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Basic Confusion: Genus, Homology (Exm of Torus)

  1. Jul 18, 2011 #1
    Hi, All:

    I'm a bit confused about this: the 1st homology of the torus T(over Z) is Z(+)Z,

    so that elements of the form (a,b) =/ (0,0) are non-trivial , meaning these are

    cycles (closed curves) that do not bound subsurfaces of the torus, and every cycle

    that does bound is

    equivalent/homologous to (0,0). Still, I don't see why, e.g., the cycle (1,1) of a meridian

    and then a parallel, is non-bounding: if we went once around meridionally and once

    longitudinally, we would disconnect the torus. Isn't (1,1) then, a cycle that bounds?

    Additionally, the genus of the torus is 1 , meaning that we can remove at most one

    SCC (simple-closed curve) from the torus without disconnecting it, so it would seem

    like any pair (a,b) would consist of 2 curves ( one going a times meridionally and b

    times longitudinally ), so that, since the genus is 1, (a,b) would disconnect the torus,

    right? But then (a,b) would be a cycle that bounds, so that it would be homologous

    to (0,0), which it is not. What am I missing?

    EDIT: I realize the obvious fact that (a,b) is not a SCCurve. Is this the issue, i.e.,

    must homology classes (at least for the torus) be represented by SCC's?
    Last edited: Jul 18, 2011
  2. jcsd
  3. Jul 18, 2011 #2


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    Nay. If you cut meridionally, you are left with a cylinder. Then when you cut longitudinally, you obtain a rectangle. rectangle = connected.

    You can realize (a,b) as a SCC. For instance, (1,1) can be realized as the diagonal of the square in the polygonal presentation of the torus. But hopefully by now, you no longer think that (a,b) disconnects.
  4. Jul 18, 2011 #3
    Thanks, Quasar, its clear now. I was wrongly focusing on the red herring that we

    were prying the space open, confusing that with the real issue of the end-space

    (after the cuts ) being disconnected. I guess I am getting a head start on senility.
    Last edited: Jul 18, 2011
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