I'm studying differential geometry using doCarmo's book, and in the chapter about Gauss-Bonnet's theorem, I got stuck in the following exercise:
Let $S \subset \mathbb{R}^3$ be a surface homeomorphic to the torus $\Rightarrow$ $S$ has a differentiable vector field without singular points.
I know that if $\xi:S \rightarrow TS$ is a differentiable field with only finite singular points (I know that I can always construct this field)
$$0= \sum_{\{x \in S;\xi(x) = 0\}} I_x = \chi (S) $$
where $I_x$ is the index of $\xi$ in the point $x$, and $\chi(S)$ is the Euler characteristic of the surface $S$. But I don't know how to use this information to build a differential vector field without singular points.
The torus is $\mathbb{T}^2 = \mathbb{S^1}\times \mathbb{S}^1$. Let $X\in \mathcal{T}(\mathbb{S^1})$ be a never vanishing vector field (is very simple to construct one with the embedding $\mathbb{S^1}\subset \mathbb{C})$. then $X\times X\in \mathcal{T}(\mathbb{S^1}\times \mathbb{S^1} )$ is a non vanishing vector field on the torus.
If you have another smooth surface $S$ such that there is a diffeo $\psi:\mathbb{T}^2\to S$, then pushing forward the non vanishing vector field you obtain $\psi_*(X\times X)$ which is a never vanishing vector field on $S$.
If $S$ is a smooth closed 2-manifold, that is homeomorphic to $\mathbb{T}^2$, then $\chi(S)=\chi(\mathbb{T}^2)$ (since the Euler characteristic is a topological invariant), therefore thanks to the classification theorem for smooth surfaces $S$ is also diffeomorphic to the torus, and the previous argument applies.
The Poincaré-Hopf theorem (and thus the Gauss-Bonnet theorem) a priori gives a necessary condition to have a never vanishing vector field ($\chi(M^2) = 0$) that a posteriori, thanks to the classification theorem of surfaces it is also sufficient. But you first have to show that the surface with Euler characteristic $0$ has a never vanishing vector field.