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

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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?

It is of course irrelevant which cohomology theory one uses

Edit: From Tom's examples, but to fix it is clear that the ideas answer is "no" if one can consider uses de Rham or even singular cohomology. As he suggests, there is some hope for a positive answer under the assumption that cohomology with any twisted coefficients is of finite type.

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# Finite type vs. finite dimensional cohomology?

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?

It is of course irrelevant which cohomology theory one uses, but to fix the ideas one can consider de Rham cohomology.