manifold with given rational homology - MathOverflow

manifold with given rational homology

2011-07-27T21:03:08Z

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?

Answer by Neil Strickland for manifold with given rational homology

2011-07-27T21:46:45Z

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

http://mathoverflow.net/questions/1922/what-is-the-cayley-projective-plane

Answer by Igor Rivin for manifold with given rational homology

2011-07-28T00:06:16Z

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

Answer by Mark Grant for manifold with given rational homology

2011-07-28T15:48:15Z

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:

$n$ is not of the form $4k$;
$n$ 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 $\sum \pm x_i^2$;
$n$ 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).

Answer by Lennart Meier for manifold with given rational homology

2011-07-28T21:21:55Z

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.