Prove that raising operator on non-highest weight is non-zero

Question

Let $\mathfrak{h}$ denote the Cartan subalgebra of some Lie algebra $\mathfrak{g}$.
Let $(\pi,V)$ be a representation of the Lie algebra. ($V$ is the vector space, $\pi : \mathfrak{g} \mapsto End(V)$).
Let $\lambda \in \mathfrak{h}^*$ be a weight such that $\pi(h)v = \lambda(h)v \;\; \forall h \in \mathfrak{h}$, $ $ for some weight vector $v \in V$.
Let $e_\alpha$ be a root vector in $\mathfrak{g}$ with root $\alpha \in \mathfrak{h}^*$ $\;$ i.e. $[h,e_\alpha] = \alpha(h)e_\alpha \;\, \forall h \in \mathfrak{h}$.

I can show that
$[h,e_\alpha] = \alpha(h)e_\alpha \\ \Rightarrow \pi(h)\pi(e_\alpha) - \pi(e_\alpha)\pi(h) = \alpha(h)\pi(e_\alpha) \\ \Rightarrow \pi(h)\,(\pi(e_\alpha)v) = (\lambda(h) + \alpha(h))\:(\pi(e_\alpha)v)$
Which means either $\pi(e_\alpha)v = 0$ or that $\pi(e_\alpha)v$ is a weight vector with weight $\lambda + \alpha$.

How do I prove that if there exists a weight $\lambda+\alpha$, then $\pi(e_\alpha)v \neq 0$ ?

Answer 1

This does not hold in general. Consider the lie algebra $\mathfrak{sl}(2,\mathbb{C})$ and the following representation of it

\begin{equation} \pi(h)= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 \end{pmatrix} ,\; \pi(e)= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\; \pi(f)= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \end{pmatrix} \end{equation}

The vector $v=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$ is a weight vector with weight $0$. The weight $0+2=2$ exists. But $\pi(e)v=0$.

$\\$

The result is true if the following conditions hold:

1. The lie algebra $\mathfrak{g}$ is semi-simple (hence $e_{-\alpha}$ exists)
2. The representation $\pi$ restricted to the subalgebra $\langle e_\alpha, e_{-\alpha}, h_\alpha = [e_\alpha, e_{-\alpha}] \rangle$, call it $\rho$, is irreducible

With the above conditions, suppose $\pi(e_\alpha)=0$, then the proper subspace spanned by $\{...,\;(\rho(e_{-\alpha}))^2v,\;\rho(e_{-\alpha})v,\;v\}$ would be stable under the action of $\rho(h_\alpha)$, $\rho(e_\alpha)$ and $\rho(e_{-\alpha})$ contradicting the fact that $\rho$ is irreducible.

Answer 2

This is not always true. The useful answer by ksnad gives a basic example, and gets to the heart of the matter in the sense that the key is how the subalgebra $\mathfrak{sl}_{2,\alpha}$ sees $V$.

The phenomenon is forced upon us on many occasions. Even when $V$ is an irreducible $\mathfrak{g}$-module. For example consider the adjoint representation $V$ of $\mathfrak{sl}_3$. The weight space $V_0$ corresponding to the weight zero is 2-dimensional (equal to the subspace of diagonal matrices). Let $\alpha=\alpha_1+\alpha_2$ be the highest root, so $V$ is the simple module of highest weight $\alpha$. Because $\dim V_{\alpha}=1$, the mapping $\pi(e_\alpha):V_0\to V_{\alpha}$ cannot be injective (by rank-nullity). So necessarily $\pi(e_\alpha)(v)=0$ for some $v\in V_0$ even though $0+\alpha=\alpha$ is a weight.

The same argument works more generally whenever the dimension of the weight space $V_{\lambda+\alpha}$ is lower than that of $V_\lambda$.

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A related observation is that if

• $\mathfrak{g}$ is semisimple,
• $V$ is finite dimensional and irreducible, and
• $\lambda$ and $\lambda+\alpha$ are both weights of $V$,

then for some $v\in V_\lambda$ we do have $\pi(e_\alpha)(v)\neq0$.

This is because to any positive root $\alpha$ we can find a subalgebra $\mathfrak{sl}_{2,\alpha}$ of $\mathfrak{g}$ that has $\alpha$ as its positive root. The subspace $$U=\bigoplus_{k\in\Bbb{Z}}V_{\lambda+k\alpha}$$ is then an $\mathfrak{sl}_{2,\alpha}$-submodule of $V$. By the well known representation theory of $\mathfrak{sl}_2$ it must have an irreducible $\mathfrak{sl}_{2,\alpha}$-submodule $W$ such that both $\lambda$ and $\lambda+\alpha$ are weights of $W$. Again, by the known structure of $\mathfrak{sl}_2$-modules we see that any non-zero $v\in W_\lambda$ will work.