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 $v_{n/2} \neq 0$.)

  • $\begingroup$ 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. $\endgroup$ – Jason DeVito May 19 '15 at 18:50
  • $\begingroup$ @Jason: I'm sort of surprised by the result! I expected it to be a 4-manifold fluke until I played with a couple examples. $\endgroup$ – user98602 May 19 '15 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.