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I'm trying to compute $\chi(T^2)$:

  • I know that the sectional curvature of $T^2$ is $\dfrac{\cos(t)}{2+\cos(t)}$ with the parametrization: $F(t,s)=((2+\cos(t))\cos(s),(2+\cos(t))\sin(s),\sin(t))$
  • Now I want to compute $\int\limits_{T^2} K \ dx$, where $K$ is the sectional curvature of the torus.

Is that the correct approach? And how does this computation work?

mathphys
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User1
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  • Do you have to compute $\chi$ through Gauß-Bonnet? There are much easier approaches like explicit triangulation or using $\chi(T^2) = \chi(T)^2$. – lisyarus Jul 05 '19 at 11:58
  • Hello, no I can compute it the way I want. So the triangulation approach would be easier? It seemed harder for me because in my lecture we've only had the basic definition for triangulation and no examples or something so I'm struggling how to start looking for a triangulation... – User1 Jul 05 '19 at 12:00

2 Answers2

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You can triangulate the torus by splitting the identification square into nine squares with diagonals in each. Counting the number of vertices faces and edges will give you the Euler characteristic after application of the definition of the euler characteristic. I will add a picture later.

One can also do this using a simple version of the Gauss-Bonnet Theorem. This states that for $S$ a closed surface we have $$ \int_S K dA = 2 \pi \chi(S) $$ where $K$ is the Gaussian curvature of $S$. Integrating the Gaussian curvature $$K= \frac{\cos(\theta)}{r(R+r\cos(\theta))}$$ of the torus (where I'm using notation from this question) gives $\int_T K dA = 0$ so we may conclude that $\chi(T) = 0$.

mathphys
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  • Thanks already for the answer! I see the identification of the torus with this square.One can also see this here https://math.stackexchange.com/questions/953586/triangulation-of-torus , right? My problem is that I can't connect this to my definition of a triangulation... In my definition, we need some covering of the Torus by sets, which are images of the triangle $((0,0),(1,0),(0,1))$ in $\mathbb{R}^2$ under an embedding s.t. the intersections is a vertex, an edge or empty. – User1 Jul 05 '19 at 12:24
  • You're welcome! Yes, the link you've provided is a useful one. I edited my original answer so that my triangulation is the proper one (i.e. the second one shown in the link). The triangulation with the nine squares with diagonals does indeed coincide with your definition; the torus (represented by the identification square) is covered by eighteen triangles (which are images on the torus of the triangles in the plane) and the intersections indeed satisfy the condition you've given. – mathphys Jul 05 '19 at 12:30
  • I can see that the intersections of the triangles satisfy the conditions and also, counting the edges etc. I get that the Euler characteristic is zero, which is what I need to show, but I still don't see yet why I get these triangles as the images of triangles in $\mathbb{R}^2$.. Is there an explicit embedding for this? – User1 Jul 05 '19 at 12:36
  • I suppose you could take the unit square in $\mathbb{R}^2$ and draw the same pattern with the nine triangles onto it. Then the embedding would take this unit square and identify the edges of the unit square as appropriate for the torus; the little triangles would be unchanged. – mathphys Jul 05 '19 at 12:58
  • I've now added a way to do it using the Gauss-Bonnet theorem in an edit. – mathphys Jul 05 '19 at 13:09
  • So I can just use the identification map, right? – User1 Jul 05 '19 at 14:20
  • And on the solution using the Gauss-Bonnet theorem, I don't get $\int\limits_0^{2 \pi} K=0$ .. – User1 Jul 05 '19 at 14:21
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The Euler characteristic can be computed through a variety of ways. The simplest way is to use the fact that the Euler characteristic is the alternating sum of the number of cells in a CW decomposition of your space. Picking the simplest one gives 1-2+1=0. An alternative and equivalent approach is to take the alternating sum of the rank of the homology, again this gives 1-2+1=0.

If you would like to use Gauss-Bonnet and you know that the torus admits a locally Euclidean metric, then you can deduce it is 0 since the curvature of the plane is 0.

Connor Malin
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  • I don't know enough to answer this question but more or less I understand it, what I don't understand nothing about is of both ways you say to compute the euler characteristics could mention some (simple and single, if I may be allowed to be more annoying) source to learn about it? – Dabed Jul 06 '19 at 00:30
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    Well more or less the first two are definitions of Euler characteristic. Most textbooks on algebraic topology will cover some form of the definition and alternate ways to compute it. Hatcher, for example, has it. However, the mention of curvature is a fact of geometry. It is called the Gauss-Bonnet theorem and something like do Carmo will have this. – Connor Malin Jul 06 '19 at 01:38
  • For me the euler polyhedra characteristic makes sense, Gauss-bonnet less, these other ways of looking at it may be just definitions but feel so advanced to me at least, glad the book is available by the author I hope to be able to go trough this, that helps thank you very much – Dabed Jul 06 '19 at 01:59