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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.

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    $\begingroup$ 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
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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 $w_3 = 0$.

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

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