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I'm looking for references that can justify to what extent is the following statement true:

Statement. Let $X$ be a smooth geometrically integral curve over a number field $k$. Then the Brauer-Manin obstruction is the only obstruction to the Hasse principle for rational points, i.e., we have $$X(k) = \emptyset \implies X(\mathbb{A}_k)^{\mathrm{Br}} = \emptyset.$$

Note that if $X$ is a projective curve, then the statement holds for genus 0 curves. Indeed, such curves $C_0$ satisfy the Hasse principle so if $C_0(k) = \emptyset$, then we have nothing to prove. On the other hand, if $C_0(k) \neq \emptyset$, then $C_0 \cong \mathbb{P}^1_k$ and so $C_0(k)$ is dense in $C_0(\mathbb{A}_k)$.

Question 1. What about genus zero affine curves with no rational points? Does it immediately follow that the statement holds for them as well?

For the case of elliptic curves $E$, a necessary and sufficient condition for the statement to hold would that the divisible part of Sha is trivial. This follows from the projective limit of the usual exact sequence involving $E(k)$, the $n$-Selmer groups and the $n$-torsion of Sha. For projective genus one curves $C_1$ without a rational point, they correspond to a class of $H^1(k,J)$, where $J$ is the Jacobian of $C_1$.

Question 2. For such a $k$-torsor $C_1$ under $J$, is there anything known about its Brauer set? Hence or otherwise, can we deduce anything about the statement in relation to the case of the affine $C_1$?

Note that the integral analogue of the statement is not true in general for the affine part of elliptic curves with a rational point. A counterexample can be found in the paper The Brauer-Manin obstruction for integral points on curves by Harari and Voloch.

Question 3. This is vague, but if any sort of information can be provided about the statement for affine curves of genus $\geq 2$, I would be glad to check them out.

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    $\begingroup$ Every genus $0$ smooth projective curve over an infinite field has a Zariski dense set of rational points. Thus, if a genus $0$ smooth affine curve has no rational points then also its projective model has no rational points. $\endgroup$ Aug 7, 2021 at 10:46
  • $\begingroup$ Minor correction: "Every genus 0 smooth projective curve over an infinite field with a rational point has a Zariski dense...." $\endgroup$ Aug 8, 2021 at 10:21

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