lie algebra $\mathfrak{g}$. Then there exists a central element $C_V \in \mathcal{Z}(\mathcal{U}(\mathfrak{g}))$ that acts
- by a nonzero constant on $V$
- and by zero on the trivial representation.
We will use this Lemma later, and not going to prove. The element $C_V$ can be choosen as follows: define a symemtric bilinear form $B_V : \mathfrak{g} \times \mathfrak{g} \to \mathbb{C}$ as $B_V(x, y) := \mathrm{Tr}(\rho(x)\rho(y))$. When $B_V$ is nondegenerate, choose a basis $x_1, \dots, x_n$ and a dual basis (with respect to $B_V$) $x^1, \dots, x^n$, then
\[C_V := \sum_{i=1}^{n} x_i x^i\]works ($C_V$ does not depend on the choice of basis). When $B_V$ is not non-degenerate, we define $C_V$ by restring it to a semisimple $\mathfrak{g}’ \subset \mathfrak{g}$ which $B_V$ is non-degenerate In fact, when $\mathfrak{g}$ is semisimple, we can just choose the Killing form $B(x, y) := \mathrm{Tr}(\mathrm{ad}(x)\mathrm{ad}(y))$, and the corresponding central element is called the Casimir element.
$\mathrm{Ext}$ groups
There are many ways to define $\mathrm{Ext}$ groups, but possible the right way is to define it as a right derived functor of the functor $\mathrm{Hom}(X, -)$ (which is left exact). One can compute it via projective or injective resolutions (if exists), but we may not compute it: the only property we are going to use is that it produces long exact sequence from a short exact sequence: from
\[0 \to A \to B \to C \to 0\]we get
\[\begin{align*} 0 &\to \mathrm{Hom}(X, A) \to \mathrm{Hom}(X, B) \to \mathrm{Hom}(X, C) \\ &\to \mathrm{Ext}^{1}(X, A) \to \mathrm{Ext}^{1}(X, B) \to \mathrm{Ext}^{1}(X, C) \\ &\to \mathrm{Ext}^{2}(X, A) \to \mathrm{Ext}^{2}(X, B) \to \mathrm{Ext}^{2}(X, C) \\ &\to \cdots \end{align*}\]Note that $\mathrm{Ext}$ can be defined for any abelian category, not necessarily $\mathbb{Z}$-modules.
Weyl’s theorem as vanishing $\mathrm{Ext}^{1}$
Now we are ready to give a proof of Weyl’s theorem Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$, and let $V$ be a representation. If $V$ is irreducible, there’s nothing to prove. If not, there exist a proper nonzero subrepresentation $V’ \subset V$ and so an exact sequence of $\mathfrak{g}$-representations
\[0 \to V' \to V \to V'' \to 0\]with $V’’ = V / V’$. If we can prove that all the short exact sequences split, then we get a decomposition $V \simeq V’ \oplus V’’$, and continue this until we get a complete decomposition as irreducibles. Hence it is enough to show that $\mathrm{Ext}^{1}(W, V) = 0$ for any $\mathfrak{g}$-representations $V, W$.
Step 0: $\mathrm{Ext}^{1}(\mathbb{C}, \mathbb{C}) = 0$
We start with trivial representations: showing that all the extensions of $\mathbb{C}$ by $\mathbb{C}$ splits (i.e. $\mathrm{Ext}^{1}(\mathbb{C}, \mathbb{C}) = 0$). Let
\[0 \to \mathbb{C} \to E \to \mathbb{C} \to 0\]be a such extension. Then $E$ is a 2-dimensional representation and can be written as a strictly upper triangular matrix, i.e.
\[\mathfrak{g} \to \mathrm{End}(E) \simeq \mathfrak{gl}_{2}(\mathbb{C}), \quad x \mapsto \left(\begin{matrix} 0 & \rho'(x) \\ 0 & 0\end{matrix}\right)\]by choosing an appropriate basis. Especially, $[\rho(\mathfrak{g}), \rho(\mathfrak{g})] = 0$ ($\rho(\mathfrak{g})$ is nilpotent). Since $\mathfrak{g}$ is semisimple, we have $\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$ ($\mathfrak{g}’ := \mathfrak{g}/[\mathfrak{g}, \mathfrak{g}]$ is semisimple and abelian, so $\mathfrak{g}’ = 0$) and get
\[\rho(\mathfrak{g}) = \rho([\mathfrak{g}, \mathfrak{g}]) = [\rho(\mathfrak{g}), \rho(\mathfrak{g})] = 0,\]i.e. $\mathfrak{g}$ acts trivially on $E$. Hence $\rho’ = 0$ and the above sequence splits.
