According to http://en.wikipedia.org/wiki/De_Rham_cohomology,

one defines the $k$-th de Rham cohomology group $H^{k}_{\mathrm{dR}}(M)$ to be the set of equivalence classes, that is, the set of closed forms in $\Omega^k(M)$ modulo the exact forms.

On the other hand, the de Rham cohomology groups of a $n$-dimensional sphere $H_{dR}^q(S^n)$ is $\mathbb{R}$ if $q=0,n$ and 0 otherwise.

I am not sure to understand the link between $\mathbb{R}$ and the equivalence groups. Does that mean that to generate the de Rham cohomology groups of $S^n$, one can take any constant function $\omega$ (case $q=0$) or non-zero $n$-differential form $\omega$ (case $q=n$), and that each equivalence class can be generated from the product of $\omega$ by a particular member of $\mathbb{R}$ ?

  • $\begingroup$ For $H^0_{dR}(S^n)$, indeed the constant functions are precisely the closed $0$-forms, and since there are no exact $0$-forms except $0$, the constant functions are the natural representatives of the cohomology classes. For $H^n_{dR}(S^n)$, take any $n$-form with nonzero integral, and its multiples are representatives of the cohomology classes. (Since we're dealing with vector spaces of dimension $1$, the multiples aren't really interesting.) $\endgroup$ – Daniel Fischer Jul 9 '13 at 22:32
  • $\begingroup$ It is worth adding that what Daniel said generalizes. Given any compact $n$-dimensional manifold $M$, $H_{dR}^n(M)=\mathbb R$ and any $n$-form with nonzero integral will be a generator for the cohomology. $\endgroup$ – Aaron Jul 9 '13 at 22:52
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    $\begingroup$ I don't have time to write a full answer, but I think the point is that abuse of notation is happening: When we say "$H^q_{dR}(S^n)$ is $\mathbb{R}$", we really mean "is isomorphic to". $\endgroup$ – Jason DeVito Jul 9 '13 at 23:04
  • $\begingroup$ I guess you nailed it Jason. The trouble is that this notation is used in every resource I came across on Internet. For someone who is self-learning it is confusing. $\endgroup$ – vkubicki Jul 9 '13 at 23:24
  • $\begingroup$ Could you put this as a short answer, so I can close the question ? $\endgroup$ – vkubicki Jul 10 '13 at 7:54

For $q=0$ the isomorphism between $H^n_{dR}(M)$ and $\mathbb{R}$ is fairly canonical, but for $q=n$ this is no longer the case. Consider for example a nontrivial fibration with fiber $M$; typically there is no natural way of identifying top-dimensional cohomology of each fiber with $\mathbb{R}$, since the determinant bundle will generally not be trivial.


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