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?

-
$k > 1$ ;-) 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

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!).

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

I will present an example involving only (non-compact) manifolds without boundary. As far as I know, the analogous problem for closed manifolds is wide open. Nevertheless, the article Configuration spaces are not homotopy invariant by Paolo Salvatore and Riccardo Longoni is very much worth looking at (it was published in Topology, volume 44, number 2, 2005, pages 375-380).$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\Conf}{\mathrm{Conf}}$$\newcommand{\To}{\longrightarrow}$

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.

-
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