I've been interested in the possible (singular) cohomology groups of complex projective algebraic varieties, and there are lots of theorems that give various restrictions on these (Hodge decomposition, Lefshetz theorem, ... ). I realized that I would like to know more about the what is true for smooth manifolds hence my questions:

1.) One can construct CW complexes that have prescribed (reduced) homology groups (coeffs in $\mathbb{Z}$), these are the Moore spaces. However, they aren't even topological manifolds in general. Can one construct compact oriented smooth manifolds that have prescribed singular cohomology groups $H^i(X, \mathbb{Z})$, provided that after we remove torsion our sequence of groups satisfy Poincare duality? Should one expect that this is "generally possible" but it may be hard to actually construct examples?

2.) If $X$ is a compact oriented smooth manifold, is there any regularity in the torsion subgroups of it's cohomology: $H^i_{sing} (X, \mathbb{Z})$? (Eg, poincare duality gives regularity between the various torsion free parts.) How about if $X$ is a nonsingular complex projective variety?

3.) For $X$ a smooth oriented manifold, it seems like compactly supported cohomology contains more information than ordinary cohomology. Can one recover ordinary cohomology $H^i_{sing}(X, \mathbb{Z})$ from compactly supported cohomology $H^i_c(X, \mathbb{Z})$? How about if we take coefficents in $\mathbb{Q}$?

I'd love to see "typical", or common examples where various phenomena appears.

  • $\begingroup$ For part 3, over $\mathbb Q$, aren't they dual? Similarly, regular integral cohomology is Pontryagin dual to compact $\mathbb R/\mathbb Z$ cohomology. $\endgroup$
    – Will Sawin
    Commented Dec 19, 2012 at 20:11
  • $\begingroup$ That's nice. I'd like to see a reference. $\endgroup$
    – LMN
    Commented Dec 19, 2012 at 20:21
  • $\begingroup$ The statement about Pontryagin dual is part of the general Poincare duality with coefficients $bR/\bZ$ coupled with the universal coefficients formula. $\endgroup$ Commented Dec 19, 2012 at 21:00

2 Answers 2


Here are some facts.


$$ P(t) = a_0+a_1t+\cdots + a_{2k} t^{2k} $$

is a polynomial with nonnegative integral coefficients such that

$$ a_0=a_{2k}=1,\;\;a_j=a_{2k-j},\;\;\forall j $$

and $\newcommand{\bZ}{\mathbb{Z}}$

$$ a_k\in 2\bZ, $$

then there exists a smooth, compact, connected, oriented manifold $M$ of dimension $2k$ whose Poincare polynomial $P_M$ is the above polynomial $P$, i.e.,

$$b_j(M)=a_j,\;\;\forall j. $$

The manifold $M$ can be found by taking connected sums of products of spheres $S^{k_1}\times \cdots \times S^{k_m}$. The result is sharp in the following sense. There do not exist oriented smooth manifolds whose Poincare polynomials are

$$1 +t^6 +t^{12},\;\; 1+ t^{10}+ t^{20}. $$

This last fact was observed by Serre and follows from Hirzebruch's signature theorem.

  • $\begingroup$ More trivially, if the dimension is $2$ mod $4$, then the middle cohomology of a manifold is even-dimensional, because Poincare duality provides a nondegenerate symplectic bilinear form. $\endgroup$
    – Will Sawin
    Commented Dec 19, 2012 at 23:01
  • $\begingroup$ Yes, that is true. But the middle Betti number in dimensions divisible by $4$ can be odd. Take for example the cade of complex projective spaces of even complex dimensions. That is why Serre's observation is so strange. He has other examples, and all have to do with the numerators of Bernoulli numbers. $\endgroup$ Commented Dec 19, 2012 at 23:34

1) You can pick the homology below the middle dimension quite arbitrarily. More precisely, given a finite complex $K$ and a number $n$, there exists a closed, parallelizable $2n$-dimensional manifold $M$ and an $n$-connected map $f:M \to K$. You begin with a constant map $S^{2n} \to K$ and make it more and more connected by surgeries.

2) as you said, a necessary condition for the homology of a manifold is Poincare duality. If you have a finite complex $X$ that satisfies Poincare duality, the question of whether there is a smooth manifold homotopy equivalent to $X$ is a basic problem in surgery theory. If $X$ is simply connected, this has largely been solved by Browder. The answer is that if $X$ is odd-dimensional, there is such a manifold; and if the dimension is divisible by $4$, there is a manifold precisely if there is a stable vector bundle on $X$ such that the Hirzebruch signature formula holds with this bundle. In dimensions $2,6,10,\ldots$, there is a subtle problem with the "Kervaire invariant". And: I forgot to say that the dimension has to be at least $5$. For nonsimplyconnected complexes, Wall gave at least a theoretical answer.

3) Poincare duality for integral coefficients (and closed oriented $M$) says that $H_i (M) \cong H^{n-i}(M)$. The universal coefficient theorem implies that the torsion subgroups (for each space with finitely generated homology) are $T H^{i+1} = T H_i$ (abstract isomorphism). Combined, these two results tie the torsion subgroups of cohomology together.

4) I would not say that compactly supported cohomology contains more information than ordinary cohomology - they contain different information. With rational coefficients, you have an isomorphism $H^i(M) \cong (H^{n-i}\_{c}(M))^{\ast}$; the other isomorphism $H^{i}_{c}(M) \cong (H^{n-i}(M))^{\ast}$ holds iff the cohomology vector space are finitely generated.

  • $\begingroup$ Johannes, I'm having a little trouble translating the language into things I am familiar with. Are you saying that the answer to my question #1 is yes (but if the dimension is odd we might not be able to control the middle cohomology group). $\endgroup$
    – LMN
    Commented Dec 19, 2012 at 21:19
  • $\begingroup$ @LMN: Even-dimensional manifolds are the ones with "middle cohomology groups." $\endgroup$ Commented Dec 19, 2012 at 21:31
  • $\begingroup$ Daniel, right - "odd" should be "even". $\endgroup$
    – LMN
    Commented Dec 19, 2012 at 21:35
  • $\begingroup$ Odd-dimensional manifolds have two middle dimensions to control, so in that case, the situation is more complicated. $\endgroup$ Commented Dec 19, 2012 at 23:26
  • $\begingroup$ @Johannes, in regards to your answer to #1, you get that the map $f$ induces an isomorphism on the first $n-1$ homotopy groups. I don't see how this allows you to deduce that the cohomology groups of $M$ are the same as those of $K$. Hurewicz's theorem doesn't seem to help. $\endgroup$
    – LMN
    Commented Dec 20, 2012 at 16:51

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