Let me discuss two possible constructions of Gaussian measures on infinite dimensional spaces. Consider the Hilbert space $l^{2}(\mathbb{Z}^{d}) := \{\psi: \mathbb{Z}^{d}\to \mathbb{R}: \hspace{0.1cm} \sum_{x\in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$ with inner product $\langle \psi, \varphi\rangle_{l^{2}}:= \sum_{x\in \mathbb{Z}^{d}}\overline{\psi(x)}\varphi(x)$. We can introduce in $l^{2}(\mathbb{Z}^{d})$ the discrete Laplacian as the linear operator: $$(\Delta \psi)(x) := \sum_{k=1}^{d}[-2\psi(x)+\psi(x+e_{k})+\psi(x-e_{k})]$$ where $\{e_{1},...,e_{d}\}$ is the canonical basis of $\mathbb{R}^{d}$. Because $(-\Delta+m^{2})$ has a resolvent for every $m\in \mathbb{R}$, we can consider its inverse $(-\Delta+m^{2})^{-1}$. It's integral Kernel or Green's function $G(x,y)$ is given by: \begin{eqnarray} G(x,y) = \frac{1}{(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}d^{d}p \frac{1}{\lambda_{p}+m^{2}}e^{ip\cdot(x-y)} \tag{1}\label{1} \end{eqnarray} where $p\cdot (x-y) = \sum_{i=1}^{d}p_{i}(x_{i}-y_{i})$ and $\lambda_{p} :=2\sum_{i=1}^{d}(1-\cos p_{i})$ is the eigenvalue of $-\Delta$ associated to its eigenvector $e^{ip\cdot x}$.

[First Approach] If $m \in \mathbb{Z}$, let $s_{m} :=\{\phi\in \mathbb{R}^{\mathbb{N}}: \hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\phi_{n}|^{2} \equiv ||\phi||_{m}^{2}<+\infty\}$, $s:=\bigcap_{m\in \mathbb{Z}}s_{m}$ and $s':=\bigcup_{m\in \mathbb{Z}}s_{m}$. Note that $s$ is a Fréchet space when its topology is given by the family of semi-norms $||\cdot||_{m}$ and $s'$ is the dual space of $s$ if $l_{\psi}$ is a continuous linear map on $s$ with $l_{\psi}(\phi) =( \psi,\phi) := \sum_{n=1}^{\infty}\psi_{n}\phi_{n}$. Let $C=(C_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' with entries $C_{xy}:= G(x,y)$. We can consider $C_{xy}$ to be a matrix $C=(C_{ij})_{i,j \in \mathbb{N}}$ by enumerating $\mathbb{Z}^{d}$. Now, let us define the bilinear map: \begin{eqnarray} s\times s \ni (\phi, \psi) \mapsto \sum_{n=1}^{\infty}\phi_{i}C_{ij}\psi_{j} \equiv (\phi, C\psi) \tag{2}\label{2} \end{eqnarray} Thus, $\phi \mapsto (\phi, C\phi)$ is a quadratic form and we can define: $$W_{C}:=e^{-\frac{1}{2}(\phi,C\phi)}$$ Using Minlos' Theorem for $s$, there exists a Gaussian measure $d\mu_{C}$ on $s'$ (or $\mathbb{R}^{\mathbb{Z}^{d}})$ satisfying: \begin{eqnarray} W_{C}(\psi) = \int_{s'}e^{i(\psi,\phi)}d\mu_{C}(\phi) \tag{3}\label{3} \end{eqnarray}

[Second Approach] For each finite $\Lambda \subset \mathbb{Z}^{d}$, set $C_{\Lambda}$ to be the matrix $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ where $C_{xy}$ are defined as before. Then, these matrices $C_{\Lambda}$ are all positive-definite, so that they define Gaussian measures $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Besides, these are compatible in the sense that if $\Lambda \subset \Lambda'$ are both finite and $E$ is a Borel set in $\mathbb{R}^{\Lambda}$ then $\mu_{\Lambda}(E) = \mu_{\Lambda'}(E\times \mathbb{R}^{\Lambda'\setminus\Lambda})$. By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_{C}$ with covariance $C$ on $l^{2}(\mathbb{Z}^{d})$ which is compatible with $\mu_{\Lambda}$ for every finite $\Lambda$.

