# A reductive group has a quasi-split inner form

Let $G$ be a connected, reductive group over a field $k$. Let $\Gamma = \textrm{Gal}(k_s/k)$. I think my question is better suited using the classical language: think of $G$ as an affine $\overline{k}$-variety with $k$-structure. This gives a continuous action of $\Gamma$ on $G$. A $1$-cocyle is a continuous map $c: \Gamma \rightarrow \textrm{Aut}(G)$ (automorphisms in the category of $\overline{k}$-group varieties) such that

$$c(\gamma \delta) = c(\gamma) \circ \gamma \circ c(\delta) \circ \gamma^{-1}$$

We say that $1$-cocycles $c$, $d$ are equivalent if there exists a $\psi \in \textrm{Aut}(G)$ such that $d(\gamma) = \psi^{-1} \circ c(\gamma) \circ \gamma \circ \psi \circ \gamma^{-1}$ for all $\gamma \in \Gamma$.

Let $\tilde{c}$ be the equivalence class of a $1$-cocycle. We say that $\tilde{c}$ is an inner form if there exists a $d \in \tilde{c}$ such that all the maps $d(\gamma) : \gamma \in \Gamma$ are inner automorphisms of $G$.

The equivalence classes of $1$-cocyles parameterize the possible $k$-structures on $G$, in the sense that if $c$ is a $1$-cocycle, then there exists an algebraic group $G_1$ over $k$ (hence a new action of $\Gamma$ on $G_1$), and an isomorphism $\phi: G \rightarrow G_1$ over $\overline{k}$, such that

$$c(\gamma) = \phi^{-1} \circ \gamma \circ \phi \circ \gamma^{-1}$$

for all $\gamma \in \Gamma$, and equivalent cocycles go to equivalent $k$-structures.

It is remarked in T.A Springer's article Reductive Groups (Proceedings in Symposia in Pure Mathematics, Vol. 33, 1979) that $G$ has a quasi-split inner form. In other words, there exists a quasi-split connected, reductive group $G_1$ over $k$, and an isomorphism $\phi: G \rightarrow G_1$ over $\overline{k}$, such that $\phi^{-1} \circ \gamma \circ \phi \gamma^{-1}$ is an inner automorphism of $G$ for all $\gamma \in \Gamma$.

Springer says that this follows from looking at the split-exact sequence

$$1 \rightarrow \textrm{Inn}(G) \rightarrow \textrm{Aut}(G) \rightarrow \textrm{Aut}(G,B,T,u_{\alpha} : \alpha \in \Delta) \rightarrow 1$$

where $u_{\alpha}$ is a "splitting," a choice of nonidentity element in each simple root subgroup $U_{\alpha} : \alpha \in \Delta$. I am familiar with the details about this exact sequence, but I don't see what this has to do with quasi-split inner forms. Is there a reference which gives more details on this? Or, even better, a hint on how to do this myself. Thank you.

Nothing is "better-suited to using the classical language"; if you cannot express things clearly via schemes then think harder about it until you can. Also, any connected reductive group over a field has a unique quasi-split inner form. (See Proposition 7.2.12 in the article Reductive Group Schemes in the Proceedings of the 2011 Luminy Summer School on SGA3 for a proof of the same result more generally over any semi-local base scheme.)

If $k_s/k$ is a separable closure and $H$ is the $k$-split form of $G$, with a split maximal $k$-torus $S$ and Borel $k$-subgroup $B\supset S$, there is a ${\rm{Gal}}(k_s/k)$-equivariant exact sequence of groups $$1 \rightarrow H^{\rm{ad}}(k_s)\to {\rm{Aut}}_{k_s}(H_{k_s})\stackrel{\pi}{\to} {\rm{Aut}}(R,\Delta)\to 1$$ where $(R,\Delta)$ is the based root datum for $(H,B,S)$. A pinning $\{u_{\alpha}\}$ identifies ${\rm{Aut}}(R,\Delta)$ with ${\rm{Aut}}_k(H,S,B,\{u_{\alpha}\})$ and thereby defines a "forgetful" homomorphic section $$\sigma:{\rm{Aut}}(R,\Delta) \rightarrow {\rm{Aut}}_k(H) \subset {\rm{Aut}}_{k_s}(H_{k_s})$$ to $\pi$.

