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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

$\displaystyle 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,

$\displaystyle [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.

$\displaystyle 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

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

so that we have a dual expression

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

Then we compute as follows.

    $\displaystyle [a,C_B]=$ $\displaystyle \sum_i [a,x_i] x^{(i)} +\sum_i x_i[a, x^{(i)}]$
    $\displaystyle =$ $\displaystyle \sum_i \sum_j c_i^{(j)}(a)x_j x^{(i)} -\sum_i \sum_j c_j^{(i)}(a)x_i x^{(j)} =0.$

$ \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

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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

$\displaystyle \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

$\displaystyle 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

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

and

$\displaystyle V\otimes_k W
$

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

$\displaystyle (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),
$

$\displaystyle (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.

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

$\displaystyle 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

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

$ \qedsymbol$


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Next: Semi direct products of Up: generalities in finite dimensional Previous: examples
2007-12-19