Manifolds with vanishing Stiefel-Whitney classes but are not stably parallelizable

Question

It is known that if a manifold is stably parallelizable, then it's Stiefel-Whitney classes must vanish. Is the converse true?

Note that we know that the converse cannot hold if stably parallelizable is replaced parallelizable; for example, spheres are stably parallelizable and so have vanishing Stiefel-Whitney classes but are not parallelizable. Also, what is known about this question if we replace Stiefel-Whitney classes with other characteristic classes?

Answer 1

The converse is not true. If a manifold is stably parallelizable, then its Pontryagin classes must also vanish, but the vanishing of the Stiefel-Whitney classes need not imply this.

For example, let $X$ be a closed simply connected smooth $4$-manifold. Since $X$ is simply connected, $w_1$ and $w_3$ automatically vanish. $w_4$ vanishes iff the Euler characteristic of $X$ is even iff $\dim H_2(X)$ is even. $w_2$ vanishes iff $X$ admits a spin structure iff the intersection form of $X$ is even. It follows that the Stiefel-Whitney classes of $X$ vanish iff its intersection form has even rank and even parity. But by the Hirzebruch signature theorem, the Pontryagin class $p_1 \in H^4(X)$ vanishes iff the signature $\sigma(X)$ of the intersection form vanishes.

Hence to give a counterexample it suffices to find a closed simply connected smooth $4$-manifold $X$ whose intersection form has even rank, even parity, and nonzero signature. For example, you can take $X$ to be a K3 surface, or more explicitly a hypersurface of degree $4$ in $\mathbb{CP}^3$ (see, for example, the calculations in this blog post), whose intersection form has rank $22$, even parity, and signature $-16$. By Rokhlin's theorem this is the smallest signature possible (in absolute value).

Edit: Incidentally, $4$ is the smallest possible dimension of a closed counterexample. If $X$ is a closed smooth $d$-manifold, $d \le 3$, then:

• When $d = 1$, $X$ is a disjoint union of circles and hence is parallelizable.
• When $d = 2$, if $w_1$ vanishes then $X$ is orientable, and this automatically implies that $w_2$ vanishes (e.g. because we know that the Euler characteristic is even in this case, but see this blog post for other arguments). But a closed orientable surface is stably parallelizable: in the usual embedding of such a surface into $\mathbb{R}^3$, the outward pointing normal trivializes the normal bundle.
• When $d = 3$, if $w_1$ vanishes then $X$ is orientable, and again this automatically implies that $w_2$ vanishes, now by some Wu class computations. It follows that the classifying map of the tangent bundle $X \to BO(3)$ lifts to a map $X \to B \text{Spin}(3)$. But $B \text{Spin}(3)$ is $3$-connected, so any map to it from a $3$-manifold is nullhomotopic. Hence $X$ is not only stably parallelizable but parallelizable.