Second Stiefel-Whitney class is a square

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?

• What has got you interested in this question? Dec 1 '16 at 18:35
• We are currently trying to get obstructions against positive scalar curvature metrics on (totally) non-spin manifolds. We can get a nice obstruction under the condition in the question, and so now we are looking for examples where we might apply our theorem. :) Dec 5 '16 at 7:09

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.

First note that the collection of orientable manifolds with $$0 \neq w_2(M) = x^2$$ for some $$x \in H^1(M; \mathbb{Z}_2)$$ is closed under products. Moreover, given two such manifolds of the same dimension, their connected sum is another; this is because the Stiefel-Whitney classes of a connected sum are the 'sum' of the Stiefel-Whitney classes of the summands (see here) and $$x^2 + y^2 = (x + y)^2$$. In addition, given such a manifold, taking a product or connected sum with a spin manifold also provides another such manifold

As $$\mathbb{RP}^5$$ is an example, the constructions mentioned above provide infinitely many examples in each dimension greater than or equal to five. In dimension four, we would also obtain infinitely many examples if we could find a four-dimensional example, but that is yet to be demonstrated. So the interesting question now becomes:

Does there exist an orientable four-manifold $$M$$ with $$0 \neq w_2(M) = x^2$$ for some $$x \in H^1(M; \mathbb{Z}_2)$$?

Added later: As is shown in this note, if $$m \equiv n \bmod 4$$, then $$w_2(\operatorname{Gr}(m, m + n)) = w_1(\gamma)^2$$ which is non-zero except when $$m = n = 1$$; here $$\gamma$$ denotes the tautological bundle. As is also shown in the previous link (alternatively, here), the grassmannian $$\operatorname{Gr}(m, m + n)$$ is orientable if and only if $$m + n$$ is even. The condition $$m \equiv n \bmod 4$$ implies $$m + n$$ is even, so the manifolds $$\operatorname{Gr}(m, m + n)$$ with $$m \equiv n \bmod 4$$, except $$\operatorname{Gr}(1, 2) = \mathbb{RP}^1 = S^1$$, are also examples. In particular, setting $$m = 1$$, we recover the examples $$\mathbb{RP}^n$$ with $$n \equiv 1 \bmod 4$$, $$n > 1$$.

This observation also provides an explicit four-dimensional example, namely $$\operatorname{Gr}(2, 4)$$.

• for what concern your last question, I believe the answer is yes, even though I'm unable to provide explicit examples. Looking at almost spin 4-manifolds, one can define the so called $w$-type of it, which basically is the preimage via the classifying map $c_* : H^2(\pi_1M;\mathbb{Z}_2)\to H^2(M;\mathbb{Z}_2)$ of $w_2(M)$ (this definition involves a Serre SS argument). Then one can compute the stable diffeomorphism classes of orientable manifolds with $w$-type $x^2$ and in some examples it turns out that the class is not empty, which means that exists the 4-manifold with the required property Jan 27 '17 at 16:32