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As I understand it, a GFF is a generalisation of Brownian motion to dimensions greater than one. However, they seem like very different objects. Brownian motion is just a continuous function (even though it is nowhere differentiable). By contrast, the Gaussian free field is not a function but only a distribution (it does not have well defined values at points, but only under integrals).

What makes these two objects 'the same'? and why does moving from dimension one to two lead to such different behaviour?

prdnr
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2 Answers2

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I would argue that the GFF is not so much a generalization of Brownian motion, but rather that the one-dimensional GFF just happens to be Brownian motion.

To explore this further, let $\varphi = \{\langle \varphi,f\rangle\}_{f\in H_0^1(\Omega)}$ be the Gaussian Free Field in some (smooth) domain $\Omega\subset \mathbb R^d$ with zero boundary conditions. Specifically, this is a mean zero Gaussian field indexed by $H_0^1(\Omega)$ such that

$$\operatorname{Cov}\Big(\langle\varphi,f\rangle,\langle\varphi,g\rangle\Big) = \int_\Omega \nabla f\cdot \nabla g \,dx =: ( f \,|\, g )$$

for all $f,g\in H_0^1(\Omega)$. The notation $\langle \varphi,\cdot\rangle$ is intentionally chosen due to its usual association with duality and linear functionals, and indeed the map $f\mapsto \langle \varphi,f\rangle$ is linear. Since Hilbert spaces are self-dual, one might reasonably try to associate $\varphi$ to a random element of $ H_0^1(\Omega)$, i.e. find a $H_0^1(\Omega)$-valued random variable $\phi$ such that $\langle \varphi,f\rangle = (\phi\,|\,f)$ for all $f\in H_0^1(\Omega)$. Indeed, if $\{\varphi_k\}_k$ is an orthonormal basis of $H_0^1(\Omega)$ and $\{\xi_k\}_k$ is a sequence of i.i.d. $\mathcal N(0,1)$ random variables, then

$$ \phi:=\sum_{k=1}^\infty \xi_k\varphi_k $$

satisfies the requirements of the GFF: if $f\in H_0^1(\Omega)$, $(\phi\,|\,f) = \sum_k \xi_k(\varphi_k\,|\,f)$, and so

$$\operatorname{Cov}\Big((\phi\,|\,f),(\phi\,|\,g)\Big) = \sum_k(\varphi_k\,|\,f)(\varphi_k\,|\,g) = \left(\sum_k(\varphi_k\,|\,f)\varphi_k\,\bigg|\,\sum_k(\varphi_k\,|\,g)\varphi_k\right) = (f\,|\,g).$$

The problem? The series in the definition of $\phi$ need not converge! Indeed, one can show that if $d \ge 2$, the series does not converge, and we cannot hope to represent $\varphi$ as a function.

However, $d=1$ is a special case. This is related to the unusually regular structure of Sobolev spaces in one dimension; in particular, $H^1(a,b)$ is simply the set of absolutely continuous functions on $(a,b)$ whose derivative is square-integrable, a property which does not carry into higher dimensions. To find that (a version of) GFF can be realized as Brownian motion in this case, observe that, if $\{\psi_k\}$ is an orthonormal basis of $L^2(0,1)$, and

$$ \phi_k(t) := \int_0^t \psi_k(s) ds, $$

then $\{\phi_k\}$ is an orthonormal basis of $V:=\{f\in H^1(0,1):f(0)=0\}$, the series $$ \phi(t):=\sum_{k=1}^\infty \xi_k\phi_k(t) $$ converges almost surely for every $t$, and $\phi$ is a Brownian motion. (In fact, if rather than an arbitrary basis $\{\psi_k\}$ we actually choose one (usually the Haar basis is used), then the convergence is uniform in $t$, so $\phi$ is an almost surely continuous Brownian motion.) Moreover, $\phi$ is precisely the realization of the GFF we were looking for earlier, with the exception that we are looking at the space $V$ instead of $H_0^1(0,1)$. This just corresponds to changing the boundary conditions to $f(0)=0$ instead of $f(0)=f(1)=0$; the latter will give you a Brownian bridge instead.

Jason
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  • A very nice answer! – zhoraster Apr 17 '20 at 10:37
  • Is it possible there is a typo in the first displayed equation defining the covariance? I am comparing with https://en.wikipedia.org/wiki/Gaussian_free_field#The_continuum_field and am missing the Greens function appearing in the covariance. On the LHS you have no gradients. The second displayed equation with covariance seems OK. Maybe I am missing something. – PPR Jan 11 '21 at 10:07
  • @PPR Sorry, I'm a little unsure what you mean. The first displayed equation is defining the GFF via its covariance, which indeed contains gradient terms. In fact, aside from notation, what I have written is the same as in the Wikipedia article you linked. – Jason Jan 29 '21 at 17:26
  • @Jason: in Wikipedia, in the definition of the covariance, both the LHS and the RHS side have the gradient-inner product, i.e., the inner product on the Sobolev space. On the other hand, in your equaiton, only the RHS of the definition of the covariance has this special inner product. Intuitively I think of the covariance of two factors of the field as yielding the Greens function. – PPR Jan 29 '21 at 22:53
  • @PPR Oh, I see the issue. The notation $\langle\varphi,f\rangle$ is purely symbolic - the GFF is simply a Gaussian field indexed by $H_0^1(\Omega)$. If you prefer, you could write $\Phi = {\varphi_f}{f\in H_0^1(\Omega)}$, where each $\varphi_f$ is a Gaussian random variable. I have intentionally used the different notation $(\cdot,|,\cdot)$ to refer to the inner product on $H_0^1(\Omega)$, since - and this is kinda the point of the post - there is no random element $\phi$ of $H_0^1(\Omega)$ such that ${(\phi,|,f)}{f\in H_0^1(\Omega)}$ is a GFF for $d>1$. – Jason Feb 04 '21 at 21:43
  • Wikipedia, on the other hand, does not use different notation, which in my opinion is not ideal. In fact, this is a perfect example of how it can lead to confusion! – Jason Feb 04 '21 at 21:43
  • @Jason I guess in your notation I would have expected the first displayed equation to rather be $ \operatorname{Cov}\Big(\langle\varphi,f\rangle,\langle\varphi,g\rangle\Big) = \langle f,(-\Delta)^{-1}g\rangle $. – PPR Feb 05 '21 at 03:53
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    No, I think you are still missing my point. In my notation, $\langle\varphi,\cdot\rangle$ is not an inner product. The inner product $(\cdot,|,\cdot)$ on $H_0^1(\Omega)$ is defined by $(f,|,g):=\int_\Omega\nabla f\cdot\nabla g dx$. I never use the standard $L^2$ inner product. – Jason Feb 05 '21 at 14:14
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A Gaussian free field, in any dimension, is a Gaussian field with covariance given by the Green's function, $G(x,y)$ satisfying $\Delta G(x,\cdot)=\delta(\cdot-x)$. In dimension $1$, $G(x,\cdot)$ is a continuous function - in fact, just a piecewise linear one. In higher dimensions, $G(x,\cdot)$ has a singularity at $x$. Therefore, the value of a field at a point cannot be well defined - on one hand, it would have to be Gaussian, but on the other hand, it must have infinite variance.

Kostya_I
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