# Quotienting $G(\mathbb{Q})_{+}$ by $G^{\text{sc}}(\mathbb{Q})$ and inner forms

Let $$G/\mathbb{Q}$$ be a connected reductive group, let $$G^{\text{ad}}$$ be the adjoint group, let $$G^{\text{der}}$$ be the derived group and let $$\rho\colon G^{\text{sc}} \to G^{\text{der}}$$ be the simply connected cover. Let $$G^{\text{ad}}(\mathbb{R})^{0}$$ be the identity component (in the real topology) of $$G^{\text{ad}}(\mathbb{R})$$, let $$G(\mathbb{R})_{+}$$ be the inverse image of $$G^{\text{ad}}(\mathbb{R})^{0}$$ under the natural map $$G(\mathbb{R}) \to G^{\text{ad}}(\mathbb{R})$$ and let $$G(\mathbb{Q})_{+}$$ be the intersection of $$G(\mathbb{R})_{+}$$ with $$G(\mathbb{Q}$$). Note that $$\rho(G^{\text{sc}}(\mathbb{R})) \subset G(\mathbb{R})_{+}$$ because $$G^{\text{sc}}(\mathbb{R})$$ is connected.

Question: Let $$H/\mathbb{Q}$$ be an inner form of $$G$$ and let the notation be as above, is there a `natural isomorphism' of abelian groups $$H(\mathbb{Q})_{+}/H^{\text{sc}}(\mathbb{Q}) \simeq G(\mathbb{Q})_{+}/G^{\text{sc}}(\mathbb{Q})$$?

When $$G^{\text{sc}}=G^{\text{der}}$$ this is true: Let $$Z$$ be the center of $$G$$, let $$\nu:G \to D$$ be the maximal abelian quotient of $$G$$, define $$D(\mathbb{R})^{\dagger}:=\operatorname{Im}(Z(\mathbb{R}) \to D(\mathbb{R}))$$ and let $$D(\mathbb{Q})^{\dagger}=D(\mathbb{R})^{\dagger} \cap D(\mathbb{Q})$$. Then Lemma 5.10 of https://www.jmilne.org/math/xnotes/svi.pdf shows that \begin{align} \nu(G(\mathbb{Q})_{+}) = D^{\dagger}(\mathbb{Q}). \end{align} Note that because $$H$$ is an inner form of $$H$$, we can identify its center with $$Z$$ and its maximal abelian quotient with $$\mu:H \to D$$. Applying the lemma again we see that \begin{align} \mu(H(\mathbb{Q})_{+})=D^{\dagger}(\mathbb{Q}) \end{align} and we are done. I have tried to follow a similar strategy in the general case, by which I mean comparing to $$Z(\mathbb{Q})/Z^{\text{sc}}(\mathbb{Q})$$, but I havent been able to get it to work.

The answer is Yes. We denote $$K(G)=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$$. We compute $$K(G)$$; see the corollary below. It is clear from the corollary that $$K(G)$$ is canonically isomorphic to $$K(H)$$.

We will use Section 3 of M. Borovoi, Abelian Galois cohomology of reductive groups. Memoirs of the AMS 132 (1998), No. 626, although all necessary results can be found in Deligne's paper Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques, Proc. Sympos. Pure Math. 33, Part 2, pp. 247–289.

We consider the crossed module $$(G^{\rm sc}\to G)$$ and the hypercohomology $$H^0_{\rm ab}({\mathbb Q},G):=H^0({\mathbb Q},G^{\rm sc}\to G),$$ where $$G$$ is in degree 0; see the Memoir. By definition $$H^0_{\rm ab}({\mathbb Q},G)$$ is a group. We consider the abelian crossed module $$(Z^{\rm sc}\to Z)$$, where $$Z=Z(G)$$ and $$Z^{\rm sc}=Z(G^{\rm sc})$$. The morphism of crossed modules $$(Z^{\rm sc}\to Z)\,\longrightarrow\,(G^{\rm sc}\to G)$$ is a quasi-isomorphism, and hence it induces a bijection on hypercohomology, permitting us to identify $$H^0_{\rm ab}({\mathbb Q},G)$$ with the abelian group $$H^0({\mathbb Q},Z^{\rm sc}\to Z)$$. We conclude that $$H^0_{\rm ab}({\mathbb Q},G)$$ is naturally an abelian group and that it does not change under inner twisting of $$G$$.

The short exact sequence $$1\to(1\to G)\to (G^{\rm sc}\to G)\to (G^{\rm sc}\to 1)\to 1$$ (where $$(G^{\rm sc}\to 1)$$ is not a crossed module) induces a hypercohomology exact sequence $$G^{\rm sc}({\mathbb Q})\to G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc}),$$ where $${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$$ is the abelianization map. This permits us to identify $$G({\mathbb Q})/\rho G^{\rm sc}({\mathbb Q})$$ with the kernel $${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$ (yes, this kernel is a subgroup of the abelian group $$H^0_{\rm ab}({\mathbb Q},G)$$ ). This kernel might change under inner twisting of $$G$$, because $$H^1({\mathbb Q},G^{\rm sc})$$ changes under inner twisting.

By definition, $$G({\mathbb R})_+=Z({\mathbb R})\cdot\rho G^{\rm sc}({\mathbb R})$$, and hence $$G({\mathbb R})_+/\rho G^{\rm sc}({\mathbb R})={\rm ab}^0(Z({\mathbb R}))\subset {\rm ker}[ H^0_{\rm ab}({\mathbb R},G)\to H^1({\mathbb R}, G^{\rm sc})].$$ We see that $$K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$$ can be identified with the preimage of $${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$$ in $${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$.

Lemma. The preimage of $${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$$ in $${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$ coincides with the preimage of $${\rm ab}^0(Z({\mathbb R}))$$ in $$H^0_{\rm ab}({\mathbb Q},G)$$.

Proof. Let $$\xi\in H^0_{\rm ab}({\mathbb Q},G)$$ lie in the preimage of $${\rm ab}^0(Z({\mathbb R}))\subset {\rm ker}[ H^0_{\rm ab}({\mathbb R},G) \to H^1({\mathbb R}, G^{\rm sc})].$$ Then the image of $$\xi$$ in $$H^1({\mathbb R},G^{\rm sc})$$ is trivial, and therefore, the image of $$\xi$$ in $$H^1({\mathbb Q},G^{\rm sc})$$ lies in the kernel of the localization map $$H^1({\mathbb Q}, G^{\rm sc})\to H^1({\mathbb R},G^{\rm sc}).$$ By the Hasse principle for simply connected groups, this kernel is trivial. Thus the image of $$\xi$$ in $$H^1({\mathbb Q},G^{\rm sc})$$ is trivial, and hence $$\xi$$ lies in the preimage of $${\rm ab}^0(Z({\mathbb R}))$$ in $${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$, as required.

Corollary. The abelianization map $${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$$ with kernel $$\rho G^{\rm sc}({\mathbb Q})$$ induces a canonical isomorphism between the abelian groups $$K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$$ and the preimage of $${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$$ in $$H^0_{\rm ab}({\mathbb Q},G)$$.

• Thank you, this is exactly what I was looking for! Mar 20, 2020 at 10:36