# Is a finite lattice determined by its Hasse diagram (as a graph)?

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).

• 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). Commented May 26 at 17:08
• 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). Commented May 26 at 17:10
• Good point! Let me rephrase the question: can we reconstruct a lattice from its Hasse diagram as an undirected graph, up to duality? Commented May 26 at 19:17
• How much information beyond just the undirected graph is enough? Commented May 28 at 16:27
• @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. Commented May 28 at 16:50

• Also the cycle graph $C_{2n+2}$ is the Hasse diagram of $n$ nonisomorphic lattices.