As is well known, a differential manifold has trivial degree-$0$ de Rham cohomology $H^0(M) = 0$ if and only if it is connected.

It seems that the degree-$1$ de Rham cohomology group $H^1(M)$ being non-trivial implies that $M$ is not simply connected. I guess the converse is not true, but I would like to know of a proof.

Do there exist similar relationships between higher de Rham cohomology groups and the homotopy groups of differential manifold?

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    $\begingroup$ You should have a look at the Hurewicz theorems: en.m.wikipedia.org/wiki/Hurewicz_theorem $\endgroup$ – Mindlack Feb 7 at 15:54
  • $\begingroup$ Great! Thanks a lot! $\endgroup$ – Pierre Dubois Feb 7 at 16:09
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    $\begingroup$ A non-simply-connected manifold with trivial first de Rham cohomology is for example the real projective plane. The Hurewicz theorem lets you relate up to two nontrivial homotopy groups to homology groups. In general obtaining information on homotopy groups from the de Rham cohomology is unlikely. But if you do not take cohomology but rather take the whole de Rham algebra, then rational homotopy theory tells you that this differential graded algebra (let's say for a simply connected manifold) determines the homotopy groups modulo torsion. $\endgroup$ – Aleksandar Milivojevic Feb 7 at 18:27
  • $\begingroup$ If the fundamental group is $G$, then the first de Rham cohomology is $\text{Hom}(G, \Bbb R)$. There are a truly massive number of groups for which this is trivial; for instance, any group whose abelianization is a torsion group. $\endgroup$ – user98602 Feb 7 at 22:17

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