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Is there a smooth compact manifold with rational homology vanishing except in dimensions $0, 8, 16$ where it is $Q$? What would be a good strategy to find such a manifold?

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$S^8 \times S^8$ works. –  Ryan Budney Jul 27 '11 at 21:15
There are simpler examples if you allow boundary. IMO a question like this is more appropriate for math.stackexchange.com. –  Ryan Budney Jul 27 '11 at 21:17
no the product above has homology $Q^2$ in dimension 8 hasnt it. Homology $Q^1$ is requested. –  Markus Ulke Jul 27 '11 at 21:39
the intend is to find something similar like cayleys projective plane in dimension 16 (at least rational). Up to now nobody i knew did find such a thing, the general opinion is that it does probably exist. (With Z replaced by Q it cant exist, but allowing torsion might render it possible) –  Markus Ulke Jul 27 '11 at 21:43
I'd say start with CP^8 and do surgery. Unfortunately, that's not very explicit, but I think you can argue this way that it exists. Alternatively, you might be able to do it with a plumbing construction. I'd recommend starting with Browder's book on surgery theory. –  Greg Friedman Jul 27 '11 at 21:44

4 Answers 4

up vote 4 down vote accepted

The following paper might be of interest: Rational Analogs of Projective Planes by Zhixu Su.

It discusses the following question: For which $n$ is there a closed $2n$-manifold $M$ with rational cohomology $H^*(M; \mathbb{Q}) \cong \mathbb{Q}[x]/x^3$ with $x$ of degree $n$? Observe that the multiplicative structure is essentially implied by the additive one by Poincare duality.

There are, of course, the classical examples $\mathbb{CP}^2$, $\mathbb{HP}^2$ and $\mathbb{OP}^2$, but are there more $n$ possible? It is clear that $n$ has to be even for this, but actually $n$ has to be divisible by $4$. Beyond the classical examples, the next example occurs in dimension $32$, where there are already infinitely many (up to homeomorphism). This is, of course, a purely rational result - by the Hopf invariant 1 problem the examples above are the only examples for the integral analogue.

The methods are those sketched in Mark Grant's answer.

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The Cayley plane $\mathbb{O}P^2$ is an example. See this question:

What is the Cayley projective plane?

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The smooth structure is discussed by John Baez here: math.ucr.edu/home/baez/octonions/node12.html –  Mark Grant Jul 27 '11 at 22:00

Here is a strategy in answer to your second question. Suppose you are given a graded vector space $H_\ast$, and you wish to realise this as $H_\ast(X;\mathbb{Q})$ for some smooth closed manifold $X$. First, you must check that the linear dual $H^\ast$ carries the structure of a finite dimensional Poincaré duality algebra. Then, assuming that $H^1=0$ (equivalently $H_1=0$) you can apply the Sullivan-Barge theorem, which essentially says that the necessary conditions that $H^\ast$ be the cohomology algebra of a smooth simply-connected closed manifold, are also sufficient.

You'll find a clean statement in Chapter 3 of this book, but let me try anyway. Suppose your Poincaré duality algebra $H^\ast$ has formal dimension $n$. Then it can be realised by a closed simply-connected manifold if, and only if, one of the following holds:

  1. $n$ is not of the form $4k$;
  2. $n$ is of the form $4k$, the signature is zero and the quadratic form on $H^{2k}$ is equivalent over $\mathbb{Q}$ to one of the form $\sum \pm x_i^2$;
  3. $n$ is of the form $4k$, the signature is nonzero, the quadratic form on $H^{2k}$ is equivalent over $\mathbb{Q}$ to one of the form $\sum \pm x_i^2$, and one can find a sequence of classes $p_i\in H^{4i}$ (the Pontrjagin classes) such that the corresponding Pontrjagin numbers satisfy certain necessary congruences.

This is of course proved by surgery theory (I'm not sure how it relates to the reference given in Igor's answer).

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A general (surgery based, surprise...) construction is give by Kalinin in:

REALIZATION OF QUADRATIC FORMS BY SMOOTH MANIFOLDS Math USSR Sbornik 62 (1989), n. 1, p 177, theorem 2.1.1

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