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 symmetric 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 page 8 of this paper). 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?

  • $\begingroup$ There are similar obstructions in dimension 20. Check my answer to this question mathoverflow.net/questions/116814/… $\endgroup$ – Liviu Nicolaescu May 31 '13 at 20:41
  • $\begingroup$ The minimal $b_6$ you seek is $2$. The manifold $S^6\times S^6$ has Poincare polynomial $1+2t^6+t^{12}$. $\endgroup$ – Liviu Nicolaescu May 31 '13 at 20:45
  • $\begingroup$ 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 $\endgroup$ – Lennart Meier Jun 3 '13 at 12:50

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.

| cite | improve this answer | |
  • $\begingroup$ 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 $b_4=b_6=b_8=1$... $\endgroup$ – Julien Marché May 31 '13 at 21:05
  • $\begingroup$ 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. $\endgroup$ – nsrt May 31 '13 at 22:24
  • 1
    $\begingroup$ I think I'm missing something here. Are we still talking about 12-manifolds? If so, what's the relation between $b_{6}$ and $b_4$? $\endgroup$ – Greg Friedman Jun 8 '13 at 2:23
  • $\begingroup$ @GregFriedman: $b_4 = 0$ is needed to show that the signature is a multiple of $62$ (in particular, even). $\endgroup$ – Michael Albanese Jul 5 '18 at 15:43

Suppose we are given non-negative integers $b_0, b_1, \dots, b_n$ with $b_k = b_{n-k}$. Is there a closed orientable manifold $M$ with $b_i(M) = b_i$? First we need $b_0 = b_n \geq 1$. It is enough to answer the question with $b_0 = b_n = 1$ as we can then just take the disjoint union with the appropriate number of spheres. So from now on, $b_0 = b_n = 1$.

If $n = 2m + 1$, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely

$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\ b_m(S^m\times S^{m+1}).$$

If $n = 2m$ and $b_m$ is even, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely

$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\ b_{m-1}(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m}{2}(S^m\times S^m).$$

Suppose now that $n = 2m$ and $b_m$ is odd. If $m$ is odd, then there is no manifold realising these Betti numbers: the intersection form on the middle dimension is a non-degenerate skew-symmetric form, so the middle Betti number is necessarily even. If $m$ is even, things are more complicated. First of all, such examples can exist, e.g. $\mathbb{CP}^2$. However, the above approach of taking connected sums of products of spheres can't work because the resulting manifolds are all nullcobordant, but a manifold with such Betti numbers is not: $b_m(M)$ and $\sigma(M)$ have the same parity and $\sigma(M)$ is a cobordism invariant.

Note that a closed smooth orientable $2m$-dimensional manifold with Betti numbers $b_0 = b_m = b_n = 1$ and all others zero has rational cohomology ring $\mathbb{Q}[\alpha]/(\alpha^3)$ where $\deg\alpha = m$. For this reason, call such a manifold a rational projective plane and denote it by $\mathbb{QP}^2$. If such a manifold exists in dimension $n$, then if $b_m$ is odd, there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely

$$M = b_1(S^1\times S^{n-1})\ \#\ \dots\ \#\ b_{m-1}(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m-1}{2}(S^m\times S^m)\ \# \mathbb{QP}^2.$$

So the natural question to ask is:

For which $n$ does there exist a rational projective plane?

Well, we have $\mathbb{CP}^2$, $\mathbb{HP}^2$ and $\mathbb{OP}^2$ which are actually simply connected and have integral cohomology ring $\mathbb{Z}[\alpha]/(\alpha^3)$; by the solution of the Hopf invariant one problem, these are the only such manifolds up to homotopy equivalence.

In Smooth manifolds with prescribed rational cohomology ring, Fowler and Su show (Theorem A) that for $n \geq 8$, a rational projective plane can only exist in dimensions of the form $n = 8(2^a + 2^b)$ for some non-negative integers $a$ and $b$.1 In Rational analogs of projective planes, Su shows (Theorem 1.1) the existence of a (simply-connected) rational projective plane in dimension $32$. Later, Kennard and Su also proved the existence of (simply-connected) rational projective planes in dimensions $128$ and $256$, see On dimensions supporting a rational projective plane, Theorem 1.1. The paper also contains non-existence results for simply-connected rational projective planes in certain dimensions.2

It may seem that the lack of a projective plane in a given dimension prevents the realisability of any collection of Betti numbers with middle Betti number odd, but that is not true. For example, when $m$ is even, $\mathbb{CP}^m$ has $b_m = 1$ regardless of whether or not a rational projective plane exists (thanks to Will Sawin for pointing this out). That is, the question of realisability of Betti numbers does not reduce to the existence of rational projective planes, it is much more complicated.

Finally, it is worth noting that if $n = 2m = 4l$ and $b_{4j} \geq 1$ for $j = 1, \dots, l - 1$ and $b_m$ is odd, then there is a closed smooth orientable manifold $M$ with $b_i(M) = b_i$, namely

$$M = b_1'(S^1\times S^{n-1})\ \#\ \dots \# \ b_{m-1}'(S^{m-1}\times S^{m+1})\ \#\ \tfrac{b_m'}{2}(S^m\times S^m)\ \#\ \mathbb{HP}^l$$

where $b_{4j}' = b_{4j} - 1$ for $j = 1, \dots, l - 1$ and $b_i' = b_i$ otherwise.

1 The definition of rational projective plane in that paper includes the hypothesis of being simply connected. However, the proof of Theorem A does not use this property, so the result holds for the definition used above.

2 I don't know the proof of these results, so I'm not sure if the simply connected hypothesis can be removed.

| cite | improve this answer | |
  • $\begingroup$ Isn't your question at the end answered in the negative by the existence of complex projective spaces, but not rational projective planes in all even dimensions? $\endgroup$ – Will Sawin Jul 6 '18 at 10:06
  • $\begingroup$ You are absolutely right. I will edit accordingly. $\endgroup$ – Michael Albanese Jul 6 '18 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.