Timeline for from affine matroid to measures
Current License: CC BY-SA 3.0
15 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 22, 2012 at 7:39 | vote | accept | Dima Pasechnik | ||
| May 22, 2012 at 7:39 | comment | added | Dima Pasechnik | OK, thanks. We will have to check the latter in detail, as it wasn't published... | |
| May 21, 2012 at 20:18 | comment | added | Gjergji Zaimi | That was a typo. The theorems are 7.4 in the first paper, and it deals with dimension 2; and theorem 4.5 in the second paper which deals with the higher values. | |
| May 21, 2012 at 20:06 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
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| May 21, 2012 at 19:46 | comment | added | Dima Pasechnik | We checked out Thm 7.3 in the first paper you cite, and we don't see how it settles our conjecture. Thm 7.3 is about the $d+2$ points in $\mathbb{R}^d$. They also have Thm 7.4, which looks more to the point, but it is about the case $d=2$ only. | |
| May 21, 2012 at 18:21 | comment | added | Gjergji Zaimi | I am familiar with that article, but notice that Pachner moves are more general than the flips considered there. What I'm having trouble with is lifting a relation in $K(S)$ to a relation in an appropriate simplicial complex. I can't find a way to do this canonically, though I can do it by ad hoc methods in any configuration I've tried by hand. | |
| May 21, 2012 at 17:08 | comment | added | Dima Pasechnik | Thanks for digging up these references, they are very useful! Regarding a possible connection with Pachner's Theorem, it does not seem likely. Indeed, in this setting, of point configurations, two triangulations of the same set need not be connected in this way, see e.g. ams.org/journals/jams/2000-13-03/S0894-0347-00-00330-1/… | |
| May 8, 2012 at 6:02 | comment | added | Gjergji Zaimi | I finally found the result in the literature, so I updated my answer accordingly. I guess my previous idea was a bit of wishful thinking. I haven't been able to turn it into a rigorous proof. | |
| May 8, 2012 at 5:57 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
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| May 5, 2012 at 6:14 | comment | added | Dima Pasechnik | Near the bottom, you write: "Now, because all the polytopes involved have vertices in $S$, the only triangulations involved are the ones with vertices in $S$." This seems to be unclear for $d\geq 3$, as then there are polytopes which cannot be triangulated without extra vertices. e.g. en.wikipedia.org/wiki/Sch%C3%B6nhardt_polyhedron Could you clarify, perhaps? | |
| May 5, 2012 at 5:01 | comment | added | Dima Pasechnik | Thanks, this is a very useful connection that we were not aware about! | |
| May 5, 2012 at 1:40 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
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| May 4, 2012 at 16:24 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
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| May 4, 2012 at 11:39 | history | answered | Gjergji Zaimi | CC BY-SA 3.0 |