# If the top Stiefel-Whitney class of a compact manifold is nonzero, must there be another non-vanishing Stiefel-Whitney class?

I was trying to collect some examples of Stiefel-Whitney class computations, just to make myself more familiar with them. It seems from my (relatively short list) that if the top Stiefel-Whitney class is nonzero, there must be another nonzero class.

Recall that, at least for oriented manifolds, the top Stiefel-Whitney class is the mod 2 reduction of the Euler class, which is just the Euler characteristic times the fundamental class. So, the top Stiefel-Whitney class is nonvanishing iff the Euler characteristic is odd.

Suppose $M$ is a compact orientable smooth manifold with non-vanishing top Stiefel-Whitney class. Must there be another nontrivial Stiefel-Whitney class?

Here are some data points:

1. A non-orientable manifold has nonvanishing $w_1$, so the answer is YES for non-orientable manifolds.

2. As proven in Milnor-Stasheff, the first nonvanishing Stiefel-Whitney class always occurs in dimension given by a power of $2$. So, the answer is YES if the dimension of the manifold is not a power of $2$.

3. In dimension $1$, the answer is YES: all manifolds are orientable so the top class vanishes. In dimension 2, the answer is YES. For, if $w_2\neq 0$, then $M$ must have odd Euler characteristic, which implies it is non-orientable.

4. In dimension 4, the answer is YES: If the Euler characteristic is odd, then, via Poincare duality, this implies the second Betti number is odd. Nonsingularity of the cup product forces there to be an element in $H^2$ which cups with itself to be a generator of $H^4$. In particular, the associated quadractic form $H^2\rightarrow H^4$ is not even, so $w_2$ is nonvanishing.

5. if $M$ is a projective spaces, the answer is YES. For, $\mathbb{R}P^n$, orientability implies $w_n = 0$. For $\mathbb{C}P^n$, $w_{2n}\neq 0$ iff $w_2\neq 0$, and for $\mathbb{H}P^n$, $w_{4n}\neq 0$ iff $w_4\neq 0$. For $\mathbb{O}P^2$, $w_8\neq 0$.

So, if there are any counterexamples, they are at least $8$ dimensional.

Is there an $8$ dimensional compact smooth manifold for which $w_8\neq 0$ but $w_1 = w_2 = ... =w_7 = 0$? Is there simply connected example?

As far as I can tell, the argument you gave for $n=4$ works just fine for all dimensions $n = 4k$. In particular this tells us that the first nonvanishing $w_j$ has $j \leq 2k$; so no 12-dimensional manifolds whose first nonvashing Steifel-Whiteny class is $w_8$.
Suppose $w_n \neq 0$, and $w_j = 0$ for all $j < n/2$. As before, Poincare duality and the assumption that $w_n \neq 0$ forces $b_{n/2}$ to be odd. Because of our assumption, $w_{n/2} = v_{n/2}$, the Wu class, and this is zero if and only if the intersection form $H^{n/2} \otimes H^{n/2} \to \Bbb Z/2\Bbb Z$ is even. But as before, because this intersection form is nonsingular and $H^{n/2}$ is odd-dimensional, this is impossible.
(To concretely see why, recall that any bilinear form $q$ over $\Bbb Z$ has a characteristic element $c$ with $q(c,x) \equiv q(x,x)\mod 2$ for all $x$; this of course reduces in our case to the Wu class $v_{n/2}$. We have the theorem: $q(c,c) \equiv \text{sign}(q) \mod 8$. Because $H^{n/2}$ is odd-dimensional, this has to be odd; and in particular, $v_{n/2} \cup v_{n/2} \neq 0$, and so $v_{n/2} \neq 0$.)
• Looks good to me. I hate it when I miss using an argument I already know. Nice catch! For some reason, I was thinking that the "$v_{n/2} = 0$ iff intersection form is even" was a $4$-manifold result. – Jason DeVito May 19 '15 at 18:50