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.



Automorphism of the Dynkin diagram
It is true that any automorphism stabilizing $B$ and $T$ induces an automorphism of $\mathcal D$. But Springer continues: *"Let $A\_0$ be the subgroup of $A$ of automorphisms of $\mathcal D$ obtained in this manner."*
So your quote is just the introduction for the definition of $A\_0$. 


