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This question is a follow-up to When does a null integral implies that a form is exact? . As mentionned in the selected answer, given certain conditions it is possible to find an isomorphism between the top de Rham cohomology and the integral $\int_M \omega$ of the members of each equivalence class $\omega$.

Moving on to de Rham cohomology below the top one. Taking the example of the torus as in , is it possible to link the rank $\binom{n}{k}$ to the values of specific integrals in a similar fashion ?

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up vote 2 down vote accepted

Let $T^{n}$ denote the $n$ torus* and $\pi_{i} : T^{n} \rightarrow S^{1}$ denote the $i-th$ projection. Let us denote the pull-back of the top from (if you do not take the normalized top form i.e. $\int_{s^{1}} d\theta = 1$, you will get some $2\pi$ factors) $\pi_{i}^{\ast}(d\theta) = \omega_{i}$. Then each $\omega_{i}\in H^{1}_{DR}(T^{n})$ and by dimension count and the fact that they are linearly independent (as can be seen by integrating on a 'suitable' component $S^{1}$) they together generate $H^{1}_{DR}(T^{n})$.

Now assume the ranks of all de-Rham cohomology of $T^{n}$ are known. The cohomologies $H^{k}(T^{n})$ are generated by wedge of $k-$ subsets of $\omega_{i}$. That they are linearly independent (in cohomology) can be seen by integrating over $k-$ cycles consisting respective component $S^{1}$ factors. Then dimension count gives isomorphism. (There is a more correct proof by showing compatability of wedge and cup products and using comparison of singular/simplicial cohomology and de-Rahm cohomology).

The inegral of each such wedge is just a product of constituent $\omega_{i}$ on $(S^{1})_{i}$.

Hence the basis element $\sum_{I \subset \{1,2,..n\}, |I| = k} (\wedge_{i \in I}\omega_{i})$ integrates on the k-cycle $\sum_{I \subset \{1,2,..n\}, |I| = k}\pm [\prod_{i \in I}(S^{1})_{i}]$ to give you the ranks. The plus minuses are important due to rule of signs in wedge products.

*Torus as per your question. Too many things are called torus in too many different contexts.

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