Timeline for Homotopy type of some lattices with top and bottom removed
Current License: CC BY-SA 3.0
26 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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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Aug 12, 2016 at 22:33 | comment | added | მამუკა ჯიბლაძე | @RichardStanley Thanks, fantastic! A pdf of the thesis can be downloaded at dtic.mil/cgi-bin/… | |
| Aug 12, 2016 at 16:17 | comment | added | Richard Stanley | @მამუკა ჯიბლაძე The earliest reference I know is the 1977 Ph.D. thesis of Scott Provan, but I don't know how to access it online. A more general result appears as Corollary 2.2 of arxiv.org/pdf/math/0505576.pdf. | |
| Aug 2, 2016 at 8:30 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
update
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| Jul 27, 2016 at 11:14 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added question about dense elements in general lattices
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| Jul 27, 2016 at 10:44 | answer | added | Uri Bader | timeline score: 1 | |
| Jul 27, 2016 at 9:38 | answer | added | მამუკა ჯიბლაძე | timeline score: 1 | |
| Jul 27, 2016 at 8:24 | comment | added | მამუკა ჯიბლაძე | If yes, then a minimal $S^1\times S^1$ would be given by the face lattice of the Császár Torus (7 vertices, 21 edges, 14 triangles) | |
| Jul 27, 2016 at 7:23 | comment | added | მამუკა ჯიბლაძე | @UriBader Somehow my brain is not functioning well enough right now, but could not actually face lattices of polytopes reflect their homotopy types? | |
| Jul 27, 2016 at 7:19 | comment | added | მამუკა ჯიბლაძე | @RichardStanley What would be an optimal reference for that in your opinion? | |
| Jul 26, 2016 at 23:28 | comment | added | Richard Stanley | The product of chains is a distributive lattice. If $L$ is any finite distributive lattice, then the geometric realization of the order complex of $L$ with the top and bottom removed is a ball, unless $L$ is a boolean algebra (product of two-element chains). In this case the order complex is a triangulation of a sphere. | |
| Jul 26, 2016 at 21:33 | comment | added | Uri Bader | მამუკა ჯიბლაძე hopefully you will get better answers to your second question in the future. I am curious myself how the lattice axiom reflects in the topology. I am not an expert. In case you don't I might try to write my very partial one. Meanwhile let's amuse ourselves with this: I see that $P([n])$ gives $S^{n-2}$. Can you find a lattice that gives $S^1\times S^1$? | |
| Jul 26, 2016 at 18:09 | comment | added | მამუკა ჯიბლაძე | @UriBader As you wish. But your proof is entirely different; and I am sure your comments contain much for the second question too. | |
| Jul 26, 2016 at 18:05 | comment | added | Uri Bader | @DanPetersen you got it. Indeed, this observation was made while trying to consider "$q\to 1$" phenomena in representation theory. With Uri Onn we considered connections between the rep theory of $\text{GL}_n(\mathbb{Z}_p)$ and $\text{O}(n)$ as well as $\text{GL}_n(\mathbb{F}_p)$ and $S_n$ and related objects. | |
| Jul 26, 2016 at 18:03 | comment | added | Uri Bader | მამუკა ჯიბლაძე I didn't post my comment as an answer originally because I assumed it is the second question in your post which is your main concern and I don't see a point in doing it now, as Goodwillie's answer is certainly satisfying. | |
| Jul 26, 2016 at 16:55 | comment | added | Dan Petersen | @UriBader it sounds like the question asked here is the "$q\to 1$" specialisation of the result in your comment! | |
| Jul 26, 2016 at 15:44 | comment | added | მამუკა ჯიბლაძე | @UriBader I believe you should make this an answer. It will give me hard time deciding which one is more beautiful but that your approach deserves to become a separate answer I have no doubt. | |
| Jul 26, 2016 at 13:57 | comment | added | Uri Bader | მამუკა ჯიბლაძე here is where I came across the lemma in my comment above: Let $V$ be vector space over a finite field and consider the lattice of all subspace. Then the homotopy type of this(chopped) lattice is given by Solomon-Tits theorem. However, if $V$ is a module over, say $\mathbb{Z}/p^2$, $x=pV$ will cause the lattice of submodules to be contractible. | |
| Jul 26, 2016 at 13:48 | comment | added | მამუკა ჯიბლაძე | @UriBader Seems like a competing answer emerges ;) | |
| Jul 26, 2016 at 13:39 | comment | added | Uri Bader | Here is a general observation: assume you have in your lattice an element $x\neq 0$ s.t for all $y\neq 1$, $x\vee y\neq 1$. Then the chopped lattice is contractible. Indeed, $y\mapsto x\vee y$ gives a deformation retract onto the sublattice $[x,1)$, which is clearly contractible as it is coned over $x$. In a product of linear orders, every element $x$ which at every coordinates is not maximal will do. | |
| Jul 26, 2016 at 13:06 | vote | accept | მამუკა ჯიბლაძე | ||
| Jul 26, 2016 at 12:07 | answer | added | Tom Goodwillie | timeline score: 11 | |
| Jul 26, 2016 at 11:44 | comment | added | Uri Bader | It is quite easy to see that the chopped product of linear orders, not all with $\leq 2$ elements, will be contractible. Pick a vertex for which (say) the first instance is intermidiate and work out the contraction to it coordinatewise, starting with the first. | |
| Jul 26, 2016 at 11:29 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added crosspost link
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| Jul 26, 2016 at 11:22 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 3.0 |