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If finite lattices $L_1,L_2$ have Hasse diagrams that are isomorphic as undirected graphs, must $L_1$ and $L_2$ be isomorphic?

NOTE: Sam Hopkins points out that the answer is “no” because there are lattices that are not isomorphic to their duals. I would like to know if this “which way is up?” ambiguity is the only obstacle to reconstructing a lattice from its Hasse diagram (viewed as an undirected graph).

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    $\begingroup$ What do you mean by "graph"? The undirected graph? Then surely not, since there are finite lattices that are not self-dual (but the Hasse diagrams of dual lattices would be isomorphic as undirected graphs). $\endgroup$ Commented May 26 at 17:08
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    $\begingroup$ On the other hand if by "graph" you mean "directed graph" then the Hasse diagram is the same information as the cover relation of the partial order, and then it is a simple fact that for a finite partial order it is the transitive, reflexive closure of its cover relation; hence this digraph does determine the finite poset (the lattice property is irrelevant here). $\endgroup$ Commented May 26 at 17:10
  • $\begingroup$ Good point! Let me rephrase the question: can we reconstruct a lattice from its Hasse diagram as an undirected graph, up to duality? $\endgroup$ Commented May 26 at 19:17
  • $\begingroup$ How much information beyond just the undirected graph is enough? $\endgroup$ Commented May 28 at 16:27
  • $\begingroup$ @MichaelHardy well, as I said, if you know the directed graph, that is enough to recover the poset (and the lattice property is irrelevant). I suppose you could ask, how many of the edge directions do you need to know. $\endgroup$ Commented May 28 at 16:50

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This is still far from true. Consider for example these lattices (just a random example):

enter image description here

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    $\begingroup$ Also the cycle graph $C_{2n+2}$ is the Hasse diagram of $n$ nonisomorphic lattices. $\endgroup$
    – bof
    Commented May 26 at 20:54
  • $\begingroup$ Instead of a "random" example, is there such a thing as a simplest example? $\endgroup$ Commented May 28 at 16:25
  • $\begingroup$ So how much information beyond just the undirected graph is enough? $\endgroup$ Commented May 28 at 16:27
  • $\begingroup$ @MichaelHardy I don't think there is any example with 5 elements, and so 6 elements is minimal (though as bof explained, there is at least one other example with 6 elements). $\endgroup$ Commented May 28 at 16:51

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