4
$\begingroup$

I've posted a question about the thermodynamic limit for Gaussian Free Fields (GFF) a couple days ago and I haven't got any answers yet but I kept thinking about it and I thought it would be better to reformulate my question and exclude the previous one, since now I can pose it in a more concrete way. The problem is basically give mathematical meaning to the infinite volume Gaussian measure associated to the Hamiltonian of the discrete GFF. In what follows, I will formulate the problem and then state the question.

A (lattice) field is a function $\phi: \Lambda_{L} \to \mathbb{R}$, where $\Lambda_{L} := \mathbb{Z}^{d}/L\mathbb{Z}^{d}$. Thus, the space of all fields is simply $\mathbb{R}^{\Lambda_{L}}$. The discrete Laplacian is the linear operator $\Delta_{L}:\mathbb{R}^{\Lambda_{L}}\to \mathbb{R}^{\Lambda_{L}}$ defined by: \begin{eqnarray} (\Delta_{L}\phi)(x) := \sum_{k=1}^{d}[-2\phi(x)+\phi(x+e_{k})+\phi(x-e_{k})] \tag{1}\label{1} \end{eqnarray} If $\langle \cdot, \cdot \rangle_{L}$ denotes the usual inner product on $\mathbb{R}^{\Lambda_{L}}$, we can prove that $\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L} > 0$ if $\langle \phi,\phi\rangle_{L}> 0$ and $m \neq 0$. Thus, $-\Delta_{L}+m^{2}$ defines a positive-definite linear operator on $\mathbb{R}^{\Lambda_{L}}$. We can extend these ideas to $\mathbb{R}^{\mathbb{Z}^{d}}$ as follows. A field $\phi: \mathbb{Z}^{d}\to \mathbb{R}$ is called $L$-periodic if $\phi(x+Ly) = \phi(x)$ for every $y \in \mathbb{Z}^{d}$. Let $\mathcal{F}_{per}$ be the set of all $L$-periodic fields, so that $\mathcal{F}_{per} \subset \mathbb{R}^{\mathbb{Z}^{d}}$. Now, using the same expression in (\ref{1}) we can define 'infinite volume' Laplacians $\Delta_{per}$ and $\Delta$ on $\mathcal{F}_{per}$ and $l^{2}(\mathbb{Z}^{d}):=\{\phi:\mathbb{R}^{d}\to \mathbb{R}:\hspace{0.1cm} \sum_{x \in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$, respectivelly. In addition, if $\phi \in \mathcal{F}_{per}$, its restriction $\phi|_{\Lambda_{L}}$ can be viewed as an element of $\Lambda_{L}$, and the action of $\Lambda_{per}$ to $\phi|_{\Lambda_{L}}$ is equivalent to the action of $\Delta_{L}$ to $\phi|_{\Lambda_{L}}$.

The Hamiltonian for the GFF in the lattice $\Lambda_{L}$ is given by: \begin{eqnarray} H_{\Lambda_{L}}(\phi) = \frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{\Lambda} \tag{2}\label{2} \end{eqnarray}

The first step is to define finite volume measures on $\mathbb{R}^{\mathbb{Z}^{d}}$. For each finite $\Lambda \subset \mathbb{Z}^{d}$, let $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ be the matrix with entries $C_{xy} := (-\Delta_{per}+m^{2})_{xy}$, where $(-\Delta_{per}+m^{2})_{xy}$ is the Kernel of $-\Delta_{per}+m^{2}$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. Because the Kernel of $-\Delta_{per}+m^{2}$ is the same as $-\Delta_{L}+m^{2}$, each $C_{\Lambda}$ is a positive-definite matrix and, thus, define a Gaussian measure $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Because this family of Gaussian measures $\mu_{\Lambda}$ is consistent, we can use Kolmogorov's Extension Theorem to obtain a Gaussian measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$ (with the product $\sigma$-algebra). Moreover, we can also obtain a family $\{f_{\alpha}\}_{\alpha \in \mathbb{Z}^{d}}$ of random variables such that $\mu_{\Lambda}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \Lambda}$. As it turns out, it is possible to prove that these random variables are given by $f_{\alpha}(\phi) = \phi(\alpha)$, $\alpha \in \mathbb{Z}^{d}$. In summary, if $A$ is a Borel set in $\mathbb{R}^{\Lambda}$, we must have: \begin{eqnarray} \mu_{\Lambda_{L}}(A) = \frac{1}{Z}\int_{A}e^{-\frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L}}d\nu_{L}(\phi) = \mu(A\times \mathbb{R}^{\mathbb{Z}^{d}\setminus \Lambda}) \tag{3}\label{3} \end{eqnarray} with $\nu_{L}$ being the Lebesgue measure on $\mathbb{R}^{\Lambda}$. The Gaussian measure $\mu$ is our a priori measure on $\mathbb{R}^{\mathbb{Z}^{d}}$ and, by (\ref{3}), it can be interpreted as a finite volume over $\mathbb{R}^{\Lambda}$.

