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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 $\pi_1$.

Francisco
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  • The same as what? Isomorphic as vector spaces? As rings? – levap Jun 08 '16 at 15:39
  • You might want to see the same question at mathoverflow: http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent – Henry Jun 08 '16 at 15:46
  • @levap As rings, thank you. I actually can't think right now of an example of manifolds with isomorphic de Rham cohomology as vector spaces but different ring structure, so that also would be appreciated. – Francisco Jun 08 '16 at 15:50

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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.)

  • You can do it in dim 2: $\mathbb{R}P^2$ and $\mathbb{R}^2$. Also, I think your second sentence needs some kind of orientability hypothesis (which you mentioned later on). – Jason DeVito - on hiatus Jun 08 '16 at 15:58
  • @JasonDeVito Ah, I was thinking of everything being closed. I'll add that in. I don't believe that result requires any sort of orientability. –  Jun 08 '16 at 16:00
  • But the covering $S^2\rightarrow \mathbb{R}P^2$ does not induce a bijection $H^\ast(N)\rightarrow H^\ast(M)$. Also, even for orientable things, you need a finite sheeted covering (which, of course, is automatic for closed manifolds). Or perhaps I'm misinterpreting what you wrote? – Jason DeVito - on hiatus Jun 08 '16 at 17:43
  • Also, I believe $S^2\times S^2$ and $\mathbb{C}P^2\sharp\overline{\mathbb{C}P^2}$ have the same rational homotoyp type. If so, their de Rham cohomology rings agree. On the other hand, I think $S^2\times S^2$ and $\mathbb{C}P^2\sharp \mathbb{C}P^2$ have different de Rham cohomology rings. – Jason DeVito - on hiatus Jun 08 '16 at 17:47
  • @JasonDeVito: Oops! I thought one thing and wrote another. I meant to say injection and finite cover... really nasty mistake, thanks for calling me out. I believe bijection is still false in the oriented case but I don't have an example on a minute's thought. –  Jun 08 '16 at 17:48
  • I agree that bijection is probably false even when oriented and finite sheeted, but I also don't know of an example. (But I do agree about the injection, and that it doesn't need orientability) – Jason DeVito - on hiatus Jun 08 '16 at 17:49
  • @Jason: They do have the same de Dham cohomology rings, but different integral cohomology rings. –  Jun 08 '16 at 17:50
  • Then I guess I misinterpreted what the OP was asking - I agree that integrally the cohomology rings differ. Sorry to be a bother. – Jason DeVito - on hiatus Jun 08 '16 at 17:51
  • @JasonDeVito No worries at all... I don't actually know how to prove that $2\Bbb{CP}^2$ have the same homotopy groups as $S^2 \times S^2$. Is that true? –  Jun 08 '16 at 17:52
  • Yes, that is true: $S^3\times S^3$ is a principal $T^2$ bundle over both of them, so one can use the LES in homotopy groups. This is clear for $S^2\times S^2$ using the Hopf bundle. For $\mathbb{C}P^2\times \mathbb{C}P^2$, the result, as far as I know, goes back to Totaro (Cheeger manifolds and the classification of biquotients). – Jason DeVito - on hiatus Jun 08 '16 at 17:54
  • Huh, that's wild. Somehow homotopy groups don't do half the job the cohomology ring does for s.c. 4-manifolds. –  Jun 08 '16 at 17:55
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    Incidentally, according to https://open.library.ubc.ca/cIRcle/collections/ubctheses/831/items/1.0079494, the covering $S^2\rightarrow S^2\rightarrow S^2\times S^2/ ( (x,y) \sim -(x,y)$ is a $2$-sheeted covering of orientable manifolds, but the second Betti number of the quotient is $0$ - so there's our counterexample to "bijection" – Jason DeVito - on hiatus Jun 08 '16 at 18:02
  • This is great, Mike and Jason. Thank you both. – Francisco Jun 11 '16 at 21:28
  • @Jason We're very silly. There are finite covers of closed oriented surfaces that give counterexamples :) –  Jun 11 '16 at 22:10
  • @Mike: Wow. Just wow. (I guess my counterexample has the added bonus of using a universal cover. Yeah, that's why I searched long and hard for it ;-) ) – Jason DeVito - on hiatus Jun 12 '16 at 02:40