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I am trying to understand how the Galois action on the $\bar{F}$-points of a connected, reductive group is related to the Galois action on the group of characters $X^*(T)$ of a maximal torus. A closely related question has already been asked at this post: Galois action on character group

The answer given there almost makes sense to me, but the Galois action on $X^*(T)$ is defined in terms of a Galois action on $\mathbb{G}_m$; is this the usual Galois action on $F$-points, or does the action on $\mathbb{G}_m$ arise from the (established) Galois action on $T(\bar{F})$? A non-trivial example (say of a unitary group) would also be greatly appreciated.

itsricobitches
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1 Answers1

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$\newcommand{\Hom}{\mathrm{Hom}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Spec}{\mathrm{Spec}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\bb}[1]{\mathbb{#1}}$

When I write $\overline{F}$ below I mean the separable closure of $F$.

So, by definition $X^\ast(T):=\mathrm{Hom}_{\ov{F}-\mathrm{grps}}(T_{\ov{F}},\mathbb{G}_{m,\ov{F}})$. The action of $\sigma\in\Gamma_F$ (I'll use $\Gamma_F$ for the absolute Galois group) is defined as follows:

$$\sigma\cdot f:= \sigma_{\bb{G}_{m,\ov{F}}}\circ f\circ \sigma_{T_{\ov{F}}}^{-1}$$

where for any $F$-scheme $X$ we define $\sigma_{X_{\ov{F}}}$ to be the natural map $X_{\ov{F}}\to X_{\ov{F}}$ which, when one writes $X_{\ov{F}}:= X\times_{\Spec(F)}\Spec(\ov{F})$, is trivial on the $X$-component and is $\Spec(\sigma^{-1})$.

So, let's do an example very explicitly. Let's set $F$ to be any field and $E/F$ to be a degree $2$ Galois extension. Then, we can form the group $U(1)_{E/F}$, or just $U(1)$, defined as the kernel of the natural map $\mathrm{Res}_{E/F}\bb{G}_{m,E}\to \bb{G}_{m,F}$ given by the norm. This is a one-dimensional torus. If we write $E=F(\sqrt{d})$ where $d\in F$, then as an affine scheme we can write $U(1)\cong \Spec(F[a,b]/(a^2-db^2-1))$ where Hopf algebra structure is the map

$$F[a,b]/(a^2-db^2-1)\to F[a_1,b_1,a_2,b_2]/(a_1^2-db_1^2-1,a_2^2-db_2^2-1)$$

(where we have made the identification of $F[a,b]/(a^2-db^2-1)\otimes_F F[a,b]/(a^2-db^2-1)$ with $F[a_1,b_1,a_2,b_2]/(a_1^2-db_1^2-1,a_2^2-db_2^2-1)$) given by

$$a\mapsto a_1a_2+db_1b_2,\quad b\mapsto ab_2+ba_2$$

Let us calculate its character lattice. Well, we have a natural isomorphism $\varphi:U(1)_{\ov{F}}\to \bb{G}_{m,\ov{F}}$ given on coordinate rings (using $t$ for the paramter of $\bb{G}_{m,F}$) by

$$\ov{F}[t]\to \ov{F}[a,b]/(a^2-db^2-1):t\mapsto a+b\sqrt{d}$$

with inverse

$$\Spec(\varphi^{-1}):\ov{F}[a,b]/(a^2-db^2-1)\to \ov{F}[t]$$

given by

$$a\mapsto \frac{1}{2}(t+t^{-1}),\quad b\mapsto \frac{1}{2\sqrt{d}}(t-t^{-1})$$

It's then not hard to see that $\Hom_{\ov{F}-\mathrm{grps}}(U(1)_{\ov{F}},\bb{G}_{m,\ov{F}})=\Z\cdot\varphi\cong \Z$. Let's now see how $\Gamma_F$ acts on $\Z\cdot \varphi$. Of course, it suffices to specify how it acts on $\varphi$. Now, by definition we have that $\sigma\cdot \varphi$ is the map $ \sigma_{\bb{G}_{m,\ov{F}}}\circ \varphi\circ \sigma_{T_{\ov{F}}}^{-1}$. Let's see how this acts on coordinate rings. Namely, this map corresponds to a map $$\ov{F}[t]\to \ov{F}[a,b]/(a^2-db^2-1)$$ which can be thought of as factorized as \begin{equation} $$\ov{F}[t]\xrightarrow{\sigma^{-1}}\ov{F}[t]\xrightarrow{\Spec(\varphi)}\ov{F}[a,b]/(a^2-db^2-1)\xrightarrow{\sigma}\ov{F}[a,b]/(a^2-db^2-1)$$

and we'll call these arrows 1, 2, and 3 (labeled left to right). Well, let's see where $t$ goes. In the first map it maps to $t$ again. Under the second map it maps to $a+b\sqrt{d}$ and under the last map it maps to $a+b\sigma(\sqrt{d})$.

