Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd like to show that the Euler characteristic ($V - E + F$) of a compact oriented surface without boundary, $S$$g$, is of the form $2 - 2g$ where $S$$g$ is a sphere with $g$ handles.

A sphere with handles is obtained by cutting $2g$ disks out of the sphere and gluing in $g$ cylinders along the boundary circles. Hence, $S$$0$ is the sphere and the torus is $S$$1$.

I'm pretty new to topology. But I think induction would make for an easy, understandable proof where the base case is the sphere.

Any help or solutions are appreciated!

share|improve this question
    
How do you define $g$? This formula is a perfectly valid definition of $g$ for a triangulable oriented surface... –  Zhen Lin Dec 8 '11 at 1:05
1  
That doesn't sound like a rigorous definition to me. Consider the one-point compactification of the surface $\{ (z, w) \in \mathbb{C}^2 : z^2 = w^3 - w \}$. What's its genus? –  Zhen Lin Dec 8 '11 at 1:22
    
I've restated the question. Hope that clarifies! –  Phil M. Dec 8 '11 at 2:39
    
There are many ways of doing this. Personally, I would work with the enlarged family of surfaces with boundary. You can compute the Euler characteristic of the disk and of the cylinder by hand. You can also find what happens to the Euler characteristic when you attach two surfaces along some of their boundary components. Once you have this, you can compute the Euler characteristic of the sphere with 2g disks removed and then use your construction. –  Sam Lisi Dec 10 '11 at 17:25

1 Answer 1

You could represent your surface with $g$ handles as a polygon with $4g$ edges suitably identified in pairs, and then compute $V$, $E$, and $F$ for that polygon, and then $V-E+F$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.