Splitting of the tangent bundle and Euler characteristic of surfaces

Question

Let $M$ be a be a closed orientable surface, and suppose that its tangent bundle $TM$ splits into a direct sum of line bundles.

How to prove that $M$'s Euler characteristic is zero?

---

Unfortunately, I am entirely unfamiliar with Characteristic Classes, but I would accept a solution based on properties of these objects;

if there is a reference for a book where I could find this claim or details of a proof, that would be great.

Answer 1

Characteristic class argument: The Euler class of a vector bundle vanishes if there is a nonvanishing section. The Euler class of the tangent bundle evaluated on the fundamental class is the Euler characteristic. If we have a trivial subbundle, we automatically have a section. We know that we have a 1-dimensional subbundle. Since we are orientable, this implies this subbundle is either trivial or both it and its complement are nonorientable. In the first case, this gives the Euler characteristic 0 by before.

In the second case, see Jason's answer.

Singularity argument: A vector field on a surface has the sum of the indices of the singularities equal to its Euler characteristic. Since a nonvanishing section of the tangent bundle is a vector field with no singularities, we deduce that if either of the bundles is trivial, the Euler characteristic of the surface vanishes.

In case both are nontrivial, we can pullback to a cover as in Jason's answer.

Answer 2

To complement Connor's answer:

Classifing spaces argument: Because $M$ is orientable, its tangent bundle is classified by a map $\phi:M\rightarrow BSO(2)=\mathbb{C}P^\infty$. The Euler class of the bundle (which gives $\pm \chi(M)$ when paired with a fundamental class) is by definition the element $\phi^\ast(z)$, where $H^\ast(\mathbb{C}P^\infty;\mathbb{Z})\cong \mathbb{Z}[z]$. Identifying $H^2(M;\mathbb{Z})$ with $\mathbb{Z}$, $\phi^\ast(z)$ can directly be interpreted as $\pm\chi(M)$.

So, we need to show that the condition that the tangent bundle splits implies that $\phi^\ast(z) = 0$.

Now a splitting of the tangent bundle implies a reduction of the structure group to $B:=BS(O(1)\times O(1))\subseteq BSO(2)$. That is, $\phi$ factors through the inclusion $B\rightarrow BSO(2)$. Since $B\simeq \mathbb{R}P^\infty$, and $H^2(\mathbb{R}P^\infty;\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$, we see that on degree $2$ cohomology, $\phi^\ast$ factors as a composition $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}$. Since any homomorphism $\mathbb{Z}/2\mathbb{Z}\rightarrow \mathbb{Z}$ is trivial, we deduce $\phi^\ast(z) = 0$ as desired.

By the way, the assumption that $M$ is orientable is not needed. For, if $M$ is non-orientable, and $\pi:\widetilde{M}\rightarrow M$ is the orientation double covering, then the fact that $T\widetilde{M}\cong \pi^\ast TM$ shows that splitting of $TM$ gives a splitting of $\widetilde{M}$. By the orientable case, $\chi(\widetilde{M}) = 0$, from which it follows that $\chi(M) = 0$ as well.