Consider the class $[G] \in {\rm{H}}^1(k_s/k, {\rm{Aut}}_{k_s}(H_s))$ corresponding to $G$. The formalism of non-abelian Galois cohomology as in section 5 of Chapter I of Serre's book Galois Cohomology shows that the set of isomorphism classes of inner forms of $G$ (i.e., the image of ${\rm{H}}^1(k_s/k, G^{\rm{ad}}(k_s))$ in ${\rm{H}}^1(k_s/k, {\rm{Aut}}_{k_s}(G_{k_s})$) is identified with the ${\rm{H}}^1(\pi)$-fiber through $[G]$.

In particular, ${\rm{H}}^1(\sigma)({\rm{H}}^1(\pi)([G]))$ is a class in the same fiber as $[G]$, so for the existence of a quasi-split inner form of $G$ it suffices to show that all classes in the image of ${\rm{H}}^1(\sigma)$ are quasi-split. But $\sigma$ is defined via the identification of ${\rm{Aut}}(R,\Delta)$ with ${\rm{Aut}}_k(H,S,B, \{u_{\alpha}\})$ and so factors through the subgroup ${\rm{Aut}}_{k_s}(H_{k_s},B_{k_s}) \subset {\rm{Aut}}_{k_s}(H_{k_s})$. Thus, any class in the image of ${\rm{H}}^1(\sigma)$ corresponds to the isomorphism class of a $k$-group $H'$ obtained through ${\rm{Gal}}(k_s/k)$-twisting of $H$ preserving its Borel $k$-subgroup $B$, so by design $H'$ admits a Borel $k$-subgroup $B'$ and thus $H'$ is quasi-split over $k$.

The remaining claim (not posed in the question, but mentioned at the start of this answer and very important in practice) is that the quasi-split inner form is unique. That is, if $G_1$ and $G_2$ are quasi-split inner forms of $G$ then they are $k$-isomorphic. More specifically, if $G_1$ is the quasi-split inner form made via the above construction and $G_2$ is any quasi-split inner form then $G_2 \simeq G_1$. This lies deeper in the sense that it rests on a more closer study of the preceding construction.

First, it is an instructive exercise to prove in a clean way that $G_2$ is necessarily an inner form of $G_1$, so we can rename $G_1$ as $G$ to reduce to showing that if $G$ admits a Borel $k$-subgroup $B$ with $(G,B)$ built from the split $k$-form via the above procedure resting on $\sigma$ then any quasi-split inner form $G'$ of $G$ is $k$-isomorphic to $G$. For a given choice of $k_s$-isomorphism $f:G'_{k_s} \simeq G_{k_s}$ corresponding to inner-twisting cocycles realizing $G'$ from $G$ via Galois descent, if we postcompose with a $G(k_s)$-conjugation we get a cohomologous 1-cocycle that has the same "inner-twisting" property. Thus, it is harmless to arrange that $f(B'_{k_s}) = B_{k_s}$, so then the inner-twisting is valued in the $B_{k_s}$-stabilizer subgroup of $G^{\rm{ad}}(k_s)$. But this stabilizer is exactly $B^{\rm{ad}}(k_s)$ for $B^{\rm{ad}} = B/Z_G$ so it suffices to prove $${\rm{H}}^1(k,B^{\rm{ad}}) = 1.$$ In this way we can replace $G$ with $G^{\rm{ad}}$ without ruining any of our running hypotheses on $G$ but gaining that we may now assume $G$ is of adjoint type.

The $k$-unipotent radical of any parabolic $k$-subgroup of a connected reductive group is always $k$-split and so has vanishing ${\rm{H}}^1$. Thus, since $B = T \ltimes U$ for a maximal $k$-torus $T$ of $B$ and $U = \mathscr{R}_{u,k}(B)$, so $B/U \simeq T$, it suffices to show ${\rm{H}}^1(k, T) = 1$ when $G$ is quasi-split of adjoint type and made from its split form via the procedure resting on $\sigma$. For this we finally have to make a serious observation that uses that $G$ is of adjoint type and is constructed from its split form via $\sigma$: the $k$-torus $T$ is "induced" (i.e., $T \simeq {\rm{R}}_{k'/k}({\rm{GL}}_1)$ for a finite etale $k$-algebra $k'$), from which it is immediate via Hilbert 90 that the desired ${\rm{H}}^1$-vanishing holds.

