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  "body": "\n## Definition\n\nLet $k$ be an [[algebraic number field|MWBDFS2KJ5]] with [[adele ring|SEH68BA1QQ]] $\\mathbb{A}_k$, and for each place $v$ let $k_v$ be the [[completion|W168DS966V]] of $k$ at $v$.\nLet $X$ be a smooth projective geometrically integral variety over $k$, and let $\\operatorname{Br} X = H^2_{\\text{ét}}(X, \\mathbb{G}_m)$ be its [[Brauer group|BEW48TA9WZ]].\nFor each place $v$ let $\\operatorname{inv}_v \\colon \\operatorname{Br} k_v \\to \\Q/\\Z$ be the local invariant map of [[class field theory|BEKCEYHCHX]].\nFor $\\alpha \\in \\operatorname{Br} X$ and $P \\in X(k_v)$ let $\\alpha(P) \\in \\operatorname{Br} k_v$ be the pullback of $\\alpha$ along $P$.\nThe *Brauer–Manin set* is\n$$X(\\mathbb{A}_k)^{\\operatorname{Br}} = \\set{(P_v)_v \\in X(\\mathbb{A}_k) : \\textstyle\\sum_v \\operatorname{inv}_v \\alpha(P_v) = 0 \\text{ for all } \\alpha \\in \\operatorname{Br} X}.$$\nThere is a *Brauer–Manin obstruction to the Hasse principle* for $X$ if $X(\\mathbb{A}_k) \\neq \\emptyset$ and $X(\\mathbb{A}_k)^{\\operatorname{Br}} = \\emptyset$.\n\n## Remarks\n\nSince $X$ is projective, $X(\\mathbb{A}_k) = \\prod_v X(k_v)$.\nFor fixed $\\alpha$, the term $\\operatorname{inv}_v \\alpha(P_v)$ is $0$ for all but finitely many $v$, so the sum is finite.\nThe reciprocity law says that $\\sum_v \\operatorname{inv}_v \\beta = 0$ for every $\\beta \\in \\operatorname{Br} k$.\nApplied to $\\beta = \\alpha(P)$ for a [[rational point|SWJYG87QQH]] $P \\in X(k)$, it gives the inclusion\n$$X(k) \\subseteq X(\\mathbb{A}_k)^{\\operatorname{Br}} \\subseteq X(\\mathbb{A}_k).$$\nSo a Brauer–Manin obstruction implies $X(k) = \\emptyset$ although $X$ has points over every $k_v$.\n\nExample: suppose that $\\operatorname{Br} X$ is the image of $\\operatorname{Br} k$, as when $X$ is a projective space.\nThen 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.\nNon-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.\n",
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