Let $\Lambda \subset \mathbb{Z}^{d}$ be finite and fixed and consider $\mathbb{R}^{|\Lambda|}$ be the vector space of all sequences $\varphi = (\varphi_{x})_{x\in \Lambda}$. We equip $\mathbb{R}^{|\Lambda|}$ with its Borel $\sigma$-algebra $\mathbb{B}(\mathbb{R}^{|\Lambda|})$. We denote by $\nu$ the Lebesgue measure restricted to $\mathbb{B}(\mathbb{R}^{|\Lambda|})$.

Now, in statistical mechanics, one usually consider an action $S_{\Lambda}: \mathbb{R}^{|\Lambda|}\to \mathbb{R}$ which is assumed to be a measurable function. This action gives rise to a finite volume Gibbs measure $\mu_{\Lambda}$, defined by means of its density with respect to $\nu$: \begin{eqnarray} d\mu_{\Lambda}(\varphi) := \mbox{const.} e^{-S_{\Lambda}(\varphi)}d\nu(\varphi) \tag{1}\label{1} \end{eqnarray} where the $\mbox{const.}$ term in (\ref{1}) is a normalization factor so that $\mu_{\Lambda}$ is a probability measure on $\mathbb{R}^{|\Lambda|}$.

From the physics point of view, we are interested in studying the behavior of the system in the infinite volume limit $\Lambda \nearrow \mathbb{Z}^{d}$. Thus, we have to first ensure that the infinite volume measure $\mu_{\mathbb{Z}^{d}} \equiv \mu$ exists in some sense. We usually take this limit as the weak-limit: \begin{eqnarray} \int d\mu e^{i\langle f, \varphi\rangle} := \lim_{n\to \infty}\int d\mu_{\Lambda_{n}}e^{i\langle f, \varphi\rangle} \tag{2}\label{2} \end{eqnarray} provided this limit exists. Here, $\Lambda_{n}$ is a sequence of increasing sets converging to $\mathbb{Z}^{d}$.

A particular case of the above scenario is when the action $S_{\Lambda}$ comes from a strictly positive quadratic form $Q_{\Lambda}: \mathbb{R}^{|\Lambda|}\times \mathbb{R}^{|\Lambda|} \to \mathbb{R}$. If we set $S_{\Lambda}(\varphi) = Q_{\Lambda}(\varphi, \varphi)$, the measure (\ref{1}) becomes Gaussian.

Now, it is a known fact that Gaussian measures are consistent in the sense of Kolmogorov Theorem, so that this very same Theorem implies the existence of a measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. (*)

My Question is whether we can interpret this measure $\mu$ obtained using Kolmogorov's Theorem as a weak limit of the form (\ref{2}) in some sense. My point here is: it seems legit to consider this $\mu$ as a infinite volume measure, but the path taken to define this limit was not by means of a limit such as (\ref{2}). Besides, the inner product $\langle f,\varphi \rangle$ only makes sense on $\mathbb{R}^{\mathbb{Z}^{d}}$ if we restrict it to some subspace, say $\mathcal{l}^{2}(\mathbb{Z}^{d})$. How to connect these two scenarios?

EDIT: Let me elaborate more on (*). Simon's book states Kolmogorov's Theorem as follows. Theorem [Kolmogorov]: Let $\mathcal{I}$ be a countable set and let a probability measure $\mu_{|I|}$ on $\mathbb{R}^{|I|}$ be given for each finite set $I\subset \mathcal{I}$, so that the family of $\mu_{I}$'s is consistent (i.e. $\mu_{I}(A) = \mu_{I'}(A\times \mathbb{R}^{|I'|-|I|})$ if $|I'|\ge |I|$). Then there is a probability measure space $(X, \mathcal{F}, \mu)$ and random variables $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$ so that $\mu_{I}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \mathcal{I}}$.

