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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 θ_{i}. Suppose that I have shown each [dθ_{i_1}[itex]\wedge[/itex]dθ_{i_2}... [itex]\wedge[/itex] dθ_{i_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 T^{2}is easy I assume T^{n}is the similiar). Infact you don't need to compute the homology groups. Since each chain C_{m}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 H_{m}(X) = H_{m}(X_{k}) for k>m. In particular here we have that X_{n}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

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# DeRham Cohomology of n-torus

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