Brauer-Manin obstruction on an open subset of an elliptic curve

Question

First a disclaimer. This is an old question that I considered years ago and that I recently remembered. Since I am no longer in active research it may be considered as 'idle curiosity', although I feel I could probably learn a good deal from an answer (and maybe others could find it instructive as well).

Now the question. Are there any examples of elliptic curves $E/\mathbb{Q}$ (or failing that over some other number field) of rank $0$, and an open subvariety $X\subset E$, such that $$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(X)} = \varnothing? $$ Here the left-hand side denotes the Brauer-Manin set of $X$.

Or more generally still, is it at all possible to use Brauer-Manin obstructions to prove the non-existence of rational points on curves outside of some closed (proper and non-empty) subset?

The main problem that I recall having is that if one deals with Brauer classes on $X$ that are ramified (i.e. that do not come from $\operatorname{Br}(E)$), the evaluation map $$ \operatorname{ev}_X : X(\mathbb{A}_\mathbb{Q}) \times \operatorname{Br}(X) \to \mathbb{Q}/\mathbb{Z} $$ is hard to describe, since for $\mathscr{A} \in \operatorname{Br}(X)$ with $X$ non-proper there are in general infinitely many primes such that the local evaluation map $\operatorname{ev}_{\mathscr{A},p} : X(\mathbb{Q}_p) \to \mathbb{Q}/\mathbb{Z}$ is not identically zero (this in contrast to the case where $X$ is projective). The usual way of computing Brauer-Manin obstructions is that one exhibits a set of Brauer classes $\mathscr{A}_i$ and shows that the images of the maps $\operatorname{ev}_{\mathscr{A}_i} :U_i \to \mathbb{Q}/\mathbb{Z}$ are disjoint from $0$, for some partition $U_i$ of $X(\mathbb{A}_{\mathbb{Q}})$. But this becomes difficult when $X$ is non-proper, since one needs to determine the images of an infinite set of local evaluation maps $\operatorname{ev}_{\mathscr{A}_i,p}$ (for each $i$). (I don't even know whether it somehow follows from this that the examples I am asking for simply do not exist.)

Answer 1

Yes you can get:

$$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(X)} = \varnothing $$

in fact we can prove

$$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)} = \varnothing $$

and we can even make this (possibly larger) set to be empty.

The Brauer Manin obstruction for Abelian varieties is nicely explained in section 4 of this article by Poonen en Voloch. And I will use the notation from this section without further explanation in this answer. Where it is useful to know that $A$ is an Abelian variety and $K$ a number field.

The key ingredients are the exact sequence

$$0 \to \widehat{A(K)} \to \widehat {\operatorname{Sel}} \to TШ \to 0$$

and the inclusions

$$\widehat{A(K)} = \overline{A(K)} \subseteq A(\mathbb{A}_K)^{\operatorname{Br}(A)}_{•} \subseteq \widehat {\operatorname{Sel}}$$ of Theorem $E$. If we take an $A=E$ an elliptic curve over $\mathbb{Q}$ of analytic rank 0. Then we know by work of Kolyvagin that it also has algebraic rank 0 and that $Ш$ is finite. If $Ш$ is finite then $TШ$ is trivial and the exact sequence and the above inclusions imply:

$$E(\mathbb Q) = \widehat{E(\mathbb Q)} = E(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}_{•}$$

If we take $S=E(\mathbb{Q})$ and $X$ to be the open subvariety $E \setminus S$ then I claim

$$ X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)} = \varnothing $$

Indeed suppose for contradiction that $x \in X(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}$ and let $x^f$ denote the finite part of $x$. Then $E(\mathbb Q) = E(\mathbb{A}_\mathbb{Q})^{\operatorname{Br}(E)}_{•}$ implies that $x^f$ comes from an element $s \in S$. However by definition of $X$ we have that the $x_v = s_v$ for at most finitely many places $v$ which is clearly incompatible with $x^f = s^f$.

For explicitness taking $X$ to be the open part obtained from the curve https://www.lmfdb.org/EllipticCurve/Q/11/a/1 by removing the origin gives an example.

This construction works more generally for $E$ over a number field $K$ whenever the rank is $0$ and $Ш$ is finite.