Constructive quantum field theory — Lecture 5 (notes)

Introduction

Lecture 4 formulated a programme (Figure 7.1 and the remark following it): solve the FBSDE

with a renormalised interaction \(V_T\), with estimates uniform in the ultraviolet cutoff, and take the limit. For \(\Phi^4_3\) the force \(F_t\) is not expected to converge to an integrable terminal value, and the naive estimates of the force diverge (Appendix A.2 of Lecture 4). In these two lectures we carry out a version of the programme on the unit torus \(\mathbb{T}^3\). The guiding principle is the one of Section 8: do not estimate \(\Phi\) and \(F\) directly, but pass to the forecast coordinate, in which the forward equation has no drift and the tree diagrams have been integrated exactly. Section 12.4 shows on one site what goes wrong with a frozen cube, and what integrating it gives.

The plan. Figure 10.1 gives the plan: this lecture builds the approximate force \(\bar{\mathcal{F}}\); Lecture 6 does the estimates and the limits. The emphasis is on the structure of the argument. We give complete proofs of the identities and of the scalar models of the estimates, and only describe the tensor estimates on the lattice. The point of view is that of [BCG26]; for other constructions of \(\Phi^4_3\) see [GJ73, FO76, BFS83, BG20, GH21].

What is proved. Theorem 10.2 holds for every coupling, every reference mass \(q_0 > 0\) and every finite part \(r\) of the counterterm, on a fixed torus, with no smallness, and for an explicit counterterm whose finite part is arbitrary. Section 18.7 lists what is not proved. Detailed proofs of what these lectures cite or condense are in the Supplement.

The reference mass. The number \(q_0 > 0\) is not an additional regulator: it is the squared mass \(m^2 = q_0\) of the Gaussian reference \(C = C_{q_0} = (q_0 - \Delta_N)^{- 1}\). It must be positive even on the torus, where there is no infrared problem at large distances, because the constants are in the kernel of \(- \Delta_N\), so that \((- \Delta_N)^{- 1}\) does not exist. The counterterm cannot supply this positivity, since its finite part \(r\) is an arbitrary real number, negative values included. The measure \(\nu_N\) depends on \(q_0\) and \(\mu_N\) only through \(q_0 + \mu_N\), so the two parameters are redundant: one may fix \(q_0\) once and for all, for instance \(q_0 = 1\), and let \(r\) be the free mass parameter. The number \(q_0\) is also where the scale starts: the Polchinski process runs over \(q \in [q_0, \infty)\) and vanishes at \(q = q_0\) (see (10.6)).

One auxiliary scale. The expansion of the force needs \(g_q\) small, and this happens above an auxiliary scale \(q_{\ast} \geqslant 1\), depending on \(\lambda\), \(q_0\) and on the finite part. Below it the coefficients are frozen (Section 12.5), so the force, the system and the results hold at all scales, and no statement about \(\nu\) depends on it.

Acknowledgements. The results presented in Lectures 4–6 are joint work with Sarah-Jean Meyer, obtained recently with the use of AI. M. Gubinelli is partly supported by UK Research and Innovation via the grant “StochFields” EP/Z534328/1 and by the Simons Collaboration on Probabilistic Paths to QFT. S.-J. Meyer is supported by the grant “StochFields” EP/Z534328/1.

10The problem and the exact FBSDE

10.1Lattice, scales and notation

Let \(N\) be an odd integer and \(\Lambda_N = (N^{- 1} \mathbb{Z}/\mathbb{Z})^3\) the periodic lattice of mesh \(N^{- 1}\) in the unit torus. Let \(H_N =\mathbb{R}^{\Lambda_N}\) with

so that the volume is \(1\) and integrals approximate continuum integrals. The nearest-neighbour Laplacian is \(\Delta_N u (x) = N^2 \sum_{i = 1}^3 [u (x + e_i / N) + u (x - e_i / N) - 2 u (x)]\); we usually drop the index \(N\). A kernel \(C (x, z)\) acts by \(Cu (x) = \int C (x, z) u (z) \mathrm{d} z\), so the identity has kernel \(N^3 \mathbb{1}_{x = z}\), and

The row norm \(\|C\|_{\text{row}}\) is the largest \(L^1\) norm of a row of the kernel. It bounds the operator norm on \(L^{\infty}\), and for a symmetric kernel on every \(L^p\), \(1 \leqslant p \leqslant \infty\), by interpolation: \(\|C\|_{p \to p} \leqslant \|C\|_{\text{row}}\). \(M_f\) is multiplication by \(f\), and \(K + \lambda y^2\) means \(K + \lambda M_{y^2}\). For a functional \(F\) on \(H_N\), \(DF\), \(D^2 F\), \(D^k F\) are its gradient and derivatives, as in Lecture 3. \(\|u\|_p\) is the \(L^p\) norm for the measure \(\int\).

