Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 102 / 2 / d / Solution

Suppose first that $x\ne 0$, and let $E$ be the real span of the root system, of dimension equal to the rank of a semisimple Lie algebra $\ell$. The images of the roots span $E/\mathbb{R}\alpha$, so select roots ${\beta }_1,\dots ,{\beta }_{\ell -1}$ whose images form a basis of this quotient.

For each $i$, take the highest endpoint ${\gamma }_i^{+}$ of the $\alpha$ root string through ${\beta }_i$, and the highest endpoint ${\gamma }_i^{-}$ of the root string through $-{\beta }_i$. By construction, ${\gamma }_i^{\pm }+\alpha$ is not a root. Moreover ${\gamma }_i^{\pm }\ne -\alpha$, since their images in $E/\mathbb{R}\alpha$ are nonzero. The root-space decomposition and its bracket rule therefore giveThese $2(\ell -1)$ roots are distinct: their quotient images are the two signs of a basis, which are distinct. Their one-dimensional root spaces consequently contribute $2(\ell -1)$ independent vectors to the Lie algebra centralizer.

In addition, the $\ell -1$ dimensional subspace $\ker \alpha \subset \mathfrak{t}$ commutes with $x$, because $[h,x]=\alpha (h)x$. The line $\mathbb{C}x\subset {\mathfrak{g}}_{\alpha }$ also commutes with $x$. These contributions are independent by the root-space decomposition; in particular none of the selected endpoint roots is $\alpha$. Thus the centralizer lower bound for a root vector isIf $x=0$, then $Z_{\mathfrak{g}}(x)=\mathfrak{g}$. The root system contains the $2\ell$ distinct roots $\pm {\alpha }_i$, so $\dim \mathfrak{g}=\ell +|\Phi |\ge 3\ell$, which also proves the desired inequality. The argument includes rank one, where the list of ${\beta }_i$ is empty.