# Hasse principle for $H^2$ of a maximal torus of a connected quasisplit group?

Let $$k$$ be a number field and let $$G$$ be a quasisplit reductive algebraic group over $$k$$. Does there exist a maximal torus in $$G$$ such that the Hasse principle in dimension $$2$$ holds, i.e., such that the map $$H^2(k,T) \to \prod_v H^2(k_v,T)$$ is injective? It is true if $$G$$ is split or if $$G$$ is adjoint. If $$G$$ is simply connected, it is true whether $$G$$ is quasisplit or not.

• Just to clarify: By "Hasse principle in dimension 2" you are asking whether the map $H^2(k,T) \to \prod_v H^2(k_v,T)$ is injective for some maximal torus $T$? Commented May 12, 2022 at 8:08
• Exactly ....... Commented May 12, 2022 at 14:53
• No. Let $G=T$ be a $k$-torus. Then $B=T$ is a Borel subgroup defined over $k$, and hence $G$ is a quasi-split reductive $k$-group. The only maximal torus in $G$ is $T=G$. I think that there exist a $k$-torus for which the Hasse principle for $H^2$ fails. Commented May 12, 2022 at 18:24
• Yes, if $G$ is semisimple, not necessarily quasisplit. Indeed, let $v$ be a nonarachimedean place of $k$. Then there exists an anisotropic maximal $k_v$-torus $T_v\subset G_{k_v}$; see Platonov and Rapinchuk, Section 6.5, Theorem 21 of the Russian edition. There exist a maximal $k$-torus $T\subset G$ that is conjugate to $T_v$ over $k_v$. Thus $T_{k_v}$ is anisotropic. It follows that $Ш^2(k,T)=1$; see formula (1.9.3) in Sansuc's paper: Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80. Commented May 12, 2022 at 18:41
• Theorem 21 in Section 6.5 of the Russian edition of Platonov and Rapinchuk is Theorem 6.21 of the English edition. Commented May 12, 2022 at 18:49