Timeline for Does the Hausdorff dimension characterise CAT(0) spaces having some bilipschitz balls?
Current License: CC BY-SA 4.0
20 events
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| May 5, 2021 at 8:05 | history | edited | Christian | CC BY-SA 4.0 |
added 421 characters in body
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| Jan 29, 2021 at 15:35 | comment | added | YCor | For sure, any tree (with the geodesic metric) is CAT(0). | |
| Jan 29, 2021 at 15:34 | comment | added | user142382 | One can take a Cantor set in R^3, say, of Hausdorff dimension 2 and build a compact (abstract) tree whose set of leaves is bi-Lipschitz equivalent to the Cantor set. (Branchings correspond to stages of the Cantor set construction.) There are details to check. Confession: I am not an expert on CAT(0), but I assume a compact (geodesic) tree will be CAT(0). I believe that Moishe Kohan's answer really takes care of the two-sided case. | |
| Jan 29, 2021 at 15:32 | comment | added | Christian | @YCor Thank you. I removed the counterexamples for now since I did not check them in all details. I will add them after checking the details. | |
| Jan 29, 2021 at 15:28 | history | edited | Christian | CC BY-SA 4.0 |
deleted 425 characters in body
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| Jan 29, 2021 at 15:23 | comment | added | YCor | By the way I'm not that confident with my answer (I wasn't aware that a compact tree can have Hausdorff dimension 2, and don't know whether there's a nonempty compact tree in which the set of branching points is dense). It's maybe very easy, but I'm not familiar enough with Hausdorff dimension. | |
| Jan 29, 2021 at 15:21 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo in title
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| Jan 29, 2021 at 14:40 | history | edited | Christian | CC BY-SA 4.0 |
Added a condition of a lower bound on the metric curvature; added the counter examples given in the comments.
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| Jan 29, 2021 at 14:10 | comment | added | Moishe Kohan | See my answer here. | |
| Jan 29, 2021 at 14:08 | comment | added | Christian | @MoisheKohan No, I was not familiar with this theorem. It looks like this is what I need. | |
| Jan 27, 2021 at 3:13 | comment | added | Moishe Kohan | Are you familiar with Nikolaev's theorem on Alexandrov spaces with 2-sided curvature bounds? | |
| Jan 26, 2021 at 20:47 | comment | added | YCor | @Christian This is not worth an answer. I'd prefer you edit the question; you might just mention that examples as a remark. | |
| Jan 26, 2021 at 20:18 | comment | added | Christian | @YCor If you post your counterexample as an answer I can accept it. I will ask a new question with both curvature conditions. | |
| Jan 26, 2021 at 20:16 | comment | added | Christian | @user142382 Thank you! I guess, I will pose a new question so that both of you can post an answer to this one which I can upvote and accept one of them (e.g. the first one). | |
| Jan 26, 2021 at 19:10 | comment | added | user142382 | Hi Christian. It can sometimes frustrate people on this website if you ask a question, get a counterexample, impose more conditions, and repeat this cycle. If you have a precise question about the case in which the curvature is bounded in two directions, I'd suggest you try to ask it explicitly, either by editing this question or starting a new one. | |
| Jan 26, 2021 at 17:26 | comment | added | YCor | I don't know, I'm not familiar with any purely metric notion of having a lower bound on the metric curvature. | |
| Jan 26, 2021 at 17:16 | comment | added | Christian | @YCor and user142382 Thank you very much for these comments. In the situation I am actually interested in this question, we also have a lower bound on the metric curvature. Does this help? | |
| Jan 26, 2021 at 15:16 | comment | added | user142382 | Or the 2-dimensional plane vs. a compact tree of Hausdorff dimension 2. | |
| Jan 26, 2021 at 15:15 | comment | added | YCor | Take the real line vs a tree in which the set of branching points is dense? | |
| Jan 26, 2021 at 15:11 | history | asked | Christian | CC BY-SA 4.0 |