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This pic below is an exploded view of a cone.

enter image description here

I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,

enter image description here

At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.

And this is actually a part of an exercise to find the total geodesic curvature of $\partial M$. I tried to figure it out using the Gauss-Bonnet theorem.

I'm sorry that the question is not clear. $M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.

Aloizio Macedo
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민광건
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  • So $M$ appears to be the region between the "top" circle on a cone and another closed geodesic meeting the first circle at a point? – Ted Shifrin Sep 05 '18 at 19:22
  • @민광건 You have chosen a cone base when an infinite number of corners $n = \infty$ at base and bottom circle are considered. Take a finite $n$ say $n=4$ and find $ V,F,E$ for a square pyramid. $\chi$ is independent of $n$ from a topological point of view. – Narasimham Sep 05 '18 at 19:41
  • I just edited the question and yes @Ted Shifrin, you are right. – 민광건 Sep 06 '18 at 04:20
  • And for @Narasimham, thanks for comment and I tried your suggestion, but I got V=7, E=11, F=4 with n=4 and still the characteristic is 0. Even when n is any natural number, I guess V=2n-1, E=3n-1, F=n and the characteristic is 0. – 민광건 Sep 06 '18 at 04:24
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    If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$. – Ted Shifrin Sep 06 '18 at 04:31
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    @민광건 I see what the problem is. If one incorrectly computes exterior angles for $\partial M$, one is led to the conclusion that $\chi(M)=1$. But $\chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane. – Ted Shifrin Sep 06 '18 at 04:51
  • @TedShifrin Oh Thanks I got what you mean. Now I'm gonna try to compute exterior angles "correctly" myself. Really thank you for commenting and helping me. – 민광건 Sep 06 '18 at 05:06
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    @민광건 Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about. – Aloizio Macedo Sep 06 '18 at 05:43
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    @AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future. – 민광건 Sep 06 '18 at 07:16
  • @민광건 For square pyramid and homeomorphs $ V=5, F=5, E=8, \chi= V+F-E =2 $ – Narasimham Sep 06 '18 at 16:44
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    I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $\partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled. – Ted Shifrin Sep 06 '18 at 16:49
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    @Narasimham: You're still not paying attention to what the region actually is. It has Euler characteristic $0$. There are various ways to prove this. – Ted Shifrin Sep 06 '18 at 16:52
  • @ Ted Shifrin: So $M$ is homeomorphic to a torus, is it ? – Narasimham Sep 06 '18 at 17:01
  • @Narasimham: No, not at all. It's homotopy-equivalent to a circle. It is a surface with boundary, for starters, and there's a non-smooth point. See the picture that was added. – Ted Shifrin Sep 06 '18 at 17:09
  • @ Ok thanks I got it what he means. Then the answer is quite simple. – Narasimham Sep 06 '18 at 17:20
  • @Narasimham: Well, that part is. Now try writing out Gauss-Bonnet and compute the integral $\int_{\partial M}\kappa_g,ds$ and sum of exterior angles. – Ted Shifrin Sep 06 '18 at 17:31
  • @TedShifrin I think I found same issue with you. – 민광건 Sep 07 '18 at 00:41
  • @민광건 Got a downvote..Not understanding what exactly is happening. Can you please comment? – Narasimham Sep 07 '18 at 23:20
  • @Narasimham First, Euler characteristic is 0 and we can all agree with that. $\partial M$ is piecewise-smooth curve. One is geodesic and the other is not. so when we calculate total geodesic curvature we would have to consider only one piece of $\partial M$. Integrate it yourself with radius of circular sector 3 and central angle ${2\pi \over 3}$. Then you can find difference between your calculation and Gauss-Bonnet. – 민광건 Sep 08 '18 at 00:17
  • I think I found something we missed. Can we apply Gauss-Bonnet? I mean, $M$ is not even orientable. – 민광건 Sep 08 '18 at 00:25
  • Sure it's orientable. It's a subset of a cone! – Ted Shifrin Sep 12 '18 at 06:21

3 Answers3

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If I'm reading that picture correctly, the Euler characteristic shouldn't exist.

The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).

A disk has $\chi=1$ , and a cylinder has $\chi=0$ . The surface $M$ is between those.

Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.

The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $\chi=5-8+3=0$.

mr_e_man
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enter image description here

We have Gauss Bonnet theorem

$$ \int K dA + \int \kappa_{g} \,ds= 2 \pi \chi $$

Solid angle $\int K dA= 0$ on the flat development of cone.

Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $\Sigma \psi_{i}= 2 \pi $ around the contour for the line integral which can be readily found as:

$$ \Sigma \psi_{i} = \int \kappa_{geodesic}\, ds = (\pi- \beta) + 2 \beta + (\pi- \beta) = 2 \pi$$

The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.

Plug these angles into GB theorem and we have

$$ 0+ 2 \pi = 2 \pi \chi\rightarrow \chi=1. $$

It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.

Narasimham
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Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short

$ \int K dA + \int \kappa_{g} \,ds= 2 \pi \chi$

where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 \pi \chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $\Sigma \psi_{i}$ around the contour for the line integral.

For a Torus with cancelling geodesic sections:

$0+ 0=2 \pi \chi \rightarrow \, \chi=0 \tag1$

For a hemisphere bounded by an equator:

$2 \pi +0 = 2 \pi \chi \rightarrow \, \chi=1 \tag2 $

For a closed convex loop on a developable surface (Gauss curvature $K=0\,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):

$0+ 2 \pi=2 \pi \chi \rightarrow \, \chi=1 \tag3 $

The matter is thus established by isometry/topological considerations.

EDIT1:

I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.

Narasimham
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    This doesn't really address the question. – Ted Shifrin Sep 07 '18 at 00:52
  • Sorry but I think you didn't get what the question is. – 민광건 Sep 07 '18 at 02:02
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    You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose. – Ted Shifrin Sep 07 '18 at 02:15