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 .

  • 2
    $\begingroup$ Surfaces are stably parallelizable, so $w_1$ and $w_2$ both vanish. In general there is no relation between $w_1 \cup w_1$ and $w_2$. An instructive example is given by a non-spinnable orientable manifold where $w_1$ vanishes but $w_2$ does not. $\endgroup$ Sep 1, 2016 at 11:27
  • 3
    $\begingroup$ This is exactly the Wu formula, see chapter 11 of Milnor and Staheff book. $\endgroup$
    – SashaP
    Sep 1, 2016 at 13:05
  • $\begingroup$ Cross-posted from math.stackexchange.com/q/1910767 $\endgroup$ Sep 1, 2016 at 14:25
  • 1
    $\begingroup$ @Jens: $w_1$ surely doesn't vanish for nonorientable surfaces... $\endgroup$ Sep 2, 2016 at 1:55
  • $\begingroup$ Sure. I meant to say "Orientable surfaces [...]" $\endgroup$ Sep 2, 2016 at 6:50

1 Answer 1


The 'natural' way to prove this is to use Wu's formula for the Stiefel-Whitney classes and basic properties of the Steenrod squares.

In more detail, if $M$ is a closed $n$-manifold (connected, but not necessarily orientable) then by mod 2 Poincaré duality there are unique classes $v_i\in H^i(M;\mathbb{Z}_2$) such that for each $x\in H^{n-i}(X;\mathbb{Z}_2)$ $$\langle Sq^i(x),[M]\rangle = \langle v_i\cup x,[M]\rangle$$ where $[M]\in H_n(M;\mathbb{Z}_2)$ is the mod 2 fundamental class. The $v_i$ are called the Wu classes of $M$ and Wu's formula expresses the $k$-th Stiefel-Whitney class as $$w_k = \sum_{i=0}^k Sq^{k-i}(v_i).$$ For $k=1,2$ this gives $$w_1=Sq^1(v_0) + Sq^0(v_1)=v_1$$ $$w_2=Sq^2(v_0) + Sq^1(v_1) + Sq^0(v_2) = w_1^2 + v_2$$ after using the basic properties that $Sq^0$ is the identity, that $Sq^i$ is the cup square on $H^i(M;\mathbb{Z}_2)$, and that $Sq^i$ vanishes on $H^k(M;\mathbb{Z}_2)$ for $i>k$.

But the latter property also tells us that $v_i$ must be zero for $i>n-i$, that is, for $i > n/2$. In particular, if $M$ is a surface, then $v_2=0$ so that Wu's formula reduces to $w_2=w_1^2$.

All of this is standard material and the standard reference is Milnor and Stasheff's "Characteristic classes".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.