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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.

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  • $\begingroup$ 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 $T$? $\endgroup$ Commented May 12, 2022 at 8:08
  • $\begingroup$ Exactly ....... $\endgroup$
    – cgb5436
    Commented May 12, 2022 at 14:53
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    $\begingroup$ 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. $\endgroup$ Commented May 12, 2022 at 18:24
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    $\begingroup$ 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. $\endgroup$ Commented May 12, 2022 at 18:41
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    $\begingroup$ Theorem 21 in Section 6.5 of the Russian edition of Platonov and Rapinchuk is Theorem 6.21 of the English edition. $\endgroup$ Commented May 12, 2022 at 18:49

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