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Timeline for A notion of 2-dimensional tree

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Apr 8, 2022 at 13:22 comment added HJRW @Agelos: Given your interest in Cannon's conjecture and the connection this question makes with npc cube complexes, you might be interested in this paper of Markovic: arxiv.org/abs/1205.5747 .
S Apr 8, 2022 at 12:15 vote accept Agelos
S Apr 8, 2022 at 12:15 vote accept Agelos
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S Apr 8, 2022 at 12:15 vote accept Agelos
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Apr 8, 2022 at 12:15 vote accept Agelos
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Apr 8, 2022 at 12:15 history edited Agelos CC BY-SA 4.0
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Apr 6, 2022 at 7:34 comment added Agelos @Sam Nead: I was thinking the same. I'll post a new question summarizing this one soon.
Apr 5, 2022 at 7:20 comment added Sam Nead It is probably time to accept Ian’s answer and post a new question. For consider the poor reader who finds this post now…
Apr 4, 2022 at 23:51 history edited LSpice CC BY-SA 4.0
Links to answers; light TeXing
Apr 4, 2022 at 22:32 comment added Geva Yashfe @Agelos Yes, I wanted stronger hypotheses for later arguments. But my comment was at least a little confused also. Anyway, see my answer for some details on my various comments so far. (Sorry for being brief, I am just very tired.)
Apr 4, 2022 at 22:29 answer added Geva Yashfe timeline score: 2
Apr 4, 2022 at 19:10 comment added Agelos @Geva Yashfe: This sounds like a nice way to exploit the fact that we can now separate any two 1-subtrees by an 1-subtree. Could you clarify your definition of collapsing? I'm willing to update Q4 to ask for collapsibility rather than contractibility. You want this in order to have a stronger inductive hypothesis to work with in higher dimensions, right?
Apr 4, 2022 at 12:26 comment added Geva Yashfe Note that your Q4 does not suffice for the argument I sketched; collapsibility is used, not just contractibility. Also, this only works as stated for simplicial complexes or regular CW complexes (not CW complexes in full generality). A similar statement may be true for higher $n$, but some more work would be required.
Apr 4, 2022 at 12:24 comment added Geva Yashfe @Agelos If my conjecture about collapsibility is correct (in one direction: if $T$ is a 2-tree, then it collapses onto a 1-dimensional subcomplex) then it seems Q5 is correct (and easy) for $n=2$. Just take the 1-dimensional subcomplex $X$ onto which $T$ collapses and pick two points $x,y$ on a nontrivial loop of $X$, if such a loop exists. The preimages of $x,y$ under the collapsing function $\pi:T\to X$ are trees; a tree in $T$ that separate these trees maps under the collapsing function to a connected subset of $X$ that separates $x,y$, contradiction, so no loop exists in $X$.
Apr 4, 2022 at 7:38 comment added Agelos I mostly agree with both last comments. At the moment we are still experimenting with the defs, and names are chosen to ease the discussion. Once there is some agreement that we have found a useful concept we can think about a good name.
Apr 3, 2022 at 20:22 comment added Sam Nead @RyanBudney - I think the point here is that we want any product of $n$ trees to be an example, and we are rather hoping that all examples sort-of-kind-of look (locally) like a product of trees. So the name is not terrible… anyway, hopefully the final name will be more reflective of the final definition.
Apr 3, 2022 at 20:16 comment added Ryan Budney For posterity, you will probably want to move away from the terminology of "n-tree". Roughly speaking, there are many different ways of extrapolating the properties of trees to high-dimensional complexes, depending on what you might want to accomplish. In analogy, Turaev gave a generalization of 3-valent graphs to 2-dimensional CW-complexes, but chose to call them "shadows". Something like this is probably the better way to go, i.e. it tells people a little more about what kind of extrapolation you are talking about. Saying n-tree could be seen as presumptuous.
Apr 3, 2022 at 17:38 history edited Agelos CC BY-SA 4.0
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Apr 3, 2022 at 17:06 comment added Agelos You are right. What if we modify the definition of n-tree, so that we require each two disjoint (n-1)-trees (instead of each two points) to be separated by a third (n-1)-tree, disjoint from both? I'm updating the question.
Apr 2, 2022 at 23:00 comment added Geva Yashfe @Agelos Regarding the new question asked in your edit, I suggest the annulus as a possible counterexample. See my comment on Sam Nead's answer (well, I hope I am not making a late-night mistake).
Apr 1, 2022 at 19:51 history edited Agelos CC BY-SA 4.0
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Apr 1, 2022 at 5:51 answer added Bob Kerns timeline score: 1
Apr 1, 2022 at 0:20 answer added Sam Nead timeline score: 8
Mar 31, 2022 at 21:48 history edited LSpice CC BY-SA 4.0
Missing TeX
Mar 31, 2022 at 20:30 history became hot network question
Mar 31, 2022 at 18:15 answer added Ian Agol timeline score: 10
Mar 31, 2022 at 17:58 comment added Geva Yashfe @Agelos Take the dunce hat (suggested in another comment above.) It is contractible but I don't think you can separate two points by a forest. My (perhaps too bold) guess is that 2-trees are precisely 2-dimensional complexes that can be geometrically collapsed onto a 1-dimensional subpolyhedron. I think it should not be hard to prove or disprove.
Mar 31, 2022 at 17:52 comment added Agelos @Geva Yashfe: good example, thanks. A pinched sphere would do as well. What if we assume X to be simply-connected?
Mar 31, 2022 at 13:27 comment added M. Winter @Geva Sorry, I misunderstood "pinched".
Mar 31, 2022 at 13:26 comment added Geva Yashfe @M.Winter There is no hole. There is just a point with a neighborhood that looks like two 2-dimensional cones glued apex-to-apex.
Mar 31, 2022 at 13:25 comment added M. Winter @GevaYashfe Topologially I would have assumed that we can replace the tiny hole by a finite size hole so the cycle does not close.
Mar 31, 2022 at 13:24 comment added Geva Yashfe I suspect changing the question to "quotients of closed surfaces that identify finite sets" is not enough: take the dunce hat. A rough (probably overly bold) guess is that the 2-trees are the 2-dimensional polyhedra that can be geometrically collapsed onto a 1-dimensional subpolyhedron.
Mar 31, 2022 at 13:17 comment added Geva Yashfe @M.Winter It seems to me that if we go back to the hole then we have created a cycle, and the separating set is no longer a forest?
Mar 31, 2022 at 13:13 comment added M. Winter @Geva I would guess that this is a 2-tree. For $x,y\in X$ choose as separating forest a path from the pinched hole to $x$, circle $x$ once, and then go back to the hole. If you do this appropriately then $y$ will be on the other side of the path, separated from $x$.
Mar 31, 2022 at 12:54 comment added Geva Yashfe What about a pinched torus?
Mar 31, 2022 at 12:30 history asked Agelos CC BY-SA 4.0