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Suppose that $X$ is a locally contractible completely metrizable topological space. Is it true that $X$ can be metrized as a (complete) CAT(1) metric space?

The only result in this direction I know is Berestovsky's theorem which settles the problem in the case of simplicial complexes:

V. Berestovskii, Borsuk's problem on metrization of a polyhedron, Dokl. Akad. Nauk SSSR 268 (1983), no. 2, 273–277.

The question itself is a spin-off this MSE question.

Edit. One more relevant result:

Theorem. Let $X$ be a separable metrizable connected locally connected locally compact topological space. Then $X$ admits a complete geodesic metric, i.e. a metric such that every two points $p, q\in X$ are connected by a geodesic, i.e. a path whose length equals the distance between $p$ and $q$.

See

A. Tominaga and T. Tanaka, Convexification of locally connected generalized continua, J. Sci. Hiroshima Univ. Ser A 19 (1955), 301–306.

Their proof builds upon the earlier work by Moise and Bing.

Since CAT(1) spaces are required to be geodesic, this and Igor Belegradek's comment suggest that in my question it makes sense to restrict to connected locally compact locally contractible separable metrizable spaces.

See also the (inconclusive) discussion here.

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    $\begingroup$ Perhaps this is naive but I would try constructing a counterexample using Kleiner's results on the topological and homological dimension of CAT(1) spaces, see theorem A in "The local structure of length spaces with curvature bounded above", math.nyu.edu/~bkleiner/locstr.pdf. $\endgroup$
    – Igor Belegradek
    Commented Dec 22, 2021 at 12:54
  • $\begingroup$ @IgorBelegradek: You are right, I forgot about Kleiner's paper. $\endgroup$
    – Moishe Kohan
    Commented Dec 22, 2021 at 16:53
  • $\begingroup$ Your link, presumably intended to point to the MSE question Existence of CAT(0)-metrics, actually pointed to your answer. I edited to point to the question, while this was on the front page. I hope that that was correct. $\endgroup$
    – LSpice
    Commented Jan 20 at 17:55

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There are examples of that type. I will construct a compact metrizable contractible and locally contractible space that does not admit a CAT(1) length metric.

Let $H$ be Bing's house (or the dunce hat).

First, let us do the CAT(0) case.

The space $H$ is compact metrizable contractible and locally contractible, but it does not admit a CAT(0) metric. Indeed, suppose it does, then $H$ has extendable geodesics. For a proof, see the solution of 9.9 in our book. But any local geodesic in a CAT(0) space is a global geodesic. The latter contradicts the compactness of $H$.

Now let us come back to the CAT(1) case.

Glue the rescalings $\tfrac 1n\cdot H$ along one point; let us call it the Hawaiian Bing's house. This space $\hat H$ is compact, metrizable, contractible, and locally contractible, but it does not admit a CAT(1) metric. Indeed, suppose it does, then the same argument shows that any geodesic in $\hat H$ is extenadable. It follows that each house in $\hat H$ has diameter at least $\pi$. This contradicts compactness of $\hat H$.

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  • $\begingroup$ I understood the CAT(0) part but not the argument in the CAT(1) setting. "The same argument" as in CAT(0) case no longer gives extendible geodesics. $\endgroup$
    – Moishe Kohan
    Commented Dec 21, 2023 at 1:05
  • $\begingroup$ @MoisheKohan It gives extension of local geodesic that has to be minimizing on any interval of length ⩽π. So, after removing one point from $\hat H$ we get countable number of connected components; each has with diameter at least π. So, $\hat H$ is not compact --- a contradiction. $\endgroup$
    – Anton Petrunin
    Commented Dec 22, 2023 at 1:19
  • $\begingroup$ You write "yes" but it seems to be a "no" answer. $\endgroup$
    – YCor
    Commented Jan 20 at 18:25
  • $\begingroup$ @YCor I will fix it. $\endgroup$
    – Anton Petrunin
    Commented Jan 20 at 18:29

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