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Coexistence implies an Ising pressure cusp

Statement

Assume Countable Choice and fix $β > 0$ and $J > 0$ with $β J \geq 2$ . For $B_{n} = [- n, n]^{2} \cap Z^{2}$ , define the free-boundary field-dependent energy and partition function by $H_{B_{n}}^{(h), f r e e} (s) = - J \sum_{\begin{array}{c} x \in B_{n}, k \in {1, 2} : \\ x + e_{k} \in B_{n} \end{array}} s_{x} s_{x + e_{k}} - h \sum_{x \in B_{n}} s_{x}, Z_{n} (h) = \sum_{s \in {- 1, + 1}^{B_{n}}} e^{- β H_{B_{n}}^{(h), f r e e} (s)} .$ The bounded finite-range pressure theorem gives the finite limit $p (h) : = \lim_{n \to \infty} ∣ B_{n} ∣^{- 1} \log Z_{n} (h)$ for each $h \in R$ . Here $h$ is the field energy per spin, while $h_{S h} (μ)$ below denotes Shannon entropy per site. The dimensionless pressure $p (h)$ is not differentiable at $h = 0$ . More precisely, if the stationary DLR states at zero field have magnetizations at least $m_{*} > 0$ and at most $- m_{*}$ , then every positive-field secant slope from zero is at least $β m_{*}$ and every negative-field secant slope is at most $- β m_{*}$ . This proves a sufficient coexistence cusp; it does not identify the critical coupling.

Facts & Assumptions

Given: The finite-range two-dimensional Ising model, $β J \geq 2$ , and Countable Choice.

[A1]

Countable Choice is retained for the DLR and stationary-state construction (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

At $β J \geq 2$ there are stationary DLR states with one-site magnetizations at least $m_{*} > 0$ and at most $- m_{*}$ (Two-dimensional Ising Gibbs coexistence at low temperature).

[F2]

For each fixed field $h$ , every stationary law satisfies $h_{S h} (μ) - β u_{h} (μ) \leq p (h)$ ; a DLR law attains equality. Here $h_{S h} (μ)$ denotes the entropy density $h (μ)$ of DLR states attain the bounded lattice pressure variational principle, and $u_{h}$ is its mean energy density with the field contribution included.

Proof

1.1A1F1F2

Under Countable Choice [A1], let $μ_{+}$ and $μ_{-}$ be the states in [F1], and let $m = μ_{+} (s_{0}) \geq m_{*} > 0$ . At zero field, [F2] gives $p (0) = h_{S h} (μ_{+}) - β u_{0} (μ_{+})$ .

2.1F2step 1.1algebra

For $h > 0$ , the energy per site of the same stationary law is $u_{h} (μ_{+}) = u_{0} (μ_{+}) - h m$ . Applying the variational inequality [F2] to $μ_{+}$ yields $p (h) \geq p (0) + β h m \geq p (0) + β h m_{*}$ .

3.1F1F2step 2.1algebra∎

For $μ_{-}$ with magnetization at most $- m_{*}$ , the same inequality gives $p (h) \geq p (0) + β h μ_{-} (s_{0})$ . If $h < 0$ , divide by $h$ to get $(p (h) - p (0)) / h \leq - β m_{*}$ . For $h > 0$ , step 2.1 gives the slope at least $β m_{*}$ . The one-sided secant slopes are separated, so no derivative at zero exists.

Coexistence implies an Ising pressure cusp

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources