I know that a real manifold $M$ is orientable iff its first Stiefel-Whitney class ($w_1(M)$) vanishes and has Spin structure iff both $w_1(M)$ and $w_2(M)$ vanish.

Is there a name given to the class of manifolds for which $w_1,w_2$ and $w_3$ vanish?

Thank you.

  • 2
    $\begingroup$ It's abusive to say that "orientable" is the "name" of the class of manifolds such that $w_1(M) = 0$. Nobody defines orientability like that, it just so happens that the two notions are equivalent... for differentiable manifolds, otherwise there's no tangent bundle. $\endgroup$ – Najib Idrissi Nov 15 '16 at 14:35
  • $\begingroup$ Point noted! I just phrased the question in this way because I wanted to specifically know what (if so) such manifolds are called. I am unsure how to rephrase it without changing the meaning. You are welcome to do so. Thank you for your comment :) $\endgroup$ – R_D Nov 15 '16 at 14:40

No there isn't because if $w_1 = 0$ and $w_2 = 0$, then $w_3 = 0$. More generally, the smallest positive $k$ such that $w_k \neq 0$ is always a power of two. This general fact follows from Wu's formula

$$\operatorname{Sq}^i(w_j) = \sum_{t=0}^i\binom{j - i + t - 1}{t}w_{i-t}\cup w_{j+t}.$$

Here $\operatorname{Sq}^i$ is the $i^{\text{th}}$ Steenrod square. In the case you are asking about, let $i = 1$, $j = 2$, then

$$\operatorname{Sq}^1(w_2) = w_1\cup w_2 + w_3,$$

so if $w_1 = 0$ and $w_2 = 0$ (or even if just $w_2 = 0$), we see that $w_3 = 0$.

  • 2
    $\begingroup$ Or it is called spin;) $\endgroup$ – Thomas Rot Nov 22 '16 at 19:39
  • 2
    $\begingroup$ Thanks for teaching us (or at least me) this remarkable result. $\endgroup$ – Georges Elencwajg Dec 30 '16 at 10:29

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.