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 .



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


