# Thread: Computing a Δ-complex homeomorphic to S^2

1. ## Computing a Δ-complex homeomorphic to S^2

I'm not very topology-savvy, and the best I could come up with was:

Identify two (2-dimensional) closed discs to be the upper and lower hemispheres of S^2, and then chop these up into four pieces each; then each sector can be identified with a 2-simplex, with the right orientations. This gives a Δ-complex with eight 2-simplices, twelve 1-simplices, and six 0-simplices (by my counting).

If I'm perfectly honest, this seems quite complicated for such a simple topological space. Even the torus has only six simplices in all. Am I missing some simpler identifying or 'gluing' procedure?

EDIT: Sorry, I'm being silly. I think there is a way to do it identifying 2-simplices (triangles) to closed discs. Can anyone just confirm that when you do this correctly, the homology groups should be $H_n(S^2) = \{0\}$ for $n \neq 2$, and $H_2(S^2) = \mathbb{Z}$ ?

2. That's right, as long as you're talking reduced homology--else H_0 is Z as well.

Actually, it's isomorphic to Z--you shouldn't write "=" here. (In case you didn't know how to write "isomorphic to" in LaTeX, it's \cong.)

3. Originally Posted by Tinyboss
That's right, as long as you're talking reduced homology--else H_0 is Z as well.

Actually, it's isomorphic to Z--you shouldn't write "=" here. (In case you didn't know how to write "isomorphic to" in LaTeX, it's \cong.)
Not sure what you mean by reduced homology. I haven't been introduced to that yet.

4. In standard simplicial homology, you declare the boundary of any 0-chain (sum of points) to be 0, and then it turns out that $H_0\cong\mathbb{Z}^n$, where n is the number of connected components. So H_0 is Z for a sphere or other connected space.

In reduced homology, you use a different map that sends every 0-simplex to 1, so that a 0-chain maps to the sum of its coefficients. It's clear that a boundary of a 1-simplex still goes to zero under this map, but now not every 0-chain is a cycle, and H_0 is Z^(n-1) instead of Z^n. (All other homology groups are identical.)

Then for the n-sphere, you have the situation you wrote, where H_n is Z and H_k is 0 for k not equal to n.