Now, It seems that these two constructions occur when the so-called thermodynamics limit is taken in QFT and Statistical Mechanics. Both Gaussian measures $\mu_{C}$ and $\nu_{C}$ are measures on $\mathbb{R}^{\mathbb{Z}^{d}}\cong \mathbb{R}^{\mathbb{N}}$. I don't know if this is true but I'd expect these two constructions to be equivalent in some sense, but it is not obvious to me that they are. For instance, the first construction provides a Gaussian measure on $s'$ and the second one on $l^{2}(\mathbb{Z}^{d})$. Are there any relation between these two measures? Are they equal? Maybe the Fourier transform of $\nu_{C}$ would give $W_{C}$, proving these two are the same. Anyway, I'm very lost here and any help would be appreciated.


The source of the confusion is not saying explicitly what are the sets and $\sigma$-algebras the measures are supposed to be on. For example, a sentence like ''By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_C$ with covariance $C$ on $l^2(\mathbb{Z}^d)$ which is compatible with $\mu_\Lambda$ for every finite $\mu_\Lambda$.'' is asking for trouble because it seems to say the measure $\nu_C$ is on the set $l^2(\mathbb{Z}^d)$, which is false.

Let's go back to basics. A measurable space $(\Omega,\mathcal{F})$ is a set $\Omega$ equipped with a $\sigma$-algebra $\mathcal{F}$. A measure $\mu$ on the measurable space $(\Omega,\mathcal{F})$ is a map from $\mathcal{F}$ to $[0,\infty]$ satisfying the usual axioms. From now on I will only talk about probability measures. For best behavior, the $\Omega$ should be a (nice) topological space and $\mathcal{F}$ should be the Borel $\sigma$-algebra for that topology. Suppose one has two topological spaces $X,Y$ and a continuous injective map $\tau:X\rightarrow Y$. Then if $\mu$ is a measure on $(X,\mathcal{B}_X)$ where $\mathcal{B}_X$ is the Borel $\sigma$-algebra of $X$, then one can construct the direct image/push forward measure $\tau_{\ast}\mu$ on $(Y,\mathcal{B}_Y)$ by letting $$ \forall B\in\mathcal{B}_{Y},\ (\tau_{\ast}\mu)(B):=\mu(\tau^{-1}(B))\ . $$ This is well defined because a continuous map like $\tau$ is also $(\mathcal{B}_X,\mathcal{B}_Y)$-measurable. Technically speaking $\mu$ and $\tau_{\ast}\mu$ are different measures because they are on different spaces. However, one could argue that they are morally the same. For example, one might be given the measure $\tau_{\ast}\mu$ without knwing that it is of that form, and only later realize that it is and thus lives on the smaller set $\tau(X)$ inside $Y$.

The first construction:

Let $s(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of fast decay $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which $$ \forall k\in\mathbb{N}, ||f||_k:=\sup_{x\in\mathbb{Z}^d}\langle x\rangle^k|f_x|\ <\infty $$ where $\langle x\rangle=\sqrt{1+x_1^2+\cdots+x_d^2}$. Put on the vector space $s(\mathbb{Z}^d)$ the locally convex topology defined by the collection of seminorms $||\cdot||_k$, $k\ge 0$. The strong dual can be concretely realized as the space $s'(\mathbb{Z}^d)$ of multi-sequences of temperate growth. Namely, $s'(\mathbb{Z}^d)$ is the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of discrete fields $\phi=(\phi_x)_{x\in\mathbb{Z}^d}$ such that $$ \exists k\in\mathbb{N},\exists K\ge 0,\forall x\in\mathbb{Z}^d,\ |\phi_x|\le K\langle x\rangle^k\ . $$ The vector space $s'(\mathbb{Z}^d)$ is given the locally convex topology generated by the seminorms $||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$ where $\rho$ ranges over elements of $s(\mathbb{Z}^d)$ with nonnegative values.

