# What are the possible Stiefel-Whitney numbers of a five-manifold?

On a compact five-manifold, the Stiefel-Whitney number $w_2w_3$ can be nonzero. An example is the manifold $SU(3)/SO(3)$, and also another example is a $\mathbb{CP}^2$ bundle over a circle where the holonomy is given by the complex conjugation automorphism of $\mathbb{CP}^2$.

My question is whether the Stiefel-Whitney number $w_1^2w_3$ can be nonzero on a compact five-manifold. I haven't found either a simple proof that it is zero, or a simple example where it is nonzero.

• Do you have a reference for the claim "the Stiefel-Whitney number $w_2w_3$ of $SU(3)/SO(3)$ is non zero?" – Andrea Mondino Dec 11 '18 at 15:10
• @AndreaMondino there's a discussion with a reference here. – Arun Debray Dec 11 '18 at 16:18

## 1 Answer

Recall that on a closed $$n$$-manifold $$M$$, there is a unique class $$\nu_k$$ such that $$\operatorname{Sq}^k(x) = \nu_kx$$ for all $$x \in H^{n-k}(M; \mathbb{Z}_2)$$; this is called the $$k^{\text{th}}$$ Wu class. The Wu classes can always be written in terms of Stiefel-Whitney classes due to Wu's theorem which states that $$w = \operatorname{Sq}(\nu)$$.

The first Wu class $$\nu_1$$ is $$w_1$$, so $$\operatorname{Sq}^1(x) = w_1x$$ for all $$x \in H^4(M; \mathbb{Z}_2)$$ where $$M$$ is a closed five-manifold. Now, by the Wu formula, we have $$\operatorname{Sq}^1(w_3) = w_1w_3$$. So $$w_1^2w_3 = \operatorname{Sq}^1(w_1w_3) = \operatorname{Sq}^1(\operatorname{Sq}^1(w_3)) = 0$$ as $$\operatorname{Sq}^1\circ\operatorname{Sq}^1 = 0$$ by the Ádem relations.

In fact, $$w_2w_3$$ is the only Stiefel-Whitney number which can be non-zero and therefore defines an isomorphism $$w_2w_3 : \Omega_5^O \to \mathbb{Z}_2$$. To see this, note that the Stiefel-Whitney numbers of a closed five-manifold are $$w_1^5, w_1^3w_2, w_1^2w_3, w_1w_4, w_1w_2^2, w_2w_3, w_5.$$

It has already been established that $$w_1^2w_3 = 0$$ and $$w_2w_3$$ can be non-zero. Note that $$w_5 = 0$$ as it is the mod $$2$$ reduction of the Euler characteristic.

One of the properties of Steenrod squares is that $$\operatorname{Sq}^k(x) = 0$$ if $$k > \deg x$$. In particular, if we consider $$\operatorname{Sq}^k : H^{n-k}(M;\mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$$, we see that if $$k > n - k$$ (i.e. $$k > \frac{n}{2}$$), it must be the zero map and hence by Poincaré duality, we must have $$\nu_k = 0$$. So on a five-manifold, the classes $$\nu_3$$, $$\nu_4$$, and $$\nu_5$$ must be zero. As can be found here (see also this note), we have

\begin{align*} \nu_3 &= w_1w_2\\ \nu_4 &= w_4 + w_1w_3 + w_2^2 + w_1^4\\ \nu_5 &= w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2. \end{align*}

As $$\nu_3 = w_1w_2 = 0$$, we have $$w_1^3w_2 = 0$$ and $$w_1w_2^2 = 0$$.

Now note that $$\nu_5 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2 = w_1w_4$$ so $$w_1w_4 = 0$$.

Finally, as $$\nu_4$$ is zero, so is $$w_1\nu_4 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^5 = w_1^5$$.