Timeline for Does every CAT(0) space embed in a measurable integral of $\mathbb{R}$-trees?
Current License: CC BY-SA 4.0
14 events
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| Nov 3, 2021 at 11:46 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| May 6, 2015 at 15:36 | comment | added | Dylan Thurston | @NicolasMonod , it's not true that the convex hull of three points in a product of trees is 2-dimensional. I worked out an explicit example with three points in a product of three trees. | |
| May 6, 2015 at 15:22 | comment | added | Dylan Thurston | I see. The point is that in $f(t) = \sqrt{\sum_i h_i(a_i t)^2}$, if one of the $h_i$ is only piecewise affine, you get a positive delta-mass in $f''(t)$ that you can't cancel. So that indeed answers Question 1, although it also suggests that the "integral" setting that @NicolasMonod suggested is more natural. | |
| May 6, 2015 at 15:20 | vote | accept | Dylan Thurston | ||
| May 6, 2015 at 13:05 | comment | added | Nicolas Monod | Other argument: In an arbitrary product of trees, I would guess that the convex hull of three points is 2-dimensional. However, in complex hyperbolic space, the convex hull of a generic (hum, hum) triple of points is 4 dimensional. Here generic means not contained in a totally real subspace. | |
| May 6, 2015 at 13:05 | comment | added | Nicolas Monod | @Yves: ...or all but in a subvariety of positive codimension? like here :P | |
| May 6, 2015 at 12:56 | comment | added | YCor | @DylanThurston: I understand the argument to work for infinite products. In this case the distance function has the form $u(t)=\sqrt{\sum_i{h_i(a_it)}}$, where $h_i$ is a (convex nonnegative) piecewise affine function with slopes in $\{-2,0,2\}$, $a_i\ge 0$ and $\sum a_i^2=1$, and some additional summability condition ensuring that $u<\infty$. Then a little argument is needed to ensure that $u$ cannot be analytic unless it's affine. | |
| May 6, 2015 at 12:45 | comment | added | YCor | (I usually understand "generic" in the sense almost all configurations, where "almost all" means something such as: all but finitely/countably many, all except measure zero, all but meager, etc.) | |
| May 6, 2015 at 11:23 | comment | added | Dylan Thurston | Also, welcome to MO! | |
| May 6, 2015 at 11:21 | comment | added | Dylan Thurston | Thanks for the nice argument! But you're addressing a question I didn't ask. In Question 1, I'm asking about embedding in an infinite product. In Question 2, I'm asking about finite products, but only about embedding finitely many points in a distance-preserving way, not their convex hull. The three corners of the triangle can trivially be embedded in a product of trees, and your argument doesn't rule out embedding a finite subset of the triangle (although it might well be impossible). As you note, it also doesn't rule out embedding inside infinite constructions. | |
| May 6, 2015 at 11:10 | comment | added | Nicolas Monod | @Yves: (1) yes that's what I meant by generic (I'd have said scalene but then I'd have gotten picked on as it's too strong) and (2) yes you're right I guess "non-analytic" would be the lazy overkill! | |
| May 6, 2015 at 9:16 | comment | added | YCor | I don't think you use "generic" anywhere, the argument works for any subset of $\mathbb{H}^2$ containing a pair of non-collinear segments. Also the argument shows that the distance $f(t)=d(u_t,v_t)$ between two speed-one geodesics $(u_t),(v_t)$ will be of the form $f(t)=\sqrt{\sum_i h_i(t)^2}$ with $h_i$ piecewise affine, which is not necessarily piecewise affine but is enough for the argument. | |
| May 6, 2015 at 8:40 | review | First posts | |||
| May 6, 2015 at 8:40 | |||||
| May 6, 2015 at 8:38 | history | answered | Nicolas Monod | CC BY-SA 3.0 |