Weyl's theorem on complete reducibility

DEFINITION 5.47   Let $L$ be a finite dimensional Lie algebra. Let $B$ be a non-degenerate invariant bilinear form on $L$ . Then we define the Casimir element $C_B\in U(L)$ with respect to $B$ by

$$C_B=\sum_i x_i x^{(i)}$$

where $\{x_i\}$ is a basis of $L$ , and $\{x^{(i)}\}$ is the dual basis of the basis $\{x_i\}$ with respect to $B$ .

PROPOSITION 5.48   Under the same assumption of the definition above, we have the following facts.

1. The Casimir operator $C_B$ is independent of the choice of the basis $\{x_i\}$ of $L$ .
2. $C_B$ commutes with $L$ . So it is in the center of $U(L)$ .

PROOF.. (1): easy exercise in linear algebra.

(2): For any $a\in L$ , let us write the adjoint action of $a$ on $L$ by using the basis $\{x_i\}$ . Namely,

$$[a, x_i]=\sum_j c_{i}^{(j)}(a) x_j \qquad (c_{i}^{(j)}(a)\in k).$$

Then the constants $\{c_{i}^{(l)}(a)\}$ (``structure constants'') are expressed in terms of $B$ as follows.

$$B(x^{(l)},[a, x_i])=\sum_j c_{i}^{(j)}(a) B(x^{(l)},x_j)= c_i^{(l)}(a)$$

We note that from the invariance of $B$ , we have

$$B([x^{(l)},a],x_i)= c_i^{(l)}(a),$$

so that we have a dual expression

$$[ x^{(l)},a]=\sum_i c_i^{(l)}(a) x^{(i)}.$$

Then we compute as follows.

$$[a,C_B]= \sum_i [a,x_i] x^{(i)} +\sum_i x_i[a, x^{(i)}]$$

$$= \sum_i \sum_j c_i^{(j)}(a)x_j x^{(i)} -\sum_i \sum_j c_j^{(i)}(a)x_i x^{(j)} =0.$$

$\qedsymbol$

DEFINITION 5.49   Let $L$ be a finite dimensional Lie algebra. Let $V$ be a finite dimensional $L$ -module. $\operatorname{Tr}_V(\bullet \bullet)$ with respect to $V$ . We assume that the Killing form $\operatorname{Tr}_V(\bullet \bullet)$ with respect to $V$ is non degenerate. Then we define the Casimir element with respect to $V$ by

$$C_V=C_{\operatorname{Tr}_V}$$

where $\{x_i\}$ is a basis of $L$ , and $\{x^i\}$ is the dual basis of the basis $\{x_i\}$ with respect to the Killing form $\operatorname{Tr}_V$ .

LEMMA 5.50   Let $k$ be a field of characteristic $p$ . Let $L$ be a $n$ -dimensional semisimple Lie algebra over a field $k$ . Let $V$ be a $m$ -dimensional $L$ -module. Let $I$ be the kernel of the representation $\rho_V$ associated to $V$ . We assume $p\in \operatorname{Ccs}(n)\cap \operatorname{Ccs}(m)$ . Then the Killing form $\operatorname{Tr}_V$ on $L/I$ is non degenerate.

PROOF.. $L$ is semisimple and $p \in \operatorname{Ccs}(n)$ so $L$ is semisimple. $L/I$ is also non-degenerate so $L/I$ is semisimple. We may thus assume $I=0$ . An ideal

$$J=L^{\perp_{\operatorname{Tr}_V}}$$

of $L$ is a solvable ideal. Since $L$ is semisimple and $p \in \operatorname{Ccs}(m)$ , we have by $J=0$ . That means, $\operatorname{Tr}_V$ is non-degenerate on $L$ .

$\qedsymbol$

LEMMA 5.51   Let $k$ be a field of characteristic $p$ (which may be 0 ). Let $(L,W\subset V)$ be a triple which satisfies the following conditions.

1. $L$ is a semisimple Lie algebra over $k$ .
2. $V$ is a finite dimensional irreducible $L$ -module.
3. $W$ is an $L$ -submodule of $V$ of codimension $1$ .
4. $p \in \operatorname{Ccs}(\dim(L)) \cap \operatorname{Ccs}(\dim(V))$ .

Then the exact sequence

$$0 \to W \to V \to V/W \to 0$$

splits. In other words, there exists a $1$ -dimensional $L$ -submodule $X$ of $V$ which is complementary to $W$ .

PROOF.. Since the question of existence of $X$ is described in terms of existence of a solution of a set of linear equations, we may assume that $k$ is algebraically closed. Let us denote by $\rho_V$ the representation of $L$ associated to $V$ . Then by replacing $L$ by $L/\ker(\rho_V)$ if necessary, we may assume that the representation $\rho_V$ is faithful.

Note that since $L$ is semisimple, it acts on $V/W$ trivially.

Let us first treat the case where $W$ is irreducible. Let $c=c_V$ be a Casimir element with respect to $V$ . Since $L$ is acts on $V/W$ trivially, $c\vert _{V/W}$ is equal to zero. Thus

$$\dim(V)=\operatorname{tr}_V(c)=\operatorname{tr}_W (c\vert _W)+\operatorname{tr}_{V/W}(c\vert _{V/W})=\operatorname{tr}_W(c\vert _W).$$

In particular, $c\vert _W$ is not equal to zero. On the other hand, by Schur's lemma, $c\vert _W$ is equal to a scalar $\lambda\in k$ . Thus $X=\operatorname{Ker}(c)$ is a required object in this case.

We next come to general case. Let $W_1$ be the maximal proper $L$ -submodule of $W$ . Then we see that $(L,W/W_1 \subset V/W_1)$ satisfies the assumption of the lemma with $W/W_1$ irreducible. By the argument above, we therefore see that there exists an $L$ -submodule $Y$ which contains $W_1$ as a submodule of codimension $1$ such that

$$V/W_1 = Y/W_1 \oplus W/W_1$$

holds. Since $(L,W_1\subset Y)$ also satisfies the assumption of the lemma with $\dim(Y)<\dim(V)$ , we deduce by induction that the lemma holds in general.

$\qedsymbol$

LEMMA 5.52   Let $L$ be a Lie algebra over a commutative ring $k$ . Then for any $L$ -modules $V,W$ , each of the vector spaces

$$\operatorname{Hom}_{k\operatorname{-linear}} (V,W)$$

and

$$V\otimes_k W$$

admits a structure of $L$ -module. Namely,

$$(x.f)(v)=x.(f(v)) -f(x.v) \qquad (\forall x \in L, \forall f\in \operatorname{Hom}_{k\operatorname{-linear}}(V,W) \forall v\in V),$$

$$(x.v \otimes_k w)=(x.v) \otimes w + v\otimes_k (x.w) \qquad (\forall x \in L, \forall v\in V, \forall w \in W).$$

PROOF.. Easy. $\qedsymbol$

THEOREM 5.53 (Weyl)   Let $k$ be a field of characteristic $p$ (which may be 0 ). Let $L$ be a non degenerate Lie algebra over $k$ . Let $V$ be a finite dimensional irreducible $L$ -module. Assume $p\in \operatorname{Ccs}(\dim(V)^2)$ . Then $V$ is completely reducible.

PROOF.. Let us define the following $L$ -modules.

$$V_1=\{ f \in \operatorname{Hom}_{k\operatorname{-linear}}(V,W); f\vert _W \in k. 1_W\}$$

$$W_1=\{ f \in \operatorname{Hom}_{k\operatorname{-linear}}(V,W); f\vert _W =0\}$$

Then it is easy to see that the triple $(L,W_1\subset V_1)$ satisfies the assumption of Lemma 5.51. We therefore have an element $f \in \operatorname{Hom}_{k\operatorname{-linear}}(V,W)$ which satisfies the following conditions.

1. $f\vert _W \in k.1_W$ .
2. $f\vert _W \neq 0$ .
3. $x.f =0 \quad (\forall x \in L)$ (In other words, $f$ is a $L$ -linear homomorphism).

by dividing by a suitable element in $k$ , we may assume $f\vert _W=1_W$ . Then $f$ gives a splitting of the exact sequence

$$0 \to W \to V \to V/W \to 0.$$

$\qedsymbol$