The resolution parameter.

It is convenient to parametrise the resolution by the squared regulator mass

Set

In the notation of Lectures 3–4, with \(T = \infty\) and using \(q\) instead of \(t\) as the resolution variable (so \(\dot{Q}\) now means \(\partial_q Q_q\)),

So \(C_q\) is the covariance of the scales not yet resolved, and \(K_q = S_q^{- 1}\) is the precision of the posterior at resolution \(q\) under the free prior \(\mu^C\). The main properties of \(C_q\) are elementary.

Proof. \(C_q = \int_0^{\infty} \mathrm{e}^{- qs} \mathrm{e}^{s \Delta} \mathrm{d} s\) and \(\mathrm{e}^{s \Delta}\) is the transition kernel of the lattice random walk. This gives positivity and the row sums. Killing the walk at rate \(\lambda y^2\) gives the comparison for \(G_q\). The symbol of \(- \Delta_N\) is comparable to \(|k|^2\) on the centred frequency cube \(n \in \{- \frac{N - 1}{2}, \ldots, \frac{N - 1}{2} \}^3\) (symbol \(4 N^2 \sum_i \sin^2 (\pi n_i / N)\), with \(k = 2 \pi n\)), and \(\sum_k (q + |k|^2)^{- 2} \leqslant C \int_{\mathbb{R}^3} (q + |k|^2)^{- 2} \mathrm{d} k = Cq^{- 1 / 2}\). Counting the \(N^3\) modes gives the other bounds.\(\Box\)

The bubble \(a_2 (q)\) is the coincident-point variance of the shell at resolution \(q\): \(\partial_q \text{Var} (X_q (x)) = a_2 (q)\) for the free field at resolution \(q\), and \(\int_{q_0}^{\infty} a_2 = \text{Tr} C_{q_0} \asymp N\) is the divergent variance of the lattice field.

10.2The cutoff measure and the goal

Fix \(\lambda > 0\). The interaction and the lattice measure are

where \(\mu^C\) is the centred Gaussian measure with covariance \(C = C_{q_0}\) on \(H_N\), and \(\mu_N \in \mathbb{R}\) is a mass counterterm to be chosen. With this normalisation \(D^2 U (y) = \lambda M_{y^2}\): the coupling is the curvature of the interaction, which is why we write \(\lambda / 12\). In Lecture 4 we wrote the interaction as

the coupling is the same, and the Wick ordering is now part of \(\mu_N\). For every real \(\mu_N\), \(\nu_N\) is a probability measure with all moments, since the quartic term dominates.

The bounds that prove (a) and (b) are uniform in \(N\): this is UV stability. They are bounds on the error \(Z\) of an approximate forecast, defined below. We now set up the exact FBSDE at fixed \(N\).

10.3The Polchinski process as a filter

At fixed \(N\) everything in Sections 7–8 of Lecture 4 is a theorem. We need it in the profile (10.2), where the regulator is \(q - q_0\) and the terminal resolution \(q = \infty\).

Proof. In this profile \(J_q = (q - q_0) C_q\) and \(S_q = C_q\), so (a) is the bridge (7.14) and (c) the filter equation of Lecture 4. For (b), the conditional law of \(\Phi_{\infty}\) is \(\nu_N\) tilted by \(\exp (\langle K_q \Phi_q, \varphi \rangle - \frac{q - q_0}{2} \| \varphi \|_2^2)\) by (7.16): an exponential family in \(K_q \Phi_q\), whose mean has derivative the covariance.\(\Box\)

The noise of the forecast is the conditional expectation of the bridge noise: taking \(\mathbb{E}_q\) in (10.6),

a function of the field, \(\eta_q = \eta_q (\Phi_q)\) with \(\eta_q (x) = K_q x - (q - q_0) m_q (x)\). It is the noise of the field equation of Lecture 4 in this profile, and it is sub-Gaussian whatever the interaction (Proposition 9.5 of Lecture 4, Lemma 13.2 of Lecture 6).

