I have a question about minuscule cocharacters, which might sound trivial to the experts:

Let $G$ be a smooth affine group scheme over $\mathbb{Z}_{p}$. Furthermore, consider a cocharacter $$\mu\colon \mathbb{G}_{m,W(k)}\rightarrow G_{W(k)},$$ where $k$ is some finite field. Consider the action of $\mathbb{G}_{m,W(k)}$ on $G_{W(k)},$ given by $x\ast g=\mu(x)^{-1}g\mu(x).$ We have the unipotent group-scheme associated to $\mu,$ denoted by $U^{+}(\mu).$ Furthermore we can consider the adjoint representation $Ad(\mu^{-1})$ of $\mathbb{G}_{m,W(k)}$ on the Lie-algebra, which induces a $\mathbb{Z}$-grading on it: $$\mathfrak{g}=\bigoplus_{n\in \mathbb{Z}} \mathfrak{g}_{n}.$$

The question: Assume that $G$ is reductive now. Is the requirement that $\mathfrak{g}_{n}=0$ for $n\geq 2$ equivalent to $\mu$ being minuscule, i.e. $\mathfrak{g}_{n}=0$ for all $|n|\geq 2?$



This has nothing to do with Witt vectors, smoothness over $\mathbf{Z}_p$ etc: formation of the $\mathfrak{g}_n$ commutes with base-extension to $\mathbf{Q}_p$ or even $\overline{\mathbf{Q}}_p$, so we may as well assume that $G$ is defined over an arbitrary alg. closed field of char 0.

The image of $\mu$ is a torus, so it is contained in a maximal torus, whose adjoint action on $\mathfrak{g}$ gives the root system of $G$. Hence the grading on $\mathfrak{g}$ is just given by the root spaces: $\mathfrak{g}_n$ is the sum of the root spaces $\mathfrak{g}_\eta$ over roots $\eta$ such that $\langle \mu, \eta\rangle = n$. Since the set of roots is preserved by $\eta \mapsto -\eta$, we have $\operatorname{dim} \mathfrak{g}_n = \operatorname{dim} \mathfrak{g}_{-n}$.

  • $\begingroup$ Ahh, I see. that happens, when one does not understand root systems... $\endgroup$ – user125861 Jun 22 '18 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy