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DeRham Cohomology of n-torus

  1. Jun 25, 2012 #1

    I want to ask for a verification of something I did. Lets say I want to compute the m-th cohomology group of the n-torus with coordinates θi. Suppose that I have shown each [dθi_1 [itex]\wedge[/itex]dθi_2 ... [itex]\wedge[/itex] dθi_m] is an independent element of this deRham cohomology (it can be shown by some integration over m-toruses that reside in the n-torus that these are closed but not exact differential forms whose differences are not exact). In particular its dimension is then atleast C(n,m)

    Now to finish I have to show that this is all. I want to a dimension arguement here. In the lattice quotient model of the Torus it is easy to show that each n-torus is a CW-complex with 1 0-cell, n 1-cells, C(n,m) m-cells. Then one can compute the homology groups and dimension of its mth homology group is C(n,m) then dimension of the mth cohomology group is also m (I know the computation of homology groups for T2 is easy I assume Tn is the similiar). Infact you don't need to compute the homology groups. Since each chain Cm has dimension C(n,m) then the homology group can have at most that dimension which is more than enough for me.

    Now we now that for simplical complices Hm(X) = Hm(Xk) for k>m. In particular here we have that Xn is the Torus itself so that cellular homology and normal homology groups can be identifies (this is the first part I am not sure). Then normal homology group is isomorphic to simplical homology groups and there fore each have the same dimension. Now simplical cohomology groups of simplical complexes are isomorphic to deRham cohomology. Thus basically m-th deRham cohomology group has the same dimension as the mth cohomology of the n-torus which is C(n,m) therefore completing the proof.

    Right? :p
    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2
    edit: finally manged to tex
    Last edited: Jun 25, 2012
  4. Jun 26, 2012 #3
    I meant " then dimension of the mth cohomology group is also C(n,m)" not m
  5. Jul 3, 2012 #4


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    i have read this a couple of times am am not sure what you are asking.

    For instance why are you not using the Kunneth formula with the reals as the coefficient group?
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