Now, the proof of this theorem shows that $X$ is actually $\dot{\mathbb{R}}^{|\mathcal{I}|}$ where $\dot{\mathbb{R}}:=\mathbb{R}\cup \{+\infty\}$ is a compactification of $\mathbb{R}$. Thus, if we take $\mathcal{I}=\mathbb{Z}^{d}$, and for each finite $\Lambda \subset \mathbb{Z}^{d}$ the associate $\mu_{\Lambda}$ to be a Gaussian measure (say, nondegenerate and associated to some positive-definite matrix $C_{\Lambda}$), the above Theorem proves the existence of a (Gaussian) measure on $\mathbb{R}^{\mathbb{Z}^{d}}$.

EDIT 2: The measures $\mu_{\Lambda_{n}}$ are, in fact, defined in $\mathbb{R}^{\mathbb{Z}^{d}}$ with the product topology. It can be constructed from (\ref{1}) by taking, instead of $S_{\Lambda}$, an action $S:\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ with free boundary conditions outside each $\Lambda_{n}$.

  • 2
    $\begingroup$ The question does not quite make sense yet. To talk about weak convergence, you need your measures to be on a fixed topological space. I guess here $\mathbb{R}^{\mathbb{Z}^d}$ with the product topology. So you have to decide what $\varphi$ is doing outside $\Lambda_n$, for each finite $n$. Also, your $\mathcal{Q}$ should be a $\mathcal{Q}_{\Lambda}$. You mention an alternate direct construction of $\mu$ in infinite volume. Explain precisely what you mean. Does it come from a $\mathcal{Q}$ in infinite volume? $\endgroup$ Commented Feb 18, 2020 at 23:34
  • $\begingroup$ related mathoverflow.net/questions/277678/… $\endgroup$ Commented Feb 18, 2020 at 23:39
  • $\begingroup$ Edited! Thanks!! $\endgroup$ Commented Feb 19, 2020 at 3:18
  • $\begingroup$ What do you mean by free boundary conditions? $\endgroup$ Commented Feb 19, 2020 at 18:47

1 Answer 1


Not quite an answer (because I still don't understand the question) but too long for a comment.

Take the following very simple example where the $\varphi_{x}$ are iid $\mathscr{N}(0,1)$ random variables. There is no problem constructing their joint law $\mu$ as a probability measure on $\mathbb{R}^{\mathbb{Z}^d}$ with the product topology. The way to do that is using the Daniell-Kolmogorov Extension Theorem. Now what about the physicists' beloved notion of action? Well, it just does not make sense.

In this case, a physicist may want to write formally $$ d\mu(\varphi)=\frac{1}{Z}\ e^{-S(\varphi)}\ \prod_{x\in\mathbb{Z}^{d}} d\varphi_x $$ with the action $S(\varphi)=\frac{1}{2}Q(\varphi,\varphi)$, (BTW please put the 1/2) where $$ Q(\varphi,\varphi)=\sum_{x\in\mathbb{Z}^d} \phi_x^2\ . $$ Now if the random variable $\varphi\in\mathbb{R}^{\mathbb{Z}^d}$ is sampled according to the (well defined) probability measure $\mu$, then almost surely $$ Q(\varphi,\varphi)=\infty $$ as can be shown say by Kolmogorov's Three Series Theorem.

This is all to say the question would make more sense if, rather than talking about actions and the matrix $Q$, it talked instead about the covariance matrix/bilinear form $C=Q^{-1}$.

  • $\begingroup$ That's a very nice comment. The point is: how to systematically define what we mean by a infinite volume limit, in order to study the system in the thermodynamic limit? The Gaussian measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$ constructed using Kolmogorov's Theorem is a legit infinite volume limit but, in general, the infinite volume measure is assumed to be that satisfying (2). It then seems that the limit is taken conveniently, depending on the model rather than a systematic approach. $\endgroup$ Commented Feb 19, 2020 at 16:49
  • $\begingroup$ although one could, one does not construct an infinite volume Gaussian measure using (2) but directly. (2) is used for constructing more physically interesting non-Gaussian measures. $\endgroup$ Commented Feb 19, 2020 at 18:45

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