I'm interested in examples of manifolds which are orientable and such that the second Stiefel-Whitney class is a square. (Of course the second Stiefel-Whitney class should be non-zero.)


An easy example is the real projective space $RP^n$ in the case $n \equiv 1 \ (\operatorname{mod} 4)$. But unfortunately, I don't know any more examples. I know that for $2$- and $3$-manifolds have the nice property $\omega\_2 = \omega\_1^2$, but it implies that orientable $2$- and $3$-manifolds have always vanishing second Stiefel-Whitney class.


If we know $\omega\_2 = x^2$, we can apply $\operatorname{Sq^1}$ to it to get $\operatorname{Sq^1}(\omega\_2) = 0$. On the other hand, using Wu's formula and since we assume $\omega\_1 = 0$, we get $\operatorname{Sq^1}(\omega\_2) = \omega\_3$. So this means that I'm especially looking for orientable manifolds with vanishing third Stiefel-Whitney class ... what does this condition mean geometrically?



Second Stiefel-Whitney class is a square
At least there are quite a lot of such manifolds: up to multiplying by powers of 2, any oriented bordism class contains such a manifold.


Proof: Let $f: X \to BSO$ be the universal map such that $w\_2$ pulled back to $X$ is a square of a class $x$. I.e. $X$ is the fiber of a map $w\_2 - x^2: BSO \times K(\mathbb{F}\_2,1) \to K(\mathbb{F}\_2,2)$. Then there is a Thom spectrum $Mf$ and a map of bordism groups $\pi\_d(Mf) \to \pi\_d(MSO)$ which becomes an isomorphism after inverting 2. If $d \geq 4$ and $M$ is some oriented manifold, then there exists an $n \gg 0$ such that $2^n [M] \in \pi\_d(MSO)$ lifts to $\pi\_d(Mf)$, i.e. there is an oriented bordism from the disjoint union of $2^n$ copies of $M$ to a manifold $N$ such that $\tau N: N \to BSO$ admits a lift to a map $N \to X$. If $d \geq 5$ (and maybe 4?), there is no obstruction to doing surgery on $N$ in order to make the map $N \to X$ 2-connected, and hence $H^2(X;\mathbb{F}\_2) \to H^2(N;\mathbb{F}\_2)$ is injective so $w\_2$ stays a non-zero square in $N$. $\square$


If you like, we can partially control the homotopy type of the manifold. For example, if $d \geq 4$ and we write $n = \lfloor d/2\rfloor$ we could use surgery to make the map $N \to X$ $n$-connected. More generally, given an $n$-dimensional finite $CW$-complex $X'$ there is only the "obvious" homotopical obstruction to $X'$ being an $n$-skeleton for a manifold with the requested property: if there exists a map $f: X' \to BSO$ with $f^\*(w\_2) \in H^2(X';\mathbb{F}\_2)$ a non-zero square, then there exists an $n$-connected map from $X'$ to a manifold with the requested property.