In particular, note that $\Gamma_F$ acts through the quotient $\mathrm{Gal}(E/F)$ and if we denote by $\sigma$ the non-trivial element of $\mathrm{Gal}(E/F)$ this satisfies that $\sigma\cdot \varphi$ sends $t$ to $a-b\sqrt{d}$. Note though that this is just $(a+b\sqrt{d})^{-1}$ and so it's not hard to see that $\sigma\cdot \varphi=-\varphi$ (our groups are multiplicatively written so it might be more natural to write $\varphi^{-1}$ but this could be confused with the inverse of $\varphi$ so I wrote it additively).

In other words, $X^\ast(U(1))$ is the $\Gamma_F$-module $\mathbb{Z}$ where the $\Gamma_F$ structure is the composition $$\Gamma_F\twoheadrightarrow \mathrm{Gal}(E/F)\to \mathrm{Aut}(\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$$

where the arrow $\mathrm{Gal}(E/F)\to\mathbb{Z}/2\mathbb{Z}$ is the obvious isomorphism.

To see the relationship between $\sigma_{T_{\ov{F}}}$ and $T(\ov{F})$ for a torus $T$ over $F$ note that $T(\ov{F})$ is nothing other than $\mathrm{Hom}_F(\mathrm{Spec}(\ov{F}),T)$. But, this is the same as $\Hom_{\ov{F}}(\Spec(\ov{F}),T_{\ov{F}})$. By the above we then know how $\Gamma_F$ should act on $f\in\Hom_{\ov{F}}(\Spec(\ov{F}),T_{\ov{F}})$:

$$\sigma\cdot f:= \sigma_{T_{\ov{F}}}\circ f\circ \sigma_{\Spec(F)_{\ov{F}}}^{-1}$$

Let's do two examples to see that this is what you should get:

