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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?

edited Jan 26 2012 at 18:39

asked Jan 21 2012 at 13:36

math3.14159
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Paul Siegel · Accepted Answer · 2012-01-21 14:28:35Z UTC

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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.

answered Jan 21 2012 at 14:28

Paul Siegel
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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 2012 at 18:44

thank you very much for the references! – math3.14159 Jan 21 2012 at 20:50

math3.14159 · Answer 2 · 2012-01-21 14:06:18Z UTC

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Yes, this is true:

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

answered Jan 21 2012 at 14:06

math3.14159
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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 2012 at 14:28
	Oops, you beat me to it! I guess that clarifies "somewhere in chapter 2" in my answer. – Paul Siegel Jan 21 2012 at 14:30

first chern class and spin structures

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