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Let $G$ be a connected reductive group over a field $k$ with separable closure $k^s$ and $\Gamma=\text{Gal}(k^s/k)$. Pick $T\subset B$ maximal split torus and Borel subgroup for $G_{k^s}$. I am aware of two ways of defining actions of $\Gamma$ on the based root datum $\Psi(G_{k^s},B,T)=(X^*(T),\Phi,\Delta,X_*(T),\Phi^{\vee},\Delta^{\vee})$.

The first is called the $*$-action (see Lemma 12.2.4 of https://math.stanford.edu/~conrad/249BW16Page/handouts/249B_2016.pdf). From what I understand, it is this: $\sigma\in\Gamma$ acts on $\chi\in X^*(T)$ by $\sigma \cdot \chi= \sigma_{\mathbb{G}_m,k^s}\circ \chi \circ \sigma^{-1}_{T}$. But this action will generally not preserve the base $\Delta$ (unless $G$ is quasi-split, i.e. has Borel defined over $k$). One can modified this by letting the Weyl group act on $X^*(T)$ to get back to the base.

The second comes from this map $\Gamma \to \text{Aut}(G_{k^s})\to \text{Aut}(\Psi(G_{k^s},B,T))$ given in section 7.3 of https://sites.duke.edu/jgetz/files/2022/04/Graduate_Text.pdf. The first map is given by sending $\sigma\in\Gamma$ to the pull-back of $\text{Spec } k^s \xrightarrow{\sigma}\text{Spec }k^s \leftarrow G_{k^s}$. For the second map (let me just do this for quasi-split $G$ first), $f\in \text{Aut}(G_{k^s})$ acts on $X^*(T)$ by $\chi \mapsto \chi \circ f|_{T}$.

I suspect these two should be the same action of $\Gamma$ on the based root datum, but I have trouble convine myself of this. For instance, when $G$ is quasi-split, the action of $\Gamma$ on $X^*(T)$ already looks so different?

Any help would be much appreciated!

Tengu
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    Dear Tengu, I am not entirely sure I follow. Can you clarify whether the following two facts, in conjunction, answer your question.

    Fact 1: There is a $\Gamma_k$-equivariant isomorphism $\mathrm{Out}(G_{\overline{k}})\xrightarrow{\approx}\mathrm{Aut}(\Psi(G_{\overline{k}}))$.

    Here $\Psi(G_{\overline{k}})$ is the canonical based root datum of $G_{\overline{k}}$ given, essentially, by $\lim \Psi(G_{\overline{k}},B,T)$ where $(B,T)$ ranges over all Borel(=Borel+Torus) pairs in $G_{\overline{k}}$, where the transition maps are given by the conjugation maps (which are independent of choices).

     – Alex Youcis Mar 05 '23 at 08:26
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    Fact 2: if one takes the natural action of $\Gamma_{\overline{k}}$ on $\Psi(G_{\overline{k}}$) which (essentially) comes by permuting the Borel pairs, then for any fixed Borel pair $(B,T)$ the projection map $\Psi(G_{\overline{k}})\to \Psi(G_{\overline{k}},B,T)$, which is an evidently an isomorphism of groups since each transition map in the system ${\Psi(G_{\overline{k}},B,T)}$ is an isomorphism, is a $\Gamma_k$-equivariant when the target is endowed with the $\ast$-action. – Alex Youcis Mar 05 '23 at 08:28
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    Not sure how helpful this is, but maybe try to apply the sesond method to the examples I went through in the answer to https://math.stackexchange.com/q/3902616/96384. – Torsten Schoeneberg Mar 05 '23 at 17:08
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    @AlexYoucis I think my question is perhaps more elementary, but thanks for telling me about based root datum with regards to $\Gamma$-action! (could I ask, where do you read about canonial based root datum? Most references I found only discussed root datum with respect to a choice of $(B,T)$. How useful is canonical based root datum in doing computation?). – Tengu Mar 06 '23 at 12:37
  • What I am confused is essentially this action $\sigma \cdot \chi = \sigma_{\mathbb{G}_m,k^s}\circ \chi \sigma_T^{-1}$ on $X^*(T)$, where when $G$ is quasi-split, I somehow cannot see this action appear in the second way. In the second way, if I read it correctly, the action is $\sigma \cdot \chi = \chi\circ \sigma_T$. – Tengu Mar 06 '23 at 12:37
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    $\chi \circ \sigma_T$ is not linear, is it? – Torsten Schoeneberg Mar 07 '23 at 02:17
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    @Tengu I am also slightly confused as to what you mean. Namely, the map $\mathrm{Aut}(G_{\overline{k}})\to \mathrm{Aut}(\Psi(B,T))$ has the source as the autormorphism of $G_{\overline{k}}$ over $\overline{k}$, and $\Gamma$ does not act in an $\overline{k}$-linear way. – Alex Youcis Mar 07 '23 at 09:44
  • @AlexYoucis TorstenSchoeneberg You two are right! Then I think my confusion is that what does this map $\Gamma \to \text{Aut}(G_{k^s})$ mean, as described in p.168 (after "$\Psi(G_{sep},B,T)$") of https://sites.duke.edu/jgetz/files/2022/04/Graduate_Text.pdf? – Tengu Mar 09 '23 at 10:26
  • @Tengu I am not sure exactly what they mean there. If you are OK with quasi-split then we don't have to introduce root datum, and presumably what they mean here is just that if $T$ is a maximally split maximal torus of $G$, so it is contained in a Borel subgroup $T\subseteq B\subseteq G$, then $\Gamma_F$ acts on $X^\ast(T_{\overline{F}})$ in the usual way I mentioned in the other post -- $\sigma\cdot \chi=\sigma\circ \chi\circ\sigma^{-1}$. This then actually presreves the roots of $G$ and $B$-positive roots, thus giving an action on the based root datum. – Alex Youcis Mar 09 '23 at 11:05
  • @AlexYoucis Great, that answers my question then. Thank you for the help! – Tengu Mar 09 '23 at 11:41

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