3
$\begingroup$

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 $\mathbb{R}$ ?

$\endgroup$
  • $\begingroup$ 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
  • 5
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.