De Rham theorem
In mathematics, more specifically in differential geometry, the de Rham theorem says that the ring homomorphism from the de Rham cohomology to the singular cohomology given by integration is an isomorphism.
But the de Rham theorem gives a more explicit isomorphism between the two cohomologies; thus, connecting analysis and topology more directly.
The key part of the theorem is a construction of the de Rham homomorphism.
It is not difficult to show that the de Rham homomorphism is a natural transformation between the de Rham cohomology functor and the singular cohomology functor.
[3] This Čech version is essentially due to André Weil.
When considering singular cohomologies with coefficients in another abelian group, for example the integers, then of course one should not expect similar isomorphism.
, and since the cohomology with real coefficients does not account for any finite (more generally, torsion) groups, we have
This indeed coincides with the corresponding de Rham cohomology group.
However, a technical result implies that the singular homology groups coincide with smooth singular homology groups.
This shows that the de Rham theorem actually implies isomorphism between de Rham cohomology and (nonsmooth) singular cohomology groups (with real coefficients).
One proof roughly follows these ideas:[4] Call a manifold "de Rham", if the theorem holds for it.
are de Rham, basically by the homotopy invariance of both cohomologies in question.
Then the result is being extended to manifolds having a basis which is a de Rham cover.
Finally, one easily shows that open subsets of
and consequently any manifold has a basis which is a de Rham cover.
Thus, invoking the previous step, finishes the proof.
Here is another proof sketch, based on sheaf-theory:[5] There are two important complexes of sheaves on our manifold
(in which categories do these complexes live is a subtle problem, let us be vague at first).
Integration over simplices give us a morphism of sheaves of complexes
Since both objects admit partition of unities, it is a standard fact that the second pages of the hypercohomology spectral sequences for both of them only have one nonzero column each, thus the hypercohomologies of the two complexes of sheaves indeed calculates the de Rham and singular cohomologies of
Therefore, to prove the de Rham theorem, it suffices to show that
To this end, we note that there are natural morphisms from the constant sheaf
Furthermore, by local contractibility of both the de Rham and singular cohomologies, the natural morphisms are indeed isomorphisms.
By the commutativity of the triangle, we have shown the desired isomorphy of
, and the proof is complete, except that we need to return to the subtle problem at the beginning: in which category of sheaves of complexes do the argument above make sense?
Since we have used local contractibility of cohomologies to conclude the isomorphy of the two morphisms going out of
There is also a version of the theorem involving singular homology instead of cohomology.
It says the pairing induces a perfect pairing between the de Rham cohomology and the (smooth) singular homology; namely, is an isomorphism of vector spaces.
[7] This theorem has the following consequence (familiar from calculus); namely, a closed differential form is exact if and only if the integrations of it over arbitrary cycles are all zero.
There is also a current (a differential form with distributional coefficients) version of the de Rham theorem, which says the singular cohomology can be computed as the cohomology of the complex of currents.
