In section 35.1 of the book "Linear algebraic groups" by Humphreys, it is stated that the quasi-split but not split semisimple groups can only arise when the root system admits a nontrivial graph automorphism.

Moreover, it seems that the relative root system in this case is obtained by adjoining the vertices of Dynkin diagram which are sent to each other by the graph automorphism.

Also in the wikipedia page on quasi-split groups, it is stated that a quasi-split groups over a field correspond to actions of the absolute Galois group on a Dynkin diagram.

In both, there is no reference about this statement. In what paper can I find some theory about this?

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    $\begingroup$ I'm not sure this is true as stated. Tori are reductive groups, and they are always quasi-split and have empty root system (hence no automorphisms), but they aren't always split. Do you maybe want your groups to be semisimple (and maybe also simply-connected), rather than reductive, here? Or perhaps you want to use root data, rather than root systems? $\endgroup$ Jul 7, 2021 at 22:28
  • $\begingroup$ Oh I think we should restrict to the semisimple case. Thanks! $\endgroup$
    – YJ Kim
    Jul 8, 2021 at 6:53

2 Answers 2


The quasi-split forms of a split reductive group $G$ over a field $k$ are classified by the elements of the first Galois cohomology group of $k$ with values in $Out(G)$ (see Theorem 23.51 of Milne's book Algebraic Groups). So no outer automorphisms means no nonsplit quasi-split forms. When $G$ is semisimple, the outer automorphisms correspond to graph automorphisms of the dynkin diagram.


The relative diagram is obtained from a decoration of the absolute diagram, called a "Tits index". Details are in Section 2 of Tits's Boulder notes "Classification of algebraic semisimple groups" which are easiest found in his collected works (article [68], p. 372 of volume 2).

  • $\begingroup$ Thank you very much! For other curious people, relevant contents is in the section 2! $\endgroup$
    – YJ Kim
    Jul 7, 2021 at 9:54

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