Filling a hole by eliminating a homology class

86 Views Asked by At

Let $S$ be a simplicial complex of dimension $m$. I read the following sentence from the book "Advances in nonlinear Geosciences" by F.C Motta :

filling an $n-$dimensional hole amounts to eliminating a class $[\gamma]\in H_n(S)$ by making the cycle $\gamma$ into the boundary of an $(n+1)-$chain so that $[\gamma]=[0]\in H_n(S)$ since $0$ and $\gamma$ differ by a boundary, namely $\gamma$.

I'm not able to understand this sentence because I don't see the meaning of filling a hole and what does it mean mathematically to eliminate a class.. and how to make a cycle into a boundary..

Thank you very much for explaining this sentence..

1

There are 1 best solutions below

1
On BEST ANSWER

I think Eric Wofsey is correct in that the statement is not meant to be construed literally.

As an illustration of the "idea" here, consider the standard "hollow" triangle constructred with $3$ one simplices and call it $X$.

If you calculate the integral homology $H_1(X)=[\alpha]$, it is generated by a single cycle. In order to make it a boundary, you can paste in a two-simplex into the interior of $X$ to create a topological disk. Now the cycle $\alpha$ bounds a higher dimensional simplex and it "fills" a one-dimensional hole, and "kills" the homology class $[\alpha]$.