10.4The exact force in the forecast coordinate

By Proposition 10.4(b) the map \(\Phi_q \mapsto \psi_q\) is invertible, and we define the exact force in the forecast coordinate \(\mathcal{F}_q\) (not to be confused with the filtration) by

So \(F_q = -\mathcal{F}_q (\psi_q)\), the force of Theorem 8.2 of Lecture 4, and the second form is the field equation in the forecast coordinate. We write \(\text{Ctr}_E T\) for the contraction of two slots of a tensor \(T\) with a kernel \(E\), and \(H_q^{\text{ex}} (y) = (K_q + D\mathcal{F}_q (y))^{- 1}\).

Proof. Differentiating (10.8) in \(\psi\) and using \(Dm_q = \text{Cov}_q K_q\) gives \(K_q + D\mathcal{F}_q (\psi_q) = \text{Cov}_q^{- 1}\). Apply Itô's formula to \(\Phi_q = \psi_q + C_q \mathcal{F}_q (\psi_q)\) with (10.5), (10.7) and \(\mathrm{d} C_q = - C_q^2 \hspace{0.17em} \mathrm{d} q\): the martingale parts agree, and the drifts differ by \(C_q\) times the left side of (10.9). At fixed \(N\) the forecast has full support, so this holds for all \(y\).\(\Box\)

This is the derivative of the Wetterich equation of Lecture 4 (Exercise 10.1). It is an identity about the unknown \(\nu_N\), not a construction. But since \(\mathrm{d} \psi_q = H_q^{\text{ex}} (\psi_q) \hspace{0.17em} \mathrm{d} W_q\), an approximate force gives approximate forecast coordinates, with an error governed by how badly it fails to solve (10.9).

10.5Approximate forecast and error

Let \(\bar{\mathcal{F}}_q\) be, for \(q \geqslant q_0\), an odd map on \(H_N\), continuous and piecewise smooth in \(q\) and \(y\), with \(D \bar{\mathcal{F}}_q\) symmetric and \(K_q + D \bar{\mathcal{F}}_q \geqslant \bar{c}_q I\) with \(\bar{c}_q > 0\). Define the approximate response, the force defect and its drift

The defect is the failure of \(\bar{\mathcal{F}}\) to solve (10.9). The map \(y \mapsto y + C_q \bar{\mathcal{F}}_q (y)\) is a bijection of \(H_N\). Define the approximate forecast and the error by

Equivalently \(Y_q\) solves the approximate coordinate equation \(K_q \Phi_q = K_q Y_q + \bar{\mathcal{F}}_q (Y_q)\), which is (10.8) with \(\bar{\mathcal{F}}\) in place of \(\mathcal{F}\). If \(\bar{\mathcal{F}} =\mathcal{F}\) then \(Y = \psi\), \(Z = 0\), \(\mathcal{D}= 0\) and \(b = 0\).

Proof. \(Y_q\) is a smooth function of \((q, \Phi_q)\), so it is an Itô process, \(\mathrm{d} Y = \beta \hspace{0.17em} \mathrm{d} q + A \hspace{0.17em} \mathrm{d} W\). Apply Itô's formula to \(Y_q + C_q \bar{\mathcal{F}}_q (Y_q) = \Phi_q\), with (10.5) and \(\mathrm{d} C_q = - C_q^2 \hspace{0.17em} \mathrm{d} q\), and multiply by \(K_q\): the martingale parts give \((K + D \bar{\mathcal{F}}) A = I\), so \(A = H\), and the drifts give

since \(CF = \psi - \Phi\) and \(C \bar{\mathcal{F}} (Y) = \Phi - Y\). Hence \(\beta = H (\psi - Y -\mathcal{D}) = HZ + b\), which is the first equation. The second is \(\mathrm{d} Z = \mathrm{d} \psi - \mathrm{d} Y\) and (10.7). The computation uses no initial value. At the reference scale \(\Phi_{q_0} = 0\), so \(Y_{q_0} = 0\) by oddness.\(\Box\)

Reading (10.12).

