# Classification Theorem of Surfaces

1. Mar 8, 2008

### Tchakra

I am having difficulties grasping the consequences of this theorem, would really appreciate a little enlightenment.

A: Well, the statement of the theorem is clear, that Every closed Surface is homeomorphic to:
1) a sphere,
2) the connected sum of g tori
3) or the connected sum of g Projective plane.

B: Closed surface means Every closed 2 dimensional manifold that is embeddable in $$R^n$$. So $$S^3$$ or in fact $$S^n$$ is homeomorphic to one of the above.

C: Another point is that $$S^n \cong R^{n+1}$$

So by transitivity we get $$R^4 \cong S^3 \cong S^2 \cong R^3$$, which is false by the invariance of dimensions.

Given A,B and C i come to this conclusion which i know to be wrong, so can someone explain to me where i went wrong.

The only conscious leap i have made is B, which i haven't read from any book, but given the definition i don't see why it should be otherwise.

2. Mar 8, 2008

### Tchakra

I just realized my mistake, my third point is wrong. The Euler characteristic is different ...

3. Mar 8, 2008

### HallsofIvy

Staff Emeritus
I was wondering why you would even think so! S1, a circle, is obviously not homeomorphic to R2, a plane. S2, the surface of a a sphere is not homeomorphic to R3. In both cases, the first is compact and the second isn't.

4. Mar 8, 2008

### Tchakra

I had a perfectly valid reason at the time :tongue2:, f(x)= x/|x| , if you discount to consider the point x=0 which i didn't.

5. Mar 11, 2008

### StatusX

Also, S^3 is clearly not a 2 dimensional manifold.

6. Mar 11, 2008

### mathwonk

i assumed this was about classifying complex surfaces, by kodaira dimension.

i.e. rational, ruled surfaces, elliptic surfaces, abelian surfaces, K3 surfaces, surfaces of general type.

7. Mar 12, 2008

### SCV

That's what I thought about too when I saw the title.

8. Mar 19, 2008

### eastside00_99

I only know the basic theorem:
every topological 2-manifold (surface by some authors) is homeomorphic to the connected sum of either a sphere or some finite combination of tori and projective planes. I also only know the basic proof which involves building your surface from simplexes, noticing that by labeling the edges you can represent each surface as an element in some free group, and then showing that each of these elements reduce to either 1 or some finite "product" of the element which corresponds to tori and the element which corresponds to the projective plane. Seems like there should be a more advanced proof via homology though.

9. Mar 19, 2008

### mathwonk

for oriented surfaces at least, i like the morse theory proof, namely there is a smooth function on the surface that has a finite number of non degenerate critical points, which can be arranged to be one max, one min, and 2g saddle points.

(all surfaces have a smooth structure.)