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Fundemental group, Torus, SVK

  1. Nov 24, 2009 #1

    So I've been using Seifert-Van Kampen (SVK) to calculate the fundemental group of the torus. I haven't done any formal group theory, hence my problem ...

    I have T^2=(S^1)x(S^1)

    If A= S^1, B=S^1, A intersection B is 0. And T^2 = union of A and B.

    Then fundemental group of (A intersection B) = 0

    And I already have fundemental group of (S^1) = Z

    Then using SVK the fundemental group of the torus is the free product of S^1 with S^1 over 0.

    Which I think is isomorphic to the ZxZ.

    In the literature this is written as the direct sum of Z+Z. Why is this the direct sum and not the cross product?

    Since I don't know group theory better I don't know if it is possible to just ask the simpler: Why is the free product of Z*Z is isomorphic to direct sum Z+Z not the product of ZxZ.
  2. jcsd
  3. Nov 24, 2009 #2


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    Three comments:

    1) If you have a finite collection of abelian groups, then their direct product and direct sum are the same.

    2) The free product of Z*Z is not isomorphic to either Z[itex]\oplus[/itex]Z or ZxZ; in fact, Z*Z is nonabelian (as most free products are).

    3) SVK is unnecessary for computing pi_1 of the torus. Instead one can just use the fact that pi_1(XxY) = pi_1(X) x pi_1(Y), which is easy to prove to directly. If you insist on using SVK, then you must be doing something wrong: pi_1 of the torus isn't Z*Z -- it's Z*Z modulo a certain normal subgroup.
  4. Nov 25, 2009 #3


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    I just noticed that you provided your working for how you applied SVK. What do you mean when you say "A= S^1, B=S^1"? I can't put any meaning to this assignment that gives us "T^2 = union of A and B".

    In any case, SVK needs a special open cover of T^2, not an arbitrary cover.
  5. Nov 25, 2009 #4
    thanks for your help on this,

    I will work on this and reply later
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