Having trouble with (simplicial) homology

Well, I've been to over two weeks' worth of lectures and reading the relevant chapters in Hatcher, but I still have no idea how homology works.

Here's what I understand

you have n-simplices like points, lines, triangles, tetrahedrons, etc.

then you have boundary maps, which take the simplex and their rotation and give the faces. For example if you had a triangle with vertices x,y,z then the boundary map of this simplex gives [x,y]-[y,z]+[z,x]...

Ok. So [x,y] would be the 1-simplex between 0 and 1. What does the sum [x,y]-[y,z]+[z,x]

The first example in the chapter is computing the homology groups of the circle. Okay, so a circle is one vertex v and one edge e.

The explanation given in Hatcher includes:

"Then $\Delta_1(S^1)$ and $\Delta_0(S^1)$ are both Z and the boundary map $\partial_1$ is zero since $\partial e = v-v$ ."

I'm lost. This is supposed to be the easiest example, and I have no clue what he's rambling on about. Could someone help?

Quote:

Originally Posted by harbottle

Well, I've been to over two weeks' worth of lectures and reading the relevant chapters in Hatcher, but I still have no idea how homology works.

Here's what I understand

you have n-simplices like points, lines, triangles, tetrahedrons, etc.

then you have boundary maps, which take the simplex and their rotation and give the faces. For example if you had a triangle with vertices x,y,z then the boundary map of this simplex gives [x,y]-[y,z]+[z,x]...

Ok. So [x,y] would be the 1-simplex between 0 and 1. What does the sum [x,y]-[y,z]+[z,x]

The first example in the chapter is computing the homology groups of the circle. Okay, so a circle is one vertex v and one edge e.

Consider 2-simplex $[v_0, v_1, v_2]$ (see Hatcher's example p105). Its boundary map is the sum of 1-simplices with signs. For instance, if you follow the counterclockwise route, $\partial_2[v_0, v_1, v_2] = [v_0, v_1] + [v_1, v_2] + [v_2, v_0] = [v_0, v_1] - [v_0, v_2] + [v_1, v_2]$ , where $[v_2, v_0] = -[v_0, v_2]$ .

Quote:

"Then $\Delta_1(S^1)$ and $\Delta_0(S^1)$ are both Z and the boundary map $\partial_1$ is zero since $\partial e = v-v$ ." I'm lost. This is supposed to be the easiest example, and I have no clue what he's rambling on about. Could someone help?

A boundary map of 1-simplex $v_0 \rightarrow v_1$ is $\partial_1[v_0, v_1]=[v_1] - [v_0]$ . In Hatcher's example, $S^1$ has a simplicial structure that consists of one edge (1-simplex) and one vertex (0-simplex). In this case, $\partial_1[v_0, v_0]=[v_0] - [v_0]=0$ .

We know that $H_n = \frac{Ker \text{ } \partial_n}{Im \text{ }\partial_{n+1}}$ .
Now consider $H_0(S^1) = \frac{Ker \text{ } \partial_0}{Im \text{ }\partial_{1}}$ . A boundary of a 0-simplex (vertex) is an empty simplex. Thus $\partial_0(v) = 0$ . Now we see that $Ker \text{ }\partial_0$ has a single generator (0-simplex v) and its group is $\mathbb{Z}$ . This generator is not an image of a boundary map $\partial_1$ , we have $H_0(S^1) = \mathbb{Z}$ .
Similary, $\partial_1(e) = 0$ as seen above. Thus $Ker \text{ }\partial_1$ has a single generator (1-simplex "e") and its group is $\mathbb{Z}$ . Similarly, this generator is not an image of a boundary map $\partial_2$ (actually we have no 2-simplex here), we have $H_1(S^1) = \mathbb{Z}$ .

this clears some things up, thank you! Expect more replies to this thread as I read on...