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  "body": "\n## Statement\n\nLet $M$ be a closed [[smooth manifold|F45VQ47GPV]] of dimension $n$, and let $[M] \\in H_n(M; \\Z/2\\Z)$ be its mod $2$ [[fundamental class|ZEA0PEM3B2]].\nFor each $k \\ge 0$, let $v_k \\in H^k(M; \\Z/2\\Z)$ be the unique class such that\n$$\\langle v_k \\smile x, [M] \\rangle = \\langle \\operatorname{Sq}^k x, [M] \\rangle \\quad \\text{for all } x \\in H^{n-k}(M; \\Z/2\\Z),$$\nwhere $\\operatorname{Sq}^k$ is the $k$-th [[Steenrod square|VR7RHP41M0]] and $\\smile$ is the [[cup product|JX47MS1R1H]].\nLet $v = 1 + v_1 + v_2 + \\cdots$ be the total Wu class.\nThen the total [[Stiefel–Whitney class|JQCK7A210W]] of the tangent bundle of $M$ is\n$$w(M) = \\operatorname{Sq}(v), \\quad \\text{that is,} \\quad w_k(M) = \\sum_{i=0}^{k} \\operatorname{Sq}^{i}(v_{k-i}).$$\n\n## Remarks\n\nThe classes $v_k$ exist and are unique by Poincaré duality with $\\Z/2\\Z$ coefficients.\nThe proof uses the Thom class of the tangent bundle of $M$, viewed as the normal bundle of the diagonal in $M \\times M$.\nSteenrod squares of a Thom class give the Stiefel–Whitney classes of the bundle.\n\nExample: for $M = \\R P^2$, $H^*(M; \\Z/2\\Z) = (\\Z/2\\Z)[a]/(a^3)$.\nSince $\\operatorname{Sq}^1 a = a^2$, one gets $v_1 = a$, and $v_2 = 0$ because $\\operatorname{Sq}^2$ vanishes on $H^0$.\nSo $w(\\R P^2) = \\operatorname{Sq}(1 + a) = 1 + a + a^2$, which agrees with $(1 + a)^3$.\n\nSince $\\operatorname{Sq}^k x = 0$ for $\\deg x \u003c k$, the class $v_k$ is $0$ for $k \u003e n/2$.\n\n## Explanations\n\n### L1 human\n\nStiefel–Whitney classes are defined from the tangent bundle, which uses the smooth structure of the manifold.\nWu's formula computes them from the mod $2$ cohomology ring and its Steenrod squares alone.\nSo these classes depend only on the homotopy type of the manifold.\n\n### L2 human\n\nThe formula has two steps: Poincaré duality turns the Steenrod squares into the Wu classes $v_k$, and then $w = \\operatorname{Sq}(v)$.\nIt implies that Stiefel–Whitney numbers are homotopy invariants of closed manifolds, which is relevant for [[Thom's cobordism theorem|HM23BV561Z]] since these numbers decide unoriented cobordism.\nFor example, $w_1(M) = v_1$, so $M$ is orientable exactly when $\\operatorname{Sq}^1 \\colon H^{n-1}(M; \\Z/2\\Z) \\to H^n(M; \\Z/2\\Z)$ is zero.\n",
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