Method for finding Euler Characteristic in Surfaces(Differential geometry)

Question

There are some problems that bothering me, "Euler Charatcteristic, $\chi(M)(= v-e+f)$"

Here are some Questions that I'm stuck.

Find the $\chi(M)$

$Q1)$ $M$ is a rotation of $y=cosx $($-\pi \leq x \leq\pi$) whose axis is $Y-axis$

$Q2$ $M$ is a rotation of $y=2+sinx $($0 \leq x \leq 2\pi$) whose axis is $X-axis$

I'm trying to find $\chi(M) = v+e-f$ by dividing surfaces by traingles or rectangulars(polygonalization) but failed. How could I found that?

Here is my trial of the Q1

I don't know what I was wrong

The answer of $Q1$ and $Q2$ in my textbook are $\chi (M) =1$ and $\chi(M) =0$ respectively.

I can't understand Why the answers are $\chi (M) =1$ and $\chi(M) =0$. :(

P.S) Please give me some method or examples for finding the $\chi (M)$ by divide as the triangle or rectangular for surfaces in $R^3$

Any help would be appreciate!

Answer 1

In Q1, it looks like you counted $f$ wrong, it should be $f=2$ and therefore $v-e+f=1$.

Perhaps you mistakenly counted the 2-dimensional round disc at the bottom of the figure as one of the faces? If so, $M$ is a surface of revolution of the graph $y = \cos(x)$ ($-\pi < x < \pi$), and that 2-dimensional disc is not part of the surface.

Added to address the question in the edit: The graph that is being rotated is the graph of the function $y = \cos(x)$ ($-\pi \le x \le \pi$), more explicitly the set $$\{(x,y) \mid y = \cos(x), -\pi \le x \le \pi\} $$ That's the set which you correctly drew in your diagram.

So, for example, when $x=0$ you get $y=1$ and the point $(0,1)$ is part of the graph that is being rotated; however the point $(0,-1)$ is NOT part of the graph that is being rotated.

For another example, when $x=\pi/2$ you get $y=0$ and the point $(\pi/2,0)$ is part of the graph that is being rotated. However, the point $(\pi/2,-1)$ is NOT part of the graph that is being rotated.

In fact, no point with $y=-1$ is on the graph that is being rotated, except for the points $(\pi,-1)$ and $(-\pi,-1)$. Those two points, when rotated, give you a circle in the plane $y=-1$ which is part of the rotated graph. However, the points on the plane $y=-1$ which are on the inside of that circle are NOT part of the rotated graph.

If you had connected the two points $(\pi,-1)$ and $(-\pi,-1)$ by a straight line segment in $y=-1$, then that entire line segment would have been rotated and your rotated graph would have contained the disc in question. But, you did NOT connect those two points by a straight line segment.

Answer 2

The good thing about Euler Characteristic is that it is invariant by homotopy. That is, you can "deform" your surfaces, and the characteristic won't change.

You don't actually have to use this fact, but having it mind will help. To picure your triangulations in your mind, you can, for instance, deform your surfaces into simpler ones, triangulate the simpler surfaces, and deform them back to the original surface. By deforming the triangulation along, you will obtain triangulations of the original surfaces.

Your first surface, for instance, if you "flatten" it, is simply a disc. Put thre points on the border of the disc, draw lines from these points to the center, and you will have a triangulation with the following data: 4 vertexes (the 3 points and the center), 6 edges, and 3 faces, then $\chi(M) = 4 - 6 + 3 = 1$, as expected.

The second one is a "tube", or equivalenty, the boundary of a cylinder, minus the two base discs (but the circles are included).

Here, it is simpler to divide it with rectangles: Put two points on each base circle, so that they are opposite, put a line between these points, you get then two rectangles (one for each "half-cilinder"). You then have 4 vertexes, 6 edges and 2 faces. Hence $\chi(M) = 4 - 6 + 2 = 0$, as expected.