Now, let $G(x,y)$ the Green function of $-\Delta+m^{2}$ in $l^{2}(\mathbb{Z}^{d})$. If $s_{m}:=\{\psi \in \mathbb{R}^{\mathbb{N}}:\hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\psi_{n}|^{2}\equiv ||\psi||_{m}^{2}\infty\}$, define $s:=\bigcup_{m\in \mathbb{Z}}$ and $s':=\bigcap_{m\in \mathbb{Z}}s_{m}$. Let u $K=(K_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' given by $K_{xy}:= G(x,y)$. If we order $\mathbb{Z}^{d}$, we can consider $K$ to be an 'infinite matrix' $K = (K_{ij})_{i,j \in \mathbb{N}}$. Now, define the following map: \begin{eqnarray} s \times s \ni (\psi,\varphi) \mapsto (\psi,K\varphi):= \sum_{i,j=1}^{\infty}\psi_{i}K_{ij}\varphi_{j}\tag{4}\label{4} \end{eqnarray} Let $W(\phi):= e^{-\frac{1}{2}(\phi,K\phi)}$. It is possible to prove that $W$ is a positive-definite function on $s$, so that, by Minlos Theorem, there exists a Gaussian measure $\tilde{\mu}_{K}$ on $s'$ such that $W$ is the Fourier transform of $\tilde{\mu}_{K}$.

[Question] I would like to establish a connection between $\mu$ and $\tilde{\mu}_{K}$ (where, here, $\mu$ is the restriction of $\mu$ to $s'\subset \mathbb{R}^{\mathbb{Z}^{d}}$ with its natural $\sigma$-algebra). It seems to me that $\tilde{\mu}_{K}$ is the infinite volume measure of $\mu$, in the sense that when we take $L\to \infty$ one should obtain $\tilde{\mu}_{K}$. In other words, $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure obtained by taking the thermodynamic limit of the measures $\mu_{L}$. But, if I'm not mistaken, to prove that $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure, I should prove that: \begin{eqnarray} \lim_{L\to \infty}\int f(\phi)d\mu_{L}(\phi) = \int f(\phi)d\tilde{\mu}_{K}(\phi) \tag{5}\label{5} \end{eqnarray} i.e. I should prove that $\mu$ converges weakly to $\tilde{\mu}_{K}$. And I don't know how to prove it.

Note: The above setup is a result of some of my thoughts about the problem during the last few days. I've been using a lot of different references and each one work the problem in a different way, with different notations and different objectives, so I'm trying to put it all together in one big picture. It is possible that my conclusions are not entirely correct or I can be going in the wrong direction, idk. But any help would be appreciated.

Note 2: I think it could be easier to prove a more particular limit such as $\lim_{L\to \infty}\int\phi(x)\phi(y)d\mu_{L}(\phi) = \int \phi(x)\phi(y)d\tilde{\mu}_{K}(\phi)$ and this would be enough to establish the existence of infinite volume correlation functions, which is one of the most important quantities in statistical mechanics. However, I don't think I could conclude that $\tilde{\mu}_{K}$ is the associate infinite volume Gibbs measure for the system just from this limit. Don't I need to prove it for a general $f$ as above?

$\endgroup$
  • $\begingroup$ I haven't read carefully, but it seems to me the $\mu_{\Lambda}$ are not consistent, so you can't use Kolmogorov. $\endgroup$ – Abdelmalek Abdesselam Jun 2 at 21:36
  • $\begingroup$ The question is not well formulated, but reading between the lines, it seems to me you are trying to obtain the infinite volume measure on $s'$ given by Bochner-Minlos as the weak limit when $L\rightarrow\infty$ of probability measures on $s'$ constructed from a finite torus of linear size $L$. Correct? If so, basically at the end of a long day (you need Levy continuity Thm on $s'$) it boils down to writing the integral for the infinite volume covariance as a limit of Riemann sums. You need also summable uniform in $L$ bounds for this limit, in order to smear with test functions in $s$. $\endgroup$ – Abdelmalek Abdesselam Jun 2 at 21:52
  • $\begingroup$ @AbdelmalekAbdesselam yes, this is what I'm trying to prove. I'm following Kupiainen's lecture notes on renormalization group (2014). As far as I understand, he difines the Gaussian measure on $s'$ and he treats the infinite volume Gibbs measure as this measure. But this measure is defined by means of the function $W$ I defined in my post, and this could be done directly in the infinite volume limit in my opinion. So I don't get how this construction connects with the thermodynamic limit and the finite volume measures. $\endgroup$ – IamWill Jun 2 at 23:01
  • $\begingroup$ Also, he writes down the Green function as a limit of Riemann sums, not the Kernel of $-\Delta+m^{2}$ itself. Well, as you may notice I'm completely lost here. $\endgroup$ – IamWill Jun 2 at 23:03
  • $\begingroup$ I don't have Antti's notes in front of me but I guess this is the calculation mentioned in my comment. I would have to write all this in a detailed answer but don't have time right now. $\endgroup$ – Abdelmalek Abdesselam Jun 2 at 23:11
3
$\begingroup$

