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In Analysis Situs, Poincaré studies the following question: which sequences of integers $b_0,\ldots,b_n$ are the Betti numbers of an orientable compact manifold of dimension $n$?.

He knows that necessary conditions are $b_k=b_{n-k}$ and if $n=4k+2$, $b_{2k+1}$ is even. Then he computes the homology of a product of spheres, reducing the problem to finding a manifold of dimension $4k$ with $b_0=b_{2k}=b_{4k}=1$ and the other are 0. He proposes the symetric product of two spheres $S^{2k}$, missing that they are singular for $k>1$.

For $n=1$ and $2$, the projective plane and the quaternionic plane answer the question and I was very surprised to learn that the Hirzebruch signature theorem implies that a manifold of dimension 12 with $b_4=0$ has signature divisible by 62 (see http://www.maths.ed.ac.uk/~aar/papers/hirzrem.pdf p.8). Hence, the smallest odd value of $b_6$ of a 12-dimensional manifold with vanishing other Betti numbers is at least 63.

Here is the question: do we know the smallest odd value of $b_6$? More generally, do we know other obstructions for realizing an arbitrary sequence of Betti numbers?

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There are similar obstructions in dimension 20. Check my answer to this question mathoverflow.net/questions/116814/… –  Liviu Nicolaescu May 31 '13 at 20:41
    
The minimal $b_6$ you seek is $2$. The manifold $S^6\times S^6$ has Poincare polynomial $1+2t^6+t^{12}$. –  Liviu Nicolaescu May 31 '13 at 20:45
    
Regarding the question when the only non-zero Betti numbers are $0$, $2k$ and $4k$: If we want the stronger statement that $H_i(M) = \mathbb{Z}$ if $i = 0,2k,4k$ and $0$ else, then the only possibilites are indeed the dimensions of the complex, quaternionic and octonionic plane, i.e. $4k = 4,8$ or $16$. For rational coefficients there are more possibilities, e.g. $4k = 32$, but a general restriction is that $k$ has to be even (and not $24$) if $M$ is simply-connected. You might have a look at this article by Su: arxiv.org/pdf/1010.3274v1.pdf –  Lennart Meier Jun 3 '13 at 12:50

1 Answer 1

up vote 1 down vote accepted

For $M^{4k}$, the intersection form on middle-dimensional homology $H_{2k}$ is non-degenerate and symmetric, so the signature equals $b_{2k} \pmod 2$. Therefore in your case the signature theorem implies $b_6$ must be even if $b_4=0$.

See also Torsion in cohomology of smooth manifolds for some related information concerning other restrictions.

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This is a nice observation, thanks! Now I have to think about the realization problem in dimension 12 with $b_4>0$. Is there a 12-manifold with $b_4=b_6=b_8=1$... –  Julien Marché May 31 '13 at 21:05
    
this is realized by CP^6, so you might want to ask more precisely whether there is a 12-manifold with b_0=b_4=b_6=b_8=b_12=1 and all other Betti numbers zero. –  nsrt May 31 '13 at 22:24
    
I think I'm missing something here. Are we still talking about 12-manifolds? If so, what's the relation between $b_{6}$ and $b_4$? –  Greg Friedman Jun 8 '13 at 2:23

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