Distinct manifolds with the same configuration spaces?

For a space $X$, let $C_k X$ denote the space of configurations of $k$ distinct unordered points in $X$.

What is an example of a pair of smooth manifolds $M$ and $N$ that are not homeomorphic but such that for all $k \in \mathbb{N}$, the spaces $C_k M$ and $C_k N$ are homotopy equivalent?

(This question is prompted by David Ayala's work on automorphisms of Ran spaces, and more generally the question of the strength of topological field theory invariants.)

Update: Qiaochu and Ricardo answered the question using examples of manifolds with boundary and non-compact manifolds, respectively, so let's now restrict attention to the (no doubt much harder) issue that is most directly relevant for TFT invariants:

Is there an example where the manifolds are closed?

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$k > 1$ ;-) Another nice question is obtained by modifying the hypothesis and asking that the configuration spaces be homotopic for all sufficienty large $k$. –  alvarezpaiva Nov 22 '13 at 19:21
@alvarez: asking that $M$ and $N$ themselves be homotopy equivalent doesn't trivialize the problem. –  Qiaochu Yuan Nov 22 '13 at 20:47
@cdouglas: Out of curiosity, what is David Ayala's work on automorphisms of Ran spaces? –  Ricardo Andrade Nov 22 '13 at 20:58
@QiaochuYuan: you're right; it was a slip on my part. –  alvarezpaiva Nov 23 '13 at 7:14


Proposition: Assume $n \geq 3$. Let $W$ be a contractible topological $n$-manifold without boundary, and $O \subset W$ an open subspace homeomorphic to $\RR^n$. Then the inclusion of $O$ into $W$ induces homotopy equivalences on all ordered and unordered configuration spaces.

A few consequences of this are:

• An embedding $V \to W$ between contractible $n$-manifolds without boundary induces a homotopy equivalence on all ordered and unordered configuration spaces.

• If $V$ and $W$ are contractible $n$-manifolds without boundary, then there are homotopy equivalences $\Conf(V,k) \simeq \Conf(W,k)$ between their ordered configuration spaces, and $C_k V \simeq C_k W$ between their unordered configuration spaces.

• In particular, the configuration spaces, both ordered and unordered, of the Whitehead manifold are homotopy equivalent to the corresponding configuration spaces of $\RR^3$.

The remainder of this post discusses the proof of the proposition above. Here is a lemma which will be used.

Lemma: Let $S$ be any finite subset of $O$. Then the inclusion $O\setminus S \to W\setminus S$ is a weak equivalence.

Proof: Apply the van Kampen theorem to the open cover of $W$ by $O$ and $W\setminus S$. Since $O\setminus S$ is simply connected (since it is equivalent to a wedge sum of $(n-1)$-dimensional spheres), it follows that $\Pi_1(W\setminus S) \simeq \Pi_1(W)$ is trivial.

Now apply the Mayer–Vietoris sequence for homology to the same open cover of $W$ by $O$ and $W\setminus S$. The contractibility of $W$ implies that the inclusion $O\setminus S \to W\setminus S$ is a homology equivalence. Since both spaces are simply connected, this inclusion is also a weak equivalence. ■

Now we prove that the map on ordered configuration spaces $\Conf(O,k) \to \Conf(W,k)$ is a weak equivalence for any $k>0$. Note that we have a map of (horizontal) fibration sequences $$\begin{matrix} O\setminus S & \To & \Conf(O,k) & \To & \Conf(O,k-1) \\ \big\downarrow & & \big\downarrow & & \big\downarrow \\ W\setminus S & \To & \Conf(W,k) & \To & \Conf(W,k-1) \end{matrix}$$ where $S$ is any subset of $O$ of size $k-1$. The previous lemma states that the map on the fibres is a weak equivalence. Thus, by induction on $k$, we conclude that the map $\Conf(O,k) \to \Conf(W,k)$ is a weak equivalence.

Now the result for unordered configuration spaces follows since we have a commutative diagram of fibration sequences $$\begin{matrix} \Sigma_k & \To & \Conf(O,k) & \To & C_k O \\ \big\downarrow & & \big\downarrow & & \big\downarrow \\ \Sigma_k & \To & \Conf(W,k) & \To & C_k W \end{matrix}$$ which is an equivalence on the fibres and total spaces. We conclude that the map on the base spaces is also a weak equivalence.

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Thanks, this example is definitely helpful! –  cdouglas Nov 24 '13 at 13:21
@cdouglas, it was my pleasure! I only thought of this because of your question, so I should thank you as well. By the way, you can use a similar argument to show, more generally, that any codimension zero embedding between manifolds which is also a weak equivalence induces a weak equivalence between all (ordered and unordered) configuration spaces. One difference is that you need to use excision in homology with local coefficients where above I used the Mayer-Vietoris sequence. –  Ricardo Andrade Nov 24 '13 at 14:14

Let $M = \mathbb{R}^2, N = D^2$. In both cases I think the configuration space of $k$ distinct unordered points has the homotopy type of $K(B_k, 1)$. More generally I think we can take $N$ to be a manifold with boundary and $M = N \setminus \partial N$, and there should be some reasonable generalization of this but I can't think of a precise statement. It would be interesting to find an example where $M$ and $N$ are both closed manifolds (or show that one doesn't exist!).

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Yes indeed, sorry: I was thinking of smooth manifolds ... I should get out more. –  cdouglas Nov 23 '13 at 1:02