For this we need: (i) a force defect \(\mathcal{D}\) integrable against the backward propagator, which is a deterministic renormalisation problem (Section 11); (ii) global coercivity of \(\bar{\mathcal{F}}\), so that \(H_q (y)\) is a good noise coefficient for all fields (Section 12); (iii) a priori estimates for the coupled system (Lecture 6). Figure 10.1 summarises the logic of the two lectures.

Figure 10.1. The structure of the construction (\(E\), \(J\): the energies of Lecture 6). The two arrows between the forward and backward estimates form the closed loop of Lecture 6: neither estimate closes on its own.

Exercises

11Renormalisation: why two of them

We now construct the force \(\bar{\mathcal{F}}_q\) on bounded fields. This section is deterministic: it uses no information on \(\nu_N\).

11.1Scales and power counting

Two scales govern everything:

\(g_q\) is the dimensionless coupling at resolution \(q\): in \(d = 3\) the coupling \(\lambda\) has dimension of mass, and \(q^{1 / 2}\) is the mass of the regulator. It is small at all scales \(q \geqslant q_{\ast}\) if \(g_{q_{\ast}} = \lambda q_{\ast}^{- 1 / 2}\) is small; this is the only large number in the construction, and it is not a physical mass. \(r_q\) is the field size at which the nonlinearity is comparable to the regulator: \(D^2 U (y) = \lambda y^2 \approx q\) when \(|y| \approx r_q\). Compare with the size of the free field at resolution \(q\):

so typical fields have size \(q^{1 / 4} = g_q^{1 / 2} r_q \ll r_q\).

We measure a \(k\)-tensor \(T (x_1, \ldots, x_k)\) (for instance a vertex with \(k\) legs, the kernel of the \((k - 1)\)th field derivative of a force) by its rooted norm

and a force \(g\) by

on the field ball

The natural size of the \(k\)-leg vertex of the loop corrections

to the bare force (\(U\prime = DU = \frac{\lambda}{3} y^3\) the bare force, \(M_q\) a linear mass term, Section 11.3) is

For \(k = 2\) this says the derivative of the correction force is \(g_q\) times the regulator \(q\) (while \(D^2 U \approx q\) at \(|y| \approx r_q\)); each extra field derivative costs \(r_q^{- 1}\). For \(k = 4\), \(g_q qr_q^{- 2} = \lambda g_q\): the four-leg vertex of the corrections is \(\lambda\) times the small coupling. The whole construction is a bookkeeping of (11.2).

The basic loop estimate.

A closed loop with \(p\) vertices, each carrying at least one external leg (vertices without external legs are resummed into \(H\)), and \(p + 1\) propagators \(H\) (two of them adjacent, \(H^2\), once the loop is closed) is estimated by rooting one external slot, using two propagators in \(L^2\) (each row has \(L^2\) norm \(\leqslant (a_2)^{1 / 2}\)) and the others in operator norm (\(\leqslant C / q\)):

Indeed, by cyclicity of the trace we may take the rooted slot \(x\) on \(T_1\). Integrating out the other external legs, each \(T_i\), \(i \geqslant 2\), is dominated by a kernel \(t_i (u, v)\) in its two internal legs whose rows and columns have \(L^1\) norm \(\leqslant s (T_i)\), so \(\|t_i \|_{2 \to 2} \leqslant s (T_i)\) by Schur's test; \(T_1\) with \(x\) fixed gives a kernel \(t_1^x\) with \(\int t_1^x \leqslant s (T_1)\). The rest of the loop is a kernel \(|H| R |H|\), where the two propagators next to \(T_1\) are kept apart and \(R\) collects the other \(p - 1\) propagators and \(t_2, \ldots, t_p\). By Cauchy–Schwarz its entries are bounded by

and integrating against \(t_1^x\) gives (11.3). We used only that the propagators are symmetric with rows of \(L^2\) norm \(\leqslant Ca_2^{1 / 2}\) and of \(L^1\) norm \(\leqslant C / q\), which also bounds their operator norm on \(L^2\). For the free resolvent these are \(\int C_q (x, z)^2 \mathrm{d} z = a_2\) and \(\int C_q (x, z) \mathrm{d} z = q^{- 1}\) (Lemma 10.1); for the propagators used below they are (12.3), and Lemma 20.1 of the Supplement.