Why is $T$ induced? By inspection of how $\sigma$ is made, $T$ is build from Galois-twisting of a split "adjoint torus" $S = {\rm{GL}}_1^{\Delta}$ through Galois action on ${\rm{X}}(S_{k_s}) = \mathbf{Z}^{\Delta}$ via a permutation action on $\Delta$. The Galois-orbits on $\Delta$ then make explicit that the associated $k$-form $T$ of $S$ is an induced torus. (Explicitly, if we pick a point in each Galois-orbit on $\Delta$ then the stabilizer in ${\rm{Gal}}(k_s/k)$ of each such base point is an open subgroup corresponding to a finite separable extension $k'_i/k$, and one shows $T \simeq \prod_i {\rm{R}}_{k'_i/k}({\rm{GL}}_1)$.)

• Thanks for your detailed answer. I am confused on a point in paragraph 4. Let's say I take the image $[c]$ of an element of $H^1(R,\Delta)$ under $H^1(\sigma)$. For any representative $1$-cocycle $c$ of the image $[c]$, I have that $c$ is a map $\textrm{Gal}(k_s/k) \rightarrow \textrm{Aut}(G,B,T)$. There exists a group $G'$ and an isomorphism $\psi: G \rightarrow G'$ over $\overline{k}$ such that $$c(\gamma) = \psi^{-1} \circ \gamma \circ \psi \circ \gamma^{-1}$$ I want to show $G'$ has a Borel which is invariant under $\gamma$.I know that $c(\gamma)(B) = B$ for all $\gamma$. – D_S May 16 '17 at 5:33
• So, I should do something to $B$ and transfer it over to a Borel subgroup $B_0$ of $G'$, and show that $\gamma B_0 = B_0$. Is this correct? – D_S May 16 '17 at 5:39
• Thinking conceptually in terms of Galois descent rather than too explicitly in the language of cocycles (in effect, using the reason that ${\rm{H}}^1(k_s/k, {\rm{Aut}}_{k_s}(X_{k_s}))$ is the set of isomorphism classes of "$k$-forms" of some "structure" $X$ over $k$) makes it obvious that such a Borel $k$-subgroup $B_0$ exists. – nfdc23 May 16 '17 at 6:16
• Sorry, I've been working on this all day and I still don't get it. You're saying that for my $k$-forms, I should be working the category of triples $(G,B_{k_s},T_{k_s})$ instead of just the single $G$? And a $k$-form would be a pair $((G',B_{k_s}', T_{k_s}'), \psi)$, where $\psi: G_{k_s} \rightarrow G_{k_s}'$ is an isomorphism which sends $B_{k_s}$ to $B_{k_s}', T_{k_s}$ to $T_{k_s}'$? – D_S May 17 '17 at 1:54
• Also, earlier you write $\sigma$ as a homomorphism of $\textrm{Aut}(R,\Delta)$ into $\textrm{Aut}_k H$. I assume this is the splitting map which is an isomorphism onto what I call $\textrm{Aut}(G,B,T, u_{\alpha})$. Are you saying that the elements of $\textrm{Aut}(G,B,T, u_{\alpha})$ are always defined over $k$? – D_S May 17 '17 at 1:57

The privileged element in $H^1(k, \mathrm{Aut}(G, B, T, (u_\alpha: \alpha \in \Delta) )$ corresponds to the rational form of $G$ which possess a Borel subgroup $B$ defined over $k$, i.e. the quasi-split form of $G$. Combining with the fact that every cocycle $z \in Z^1(k, \mathrm{Aut}(G))$ which lies in the image of the mapping $Z^1(k, G_{ad}) \to Z^1(k, \mathrm{Aut}(G) )$ is mapped to $1 \in Z^1(k, \mathrm{Aut}(G, B, T, (u_\alpha: \alpha \in \Delta) )$. So every connected reductive algebraic $k$-group $G$ has an inner form which is quasi-split.