Trivial tangent bundle of sphere with handles 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.
 A: Let us suppose that the tangent bundle to a compact, orientable surface $M$ with genus $g\geq2$ is trivial. In particular, there is a vector field $X$ on $M$ which is nowhere zero.
We can pick a Riemannian metric on $M$ such that $X$ is everywhere of norm $1$, so that its integral curves are all parametrized by arc length. Let $\phi:\mathbb R\times M\to M$ be the flow of $X$, and for each $t\in\mathbb R$ let $\phi_t=\phi(t,\mathord-):M\to M$. It is clear that all the maps $\phi_t$ are homotopic to the identity, so that their Lefschetz number $\Lambda(\phi_t)$ is equal to $\Lambda(\mathrm{id}_M)=\chi_M$, the Euler characteristic of $M$. Since $g\geq2$,we have $\chi_M\neq0$ and therefore Lefschetz's fixed point theorem tells us that for all $t>0$ the map $\phi_t$ has a fixed point.
Now pick $\epsilon>0$ so small that each ball of radius $\epsilon$ (with respect to the metric derived from the Riemannian metric me picked above) is contained in a coordinate chart, and let $t=\epsilon/2$ and $x\in M$ be a fixed point of $\phi_t$. The integral curve of $X$ through $x$ is a circle of diameter at most $\epsilon$, so it is contained in a coordinate patch. It therefore bounds a (closed,smooth) disc $D$. The restriction of the vector field $X$ to this disc is nowhere zero and it is tangent to the boundary of $D$.
This is impossible. One way to see it is to use thw following easy result:

Lemma. Let $D$ be the unit disc in the plane. There is no nowhere zero vector field on $D$ which is tangent to the boundary of $D$.

Proof. Such a vector field can be normalized to give a map $X:D\to S^1$. The restriction $X|_{S^1}:S^1\to S^1$ of $X$ to $S^1=\partial D$ has no fixed points, so that the Lefchetz number $\Lambda(X|_{S^1})$ is zero. It follows at once from this that $X|_{S^1}$ has to act like the identity on $H_1(S^1)$. This is impossible, since $X|_{S^1}$ is homotopic to a constant map.
A: The only proof I have every seen is the Poincare-Hopf theorem, or a very slightly weaker version which says that the existence of a non-vanishing vector field is equivalent to the vanishing of the euler characteristic. Any other proof you might see will just be a disguised version of this statement.
A: This is of course not a proof, but it is just intended to give an intuitive picture (sorry for the very bad drawings).
First, draw a picture of the torus $T^2$, with $g=1$. It's easy to see that you can draw on it a constant tangent bundle, putting the same vector on every point.

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Next, consider $S^2$, the usual sphere with $g=0$. Here things are different... try to start from the equator and draw tangent vectors which are perpendicular to the equator, say pointing up, and then try to extend the bundle on the upper emisphere. You'll see that at the north poles the vectors need to converge, thus producing a singularity. Not allowed! (This is the hairy ball theorem)

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Finally, consider a $2$-torus with $g=2$. You can produce it by glueing two $1$-tori, and you'll easily see that there's no way to make the constant bundles on those to fit together.

A: The Poincare-Hopf theorem is not needed to prove the theorem. 
The sum of the indices of a vector field with isolated zeros is the same for all vector fields. This follows because such a vector field is just a smooth map of the surface into its tangent bundle which is transverse to the zero section. Given this one needs a vector field whose indices are easy to compute. One example is the vector field that flows from the barycenters of the triangles in a triangulation of the surface towards the vertices. This vector field will have zeros of index 1 at the vertices and at the barycenter of each triangle  and zeros of index -1 at the barycenter of each edge.
Here is another way to avoid the Poincare-Hopf Theorem
The tangent bundle of an oriented surface has a natural complex structure. Given a Levi-Civita connection, the first Chern class of the tangent bundle is represented by the curvature 2 form (or Gauss-Bonnet 2 form) divided by 2pi. Therefore the integral of the curvature 2 form over the surface is the same for all connections since they all give the evaluation of the first Chern class on the fundamental cycle of the surface.
For the sphere with the standard connection, the integral is 2, For the torus it is zero since the torus can be given a flat connection. For surfaces of higher genus the integral is a negative integer since they all can be given connections with constant negative curvature.  
But the tangent bundle can not be trivial if the first Chern class is not zero so the torus is the only one.
A: 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 $g \ne 1$.
