algebraic-topology
How is it calculate the Euler characteristic of sphere using its fundamental polygon?
It is $V = 2, E = 2, F = 1$, so $\mathcal{X}(S^2) = 1$? (I know its 2, so there is something wrong)
And for the projective plane?
It is $V = 2, E = 2, F = 1$, so $\mathcal{X}(\mathbb{R}P^2) = 1$?
I saw this question, but it mention a graph: Euler characteristic of the projective plane (using embedding diagram)
Regards!
If you're ok with the fundamental polygon being a polygon on a surface which isn't $\mathbb{R}^2$, then you can get a polygon with two edges on the sphere $S^2$.
The polygon is given by taking the northern hemisphere, and the edges are given by each a half of the equator, with the identification being the antipodal map on the circle defining the equator.
If you know how covering spaces work, this region is a fundamental region of the $2$-fold covering of the sphere on to the real projective plane.
You can see four squares in the figure. The two diagonally opposite squares form a face, as you can see from the gluing. That is why it has two faces in total.
Best Answer
Your image https://i.sstatic.net/oG7eW.png is $\mathbb{R}P^2$: 2 vertices, 2 edges, 1 face, and its Euler characteristic is $2-2+1=1$.
The corresponding picture for $S^2$ could look like the image https://i.sstatic.net/lbGOz.png. There are 3 vertices, 2 edges, and 1 face, and its Euler characteristic is $3-2+1=2$. (How do I know there are three vertices and two edges? When you glue this, you are gluing the bottom edge to the left edge, and the right edge to the top edge. So the resulting "sphere" will have a blue edge and a red edge, a vertex at the start of the blue edge, a vertex between the two, and a vertex at the end of the red edge. There is no more gluing among the vertices.)