The vertices of the bare force have

on the ball (\(D^3 U = 2 \lambda y\), \(D^4 U = 2 \lambda\)). With \(a_2 \leqslant Cq^{- 1 / 2}\), a loop with \(k\) external legs built from them is \(\leqslant Cq^{1 / 2} r_q^{- k}\), and after the integration in \(q\) that produces the force \(\mathcal{F}\) (which gains a factor \(q\)) it has size \(q^{3 / 2} r_q^{- k} = g_q qr_q^{2 - k}\): this is the origin of (11.2), the size of one loop. Replacing a bare vertex by a vertex of a correction of relative size \(\varepsilon\) multiplies the loop by \(\varepsilon\) (Exercise 11.1). Every further loop gains one power of \(g_q\).

11.2The seed: one loop

The first approximation is the one-loop formula of Exercise 8.6(b): replace \(D\mathcal{F}\) by \(DU\prime = \lambda y^2\) in the force equation and integrate from \(q = \infty\). With the linear part subtracted this gives the seed

with \(G_q (y) = (K_q + \lambda y^2)^{- 1}\). It is odd, and its linear Taylor coefficient at \(y = 0\) vanishes. Its expansion in powers of the field,

is the sum of one-loop diagrams with \(2 j\) external legs, one of them the output \(x\). The leading term is the bubble

with

as predicted by (11.2).

Figure 11.1. The seed (11.4) as one-loop graphs: lines \(C_q\), and a vertex \(\lambda y^2\) with its two external fields at each corner. The \(j\)-th ring has \(2 j\) legs and coefficient \((- 1)^{j + 1} / (2 j)\). Marking one leg of a ring gives its term in the seed force. The series sums to a logarithm and has radius of convergence \(r_q\). The quartic ring is the bubble.

11.3Corrections and the mass

The seed solves the force equation only with \(DU\prime\) inside the loop. Its defect is of relative size \(g_q\) compared with the one-loop flow \(q^{1 / 2} r_q^{- k}\) (two loops). To improve it, let

remove the linear Taylor coefficient (the low jet) of an odd force, let

be a linear operator path (a running mass \(\mu_q\) and an even, translation-invariant, centred kernel \(\Theta_q\), \(\Theta_q \mathbb{1} = 0\)), and set

The approximate force on the field ball is

with the nonlinear part built by two integrated steps of the force equation,

so that

each step inserts the previous approximation into the loop. The integrals run from \(\infty\): the nonlinear part is fixed by the requirement that it vanishes at infinite resolution, where \(\mathcal{F}_q\) must become the bare force (Theorem 8.2).

Where the iteration comes from.

The projections \(\Pi\) and \(1 - \Pi\) split it into two equations of different nature:

The first is the nonlinear part of the flow. By power counting (11.2) its vertices with \(k \geqslant 4\) legs have integrands \(O (s^{- k / 2})\), integrable at \(s = \infty\), so it can be integrated from infinite resolution with the boundary value \(\ell_{\infty} = 0\) required by the bare force:

This is a fixed-point equation: \(\ell\) appears on both sides, inside the propagator \(R\) and in \(D^2 \ell\). The second equation is the linear part of the flow. It is not integrable at \(q = \infty\) uniformly in \(N\), since the tadpole and the sunset diverge (Section 11.4). It is solved exactly, as the mass equation (11.8) with a boundary value at \(q_{\ast}\), and its value at \(q = \infty\) is the bare mass. This is the reason for the subtraction \(\Pi\): it separates the convergent nonlinear flow, which we approximate, from the divergent linear flow, which we solve exactly and which produces the counterterm.

The iteration as Picard steps.

Why the steps approximate the flow.

the loop difference between the last two steps, of order \(g_q^3\) by (11.12). This is what the FBSDE sees: by Theorem 10.6 the error \(Z\) is driven only by the defect, and Section 11.5 shows that two steps are the minimum for a source that is integrable in the backward equation. The iteration cannot be continued indefinitely: each step consumes two derivatives (see the end of this section), so the number of steps is fixed in advance.

Figure 11.2. The iteration (11.7), schematically, with only two external legs drawn per vertex. Each step inserts the previous correction into the loop and gains one loop and one factor \(g_q\), as in (11.11). After two steps the defect is the loop difference of order \(g_q^3\), which gives (11.12) and makes the source integrable in the backward equation.

The linear part.

where \(R_{0, q} = (K_q + M_q)^{- 1}\). Requiring that \(M\) solves this part of the flow exactly gives the mass equation

(the source \(B_q [M]\) is not the sampling Brownian motion \(B\)), with the mixed boundary conditions

(the latter in first spatial moment). The running mass is pinned at the auxiliary scale \(q_{\ast}\), at a value \(b\) with \(|b| \leqslant B \lambda q_{\ast}^{1 / 2}\), while the nonlocal kernel is pinned at infinity. The bare mass is then

and it is this number that enters the Gibbs measure (10.4).

Note the order: the counterterm is an output of the deterministic flow, not an input.

Idea of the proof (complete proof: Proposition 20.6 of the Supplement). (a) Loop Lipschitz estimate. If two corrections \(\ell, \hat{\ell}\) obey the individual bounds (11.2) up to order \(k + 2\) and their difference obeys them with \(g_q\) replaced by \(\varepsilon_q\), then \(\mathcal{N}_k (\mathcal{T}[M, \ell] -\mathcal{T}[M, \hat{\ell}]) \leqslant C \varepsilon_q q^{1 / 2} r_q^{- k}\). Expand \(D^{k - 1}\) of the loop force into loops; in each, telescope the difference onto one marked vertex or propagator and apply (11.3). (b) Integration. Each step gains one \(g\); the integrand of (11.7) is \(O (\lambda^{1 + k / 2} s^{- k / 2})\) in the \(k\)-leg vertex, and that of the increment \(\ell^{[j + 1]} - \ell^{[j]}\) is \(O (\lambda^{1 + k / 2} s^{- k / 2} g_s^j)\), integrable at \(s = \infty\) for \(k \geqslant 4\). The low jets have been removed, so the lower derivatives are controlled by Taylor's formula from the four-leg vertex. (c) Mass path. Given \(M\) in the ball, the right side of (11.8) is computed from the corrections, and the map \(M \mapsto (\mu, \Theta)\) is a contraction for large \(q_{\ast}\): differences gain a factor \(g_{q_{\ast}}\) (for \(\mu\) after integrating from \(q_{\ast}\), using \(\sup_x \log x / \sqrt{x} < \infty\)). (d) Defect. The mass equation supplies exactly the linear jet of the flow, and \(\partial_q \ell^{[2]} = \Pi \mathcal{T} [M, \ell^{[1]}]\), so the defect is the loop difference between the last two approximations, of order \(g^3\): by (a) with \(\varepsilon_q = g_q^3\), \(\textsf{S}_4 (\mathcal{D}^{\text{o}}) \leqslant Cg_q^3 q^{1 / 2} r_q^{- 4} = C \lambda^5 q^{- 3}\).\(\Box\)

Two technical points are hidden in this sketch. First, the loop estimate (a) needs more derivatives than it controls: the \(k\)-leg vertex of the loop involves vertices with \(k + 2\) legs. Ten legs of the seed leave eight after one step and six after two; this is why the number of steps must be finite and fixed in advance (an infinite iteration loses all derivatives). Second, the nonlocal kernel \(\Theta_q\) must be measured in a first spatial moment \(v_1\), not in a second one: the estimate \(\| \delta_z^2 C_q \|_{\text{row}} \lesssim d (0, z)^2\), which a second moment would require, fails by a logarithm, and one uses \(\| \delta_z^2 C_q \|_{\text{row}} \lesssim q^{- 1 / 2} d (0, z)\) instead (Lemma 19.3 of the Supplement). This is also why \(\mathcal{N}_k\) contains a spatial moment at \(y = 0\).

11.4Two mass renormalisations

The bare mass \(\mu_N\) diverges as \(N \to \infty\). Its two divergent parts are the two divergent self-energy diagrams of Appendix A.2 of Lecture 4.

• The tadpole. The first term of (11.8) integrates to \(- \lambda \int_{q_{\ast}}^{\infty} \text{Tr} R_{0, q}^2 \hspace{0.17em} \mathrm{d} q \approx - \lambda \text{Tr} C \sim - N\) (up to a term bounded uniformly in \(N\)). This is Wick ordering: \(\frac{\lambda}{12} \varphi^4 - \frac{1}{2} \lambda c \varphi^2\) agrees, up to a constant, with \(\frac{\lambda}{12} H_4 (\varphi ; c)\), with \(c = C (x, x)\) ((A.5) of Lecture 4).

