manifold with given rational homology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:03:34Z http://mathoverflow.net/feeds/question/71439 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71439/manifold-with-given-rational-homology manifold with given rational homology Markus Ulke 2011-07-27T21:03:08Z 2011-07-28T21:21:55Z <p>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?</p> http://mathoverflow.net/questions/71439/manifold-with-given-rational-homology/71442#71442 Answer by Neil Strickland for manifold with given rational homology Neil Strickland 2011-07-27T21:46:45Z 2011-07-27T21:46:45Z <p>The Cayley plane $\mathbb{O}P^2$ is an example. See this question: </p> <p><a href="http://mathoverflow.net/questions/1922/what-is-the-cayley-projective-plane" rel="nofollow">http://mathoverflow.net/questions/1922/what-is-the-cayley-projective-plane</a></p> http://mathoverflow.net/questions/71439/manifold-with-given-rational-homology/71450#71450 Answer by Igor Rivin for manifold with given rational homology Igor Rivin 2011-07-28T00:06:16Z 2011-07-28T00:06:16Z <p>A general (surgery based, surprise...) construction is give by Kalinin in:</p> <p>REALIZATION OF QUADRATIC FORMS BY SMOOTH MANIFOLDS Math USSR Sbornik 62 (1989), n. 1, p 177, theorem 2.1.1</p> http://mathoverflow.net/questions/71439/manifold-with-given-rational-homology/71503#71503 Answer by Mark Grant for manifold with given rational homology Mark Grant 2011-07-28T15:48:15Z 2011-07-28T15:48:15Z <p>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. </p> <p>You'll find a clean statement in Chapter 3 of this <a href="http://www.amazon.com/Algebraic-Models-Geometry-Graduate-Mathematics/dp/019920652X" rel="nofollow">book</a>, 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:</p> <ol> <li>$n$ is not of the form $4k$;</li> <li>$n$ 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 $\sum \pm x_i^2$;</li> <li>$n$ 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.</li> </ol> <p>This is of course proved by surgery theory (I'm not sure how it relates to the reference given in Igor's answer). </p> http://mathoverflow.net/questions/71439/manifold-with-given-rational-homology/71519#71519 Answer by Lennart Meier for manifold with given rational homology Lennart Meier 2011-07-28T21:21:55Z 2011-07-28T21:21:55Z <p>The following paper might be of interest: <a href="http://arxiv.org/PS_cache/arxiv/pdf/1010/1010.3274v1.pdf" rel="nofollow">Rational Analogs of Projective Planes</a> by Zhixu Su. </p> <p>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. </p> <p>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. </p> <p>The methods are those sketched in Mark Grant's answer. </p>