Non-homotopical manifolds with same de Rham cohomology

Question

I am searching for manifolds $M$ and $N$ with different homotopy type such that their de Rham cohomology is isomorphic as rings. It would, of course, be enough to find $M$ and $N$ with different $\pi_1$.

Answer 1

All of the examples here are closed manifolds (compact without boundary); as Jason DeVito points out, you can simplify this a bit if you allow noncompact manifolds.

You can do this as soon as dimension 3. One nice fact is that if $M \to N$ is a finite covering map, then it induces an injection $H^*(N) \to H^*(M)$; applying this to $M = S^3$, we see that any orientable quotient of $M$ has the same de Rham cohomology of $M$. One simple case is $\Bbb{RP}^3$, but there are also the many lens spaces.

One particularly worthwhile example is the Poincare sphere, defined to be $SO(3)/I$, where $I$ is the isometry group of the icosahedron. This doesn't just have the same de Rham cohomology as $S^3$, it has the same singular homology - but is still not homotopy equivalent.

As mentioned in the comment above, the simplest example of two non-homotopy-equivalent manifolds with the same homology, higher homotopy groups, and fundamental group are $S^2 \times S^2$ and $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$; one can distinguish them via the ring structure on cohomology. (But the de Rham cohomology rings are isomorphic, like you want.)