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I am not quite sure about the terminology, but let's call a $n$-dimensional manifold of finite type if it has a finite open cover $U_1,\ldots,U_k$ such that all intersections are either empty or diffeomorphic to ${\mathbb R}^n$. Every compact manifold is of finite type (endow it with a Riemannian metric and use geodesically convex neighbourhoods), while $\mathbb{R}^2\setminus \mathbb Z$ isn't.

A straightforward argument using the Mayer-Vietoris sequence shows that the cohomology of a manifold of finite type is finite dimensional. My question is about the converse statement:

If the cohomology algebra of a smooth manifold $M$ is finite dimensional, does this imply that $M$ is of finite type?

Edit: From Tom and Charles' examples, it is clear that the answer is "no" if one uses de Rham or even singular cohomology. As Tom suggests, there is some hope for a positive answer under an appropriate assumption on cohomology with twisted coefficients, but not being an expert on these matters I will not take the risk of further conjectures.

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What quotient is $\mathbb{R}^2/\mathbb{Z}$? Is it a cylinder? If so it is of 'finite type', so I'm guessing it's not. I suggest comparing deRham to Cech cohomology, which is natural given you are dealing with open covers. Also, what you call 'finite type' is called having a finite good (open) cover. Usually finite type means having finitely-generated cohomology in each dimension, or something similar. And finally, when you say cohomology algebra, do you mean with real coefficients? (It may not matter for manifolds, but it would for spaces, which is a natural extension of your question) –  David Roberts Dec 6 '10 at 22:55
@ David: I thought Andrei meant $\mathbb{R}^2-\mathbb{Z}$, i.e., the complement of a discrete subset. –  Charles Rezk Dec 6 '10 at 22:58
I think the blackslash means complement, not quotient. It has infinite first Betti number. Anyway, I suspect there is a counterexample to the question, but I can't think of one. –  Donu Arapura Dec 6 '10 at 23:00
My goodness, that line does lean to the left. In that case I hang my head in shame. :) –  David Roberts Dec 7 '10 at 0:50
What does it mean to say that "the cohomology with any twisted coefficients is of finite type"? The category of coefficient systems is closed under arbitrary direct sums. –  Charles Rezk Dec 7 '10 at 15:17
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2 Answers

up vote 11 down vote accepted

It does matter what kind of cohomology you use:

First think of the connected sum of an infinite number of real projective spaces end to end. De Rham cohomology cannot tell that these were not spheres, but mod 2 cohomology shows that the thing is not of finite type.

Now take it further. Take a (necessarily non simply connected) closed manifold whose integral homology is the same as that of a sphere (for example Poincare's 3-dimensional example whose fundamental group is the binary icosahedral group). Just as this looks like sphere to integral cohomology, the connected sum of infinitely many of these appears is indistinguishable from the connected sum of infinitely many spheres.

If you allow cohomology with twisted coefficients, then things get more hopeful. I would be inclined to separate the problem into two parts: (1) if a smooth manifold has the homotopy type of a finite CW complex, does this imply that it is of finite type in your sense? (2) What finiteness conditions on cohomology suffice to guarantee that a CW complex (space) has the (weak) homotopy type of a finite CW complex? (2) has the expected affirmative answer if you confine yourself to simply connected spaces, but the general case is more subtle.

EDIT What I really mean is: First assume that there are finitely many components, and that each of these has a finite presented fundamental group. Then use Wall's finiteness obstruction (which has to do with modules over the group ring of the fund gp -- it's a long story).

RE-EDIT Some more details: There are finite groups $G$ for which the ring $\mathbb ZG$ has some projective modules that are not stably free. For any such $G$ there are examples of finite CW complexes $K$ and $L$ with fundamental group $G$ such that $L$ is finite and $K$ is a retract of $L$ up to homotopy (i.e. there are maps $K\to L\to K$ whose composition is homotopic to the identity) but $K$ is not homotopy equivalent to any finite complex. The other composition $p:L\to L$ is then such that $p\circ p$ is homotopic to $p$, and $K$ is equivalent to the mapping telescope or homotopy colimit of $L\to L\to L\to\dots$. Take any such example and realize it by manifolds: Choose $L$ to be a smooth compact parallelizable manifold with boundary and (by taking the dimension big enough) take $p$ to be a smooth embedding of $L$ in the interior of $L$, and take $K$ to be the increasing union of infinitely many nested copies of $L$. So $K$ is a noncompact manifold without boundary whose (co)homology is in any sense no bigger than that of $L$, but which is not homotopy equivalent to any finite complex. In particular it is not the interior of any compact manifold with boundary. I can't see why it cannot be of finite type in the sense of the problem, but it looks unlikely.

It just struck me that this example can be improved: By doing surgery on a finite number of embedded $1$-spheres, one can arrange for the noncompact manifold $K$ to be simply connected. In this case it is homotopy equivalent to a finite complex (basically because $\mathbb Z$-modules are (stably) free), but it still is not the interior of any smooth compact manifold (because outside a compact set it's still the same example).

Background to this (but I am not using the full strength of it) is Wall's finiteness obstruction for CW complexes and Siebenmann's thesis in which Wall's obstruction is adapted to give an obstruction to putting a boundary on an isolated end of a manifold.

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Great answer, Tom! I now realize that it does matter which cohomology one uses and edited my question correspondingly. –  Andrei Moroianu Dec 7 '10 at 9:44
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How about the unbounded complement $X$ of the Alexander horned sphere? Its cohomology is certainly finite dimensional (by the usual generalization of the Jordan curve theorem). The fundamental group is a direct limit $G=\mathrm{colim} F_{2^n}$ of of an increasing sequence free groups, where a generator in $F_{2^n}$ is sent to the commutator of a pair of generators of $F_{2^{n+1}}$; in particular, the group $G$ is not finitely generated. This means that $X$ cannot have a nice finite cover of the type you describe (by application of Van Kampen's theorem, a space with such a finite cover must have finitely generated fundamental group in each path component.)

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