# A proof for this equivalent version of the Infrared Bound/Gaussian Domination

Consider the Ising Model in the $d$-dimensional discrete torus with side lengh $L$, denoted by $\mathbb{T}_L$, with nearest neighbors interaction (with interaction parameter $J$, no magnetic field, and inverse of temperature $\beta$

The Gaussian Domination Bound/ Infrared Bound states that for every non-zero $p \in \mathbb{T}_L^*:=(\frac{2\pi}{L} \mathbb{Z}^d) \cap (-\pi,\pi]$, we have:

$$\sum_{x \in \mathbb{T}_L} e^{ip \cdot x} \langle \sigma_0 \sigma_x \rangle_{L,\beta} \le \frac{1}{2\beta E(p)},$$

where $\langle \cdot \rangle^{0}_{L,\beta}$ denotes the expected value of a random variable with respect to finite volume Gibbs Measure in $\mathbb{T}_L$; $i$ is the imaginary unit; $ip \cdot x$ denotes the inner product of $ip$ and $x$; and $$E(p):= J \sum_{x: \|x\|_1=1} (1-e^{ip\cdot x}).$$

Multiple articles point out that the following statement is equivalent to that bound, $\forall (v_x) \in \mathbb{C}^{\mathbb{T}_L}$, we have $$\sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 }v_x \bar{v}_y \langle \sigma_x \sigma_y \rangle_{L,\beta} \le \frac{1}{2\beta}\sum_{x,y \in \mathbb{T}_L: \|x-y\|_1=1 } v_x \bar{v}_y G_L(x,y) + \frac{1}{L^d} \big| \sum_{x \in \mathbb{T}_L} v_x \big|^2 \sum_{x \in \mathbb{T}_L} \langle \sigma_0 \sigma_x \rangle_{L,\beta},$$ where $$G_L(x,y) := \sum_{p \in \mathbb{T}_L^* \backslash \{0\} } \frac{1}{L^d} \frac{e^{ip \cdot (x-y)}}{E(p)}.$$

Well, I tried to use the usual proof as in the Chapter 10 of the book Statistical Mechanics of Lattice Systems:a Concrete Mathematical Introduction. Following the proof in page 465 until the moment they choose the values of $\alpha_i$, I believe that this equivalence will come choosing new values for $\alpha_i$, but to this moment, I don't know what to choose.

• Are you asking about the proof of the infrared bound (which is what is on p. 465) or about the equivalence between the two formulations?
– Ben
May 6, 2017 at 20:31
• My question is about the equivalence. I understand the proof in p. 465 and I was trying to somehow adapt it to the second statement. May 7, 2017 at 13:21

So what you actually prove is that $$\langle (\sigma (-\Delta h))^2 \rangle \le \frac{1}{2\beta} (-h\Delta h)$$ When you take $$h=e^{ikx}$$, your first statement pops up. However, for any $$v$$ with $$\sum_x v_x =0$$, we can always find $$h$$ such that $$-\Delta h =v$$ (the kernel of $$-\Delta$$ is all constant functions). Plug in $$h$$, and your second statement follows for "neutral" $$v$$. If $$\sum_x v_x\ne0$$, we can always define $$u=v-1/L^d\sum _x v$$ and use $$u$$ instead.