= Solution
For a finite graph $\Lambda=(V,E)$ with <free boundary condition>[free boundary conditions], write $\sigma_x=(\cos\theta_x,\sin\theta_x)\in S^1$. The ferromagnetic <O(2) model> is
$$
d\mu_{\Lambda,\beta,h}(\theta)
=\frac1Z\exp\left\{\beta\sum_{xy\in E}\cos(\theta_x-\theta_y)
+h\sum_{x\in V}\cos\theta_x\right\}
\prod_{x\in V}\frac{d\theta_x}{2\pi}.
$$
The <Ginibre inequality> says, in particular, that for $a,b\in\mathbb Z_{\geq0}^V$,
$$
\langle\cos(a\cdot\theta)\cos(b\cdot\theta)\rangle
\geq
\langle\cos(a\cdot\theta)\rangle
\langle\cos(b\cdot\theta)\rangle.
$$
For the proof, take two independent replicas $\theta,\theta'$ and write the covariance as one half of the expectation of
$$
\{\cos(a\cdot\theta)-\cos(a\cdot\theta')\}
\{\cos(b\cdot\theta)-\cos(b\cdot\theta')\}.
$$
Set $u=(\theta+\theta')/2$ and $v=(\theta-\theta')/2$. Product-to-sum identities turn each difference into $-2\sin(a\cdot u)\sin(a\cdot v)$, while every replicated interaction becomes
$$
\cos(\theta_x-\theta_y)+\cos(\theta'_x-\theta'_y)
=2\cos(u_x-u_y)\cos(v_x-v_y).
$$
Expand every exponential in a power series and then every cosine power into Fourier modes. Integration over each angle kills all unmatched modes. Because the couplings, field, and entries of $a,b$ are nonnegative, every surviving paired coefficient in the covariance is nonnegative. Their sum is therefore nonnegative, proving the inequality. The same replica expansion proves the usual product version.
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