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How to show that the Stiefel Whitney class for all closed 3-manifolds obey $$ w_1w_2 = w_3 = {w_1}^3 = 0 $$ for any closed 3-manifolds?

  • In this case, can one relate the Spin, Pin$^+$, and Pin$^-$ manifolds to each other?

  • Do we still have distinctions of Spin, Pin$^+$, and Pin$^-$ manifolds in 3$d$?

  • Do we have any unorientable manifold that are neither of them (Spin, Pin$^+$, and Pin$^-$ )?
annie marie cœur
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  • This is the Wu formula, see https://math.stackexchange.com/questions/141213/the-wu-formula-and-steenrod-algebras – Tyrone Jul 14 '18 at 09:44
  • The easiest way to do this is via the Wu formula, but note also that any closed $3$-manifold bounds (by the classical computation of $\Omega_3^O$, for example), and thus all of its Stiefel-Whitney numbers vanish. – anomaly Jul 15 '18 at 15:33
  • @anomaly: What is the classical computation of $\Omega^O_3$? – Michael Albanese Jul 15 '18 at 15:54
  • @MichaelAlbanese: Thom's computation in the 50s. – anomaly Jul 15 '18 at 17:00

1 Answers1

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First of all, the top Stiefel-Whitney class of a manifold is the mod $2$ reduction of the Euler characteristic. In particular, for a three-manifold $w_3 = 0$ (odd-dimensional manifolds have Euler characteristic zero by Poincaré duality). The vanishing of the other two expressions requires more work.

As I explain in this answer, if $M$ is a closed $n$-dimensional manifold, then there is a unique class $\nu_m \in H^m(M; \mathbb{Z}_2)$ called the $m^{\text{th}}$ Wu class such that $\operatorname{Sq}^m(x) = \nu_m x$ for all $x \in H^{n-m}(M; \mathbb{Z}_2)$; here $\operatorname{Sq}^m$ denotes the $m^{\text{th}}$ Steenrod square. Furthermore, if $M$ is smooth, then $\nu_m$ can be expressed in terms of Stiefel-Whitney classes. For example, in this answer I show that $\nu_2 = w_1^2 + w_2$ and $\nu_3 = w_1w_2$.

Now suppose $M$ is a closed three-dimensional manifold. As $\operatorname{Sq}^3 : H^0(M; \mathbb{Z}_2) \to H^3(M; \mathbb{Z}_2)$ is necessarily zero ($\operatorname{Sq}^i(x) = 0$ if $i > \deg x$), we see that $\nu_3 = w_1w_2 = 0$ by Poincaré duality. Likewise, $\nu_2 = w_1^2 + w_2 = 0$. Multiplying by $w_1$ and rearranging, we see that $w_1^3 = w_1w_2 = 0$.

For a manifold $M$, we have

  • $M$ is spin $\,\Leftrightarrow$ $w_1 = 0$, $w_2 = 0$,
  • $M$ is pin${}^+$ $\Leftrightarrow$ $w_2 = 0$, and
  • $M$ is pin${}^-$ $\Leftrightarrow$ $w_1^2 + w_2 = 0$.

As $\nu_2 = w_1^2 + w_2 = 0$, every closed three-manifold is pin${}^-$. Also, it follows that a closed three-manifold is spin if and only if it is orientable. Every spin manifold is pin${}^+$, but the converse is not true as $K\times S^1$ demonstrates where $K$ is the Klein bottle. Finally, $\mathbb{RP}^2\times S^1$ is an example of a closed three-manifold which is not pin${}^+$.

Michael Albanese
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