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.

We know that the sphere $S^n$ has $n$-th singular homology $H_n(S^n)= \mathbb{Z}$. A generator is given by a fundamental class, which is nothing else than the sum of the simplices in some triangulation. Thus my question is: Is there any intuition behind the fact that such a sum of simplices generates the whole homology? Is there any picture one can have in mind to see every element of the singular homology is a power of this fundamental class?

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

The idea is simply this. Suppose you have (a representative of) a nonzero top homology class. Then it has $k\not= 0$ copies of some n-simplex. Since this is a manifold, each boundary face is attached to the boundary face of exactly one other n-simplex, so in order for our chain to be a cycle, we must have $k$ copies of those simplices too. These, in turn, force their neighbors to be in our chain with multiplicity $k$, and so everything propagates around the entire (connected component of) the manifold until we see that this is just $k$ copies of the fundamental class...

Or we get a contradiction. Precisely in the case that we're working with a nonorientable manifold, we'll be able to come back around to our original simplex and we'll want to say that it has multiplicity $-k$, which is impossible unless $k=0$. This illustrates that the top integral homology of a nonorientable manifold is 0, and also suggests why you get a fundamental class for any manifold when you use $\mathbb{Z}/2$ coefficients.

share|improve this answer
1  
This is more of a proof than a piece of intuition :) –  Mariano Suárez-Alvarez Dec 4 '11 at 1:27
    
Haha. Well, math makes the most sense when the proof gives intuition! (And, I suppose one could argue, if your proof doesn't give intuition then you've got the wrong proof!) –  Aaron Mazel-Gee Dec 4 '11 at 2:12
add comment

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.