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Consider a sphere with two handles. If I don't make a mistake it is torus with one handle: enter image description here

Can you give me an example of vector field on it with one singular point.

Thanks.

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Did you try to get the surface by gluing sides of an octagon? Then it should be clear how to do it. –  t.b. Sep 12 '11 at 4:34
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up vote 13 down vote accepted

The key idea behind this problem is the Poincaré–Hopf Theorem, which states that the sum of the degrees of singular points of a vector field on a compact, orientable differentiable manifold is equal to the Euler characteristic of the manifold. This is a really pretty theorem; I first learned it from Alexei Sossinski during a semester in Moscow.

You are correct that the sphere with two handles is the same thing as the torus with one handle; this is the surface of genus 2. (A sphere has genus 0, and a torus has genus 1.) The Euler characteristic of a genus $g$ surface is $2-2g$, so we have Euler characteristic $-2$. Thus, by the theorem, if there is only one singular point; it has degree $-2$.

Singularity Degrees

The degree of a singularity on a 2D vector field is the winding number of the vector field along a small circle around the singularity. That is, we travel around the singularity counter-clockwise, and count the rotations made by the directions of vectors we encounter (counting clockwise rotations as negative and counter-clockwise rotations as positive).

Your diagram does give a genus 2 surface topologically, but we need to work with differentiable manifolds. If we look at the "hole" you use to make the handle, all four corners map to the same point. But that means the neighborhood of this point has 4*270 degrees = 1080 degrees! So to make your diagram a differentiable manifold, we need to use a different coordinate system at that point, so that it has a normal 360 degree neighborhood. Below is a vector field with just one singularity on your surface, taking all of this into account. I know I have to start with a degree $-2$ singularity, and I place vectors around the glued edges and points, taking into account the coordinate change required at your handle.

enter image description here

With that starting point, I found a nice continuous, non-zero vector field to fill the rest of the space. (I drew curved flow lines instead of just the vector field, because it's a lot easier to see what's going on that way.)

Theo suggested using the octagon representation of the genus 2 surface, which is generally easier to work with and to generalize. (For this problem, I don't think it's actually any easier.) Notice that all 8 corners are really the same point, so each corner is drawn as 45 degrees. In my picture, there is no need to adjust angles. (If you're familiar with the hyperbolic plane, that's where my drawing comes from. We use the hyperbolic plane, because this surface has negative curvature.) I labeled the corners so that you can see how they all get glued together to form a nice 360 degree neighborhood (and so I could keep track of them for drawing my vector field).

gluing diagram

Here's the vector field with just one singularity drawn on my octagon. I started with a vector pointing in the "1" direction at the corners; from the picture above, you can see that the vector field at the corners lines up perfectly.

vector field on octagon

The edges are a bit harder to check, but if we straighten out each edge, we will see that the vector field on both sides lines up as well. (There are four edges, but they all give essentially the same picture.) I drew this picture to show how the vector field along the top edge lines up with the field along the right edge; think about where my picture has vectors going into, parallel to, or out of the edge.

edge arrows

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Best pictures evar! –  Mariano Suárez-Alvarez Sep 12 '11 at 18:44
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Yes, very nice! I still think using $4g$-gons is the way to go :) –  t.b. Sep 12 '11 at 18:59
    
@Theo: Thanks! And apologies for misspelling your last name; I just read it carelessly (and apparently have been doing so in my head for a while). –  Jonas Kibelbek Sep 12 '11 at 19:03
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