Let $k$ be a number field and $X$ be a $k$-scheme. Let $G$ be a linear algebraic group over $k$ and let $f: Z \to X$ be a $G\_X$-torsor ($G\_X = G \times\_k X)$. We can twist the torsor $f$ by 1-cocycles in $Z^1(k,G)$ (see Skorobogatov's "Torsors and rational points" for more details), and we denote the twist of $f$ by $[\tau] \in H^1(k,G)$ by $f^\tau : Z^\tau \to X$, which is a $G\_X^\tau$-torsor. Now to my question.



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> Is it true that $X(\mathbb{A}\_k) = \bigcup\_{[\tau] \in H^1(k,G)} f^\tau(Z^\tau(\mathbb{A}\_k))$?
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I remark the level of generality of $X$ in the above.



Decomposing adelic points using torsors
The answer is **No.** Here is a counterexample.


Take $k = \mathbb Q$, $X = {\mathbb G}\_{\text{m}}$, $G = \mu\_2$ and $f \colon Z \to X$ the squaring map ${\mathbb G}\_{\text{m}} \to {\mathbb G}\_{\text{m}}$. Then $H^1(k, G) = {\mathbb Q}^\times/\text{squares}$, and its elements can be represented by squarefree integers $d$. If $P = (P\_v) \in f^d(Z^d({\mathbb A}\_{\mathbb Q}))$, then $P\_v = d a\_v^2$ for some $a\_v \in {\mathbb Q}\_v^\times$, for all places $v$. If you consider $P\_p = p$ for primes $p$ and $P\_\infty = 1$, then to have $P\_p = d a\_p^2$, you need $d$ to be divisble by $p$, for all $p$, which is impossible.


**Edit:** I just realized that this is actually not a counterexample, since the point is not an adelic point on ${\mathbb G}\_{\text m}$ (it is not the case that $P\_p \in {\mathbb Z}\_p^\times$ for all but finitely many $p$). But one can construct a counterexample using a similar idea. Enumerate the squareferee integers in some way as $(d\_n)\_{n \in {\mathbb N}}$. For each $n$, pick a prime $p\_n$ such that the sequence $(p\_n)$ is increasing and $d\_n$ is not divisible by $p\_n$. Then let $P\_{p\_n}$ be $d\_n$ times a quadratic nonresidue mod $p\_n$. Fill the other components with $1$'s. Then each possible $d\_n$ is ruled out by some place $p\_n$, so the point is not in the image of the adelic points of any twist.


