A Negative Answer to Bäumler's Question on the Number of Spin-Glass Ground States

Consider the Edwards–Anderson spin glass at zero temperature on an infinite, connected, locally finite graph, with i.i.d. absolutely continuous couplings, and let G(J) be its set of ground states. Bäumler proved that on every locally finite tree, for coupling laws of linear growth, |G(J)| is almost surely 2 or ∞, and that |G(J)| = 2 if and only if simple random walk on the tree is recurrent. He asked whether |G(J)| ∈ {2, ∞} holds for all graphs and all distributions of linear growth. We show that it does not. Join each pair of consecutive integers k, k+1 by m_k internally disjoint paths of length two. For couplings uniform on (−1, 1), the resulting graph has |G(J)| = 4 almost surely if Σ_k m_k^(−1/2) < ∞, and |G(J)| = 2 almost surely otherwise; the first conclusion holds for every absolutely continuous coupling law. We also give a planar graph of maximum degree 4 on which simple random walk is recurrent and |G(J)| = 4 almost surely for couplings uniform on (0, 1), a recurrent example with couplings of both signs, and a recurrent graph on which |G(J)| is a non-degenerate random variable. The last example answers the other half of the question in the form Bäumler posed it in 2019. The examples show that several implications of the tree theorem, between uniqueness, vanishing maximal flow and recurrence, fail on general graphs. The mechanism, a unique cheapest finite domain wall on a two-ended graph, is elementary. The question remains open for bounded-degree graphs with a symmetric coupling law and for quasi-transitive graphs such as ℤ^d. This is an unrefereed note.