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One possible approach to constructive field theory is to define it on a lattice and take the scaling limit, and there are famous results stating that in $d\geq4$ this cannot lead to a non-trivial theory.

What is the status of approaches using the Gaussian free field?

What I mean is this: let $\Omega\subseteq\mathbb{R}^d$ be a smooth domain and consider the Gaussian free field $\{\langle f,\varphi\rangle\}_{f\in H_0^1(\Omega)}$ in $\Omega$ with zero boundary conditions. Let $F$ be a nice functional of $\varphi$, e.g., a mollified two-point function between $x,y\in\Omega$.

Why doesn't the following definition of $\varphi^4$ theory work?

$$ \langle F \rangle_\lambda := \lim_{\Omega\to\mathbb{R}^d} \frac{\mathbb{E}\left[F(\varphi)\exp(-\lambda\int_\Omega\varphi^4)\right]}{\mathbb{E}\left[\exp(-\lambda\int_\Omega\varphi^4)\right]} \,. $$

I suppose the first question one should ask about this is how to make sense of $\int_\Omega\varphi^4$, and then, if the infinite volume limit exists.

Do people in constructive field theory work on this approach? What are some of the major hurdles in such or similar frameworks?

PPR
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  • Related: https://mathoverflow.net/questions/260854/a-roadmap-to-hairers-theory-for-taming-infinities/260941#260941 the part at the end. As Martin said, trying to use the free field as your underlying fixed (anchored) probability space and construct (even by a limiting procedure) $\int \phi^4$ as a functional of that free field, is okay in 2d and finite volume but will not work in 3d. There you need to see the problem as that of weak convergence of a sequence of cut-off measures (fixed measurable space, but ton of probability spaces). – Abdelmalek Abdesselam Feb 05 '21 at 18:28
  • For a proof of the singularity of measures mentioned by Martin you can see, e.g., https://arxiv.org/abs/2004.01513 – Abdelmalek Abdesselam Feb 05 '21 at 18:28
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    A sketch of a slightly different proof of that singularity can be found here (this is the unpublished proof mentioned on p.3 of the above preprint). – Martin Hairer Feb 05 '21 at 21:43
  • @MartinHairer: I guess that's what you showed us at the Imperial College conference in 2019. Thanks for posting it, and making it available. – Abdelmalek Abdesselam Feb 06 '21 at 00:16
  • @MartinHairer, thanks for the reference to the proof. Is it obvious from the fact the $d=3$ measure is singular w.r.t. the GFF that the same is true also for $d>3$? If not, are there analogous proofs for $d\geq4$? – PPR Feb 06 '21 at 00:34
  • @PPR There exists no $\Phi^4$ measure in $d\ge 4$, see this recent paper by Aizenman and Duminil-Copin. – Martin Hairer Feb 06 '21 at 08:26
  • @AbdelmalekAbdesselam Yes, that's the one. – Martin Hairer Feb 06 '21 at 12:18
  • @MartinHairer, Thanks. My understanding was that what A.-D.C. prove in $d=4$ is that any scaling limit of a lattice $\Phi^4$ theory converges to a Gaussian limit. Is it clear that that implies there is no $\Phi^4$ measure in $d=4$, rather than just that if $\Phi^4_4$ exists, it cannot be obtained via a scaling limit of a discrete theory? – PPR Feb 06 '21 at 15:39
  • @PPR What would be the precise meaning of the statement "$\Phi^4_4$ doesn't exist"? Yes, there are reasonable looking approximations of $\Phi^4$ to which A-DC doesn't apply and this will be always be true, whatever more general statement of this form they come up with. I don't think anyone would expect any other approximation to behave in a substantially different way. – Martin Hairer Feb 06 '21 at 16:03

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When $d=2$, this works fine and this is precisely how Nelson originally constructed the $\Phi^4$ measure (in finite volume). Already for $d = 3$, the $\Phi^4$ measure is singular with respect to the free field, even in finite volume, so this approach is bound to fail. The reason why it is singular is subtle, but you can (kind of) convince yourself that this should be the case from the fact that the correct renormalisation in $d=3$ has an additional logarithmic correction on top of Wick renormalisation.

Martin Hairer
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