Step 1: $\mathrm{Ext}^{1}(\mathbb{C}, V) = 0$ for irreducible $V$
In other words, we need to prove that any extension
\[0 \to V \to E \to \mathbb{C} \to 0\]splits. Recall that the Casimir element $C_V \in \mathcal{U}(\mathfrak{g})$ of $V$ acts on $V$ as a nonzero scalar $\lambda$ and trivially on $\mathbb{C}$. If we consider eigenspace decomposition of $C_V$ on $E$, $V \subset E$ becomes $\lambda$-eigenspace of multiplicity $\dim V$, and 0-eigenspace $W$ of multiplicity 1, with $E = V \oplus W$. This gives a splitting with $W \simeq \mathbb{C}$.
Step 2: $\mathrm{Ext}^{1}(\mathbb{C}, V) = 0$ for any $V$
We can prove the same statement for any $V$ via induction on dimension of $V$. If $V$ is not irreducible, we can find a nonzero proper subrepresentation $V’ \subset V$ that gives a short exact sequence
\[0 \to V' \to V \to V'' \to 0\]Here both $V’, V’’$ are nonzero and have smaller dimensions. Now take $\mathrm{Hom}(\mathbb{C}, -)$ and we get a long exact sequence
\[\cdots \to \mathrm{Hom}(\mathbb{C}, V'') \to \mathrm{Ext}^{1}(\mathbb{C}, V') \to \mathrm{Ext}^{1}(\mathbb{C}, V) \to \mathrm{Ext}^{1}(\mathbb{C}, V'') \to \cdots\]and by inductive hypothesis, $\mathrm{Ext}^{1}(\mathbb{C}, V’) = 0 = \mathrm{Ext}^{1}(\mathbb{C}, V’’)$, so we get $\mathrm{Ext}^{1}(\mathbb{C}, V) = 0$.
Step 3: $\mathrm{Ext}^{1}(W, V) = 0$ for any $W, V$
For a representation $V$, define $V^{\mathfrak{g}}$ as a $\mathfrak{g}$-trivial part of $V$, i.e.
\[V^{\mathfrak{g}} := \{v \in V: x.v = 0, \,\forall x \in \mathfrak{g}\}.\]One can check that this is naturally isomorphic to
\[\mathrm{Hom}(\mathbb{C}, V)\]where $\mathbb{C}$ is the trivial representation (observe where $1 \in \mathbb{C}$ goes). Also, by the definition of $\mathfrak{g}$-action on $\mathrm{Hom}_{\mathbb{C}}(V, V’)$, one can check that
\[\mathrm{Hom}_{\mathbb{C}}(V, V')^{\mathfrak{g}} = \mathrm{Hom}(V, V')\](the later Hom is $\mathfrak{g}$-linear homomorphisms as before).
Recall that our final goal is equivalent to proving all $\mathrm{Ext}^{1}$ group vanishes. Let
\[0 \to V \to E \to W \to 0\]be an arbitrary short exact sequence (of $\mathfrak{g}$-representations). We first apply the functor $\mathrm{Hom}_{\mathbb{C}}(W, -)$, $\mathbb{C}$-linear hom (not $\mathfrak{g}$-linear hom), which is exact (any short exact sequence of vector spaces splits):
\[0 \to \mathrm{Hom}_{\mathbb{C}}(W, V) \to \mathrm{Hom}_{\mathbb{C}}(W, E) \to \mathrm{Hom}_{\mathbb{C}}(W, W) \to 0\]Now apply $(-)^{\mathfrak{g}} = \mathrm{Hom}(\mathbb{C}, -)$ to this exact sequence. By Step 2, $\mathrm{Ext}^{1}(\mathbb{C}, \mathrm{Hom}_{\mathbb{C}}(V, W)) = 0$ and get an another short exact sequence:
\[0 \to \mathrm{Hom}(W, V) \to \mathrm{Hom}(W, E) \to \mathrm{Hom}(W, W) \to 0\]Especially, we have $\mathrm{id}_{W} \in \mathrm{Hom}(W, W)$ and there exists $s \in \mathrm{Hom}(W, E)$ that maps to $\mathrm{id}_{W}$. In particular, we get a section $W \to E$ of $E \to W$, which shows that the original exact sequence splits. This completes the proof. $\square$
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