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Math Help - Having trouble with (simplicial) homology

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by harbottle View Post
    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}.
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  3. #3
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    this clears some things up, thank you! Expect more replies to this thread as I read on...
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