I have a question about a passage from Springer's Linear Algebraic Groups (Birkhauser). The setup is: $F$ is an arbitrary field, $G$ is an $F$-split group, and $B$ is a Borel subgroup of $G$ containing a maximal $F$-torus $T$. This corresponds to a choice $R^+$ of positive roots. Let $D$ be the root basis. Let $\mathcal{D}$ be the Dynkin diagram defined by $D$. On page 276, Springer writes, "An automorphism of $G$ stabilizing $B$ and $T$ and fixing the elements of the connected center $C$ induces an automorphism of $\mathcal{D}$."

My question is, if an automorphism of $G$ stabilizes $B$ and $T$, why doesn't it already induce an automorphism of the Dynkin diagram? My reasoning was that if $\sigma$ is the automorphism, then $\sigma$ preserves the minimal $T$-stable subspaces of the unipotent radical $R_u(B)$, namely the root subgroups $U_\alpha$ for $\alpha \in R^+$. This induces a permutation of the positive roots, which stabilizes the set of simple roots. Therefore one gets an automorphism of the Dynkin diagram.

  • $\begingroup$ One reason i could think of is the following. Dynkin diagram and hence their automorphisms depend only on the root system and not the root datum. Thus this root system doesnot see any isogeny or connected centre of a reductive group. $Z(G) \times \mathcal{D}G \rightarrow G$ is an isogeny and thus root system of $G$ depends only on $\mathcal{D}G$(more precisely $G_{ad}$). Just consider the case when this isogeny is an isomorphism and it is clear that automorphisms of $Z(G)$ has no effect on dynkin diagrams. $\endgroup$ – random123 Sep 21 '17 at 12:03

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