For $x\in\mathbb{Z}^d$ I will denote by $\bar{x}$ the corresponding equivalence class in the discrete finite torus $\Lambda_{L}=\mathbb{Z}^d/L\mathbb{Z}^d$. I will view a field $\phi\in\mathbb{R}^{\Lambda_L}$ as a column vector with components $\phi(\bar{x})$ indexed by $\bar{x}\in\Lambda_L$. The discrete Laplacian $\Delta_L$ then acts by $$ (\Delta_L\phi)(\bar{x})=\sum_{j=1}^{d}\left[ -2\phi(\bar{x})+\phi(\overline{x+e_j})+\phi(\overline{x-e_j}) \right]\ . $$ Now take the column vectors $$ u_k(\bar{x})=\frac{1}{L^{\frac{d}{2}}} e^{\frac{2i\pi k\cdot x}{L}} $$ for $k=(k_1,\ldots,k_d)\in\{0,1,\ldots,L-1\}^d$. They give an orthonormal basis in $\mathbb{C}^{\Lambda_L}$ which diagonalizes the Laplacian matrix. Let $C_L=(-\Delta_L+m^2{\rm I})^{-1}$ and denote its matrix elements by $C_L(\bar{x},\bar{y})$. We then have, for all $x,y\in\mathbb{Z}^d$,

$$ C_L(\bar{x},\bar{y})= \frac{1}{L^d}\sum_{k\in\{0,1,\ldots,L-1\}^d} \frac{e^{\frac{2i\pi k\cdot(x-y)}{L}}}{m^2+ 2\sum_{j=1}^{d}\left[1-\cos\left(\frac{2\pi k_j}{L}\right)\right]} =:G_L(x,y) $$

where we used the formula to define $G_L$ on $\mathbb{Z}^d\times\mathbb{Z}^d$. Because we assumed $m>0$, we have the trivial uniform bound in $L$ saying $$ |G_L(x,y)|\le \frac{1}{m^2}\ . $$ Now let $\nu_L$ denote the centered Gaussian probability measure on $\mathbb{R}^{\Lambda_L}$ with covariance matrix $C_L$. We also define an injective continuous linear map $$ \tau_L:\mathbb{R}^{\Lambda_L}\longrightarrow s'(\mathbb{Z}^d) $$ which sends $\phi\in\mathbb{R}^{\Lambda_L}$ to $\psi\in s'(\mathbb{Z}^d)$ defined by $\psi(x)=\phi(\bar{x})$ for all $x\in\mathbb{Z}^d$. Of course $\mathbb{R}^{\Lambda_L}$ has its usual finite-dimensional space topology, whereas $s'(\mathbb{Z}^d)$ is given the strong topology and the resulting Borel $\sigma$-algebra.

As I explained in my answer to the previous MO question we can use such a map to push forward probability measures. We thus go ahead and define $\mu_L=(\tau_L)_{\ast}\nu_L$ which is a Borel probability measure on $s'(\mathbb{Z}^d)$.

Now we switch gears and consider the Green's function $G_{\infty}(x,y)$ for $-\Delta+m^2$ on $\mathbb{Z}^d$. More explicitly, $$ G_{\infty}(x,y)=\frac{1}{(2\pi)^d} \int_{[0,2\pi]^d}d^d\xi\ \frac{e^{i\xi\cdot(x-y)}}{m^2+ 2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)}\ . $$ The function $$ \begin{array}{crcl} W_{\infty}: & s(\mathbb{Z}^d) & \longrightarrow & \mathbb{C} \\ & f & \longmapsto & \exp\left( -\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y) \right) \end{array} $$ satisfies all the hypotheses of the Bochner-Minlos Theorem for $s'(\mathbb{Z}^d)$. Therefore it is the characteristic function of a Gaussian Borel probability measure $\mu_{\infty}$ on $s'(\mathbb{Z}^d)$.

Finally after all these preliminaries we can state the main result which the OP asked for.

Theorem: When $L\rightarrow\infty$, the measure $\mu_L$ converges weakly to $\mu_{\infty}$.

