Why is the gaussian free field a distribution but Brownian motion is a function? 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?
 A: 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. 
A: 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.
