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Timeline for A notion of 2-dimensional tree

Current License: CC BY-SA 4.0

17 events
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Apr 8, 2022 at 12:15 vote accept Agelos
Apr 8, 2022 at 12:15
Apr 3, 2022 at 17:39 comment added Agelos I wrote another update.
Apr 3, 2022 at 17:05 comment added Geva Yashfe @SamNead No problem. Regarding the statement for higher dimensions: I made a conjecture about collapsibility for 2-trees (namely that they should collapse onto a one-dimensional subcomplex, in other words a graph). For $(n+1)$-trees, you could generalize my guess to the statement that they should collapse onto an $n$-dimensional CW complex. In particular if they are $n$-connected then they are contractible. In this sense $(n+1)$-manifolds with boundary are an example, but not a problematic one.
Apr 3, 2022 at 13:39 history edited Sam Nead CC BY-SA 4.0
pointed out even worse problems.
Apr 3, 2022 at 13:30 comment added Sam Nead @GevaYashfe - Your technique applies to all manifolds with boundary. For example, properly embedded disks (that is, two-trees) separate points in three-manifolds. For example, in $S^2 \times I$. So all hope is lost. I'll add this to my answer.
Apr 3, 2022 at 13:19 comment added Sam Nead @GevaYashfe - Fixed, and my apologies.
Apr 3, 2022 at 13:18 history edited Sam Nead CC BY-SA 4.0
fix misspelt name.
Apr 3, 2022 at 9:52 comment added Geva Yashfe @SamNead By the way, please fix my name's spelling in your edit, if you can. Mathoverflow will not let me make the 1-character change.
Apr 3, 2022 at 9:49 comment added Geva Yashfe @SamNead If my conjecture on collapsibility (in comments on the question) is correct, then the fundamental groups of these complexes are free, and simple-connectedness is a sufficient strengthening to imply contractibility.
Apr 3, 2022 at 8:21 history edited Sam Nead CC BY-SA 4.0
Adde
Apr 3, 2022 at 8:15 comment added Sam Nead @GevaYashfe - Very nice counter-example. I’ll edit my post… as I now think that there is no obvious strengthening to give what the OP wants (other that CAT(0) cube complexes, I suppose?).
Apr 2, 2022 at 22:51 comment added Geva Yashfe How about an annulus, e.g. $\{p \in \mathbb{R}^2 \mid 1\le \Vert p \Vert \le 2\}$? Any two points can be separated by a segment that has both endpoints on the same component of the boundary. In particular it is a 2-tree in the sense suggested above. But it is not simply connected.
Apr 1, 2022 at 9:02 comment added Sam Nead That is a good question! I would hope that, with the correct set-up, these spaces are contractible…
Apr 1, 2022 at 8:47 comment added Agelos Yes, I do like this variant! A nice feature is that 1-trees now coincide with standard trees. A proof of the Lemma would be very welcome. Do higher homotopy groups vanish as well?
Apr 1, 2022 at 0:33 history edited Sam Nead CC BY-SA 4.0
added lemma
Apr 1, 2022 at 0:27 history edited Sam Nead CC BY-SA 4.0
wording
Apr 1, 2022 at 0:20 history answered Sam Nead CC BY-SA 4.0