# Thread: Having trouble with (simplicial) homology

1. ## 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?

2. 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 $[v_2, v_0] = -[v_0, v_2]$.

"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 $H_1(S^1) = \mathbb{Z}$.

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