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

Let $k$ be a number field, and let $G$ be a split simply connected algebraic group over $k$. Let $\Omega_k$ denote the set of places of $k$. Let $T$ be a maximal torus of $G$ (defined over $k$). Consider the second Tate-Shafarevich group $$Ш^2(k,T)=\ker\left[H^2(k,T)\to\prod_{v\in\Omega_k} H^2(k_v,T)\right].$$

Question. Is it true that $Ш^2(k,T)=1$ for any maximal torus $T$ of any split simply connected $k$-group $G$?

Note that the answer is YES for $G=SL_n$. Indeed, then $T=\ker[N: S\to {\mathbb G}_m ]$, where $S$ is a maximal torus in $GL_n$. We have a cohomology exact sequence $$1\to H^2(k,T)\to H^2(k,S)\to H^2(k,{\mathbb G}_m)$$ and similar exact sequences for any place $v$ of $k$. We see that $Ш^2(k,T)$ embeds into $Ш^2(k,S)$. Since $S$ is a product of tori of the form $R_{L_i/k}\, {\mathbb G}_{m,L_i}$ where each $L_i$ is a finite extension of $k$, we have $$Ш^2(k,S)=\prod_i Ш^2(k,R_{L_i/k}\,{\mathbb G}_{m,L_i})=\prod_i Ш^2(L_i,{\mathbb G}_m)=1,$$ hence $Ш^2(k,T)=1$.

• Presumably the same argument holds for $G=\text{Sp}_{2n}$. Aug 2 '16 at 18:22
• @JasonStarr: Could you please elaborate? Aug 2 '16 at 18:26
• "Could you please elaborate?" No, probably I cannot. I just wrote something that came into my head. I will think it through more carefully and write more if I have an actual argument. Aug 2 '16 at 18:34
• Here is my line of thinking, probably wrong. Let $(\bullet)^t:\text{GL}_{2n}\to \text{GL}_{2n}$ be the "transpose" involution associated to the perfect nondegenerate pairing, so that $\text{Sp}_{2n}$ is the fixed point locus of the involution $\iota(B) = (B^t)^{-1}$. Let $T$ be a maximal torus in $\text{Sp}_{2n}$. Denote by $C_{GL}(T)$ the centralizer of $T$ in $\text{GL}_{2n}$. Consider the morphism $f:C_{GL}(T)\to C_{GL}(t)$ by $f(B) = \iota(B)B$. Argue (geometrically) that $f$ is a group homomorphism onto $T$ that restricts to the squaring homomorphism on $T$ . . . Aug 2 '16 at 19:14
• @JasonStarr: Yes, it seems that any maximal torus $T$ of $\mathrm{Sp}_{2n}$ is a product $\prod T_i$, where each $T_i$ is a Weil restriction of a one-dimensional torus. The argument above shows that $Ш^2$ is trivial for a one-dimensional torus, and hence, for $T$. Aug 2 '16 at 19:19

This is a partial answer inspired by comments of Jason Starr. I show that the answer is YES for $G=\mathrm{Sp}_{2n}$ (and also for the classical non-simply-connected groups $\mathrm{SO}_{2n+1}$ and $\mathrm{SO}_{2n}$).
Let $T\subset G$ be a maximal torus. The Galois group ${\Gamma}=\mathrm{Gal}(\bar k/k)$ acts on the character group $X(T)$ via $\mathrm{Aut}\, R(G_{\bar k},T_{\bar k})$, and hence, via the Weyl group $W$. We know that $$W=(\pm 1)^n\rtimes S_n.$$ We have $G\subset \mathrm{GL}(V)$, where $V=k^{2n}$. Our torus has $2n$ different eigenspaces in $V$ with characters $\chi_1,-\chi_1,\chi_2,-\chi_2,\dots$, which are naturally divided into pairs. The Galois group ${\Gamma}$ acts on these characters and on the pairs. Let ${\Gamma}_1$ denote the stabilizer of the pair $(\chi_1,-\chi_1)$ in ${\Gamma}$, and let ${\Gamma}'_1$ denote the stabilizer of $\chi_1$ in ${\Gamma}$. Let $K_1\subset{\bar k}$ and $K'_1\subset {\bar k}$ denote the subfields corresponding to ${\Gamma}_1$ and ${\Gamma}'_1$, respectively, then either $K'_1=K_1$ or $K'_1$ is a quadratic extension of $K_1$. Let $X=X(T_{\bar k})$ denote the character group of $T$, and let $X_1$ denote the subgroup of $X$ generated by the characters $\sigma\chi_1$ for $\sigma\in\Gamma$. Then $X_1$ is a $\Gamma$-invariant direct factor of $X$, and we obtain a direct factor $T_1$ of $T$, where $$T_1=R_{K_1/k}\, S_1,$$ and $S_1$ is a certain one-dimensional torus over $K_1$. Namely, if ${\Gamma}'_1={\Gamma}_1$, then $S_1$ is a split one-dimensional $K_1$-torus; otherwise it is the nonsplit one-dimensional $K_1$-torus that splits over the quadratic extension $K'_1/K_1$. We see that $T$ is a product of the Weil restrictions $T_i=R_{K_i/k}\, S_i$ of one-dimensional tori $S_i$ corresponding to the orbits of the Galois group in the set of pairs of characters of $T$. Since $Ш^2$ is trivial for a one-dimensional torus, it is trivial for $T$.