The measure $\mu_C$ obtained via the Bochner-Minlos Theorem is a measure on $X=s'(\mathbb{Z}^d)$ with its Borel $\sigma$-algebra $\mathcal{B}_X$.

The second construction:

Let $s_0(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of finite support $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which $f_x=0$ outside some finite set $\Lambda\subset\mathbb{Z}^d$. Put on the vector space $s_0(\mathbb{Z}^d)$ the finest locally convex topology. Namely, this is the locally convex topology generated by the collection of all seminorms on $s_0(\mathbb{Z}^d)$ . Note that $s_0(\mathbb{Z}^d)\simeq \oplus_{x\in\mathbb{Z}^d}\mathbb{R}$. Let $s'_0(\mathbb{Z}^d)$ be the strong topological dual realized concretely as $\mathbb{R}^{\mathbb{Z}^d}$. One can also define the topology by the seminorms $||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$ where $\rho$ ranges over elements of $s_0(\mathbb{Z}^d)$ with nonnegative values. However, this is the same as the product topology for $s'_0(\mathbb{Z}^d)=\prod_{x\in\mathbb{Z}^d}\mathbb{R}$.

The measure $\nu_C$ constructed via the Daniell-Kolmogorov Extension Theorem is a measure on $Y=s'_0(\mathbb{Z}^d)=\mathbb{R}^{\mathbb{Z}^d}$ with its Borel $\sigma$-algebra for the product topology a.k.a strong dual topology.

The precise relation between the two measures:

We simply have $\nu_C=\tau_{\ast}\mu_C$ where $\tau$ is the continuous canonical injection due to $X=s'(\mathbb{Z}^d)$ being a subset of $Y=\mathbb{R}^{\mathbb{Z}^d}$.

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    $\begingroup$ Thank you SO much for this amazing answer! This was exactly what I was searching for! $\endgroup$ – IamWill May 31 '20 at 18:55
  • $\begingroup$ Let me see if I understand the application of your explanations. Suposse a field is $\phi: \mathbb{Z}^{d}\to \mathbb{R}^{d}$ so that the space of all fields is $\mathbb{R}^{\mathbb{Z}^{d}}$. For each $x \in \mathbb{Z}^{d}$ I can define random variables $f_{x}: \Omega=\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ by $f_{x}(\phi) :=\phi(x)$. Thus, I can calculate, e.g. correlations $\mathbb{E}_{\nu_{C}}[f_{x}f_{y}] = \mathbb{E}_{\mu_{C}}[\phi(x)\phi(y)]$ both ways. That's why these are "basically the same"? $\endgroup$ – IamWill May 31 '20 at 19:12
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    $\begingroup$ You are moving the goalposts again and replying to your question requires making corrections as a preliminary step. I suppose you mean $\phi$ is a function from $\mathbb{Z}^d\rightarrow\mathbb{R}$ and not $\mathbb{Z}^d\rightarrow\mathbb{R}^d$. So $\phi$ means an element of $Y=\mathbb{R}^{\mathbb{Z}^d}$. Then, technically $\mathbb{E}_{\mu_C}[\phi_x \phi_y]$ is not well defined because $\mu_C$ is a measure on $X$ and not $Y$. $\endgroup$ – Abdelmalek Abdesselam May 31 '20 at 19:51
  • $\begingroup$ I must think more about all this. I'll possibly post another question on applications of these ideas. It might be better. $\endgroup$ – IamWill May 31 '20 at 19:57
  • $\begingroup$ No need. All you have to do is say $\phi$ designates an element of the space of fields $X$ and not $Y$. Then the equation $\mathbb{E}_{\nu_C}[f_x f_y]=\mathbb{E}_{\mu_C}[\phi_x \phi_y]$ is correct. This is just a trivial consequence of the abstract change of variable theorem relating expectations for the direct image measure to expectations of the original measure. $\endgroup$ – Abdelmalek Abdesselam May 31 '20 at 20:32

I think that what you are looking for is the link between the white noise measure $\mu_C$ and the isonormal process indexed by $\ell^2(\mathbb{Z}^d)$ with covariance structure given by $C$. The white noise measure $\mu_C$ is a Gaussian measure on $s'$ so that for all $\varphi \in s$, $\langle ;\varphi\rangle_{s',s}$ is a centered Gaussian random variable with variance $\langle \varphi ; C \varphi\rangle$. By an approximation argument, you should be able to give some sense to $\langle ; f\rangle$ with $f \in \ell^2(\mathbb{Z}^d)$ so that it is a centered Gaussian random variable under $\mu_C$ with variance $\langle f;C f\rangle$. Now, your second construction gives rise to a Gaussian stochastic process indexed by $\mathbb{Z}^d$ with covariance structure given $C$. By re-indexing, each element $X_j$ of this Gaussian stochastic process admits the representation $\nu_{C}(e_j)$ where $e_j=(0,\dots,0,1,0,\dots)$. Now, again by approximation, you can extend $\nu_C$ to all $\ell^2(\mathbb{Z}^d)$ and it is completely defined, for all $f,g \in \ell^2(\mathbb{Z}^d)$, by $$ \mathbb{E}\left(\nu_{C}(f)\nu_{C}(g)\right)= \langle f;Cg\rangle ,$$ and $\mathbb{E}(\nu_C(f))=0$. Now, the link is clear and you have the following equality in law under $\mu_C$, for all $f \in \ell^2(\mathbb{Z}^d)$ $$\nu_c(f) = \langle ; f\rangle.$$ This is completely similar to the classical construction of the white noise probability measure on the space of tempered distributions on $\mathbb{R}$ ($S'(\mathbb{R})$) and the classical isonormal Gaussian process indexed by $L^2(\mathbb{R})$.

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    $\begingroup$ The underlying probability space is $s'$ then $\langle ;f\rangle$ is the random variable $X$ defined, for all $\omega \in s'$, by $$X(\omega) = \langle \omega; f \rangle$$. The ``difficulty" is that the bilinear pairing is defined for $\omega \in s'$ if $f$ belongs to $s$ and only $\mu_C$ a.e.if $f$ belongs to $\ell^2(\mathbb{Z}^d)$. $\endgroup$ – user69642 May 31 '20 at 14:29
  • $\begingroup$ I'm still having trouble understanding the origin of such differences. My main interest is statistical mechanics. I've been reading the section Gaussian Free Field (GFF) of two different books and the Hamiltonian on both books is the same. But when the time comes to discuss the thermodynamic limit, one book goes in the $l^{2}(\mathbb{Z}^{d})$ direction and the other goes in the $s'$ direction. But it seems that the target here is the same, but the constructions are differente and the results are different. Do you know why? $\endgroup$ – IamWill May 31 '20 at 15:05
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    $\begingroup$ I understand your question this way: what is the difference between white noise distribution theory and Malliavin calculus ? In the first theory, the white noise (derivative of Brownian motion) is the central object of interest whereas in the second it is Brownian motion. Both theories have several similarities but they are intrinsically different by nature. $\endgroup$ – user69642 May 31 '20 at 17:03
  • $\begingroup$ I see. I think this confusion of mine is worth a new post, to put it some context. Anyway, you've been really helpful! Thanks a lot! $\endgroup$ – IamWill May 31 '20 at 17:08

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