TITLE: Vanishing of the second Stiefel–Whitney classes of orientable surfaces
QUESTION [10 upvotes]: 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!

REPLY [10 votes]: 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.