It's confusing to calculate Euler characteristic of this surface

Question

This pic below is an exploded view of a cone.

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

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.

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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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\chi=5-8+3=0$.