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This post here is a specification of this post.

Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of complete geodesic metric spaces verifying :

  • $X_{n}$ is a $n$-dimensional regular CW complex.
  • $X_{n}$ is boundaryless and unbounded.
  • $X_{n+1}$ is n-connected.
  • $X_{n} \subset X_{n+1}$
  • The distance $d_{n}$ and $d_{n+1}$ generate the same topology on $X_{n}$.
  • $\forall x,y \in X_{n}$ : $d_{n+1}(x,y) \le d_{n}(x,y)$.
  • $(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$

Definition : Let $d$ be a distance on $ \bigcup{X_{n}}$, defined as follows : $d(x,y) = lim_{n \to \infty} d_{n}(x,y)$.

Remark : There is a small abuse in the previous definition because $d_{n}(x,y)$ is defined only for $x, y \in X_{n}$. But because we take $n \to \infty$, there is no problem.

Definition : Let $X:=\overline{\bigcup{X_{n}}}$, the complete metric space obtained as a completion of $\bigcup{X_{n}}$ for $d$.

Question : Is $X$ weakly contractible ?

Remark : If yes, some of these conditions could be useless for a proof, and others, highly generalized.

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Sebastien Palcoux
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1 Answers1

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No, it doesn't have to be weakly contractible. The idea of the counter-example is similar as in my previous answer, just some tricks need to be done to satisfy more assumptions.

Fix $d\geq 2$ and take $d$-disks approximating the sphere $S^d$: $$D_n' = \left\{x\in S^d\ \big|\ x_1\leq 1-\frac{1}{n} \right\} \subseteq \mathbb R^{d+1}.$$ Now let (trick to remove boundary and preserve completeness): $$D_1=D_1' \times \{0\} \cup \partial D_1'\times [0,\infty) \subseteq \mathbb R^{d+2}\\ D_n = D_{n-1} \cup D_n' \times \{0\} \cup \partial D_n' \times [0,\infty)$$ with natural geodesic metrics. What is going on? For $d=2$ to each circle being a border of some $D_m' \subseteq D_n'$ we glue an infinite cylinder (and analogously in higher dimensions).

Each $D_n$ is contractible and in the limit we get a space that is homotopy equivalent to the sphere $S^d$. So we have the desired counter-example up to a few details. Btw. note that there is no cylinder or interval over the north pole: it's completion with respect to the geodesic metric, not the euclidean metric.

One of the assumptions not satisfied yet is the condition of quasi-isometry. Fortunately we can easily correct it. Attach to the south pole the following set looking like an infinite comb: $$C=\{x\ |\ x_1=-1 \lor (x_1\leq -1 \land x_2\in \mathbb Z),\ x_3=\ldots=x_{d+2}=0\}.$$ I claim that all our spaces ($X_n=D_n\cup C$) are quasi-isometric with $C$ (with the geodesic metric). In an infinite comb we can always spread the teeth apart to make some free space in the middle - that's where the cylinders go via the quasi-isometry (I skip formal details). Note that the inclusion map is not a quasi-isometry.

Similarly with the dimension assumption - to lower dimension for $n=0$ we can take a discrete net in $C$ as $X_0$ and for $0<n<d$ we can set $X_n=C$. And to increase dimension, we can always take a wedge sum with the $n$-sphere. These changes don't affect other assumptions.

savick01
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  • instead of deleting your answer, do you see a manner for improving it ? But perhaps this time the answer is "yes"... What do you suspect now ? – Sebastien Palcoux Aug 29 '13 at 09:54
  • I was thinking about fixing $d=2$ and glueing something compact and boundaryless (like the $n$-th projective space) instead of the cylinders. If we have something like van Kampen theorem for higher homotopy groups, it would work: homotopy groups of disk are trivial and the same for low homotopy groups of the projective space - except for the first group, that we kill with the disk. – savick01 Aug 29 '13 at 10:03
  • By analising the example with the projective spaces I can say, that it may be possible to construct a counterexample by Hurewicz and long exact sequence of homology, if we have a space with trivial middle ($1<i<n$) homology groups, just like the middle homotopy groups of projective spaces are trivial... P.S. The projective spaces don't work by the same argument (Hurewicz and long exact sequence). – savick01 Aug 29 '13 at 11:24
  • P.P.S. The spheres are not OK, because they have trivial fundamental groups and that leads to nontrivial second homology group. – savick01 Aug 29 '13 at 11:36
  • For information : I have added in the post the assumption "regular" for the CW complexes. – Sebastien Palcoux Aug 29 '13 at 13:32
  • Actually my counterexample needs spaces that have homology groups of the circle (up to $n-1$), I really doubt if we can found such examples. P.S. You told me about regularity before and the above example is regular. – savick01 Aug 29 '13 at 21:30
  • You could just ask a question on MSE about such examples ? (Another politics on MSE (MO also) would be to ask questions as less as possible, why ?) – Sebastien Palcoux Aug 29 '13 at 22:03