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Let M be a compact complex manifold. Then is it true that if the first Chern class of M is even, then M admits a spin structure?

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up vote 8 down vote accepted

Yes. An oriented real vector bundle is spin if and only if its second Stiefel-Whitney class vanishes. If $E_\mathbb{C}$ is a complex vector bundle and $E_\mathbb{R}$ is the underlying real bundle then the second Stiefel-Whitney class is given by $w_2(E_\mathbb{R}) = c_1(E_\mathbb{C})$ mod 2. The details appear somewhere in chapter 2 of Spin Geometry by Lawson and Michelsohn.

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More generally, the total Stiefel-Whitney class of $E_{\mathbb R}$ is the reduction mod $2$ of the total Chern class of $E_{\mathbb C}$. This is Problem 14-B on page 171 of Characteristic Classes, by John Milnor and James Stasheff –  Alex Suciu Jan 21 '12 at 18:44
thank you very much for the references! –  314159. Jan 21 '12 at 20:50
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Yes, this is true:

See for example Spin geometry -- H. Blaine Lawson, Marie-Louise Michelsohn Remark 2.2/ page 87.

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It's what you'd expect if the squaring map on Pic(M) has the image you'd first guess. The connected component being divisible, it's about the discrete part ... –  Charles Matthews Jan 21 '12 at 14:28
Oops, you beat me to it! I guess that clarifies "somewhere in chapter 2" in my answer. –  Paul Siegel Jan 21 '12 at 14:30
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