Torsion in cohomology of smooth manifolds

Question

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.

Answer 1

Here are some facts.

If

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

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.

Answer 2

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.