TITLE: Trivial tangent bundle of sphere with handles
QUESTION [5 upvotes]: I am wondering if there is a simple proof of this statement:

A sphere with $g$ handles has trivial tangent bundle iff $g=1$

I know that it is a corollary of Poincaré-Hopf theorem, but it seems to be too hard for this problem.

REPLY [2 votes]: I found a school-level solution in booklet Skopenkov A.B. Characteristic classes for beginners (in Russian), there is a reduced version of this proof.
Choose a triangulation, a point inside each of triangles, equiped with non-zero vector, and paths beetween centers of adjacent triangles. Let us call the union of these paths as skeleton.

Deform our supposed vector field to coincide with vectors in the vertices of skeleton. As this field has no zeroes, rotation of vector along any of the curves of triangulation, surrounding a vertex of triangulation, is zero.
But for any vector field on skeleton the sum of this numbers is the same (in fact, it equals to Euler characteristic, as it is special case of Euler class of a tangent bundle), because a change of vector field along one of paths will add $k$ to one of adjacent triangles and $-k$ to another.
It is easy to calculate it explicitly for spheres with $g$ handles from polygonal model, get Euler characteristic, and prove that there does not exist non-vanishing vector field on sphere with $g$ handles for $g \ne 1$.