  • If $T$ is $\mathbb{G}_{m,F}$ then $T(\ov{F})=\ov{F}^\times$ and for $\alpha\in T(\ov{F})$ you expect that $\sigma_{T_{\ov{F}}}(\alpha)=\sigma(\alpha)$. Well, note that in scheme world this $\alpha\in\ov{F}$ corresponds to the map $\Spec(\ov{F})\to T_{\ov{F}}$ that corresponds to $\ov{F}$-algebra map $\ov{F}[t,t^{-1}]\to \ov{F}$ that takes $t$ to $\alpha$. Note then that $\Spec(\sigma\cdot f)=\Spec(\sigma_{\Spec(F)_{\ov{F}}}^{-1})\circ \Spec(f)\circ\Spec(\sigma_{T_{\ov{F}}})$ takes (using the sequence of maps on the right hand side) $t$ iteratively to $t$, then $\alpha$, then finally $\sigma(\alpha)$. So you get what you expect.
  • If $T=U(1)_{E/F}$ then $T(\ov{F})=\{(\alpha,\beta)\in\ov{F}^2:\alpha^2-d\beta^2=1\}$. You expect that $\sigma$ acts on $(a,b)$ as $(\sigma(a),\sigma(b))$. To see this, note that $(\alpha,\beta)$ corresponds to the $\ov{F}$-algebra map $\ov{F}[a,b]/(a^2-db^2-1)\to \ov{F}$ taking $a$ to $\alpha$ and $b$ to $\beta$. Note then that if you consider $\Spec(\sigma\cdot f)=\Spec(\sigma_{\Spec(F)_{\ov{F}}}^{-1})\circ \Spec(f)\circ\Spec(\sigma_{T_{\ov{F}}})$ this takes $a$ to $\sigma(\alpha)$ and $b$ to $\sigma(\beta)$, just as you'd expect.
Alex Youcis
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  • Thank you! It took me a bit to crack the AG approach but this mostly answers my question. One lingering query - does this approach mean, essentially, that every connected, reductive group can be realized in some basis as the Galois-fixed points, where $\mathrm{Gal}(\bar{F}/F)$ acts on the right factor of the fibre product as you have described? – itsricobitches Aug 12 '19 at 06:56
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    $\newcommand{\ov}[1]{\overline{#1}}$In general, there is an equivalence of categories between quasi-projective varieties over $k$ and quasi-projective varieties over $\ov{k}$ with a 'continuous' $\mathrm{Gal}(\ov{k}/k)$-action given by sending $X$ to $X_{\ov{k}}$ with the action of $\mathrm{Gal}(\ov{k}/k)$ I gave above. If you specialize this to finite type groups (which are all quasi-projective by Chevalley's extension theorem) you get an analagous statement for groups. – Alex Youcis Aug 12 '19 at 08:04
  • Hi, could I ask for the references for this? I tried to read about L-groups and at all the places I know discussing about this, they never explicitly write down the action of $\Gamma$ on $X_(T)$ or $X^(T)$. Thanks! – Tengu Feb 19 '23 at 11:10
  • And when I tried to make a guess about it, I thought it should be $\sigma \cdot f= f\circ \sigma_{T}^{-1}$ ... – Tengu Feb 19 '23 at 11:17
  • @Tengu 天狗様へ、do you mean literally the action of $\Gamma$ on $X_\ast(T)$/$X^\ast(T)$ for an abstract torus, or do you mean the action on $X^\ast(T)$ that is related to the action on the based root datum that you pick when defining the action of $\Gamma$ on the $L$-group (they only coincide when $G$ is quasi-split and $T$ is a maximally split maximal torus in $G$)? – Alex Youcis Feb 19 '23 at 11:24
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    @Tengu What you wrote down will not be a morphism over $\overline{F}$ and so not be an element of $X^\ast(T)$. For instance, if $f=\mathrm{id}$ on $\sigma$ then your map $\sigma\cdot f$ will send $\alpha$ to $\sigma^{-1}(\alpha)$ -- that's not a morphism of algebaric groups. – Alex Youcis Feb 19 '23 at 11:25
  • I meant the action of $\Gamma$ on $X_(T)$ and $X^(T)$ for abstract torus. But I see now why it is the 'canonical' action. – Tengu Feb 19 '23 at 11:28
  • I just want to make sure, the action of $\Gamma$ on $X_(T)$ as abstract torus should be the same action on the based root datum $(X_(T),\Phi^{\vee}, X^*(T), \Phi)$ for quasi-split $G$ with split maximal torus $T$ of $G$ over separable closure right? – Tengu Feb 19 '23 at 11:34
  • @Tengu Is your torus chosen over the algebraic closure? There is a so-called 'canonical based root datum' which is given by taking the limit over the based root data associated to all Borel pairs $(\mathcal{B},\mathcal{T})$ in $G_{\overline{F}}$. One might then ask the following question: if I project from $(\mathcal{B},\mathcal{T})$ to $(B,T_{\overline{F}})$, then how does the Galois action on the canonical based root datum relate to the action on based root datum of $G_{\overline{F}}$ relative to $(B,T_{\overline{F}})$? – Alex Youcis Feb 19 '23 at 11:39
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    In particular, do you get the right action when you take the usual action on $X^\ast(T)$ (and $X_\ast(T)$)? The answer is that you get the usual action when $G$ is quasi-split, and $T$ is a maximally split maximal torus of $G$ (i.e. is contained in a Borel of $G$!). In general the two differ, and what you instead get is what is called the $\ast$-action. See this note: http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/*-action.pdf – Alex Youcis Feb 19 '23 at 11:40
  • Hi Alex, thanks for the explanation! So here is what I understand so far: For quasi-split $G$, we have Borel $B$ defined over $k$, so the usual action of $\Gamma$ on $X^(T_{k^s})$ (i.e. the one you described in this post) will preserve the positive roots (or the base), which will induce a map on the based root datum. But if $G$ is not quasi-split, i.e. has no Borel defined over $k$, the usual action on $X^(X_{k^s})$ will not preserve the base, so one needs to compose it with the action of the Weyl group in order to preserve the base, and this is the $*$-action. – Tengu Mar 03 '23 at 14:40
  • I have some another questions, if you don't mind. In Getz, Hahn book Intro to Automorphic rep, in the section of Langlands dual group, they describe the action of $\Gamma$ on the based root datum $\Psi(G_{k^s},B_{k^s},T_{k^s})$ for (connected) reductive $G$ over $k$ as the map $\Gamma \to \text{Aut}(G_{k^s})\to \Psi(G_{k^s},B_{k^s},T_{k^s})$. I suspect this should be the same as the $*$-action but I just couldn't figured out why. – Tengu Mar 03 '23 at 15:16
  • In particular, $\sigma\in \Gamma$ induces action $\sigma_{G_{k_s}}$ on $G_{k^s}$, whose restricts to an action on $T_{k_s}$ (suppose $G$ quasi-split), which gives an action on $X^*(T_{k_s})$ by sending $\chi\mapsto \chi\circ \sigma_{T_{k_s}}$. And here this is not the usual action of $\Gamma$ on $T_{k^s}$ (in your post)? – Tengu Mar 03 '23 at 15:16
  • @Tengu You should probably post this a separate question, the comments of another post are not a good place to discuss this. If you post it, I will try to answer. – Alex Youcis Mar 04 '23 at 02:58
  • Yes, you are right, I should have done that. I have relocated my question to https://math.stackexchange.com/q/4651837/58951. – Tengu Mar 04 '23 at 10:37