Semisimple complex Lie Algebra and decomposition into weight spaces.

Question

So I was wondering why a semisimple complex Lie Algebra $L$ is a direct sum of its weight spaces.

Given a Cartan Subalgebra of $H$ of $L$ then since $L$ is a semisimple complex Lie Algebra, then every element of ad(H) is semisimple (diagonalizable) . And since H is abelian, then all elements of H can be simultaneously diagonalised w.r.t some basis $x_1, \ldots , x_n$

But then consider $\alpha _i: H \rightarrow \mathbb{K}$ where $\alpha _i(h)$ specifies the $i^{\text{th}}$ diagonal element of $ad_h$.

Then for $L_{\alpha _i} := \{x\in L\space | \space ad_h(x) = \alpha _i(h)x \space \forall h \in H\}$ we know that $x_i \in L_{\alpha _i}$.

So if L is indeed a direct sum of it's weight spaces $L_{\alpha _i}$, then it must be the direct sum of these. Implying each weight space has dimension $1$. But this implies inturn that $H$ has dimension $1$ as $H$ = $L_{0}$. Which surely is false!?

What exactly have I done wrong?

Answer 1

There are a few places where I think things could have gone wrong.

1. The definition of the weight space (of $L$ as an $H$-module) is $$ L_{\lambda}=\lbrace y\in L \mid ((\operatorname{ad}(x)-\lambda(x))^ny=0\text{ for all }x\in H\rbrace $$ where $\lambda:H\rightarrow\mathbb{C}$ is a $1$-dimensional representation, which isn't exactly what you've written: you need that some power of $\operatorname{ad}(x)-\lambda(x)$ annihilates $y\in L$, for all $x\in H$, and so it isn't necessarily true that $\operatorname{ad}(x)=\lambda(x)y$.

2. You haven't necessarily shown that $L_0$ is $1$-dimensional. You've taken some $\alpha_i:H\rightarrow\mathbb{C}$, but none of these are the $0$ weight. Whilst it's true that every non-zero weight gives rise to a $1$-dimensional weight space, the same is not true of the Cartan subalgebra.

3. Why is $x_i\in L_{\alpha_i}$?

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As to the reason why: every module over a nilpotent Lie algebra can be decomposed into the direct sum of its weight spaces.

Let $V$ be a module over a nilpotent Lie algebra $L$, so $V=V_1\oplus\cdots\oplus V_k$, with $V_j$ an indecomposable submodule inducing a weight $\lambda_j$. Let $W_\lambda=\bigoplus_{\lambda_j=\lambda}V_j$. Then $V=\bigoplus_\lambda W_\lambda$. Clearly $W_\lambda\subset V_\lambda$, so we try to show the reverse inclusion.

Let $v\in V_\lambda\subset V=\bigoplus_\mu W_\mu$. So $v=\sum_\mu v_\mu$. Then $v(x-\lambda(x))^n=0$, and so $\sum_\mu v_\mu(x-\lambda(x))^n=0$. So each $v_\mu(x-\lambda(x))^n=0$. Suppose $\lambda\ne\mu$, so $\lambda(x)\ne\mu(x)$ for some $x\in L$. Then $v\mapsto vx$ has matrix $$ \left(\begin{matrix} \lambda(x) & & * \\ & \ddots & \\ 0 & & \lambda(x) \end{matrix}\right) $$ on $W_\mu$, which is non-singular. So $v\mapsto v(x-\lambda(x))$ is non-singular on $W_\mu$, and $v\mapsto v(x-\lambda(x))^n$ is non-singular on $W_\mu$. So $v_\mu(x-\lambda(x))^n=0\Rightarrow v_\mu=0$. So $v=\sum_\mu v_\mu=v_\lambda\in W_\lambda$.