# Algebraic relationships between the Stiefel-Whitney classes of a manifold?

I've noticed that in a 2D manifold, the second Stiefel-Whitney class can always be obtained as the cup product of the first one with itself.

In other words $w_2 = w_1 \smile w_1$ .

Is there a 'natural' way to prove this? Does it appear as a consequence of some deeper relationship between the Stiefel-Whitney classes of a manifold? I can't think of a proof that doesn't involve the tedious explicit construction of classes and the classification theorem for 2D manifolds .

• For orientable manifolds $w_1=0$ and for a surface $w_2([M])$ is the Euler characteristic mod 2 which also always vanishes for orientable surfaces which solves the question for the orientable surfaces. – Thomas Rot Sep 1 '16 at 10:54
• For non-compact surfaces $H^2$ is trivial so that also works. But that's not where the 'meat' of the problem is . Haven't yet found a counterexample in 3D manifolds either. – user3257842 Sep 1 '16 at 11:24
• But sw classes live live cohomology with $Z_2$ Coefficients – Thomas Rot Sep 1 '16 at 11:28
• Yes, meant to say $H^2(Z_2)$ . Regardless of group, the top cohomology always vanishes in a non-compact manifold. – user3257842 Sep 1 '16 at 11:44
• Hi, i'm sorry misread this. I thought you said non-orientable for some reason. – Thomas Rot Sep 1 '16 at 11:57

We have Wu's formula. Let $X$ be a closed $n$-manifold. Define the $k$-th Wu class $v_k \in H^k(X;\mathbb{F}_2)$ by the following property: for any $x \in H^{n-k}(X;\mathbb{F}_2)$, we have $$v_k \smile x = \operatorname{Sq}^k x.$$ (The existence and uniqueness of Wu classes is a consequence of Poincare duality.) Then Wu's formula states that $$\operatorname{Sq}(v) = w,$$ where $\operatorname{Sq}$ is the total squaring operation, $v$ is the total Wu class, and $w$ is the total Stiefel-Whitney class.
Writing out the first few terms, we find that $w_1^2 + w_2 = v_2$. So to show that $w_1^2 = w_2$, we just need to show that $v_2 = 0$. For surfaces, this is clear from the defining property of Wu classes, since we need $v_2 \smile x = \operatorname{Sq}^2 x = 0$ for any $x \in H^0(X;\mathbb{F}_2)$.