• The sunset. The second term contracts the bubble contained in \(D^3 \ell^{[2]} (0)\) (the cubic coefficient of the seed force) with \(R_0^2\), and integrating in \(q\) produces the two-loop self-energy, which diverges like \(\lambda^2 \int_{|z| > 1 / N} |z|^{- 3} \mathrm{d} z \sim \lambda^2 \log N\). This is the sunset counterterm \(\gamma_T \sim \log T\) of (A.7) of Lecture 4: with \(\gamma_T \int H_2 = \gamma_T \int \varphi^2 +\)const, the sunset part of \(\mu_N\) is \(2 \gamma_T = \frac{2}{3} \lambda^2 \int C^3 \approx \frac{\lambda^2}{24 \pi^2} \log N\), positive. (The double-tadpole logarithms in the two terms of (11.8), the bubble with a tadpole insertion and the mass insertion in \(\text{Tr} R_{0, q}^2\), cancel each other.)

Figure 11.3. The mass equation (11.8) and the two divergent parts of the bare mass. The cross is the insertion \(R_{0, q}^2\). Since \(D^3 \ell^{[2]} (0)\) contains the bubble of Figure 11.1, its contraction contains the sunset. Integrating in \(q\) turns the crossed line into a line \(C\): the tadpole is linear in \(N\) and the sunset logarithmic.

Power counting (Exercise A.3 of Lecture 4) says that no other counterterm is needed. In the flow this is the statement that the integrals (11.7) of all vertices with \(k \geqslant 4\) legs converge at \(q = \infty\): only the linear jet must be fixed by a boundary condition.

There is a subtlety: the running mass \(\mu_q\) at fixed \(q\) has a limit as \(N \to \infty\), but \(\mu_N = \mu_{\infty}\) does not. One never interchanges these two limits. Also, the whole running kernel \(\Theta_q\) is kept: replacing the sunset by its local part would leave a nonlocal remainder of the wrong size.

11.5Two correction steps

Why two corrections, and not one or three? The answer is in the backward equation of (10.12). Freeze the noise and the large fields, and estimate the size of \(Z\) produced by a defect on the ball. The propagator of \(\mathrm{d} Z = - HZ \hspace{0.17em} \mathrm{d} q + \ldots\) from \(q\) to \(s\) is about \(s / q\) (since \(H \approx C_q\) has norm \(q^{- 1}\)), so

where we used that the low jet of \(\mathcal{D}\) vanishes, so on the ball \(|\mathcal{D}| \leqslant \textsf{S}_4 (\mathcal{D}) |y|^3 / 6\). After \(n\) correction steps the defect is the loop difference of the last two approximations, of relative size \(g^{n + 1}\): by the loop estimate,

and therefore

With one step the bounded-field source produces a logarithmically divergent error; two steps are the minimum, and they suffice: \(Z^{\text{b}}_q = O (q^{- 3 / 2})\), and the backward energy \(\lambda^{- 1} \int q^{1 / 2} |Z^{\text{b}} |^2 \mathrm{d} q\) is finite (Lecture 6). The general (fractional) theory gives the depth criterion \((n + 2) \varkappa > (3 + \varkappa) / 2\), with subcriticality \(\varkappa = \frac{1}{2}\) for \(\Phi^4_3\): \((2 + 2) \cdot \frac{1}{2} = 2 > \frac{7}{4}\), while \(n = 1\) gives \(\frac{3}{2} < \frac{7}{4}\). This is a sufficient condition, which loses \(\varkappa / 2\); for \(\Phi^4_3\) the heuristic above shows that \(n > 1\) suffices, with \(n = 1\) borderline.

The two senses of “two renormalisations” are related. The mass counterterm must contain the sunset because the quartic vertex of the corrections, already of the seed, contains the bubble, and contracting it produces the sunset; and the second step is needed because the defect of the first one, the three-loop remainder, is borderline in the backward equation, exactly as the sunset is borderline in the mass.

11.6The explicit counterterm

Formula (11.9) gives \(\mu_N\) as an output of the flow, from the boundary value \(b\). The input can be made explicit, and this also fixes what the finite part is.

