# Vanishing of the second Stiefel–Whitney classes of orientable surfaces

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!

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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.
@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
(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