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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.

Matheus Manzatto
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  • More generally, if $M$ is a smooth compact connected manifold with $\chi(M)=0$ then $M$ admits a nonvanishing vector field: https://math.stackexchange.com/questions/47370/if-a-manifold-m-has-zero-euler-characteristic-there-is-a-non-vanishing-vector-f, https://mathoverflow.net/questions/129752/does-a-connected-manifold-with-vanishing-euler-characteristic-admit-a-nowhere-va – Moishe Kohan Nov 24 '17 at 14:14

1 Answers1

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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.

Overflowian
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    But the hypothesis only says that $S$ is homeomorphic to the torus, not diffeomorphic.. – Eduardo Longa Nov 22 '17 at 18:11
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    You ask for the vector field on the surface to be smooth. So I assumed that $S$ has a smooth structure. Then we know from theory that if $S$ is homeomorphic it is also diffeomorphic (closed 2-manifolds) – Overflowian Nov 22 '17 at 18:15
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    I see. Maybe you could add this to your answer, because the last sentence is a nontrivial theorem, right? – Peter Franek Nov 22 '17 at 18:16
  • @WarlockofFiretopMountain Do you know how I prove that if there is a homeomorphism between $S$ and $\mathbb{T}^2$, and $S$ is a smooth surface then there is a diffeomorphism between $S$ and $\mathbb{T}^2$?. – Matheus Manzatto Nov 22 '17 at 18:21
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    I would use the classification theorem for closed 2 manifolds. – Overflowian Nov 22 '17 at 18:28
  • @WarlockofFiretopMountain In doCarmo's Book, we only have this Proposition about classification of surfaces "PROPOSITION 4. Let $S ⊂ \mathbb{R}^3$ be a compact connected surface; then one of the values $2,0,−2,...,−2n,.... $ is assumed by the Euler-Poincaré characteristic $χ(S)$. Furthermore, if $S′ ⊂ \mathbb{R}^3$ is another compact surface and $χ(S) = χ(S′)$, then $S$ is homeomorphic to $S′$." There is no mention about the classification of surfaces up to diffeomorphism (only homeomorphism). Do you know a reference where I can learn how to demonstrate the part concerning the diffeomorphism. – Matheus Manzatto Nov 22 '17 at 18:33
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    Hirsch - Differential Topology – Overflowian Nov 22 '17 at 18:46