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I have seen this fact stated as obvious a number of times, but I can't see it. Let $\alpha$ and $\beta$ be roots of a Lie algebra $\mathfrak{g}$ with CSA $H$ and Let $\mathfrak{g}_{\alpha}$ be the set of generators satisfying the eigenvalue equation $[H,E_\alpha]=\alpha (H) E_\alpha$. Then if $\alpha+\beta \ne 0\text{ and } X \in \mathfrak{g}_{\alpha}\text{ and } Y \in \mathfrak{g}_{\beta}\text{ ,then } ad(X)\circ ad(Y)$ is nilpotent. As in p.490 of Fulton and Harris.

I don't see this, could anyone explain it?

лаетековский чайник
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1 Answers1

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The point is that $ad(X)\circ ad(Y)$ maps $\mathfrak h$ to $\mathfrak g_{\alpha+\beta}$ (which may be $\{0\}$) and $\mathfrak g_\gamma$ to $\mathfrak g_{\gamma+\alpha+\beta}$ (which may be $\{0\}$) . If $\alpha+\beta$ is positive, then you can find $N\in\mathbb N$ such that for the lowest root $\nu$, the functional $\alpha+\beta+\nu$ is larger than the largest root $\mu$. Then $(ad(X)\circ ad(Y))^N$ has to be the zero map. If $\alpha+\beta$ is negative, you argue simililarly exchanging the roles of $\mu$ and $\nu$. (BTW: Usually, you only need that $ad(X)\circ ad(Y)$ is tracefree, which is obvious for a basis of joint eigenvectors.)

Andreas Cap
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