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Finding the order dimension of a set of lattices

asked 2018-01-11 16:35:23 -0500

maresage gravatar image

updated 2018-01-11 16:51:41 -0500

Let $L_n$ denote the set of all lattices with $n$ points in SAGE. I want to find for given $n$ (lets say $n$ is smaller than 10 or so) all lattices with order dimension at least 3. For the definition of order dimension see I can do it for a concrete example, but I am not sure how to obtain a (graphical if possible) output of the result for a given $n$ (that also possibly includes the value of the order dimension in case it is at least 3). Here the concrete example for the boolean lattice with 3 elements:

P=posets.BooleanLattice(3) dimension(P)

I also have two small mathematical questions:

  1. I hope I see it correctly that the order dimension of a lattice is 1 if and only if this lattice is a chain? In case this is true, this motivates to only look at order dimension at least 3 since then it is clear which lattices have order dimension equal to two (knowing that only chains have order dimension equal to one).
  2. Are there tables on the internet or the literature on known results of what the order dimension is for lattices coming from combinatorics such as the divisor lattice. Do I see it correcty that the order dimension of the divisor lattice of a natural number n is equal to the number of prime divisors of n?
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answered 2018-01-12 01:54:16 -0500

vdelecroix gravatar image

You can do a 2 dimensional table using just a tiny bit of Python, though this is dramatically slow

sage: for n in range(1,8):
....:     d = [0] * (n+1)
....:     for P in Posets(n):
....:         d[P.dimension()] += 1
....:     for i in d:
....:         print "%4s" %i,
....:     print
   0    1
   0    1    1
   0    1    4    0
   0    1   15    0    0
   0    1   62    0    0    0
   0    1  314    3    0    0    0
   0    1 1955   89    0    0    0    0

As you see, up to 7 vertices there is no such lattice.

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Asked: 2018-01-11 16:35:23 -0500

Seen: 20 times

Last updated: Jan 12