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If you replace spheres in the Poincaré conjecture with objects-with-some-kinds-of-holes can you say that every manifold with the same number and type of holes is homeomorphic to every other such manifold ?

Motivation: mainly interested in shapes that answer the question: Topological shape of sphere-like equations with complex radius to see if that would lead to a Holomorphic Poincaré conjecture for such shape.

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That depends on what you mean by "some kinds of holes." – Qiaochu Yuan May 30 '12 at 21:59
Whatever kinds of holes that would make the conjectures true. – vtt May 30 '12 at 22:15
Then yes, by definition! – Qiaochu Yuan May 30 '12 at 22:16
Perhaps this question can be phrased as "What closed manifolds are characterized by their homotopy type?" – MartianInvader May 30 '12 at 22:53
up vote 5 down vote accepted

Here is my interpretation of your question:

Given two homotopy equivalent (perhaps smooth) $n$-manifolds $M$ and $N$, are $M$ and $N$ homeomorphic (or even diffeomorphic)?

For $n = 1$ the answer is of course always yes, and the classification of surfaces shows that the answer is also always yes for $n = 2$.

In higher dimensions, however, the answer to the question is in general no.

Already in dimension $3$, we can find counterexamples. For example, one can use Reidemeister torsion to find homotopy equivalent lens spaces that are not homeomorphic.

In dimension $4$, the answer is again no in general; for example by Freedman's classification theorem, there is a $4$-manifold $\ast \mathbb{C}P^2$ homotopy equivalent to $\mathbb{C}P^2$ but not homeomorphic to it (although $\ast\mathbb{C}P^2$ doesn't admit a smooth structure). In fact, dimension $4$ is where you first see a more striking phenomenon: many closed smooth $4$-manifolds $M$ admit infinitely many exotic smooth structures, i.e. there are infinitely many smooth $4$-manifolds homeomorphic to $M$ but not diffeomorphic to $M$. I believe the current conjecture is that all closed smooth $4$-manifolds admit infinitely many exotic smooth structures. The existence or non-existence of exotic smooth structures on $S^4$ is still open and is known as the smooth Poincaré conjecture.

Higher dimensional examples can be exhibited as well, for example see Novikov's paper Rational Pontrjagin Classes, Homeomorphism and Homotopy Type of Closed Manifolds, I.

Note that my interpretation of your question does not admit a complete answer for $n \geq 4$; for such $n$, any finitely presented group is isomorphic to the fundamental group of some $n$-manifold. Since it is impossible to construct an algorithm to tell whether two finitely presented groups are isomorphic (look up the "word problem for groups"), we see that we cannot even classify $n$-manifolds up to homotopy type for $n \ge 4$.

Because of this, topologists ask the following easier question:

Let $f: M \longrightarrow N$ be a homotopy equivalence between smooth $n$-manifolds. Is $f$ homotopic to a diffeomorphism?

The field of geometric topology known as surgery theory provides tools with which to approach this reformulated question. This is a large field that I don't think I should go into much detail about here. See Chapter 1 of Ranicki's Algebraic and Geometric Surgery (available here) for some basic background on surgery and what it can do.

You may also be interested in reading about the Borel conjecture, which claims that any homotopy equivalence between aspherical manifolds is homotopic to a homeomorphism; this is an important open problem in topology.

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My own interpretation of the question is "In sufficiently high dimensions, is the sphere the only manifold $M$ with the property that if $M$ and $N$ are homotopy equivalent, then they are homeomorphic." Would you happen to know the answer to this question? +1 either way! – Jason DeVito May 30 '12 at 23:14
@JasonDeVito: The answer is no. For example, any homotopy equivalence of closed hyperbolic $3$-manifolds is homotopic to a homeomorphism (in fact, an isometry) by Mostow rigidity. – Henry T. Horton May 30 '12 at 23:17
I'd forgotten about Mostow rigidity. Doesn't Mostow also work in higher dimensions? – Jason DeVito May 30 '12 at 23:18
Yes, Mostow rigidity holds for any complete, finite-volume hyperbolic $n$-manifolds with $n \ge 3$, so that does in fact give us a negative answer in higher dimensions too. – Henry T. Horton May 30 '12 at 23:19
@JasonDeVito: Good point... Perhaps we should stick to closed manifolds then. A closed $n$-manifold $M$ with $n \ge 5$ admits a hyperbolic metric if and only if it is aspherical and $\pi_1(M)$ is isomorphic to a discrete cocompact subgroup of $\mathrm{O}(1,n)$. So in the closed case, if $M^n$ ($n \ge 5$) admits a hyperbolic metric and is homotopy equivalent to $N$, then $N$ admits a hyperbolic structure too. See this paper: – Henry T. Horton May 31 '12 at 1:14

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