2

It is said that the boundary operator $\partial_k$ maps a $k$-chain to a $(k-1)$-chain. I've also seen that this operator can be represented with a matrix of dimension $|K^{k-1}|\times|K^k|$. I can't figure out with a simple example how I can multiply this matrix with a vector of a $k$-chain to obtain the $(k-1)$-chain... Could someone show me a simple example on a single $2$-simplex or $3$-simplex ?

I'm a computer scientist so be easy on me, math is not my strongest point. Thanks a lot.

MITjanitor
  • 2,690
Rodolphe Vaillant
  • 23
  • 3

1 Answers1

1

If you didn't know this already, Abelian groups are $\Bbb{Z}$-modules.

So, I will do this treating the boundary operator as a homomorphism, not a matrix. The $k$-chains are in one $\Bbb{Z}$-module $C_k$, generated by the k-simplices, and the (k-1)-chains are in another $C_{k-1}$, generated by the (k-1)-simplices. The boundary operator is a linear map, which is is a group homomorphism $\partial_k(\mathbf{x}+\mathbf{y})=\partial_k(\mathbf{x})+\partial_k(\mathbf{y})$ such that scaling can go before or after the map $\partial_k(\lambda \mathbf{x})=\lambda\partial_k(\mathbf{x})$ for $\lambda$ a scalar.

Now, if you have a basis for $C_k$ and $C_{k-1}$, Wikipedia will tell you how to convert the linear map into a matrix over the integers.

If you don't, I would watch someone compute one for a homology group.

Loki Clock
  • 2,193
  • I will annotate that video's calculation, since he starts off with matrices to begin with. He starts off with four edges to begin with, which gives a map into $\Bbb{Z}^4$. The trouble is, $C_{k-1}$ is not necessarily $\Bbb{Z}^4$, but could be a lower-dimensional submodule. So you need to find the dimension of the image, which is $C_{k-1}$, and is spanned by those vectors. Eliminating dependencies, you get a basis for the image, $C_{k-1}$. In other words, compute the rank of the matrix form of the linear system of equations produced by the boundary operator. He chooses Gaussian elimination. – Loki Clock Mar 09 '13 at 21:25
  • Thanks a lot, took me some time but I think I finally understood how this matrix mapping works. Video link is great I recommend watching the other videos for anyone trying to learn algebraic topology – Rodolphe Vaillant Mar 10 '13 at 13:23
  • Hey, so I did the calculations, and there are two things: 1) The image is not $C_{k-1}$. $C_{k-1}$ is $\Bbb{Z}^4$. The image is just the set of cycles that you can obviously contract to a point, because you can contract it through the thing they're a boundary of. 2) You don't need a basis for the image to have a basis to do the linear map with. Just use the basis of k-simplices and (k-1)-simplices. However, you'll want a basis for the kernel of the next map in terms of a basis for the image to compute the homology group. Also, it lets you factor the map into quotient and inclusion matrices. – Loki Clock Mar 13 '13 at 14:16