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Let $T$ be a revolution Torus. My objective is to prove that $\chi(T)=0,$ but without computing $\iint Kd\sigma$. I already know that this integral is zero, but I can't use this.

I parametrized $T$ by

$$\mathbb{x}(u,v)=((a+r\cos u)\cos v,(a+r\cos u)\sin v,r\sin u),\; 0<u<2\pi, 0<v<2\pi. $$ I have two ideas to compute $\chi (T).$

1) Finding some homeomorphic surface $S$ to $T$ and computing $\chi(S)=0.$

2) Explicit computing $\chi(T)=0$

On 1), I know that $S$ is homeomorphic to $S^{1}\times S^{1},$ but I don't know if this could help.

On 2), I actually don't know how to take an easy triangulation for $T$ to compute $\chi(T)$. The easiest triangulations I taught for $T$ has $V=4$, $F=8$ and $E=12$, but write this triangulation explicit would be laborious. Can I do it in another way?

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  • $\begingroup$ Do you know $\chi(S^1)$ and the fact that $\chi(X\times Y)=\chi(X)\chi(Y)$? That's the east (fancy) way. There is no triangulation with so few vertices, edges, and faces. It takes a surprisingly larger number of all of them. $\endgroup$ – Ted Shifrin Nov 8 at 18:42
  • $\begingroup$ I didn't know that "product" rule for Euler-Poincaré characteristic. But in my Differential Geometry class we defined it just for regular surfaces in $\Bbb{R}^{3}$. Using my intution, I split $S^{1}$ in two equal sides, so $F=2$. The two opposites podes divinding the circle leads me to $V=2.$ Also, we have the two arcs of $S^{1}$ with the middle edge, so $E=3.$ Then, $\chi(S^{1})=1$, which is impossible, since the characteristic only assumes values for $2,0$ and negative even numbers. What was my mistake? $\endgroup$ – Mateus Rocha Nov 8 at 18:53
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    $\begingroup$ No faces for the circle $S^1$. Just $V=E=2$. $\endgroup$ – Ted Shifrin Nov 8 at 21:13
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The formula $\chi(T) = V-E+F$ is true not just for triangulations, but for more general cell decompositions of $T$.

For example, let's consider the map $\mathbb{x}$ in your question, but let's extend that map to the entire closed square $[0,2\pi] \times [0,2\pi]$. I can see that you used the open square in your parameterization, not the closed square. But as I'm sure you understand, parameterizing $T$ with the open square misses some points of the revolution torus, namely points that are parameterized by the sides and corners of the square. One might object that the sides and corners are not uniquely parameterized, but that non-uniqueness becomes a tool you can use.

So, let's consider the map $\mathbb{x} : [0,2\pi] \times [0,2\pi] \to T$. This suggests a cell decomposition of $T$ whose unique 2-cell is a square parameterized by $[0,2\pi] \times [0,2\pi]$, and so $F=1$.

The two vertical sides of that square become a single edge of $T$; that edge is parameterized by $\{0\} \times [0,2\pi]$, and it is also parameterized by $\{2\pi\} \times [0,2\pi]$, but it's just a single edge of $T$ itself. Similarly, the two horizontal sides of that square become a single edge of $T$. That's two edges altogether, and so $E=2$.

Finally, the four corners of the square come together to form a single vertex of $T$, the vertex of the cell decomposition, and so $V=1$.

Thus $$\chi(T)=1-2+1=0 $$

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  • $\begingroup$ Nice answer. Thank you :) $\endgroup$ – Mateus Rocha Nov 8 at 18:46

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