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.
Best Answer
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.