Timeline for Lattice Flatness Measure
Current License: CC BY-SA 3.0
9 events
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Sep 2, 2015 at 16:37 | comment | added | Gerhard Paseman | It should be clear. M_n is flat with one maximal antichain of n elements (and ratio of n/n+2). The sublattice induced by taking the Boolean lattice of subsets of a (2k+1) element set and identifying all the subsets of size less than k, and then identifying all the subsets of size greater than k+1, is pretty flat, with all but two of its elements belonging to one of two disjoint maximal antichains, giving a measure of m/m+2 for m a quantity nearly exponential in k. You need to think about what flat "means" . Gerhard "Go And Play With It" Paseman, 2015.09.02 | |
Sep 2, 2015 at 16:26 | comment | added | malaverdiere | @GerhardPaseman mind elaborating on 'disjoint union of all disjoint antichains of maximal length?' | |
Sep 2, 2015 at 16:25 | comment | added | malaverdiere | @ThomasKahle The difference is that a lattice composed of n antichains of size 1 and 1 complex sublattice of size s gives us a height of s. I want a measure that shows that (for a n > s), the lattice is flat. | |
Aug 31, 2015 at 21:26 | comment | added | Gerhard Paseman | For finite lattices, one could consider a ratio such as length of a maximal antichain to the number of all lattice members, or a similar ratio involving a disjoint union of all disjoint antichains of maximal length. The latter means that a chain is also very flat. Gerhard "Lay It On Its Side" Paseman, 2015.08.31 | |
Aug 31, 2015 at 19:47 | history | edited | Eric Wofsey |
edited tags
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Aug 31, 2015 at 19:01 | comment | added | Thomas Kahle | Can you give some more examples: How should that flatness measure be different from just height? | |
Aug 31, 2015 at 18:51 | history | edited | R W |
the measure-theory tag is out of place here
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Aug 31, 2015 at 18:48 | review | First posts | |||
Aug 31, 2015 at 20:04 | |||||
Aug 31, 2015 at 18:42 | history | asked | malaverdiere | CC BY-SA 3.0 |