Wu's formula

Remarks

The classes $v_k$ exist and are unique by Poincaré duality with $\Z/2\Z$ coefficients. The proof uses the Thom class of the tangent bundle of $M$, viewed as the normal bundle of the diagonal in $M \times M$. Steenrod squares of a Thom class give the Stiefel–Whitney classes of the bundle.

Example: for $M = \R P^2$, $H^*(M; \Z/2\Z) = (\Z/2\Z)[a]/(a^3)$. Since $\operatorname{Sq}^1 a = a^2$, one gets $v_1 = a$, and $v_2 = 0$ because $\operatorname{Sq}^2$ vanishes on $H^0$. So $w(\R P^2) = \operatorname{Sq}(1 + a) = 1 + a + a^2$, which agrees with $(1 + a)^3$.

Since $\operatorname{Sq}^k x = 0$ for $\deg x < k$, the class $v_k$ is $0$ for $k > n/2$.

Going deeper

For a student

The formula has two steps: Poincaré duality turns the Steenrod squares into the Wu classes $v_k$, and then $w = \operatorname{Sq}(v)$. It implies that Stiefel–Whitney numbers are homotopy invariants of closed manifolds, which is relevant for Thom's cobordism theorem since these numbers decide unoriented cobordism. For 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.