The proof uses the Lévy Continuity Theorem for $s'(\mathbb{Z}^d)$ which is due to Xavier Fernique. One only has to prove that for all discrete test function $f\in s(\mathbb{Z}^d)$, $$ \lim\limits_{L\rightarrow \infty} W_L(f)\ =\ W_{\infty}(f) $$ where $W_L$ is the characteristic function of the measure $\mu_L$. By definition, we have $$ W_L(f)=\int_{s'(\mathbb{Z}^d)} \exp\left[ i\sum_{x\in\mathbb{Z}^d}f(x)\psi(x) \right]\ d[(\tau_L)_{\ast}\nu_L](\psi)\ . $$ By the abstract change of variable theorem, $$ W_L(f)=\int_{\mathbb{R}^{\Lambda_L}} \exp\left[ i\sum_{x\in\mathbb{Z}^d}f(x)\phi(\bar{x}) \right]\ d\nu_L(\phi) $$ $$ =\exp\left( -\frac{1}{2}\sum_{\bar{x},\bar{y}\in\Lambda_L} \tilde{f}(\bar{x})\ C_L(\bar{x},\bar{y})\ \tilde{f}(\bar{y}) \right) $$ where we introduced the notation $\tilde{f}(\bar{x})=\sum_{z\in\mathbb{Z}^d}f(x+Lz)$. Hence $$ W_L(f)= \exp\left( -\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y) \right)\ . $$ Since the function $$ \xi\longmapsto \frac{1}{(2\pi)^d} \frac{e^{i\xi\cdot(x-y)}}{m^2+ 2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)} $$ is continuous on the compact $[0,2\pi]^d$ and therefore uniformly continuous, we have that, for all fixed $x,y\in\mathbb{Z}^d$, the Riemann sums $G_L(x,y)$ converge to the integral $G_{\infty}(x,y)$. Because of our previous uniform bound on $G_L(x,y)$ and the fast decay of $f$, we can apply the discrete Dominated Convergence Theorem in order to deduce $$ \lim\limits_{L\rightarrow\infty} \sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y)\ =\ \sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y)\ . $$ As a result $\lim_{L\rightarrow \infty}W_L(f)=W_{\infty}(f)$ and we are done.

Note that we proved weak convergence which as usual means $$ \lim\limits_{L\rightarrow \infty} \int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_L(\psi)\ =\ \int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_{\infty}(\psi) $$ for all bounded continuous functions $F$ on $s'(\mathbb{Z}^d)$. One also has convergence for correlation functions or moments because of the Isserlis-Wick Theorem relating higher moments to the second moment and the previous argument where we explicitly treated the convergence for second moments. Finally, note that the extension map $\tau_L$ used here is the periodization map, but there are lots of other choices which work equally well. A good exercise, is to construct the massive free field in the continuum, i.e., in $\mathscr{S}'(\mathbb{R}^d)$, as the weak limit of suitably rescaled lattice fields on $\mathbb{Z}^d$ with a mass adjusted as a function of the (rescaled) lattice spacing.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It couldn't be more clear! Perfect answer again! Thank you so much for always saving me! $\endgroup$ – IamWill Jun 3 at 17:07
  • $\begingroup$ Thanks. To better understand the subject, see also the other things I have written in particular about the topology of spaces like $S,S',D,D',s,s',s_0,s'_0$. Presentations in mathematical physics (even the book by Glimm and Jaffe) typically do not give a proper treatment of these issues. For example, the $\sigma$-algebra used is often defined as a cylinder $\sigma$-algebra instead of emphasizing it is a Borel $\sigma$-algebra for the canonical topology these spaces must be endowed with and which makes them as good as finite dimensional spaces. $\endgroup$ – Abdelmalek Abdesselam Jun 3 at 17:13
  • $\begingroup$ You can start here: math.stackexchange.com/questions/2623515/… $\endgroup$ – Abdelmalek Abdesselam Jun 3 at 17:15
  • $\begingroup$ Thanks! I'm always studying your previous answers and they always help me a lot. Out of curiosity: why do some references prefer to address these problems as Gaussian processes? Is it just a matter of taste? I think this is where my confusion begun. If you check, say, Velenik's book, this very same problem is addresses in terms of random variables, Gaussian processes and Kolmogorov. I prefer to deal with it as you did, but these other references treat these problems in a way that they seem to be completely different. $\endgroup$ – IamWill Jun 3 at 17:23
  • $\begingroup$ I think it's just due to widespread discomfort with topological vector spaces. Indeed, in this point of view one would talk about a process, i.e., the collection of real-valued random variables $\psi(x)$, indexed by $x\in\mathbb{Z}^d$ on a fixed probability space, e.g., $s'(\mathbb{Z}^d)$ or $\mathbb{R}^{\mathbb{Z}^d}$. I prefer to talk about a single random variable $\psi$ which is $s'(\mathbb{Z}^d)$-valued. Strictly speaking my point of view is more powerful, because from a random $\psi$ you trivially get a process by post-composition with the coordinate projection functions... $\endgroup$ – Abdelmalek Abdesselam Jun 3 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.