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How does one see that the second Stiefel-Whitney class is zero for all orientable surfaces. For $S^2$ this can be seen by $TS^2$ being stably trivial, and for $S^1 \times S^1$ one can use $T (S^1 \times S^1) = TS^1 \times TS^1$, which gives the class in terms of the classes on $TS^1$ (which are all trivial). What about higher genus?

Thanks!

Eric O. Korman
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2 Answers2

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Tangent bundle is stably trivial for any orientable surface (because the normal bundle is trivial).

Grigory M
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The second Stiefel-Whitney class of a surface is the mod 2 reduction of the Euler class. Since the Euler characteristic (and hence number) is divisible by 2, $w_2$ is zero.

Grigory M
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Adam
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  • @Grigory Can you give some reference to read this proof - hopefully in the differential forms language? This is important for physics since there is some theorem which states that to have a spin structure on a manifold its necessary that the second Stiefel-Whitney class be 0. Now if for all orientable surfaces the second Stiefel-Whiteny class is 0 then in particular it means that it is so for all Riemann surfaces. – Student Nov 26 '11 at 17:45
  • @Anirbit Are you sure, you need this? Spin structures on a Riemann surface are in bijection with "halves" of the canonical class. So there are always exactly $2^{2g}$ (think of 2-torsion in the Jacobian) spin structures on a Riemann surface of genus $g$. – Grigory M Nov 26 '11 at 18:26
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    (Anyway, $\chi=w_n\mod 2$ is the Property 9.5 in Milnor-Stasheff's «Characteristic classes».) – Grigory M Nov 26 '11 at 18:33