Brauer–Manin obstruction

Remarks

Since $X$ is projective, $X(\mathbb{A}_k) = \prod_v X(k_v)$. For fixed $\alpha$, the term $\operatorname{inv}_v \alpha(P_v)$ is $0$ for all but finitely many $v$, so the sum is finite. The reciprocity law says that $\sum_v \operatorname{inv}_v \beta = 0$ for every $\beta \in \operatorname{Br} k$. Applied to $\beta = \alpha(P)$ for a rational point $P \in X(k)$, it gives the inclusion $$X(k) \subseteq X(\mathbb{A}_k)^{\operatorname{Br}} \subseteq X(\mathbb{A}_k).$$ So a Brauer–Manin obstruction implies $X(k) = \emptyset$ although $X$ has points over every $k_v$.

Example: suppose that $\operatorname{Br} X$ is the image of $\operatorname{Br} k$, as when $X$ is a projective space. Then the sum vanishes for every adelic point by reciprocity, so $X(\mathbb{A}_k)^{\operatorname{Br}} = X(\mathbb{A}_k)$ and there is no obstruction. Non-example of a sufficient test: Skorobogatov constructed a surface with $X(\mathbb{A}_k)^{\operatorname{Br}} \ne \emptyset$ and $X(k) = \emptyset$, so the Brauer–Manin obstruction does not explain all failures of the Hasse principle.