Proof. Expansion. Expand \(R_0 = C - CMC + CMCMR_0\) with \(C = C_q\). Since \(a_2' = - 2 \text{Tr} C^3\), the first term of (11.8) is \(- \lambda a_2 - \lambda \mu a_2' + 2 \lambda \text{Tr} (C^3 \Theta)\), up to \(O (\lambda a_2 g_q^2)\), and \(| \text{Tr} (C^3 \Theta) | \leqslant Cq^{- 1} v_1 (\Theta)\) because \(| \hat{\Theta} (k) | \leqslant 2 |k| v_1\). The corrections in \(\ell^{[2]}\) give \(O (\lambda^3 q^{- 3 / 2})\) after contraction, by (11.11). The cubic part of the seed is \(- \lambda^2 y (x) \int C^2 (x, z) y (z)^2\), and contracting its third derivative at zero with \(R_0^2\) gives the operator \(\lambda^2 a_2 \text{diag} (R_0^2) + 2 \lambda^2 C^2 \cdot R_0^2\), where \(\cdot\) is the pointwise product of kernels. Its constant-mode eigenvalue, with \(R_0^2\) replaced by \(C^2\) at a relative cost \(g_q\), gives the remaining two terms.

Integration. Use \(\partial_q \text{Tr} C_q = - a_2\), \(\partial_q \int C_q^3 = - 3 \int C_q^2 C_q^{(2)}\), and \(- \lambda \mu a_2' + \lambda^2 a_2^2 = - \lambda \partial_q (\mu a_2) + \lambda (\beta + \lambda a_2) a_2\). This is the cancellation of the double-tadpole logarithms: \(\beta + \lambda a_2 = O (\lambda^2 / q)\). At fixed \(N\) the boundary terms at \(q = \infty\) vanish, which gives (11.15).

Response. For two boundary values with difference \(\delta b\), put \(D = \sup_q | \delta \mu_q |\) and \(X = \sup_q q^{3 / 2} v_1 (\delta \Theta_q)\), which are finite at fixed \(N\). The loop comparison gives \(| \delta \beta_q | \leqslant C \lambda (Dq^{- 3 / 2} + Xq^{- 5 / 2})\) and \(v_1 (\delta B_q - \delta \beta_q I) \leqslant C \lambda^2 (Dq^{- 5 / 2} + Xq^{- 7 / 2})\), so \(X \leqslant C \lambda^2 D + C \lambda^2 q_{\ast}^{- 1} X\) and \(D \leqslant | \delta b| + C \lambda q_{\ast}^{- 1 / 2} D + C \lambda q_{\ast}^{- 3 / 2} X\). Hence \(D \leqslant 2 | \delta b|\), and integrating \(\delta \beta\) gives the claim.

Limit and inversion. In (11.15), \(\text{Tr} C - \text{Tr} C_{q_{\ast}}\) has summable terms \(\lesssim (1 + |k|)^{- 4}\); \(C - C_{q_{\ast}}\) converges uniformly and \(C^2 + CC_{q_{\ast}} + C_{q_{\ast}}^2\) in \(L^1\); and the integrand converges at fixed \(q\) by consistency of the coefficients, dominated by \(C \lambda^3 q^{- 3 / 2}\). As \(\text{Tr} C - \text{Tr} C_{q_{\ast}} \leqslant C + q_{\ast}^{1 / 2} / 10\) uniformly in \(N\), the range of \(f_N\) contains \(r\) for \(B\) above a universal constant and \(q_{\ast}\) large. Monotonicity and pointwise convergence give the convergence of \(b_N\).\(\Box\)

The boundary value \(b\) is not the finite part: by (11.15), \(f_N (b) \approx b + \lambda q_{\ast}^{1 / 2} / (4 \pi)\) at leading order, so at fixed \(b\), raising \(q_{\ast}\) changes the theory.

Exercises

12The flow at large fields

12.1The large-field problem

All the estimates of Proposition 11.1 hold on the field ball \(\|y\|_{\infty} < 2 r_q\), and only there. But the forecast \(Y_q\) is a random field on \(N^3\) sites, and nothing prevents it from being larger than \(r_q\) somewhere. The free field has typical size \(q^{1 / 4} \ll r_q\), but it exceeds \(r_q\) with small, nonzero probability, and nothing a priori controls the interacting forecast. We need a force \(\bar{\mathcal{F}}_q\) defined and uniformly coercive for all fields, which agrees with \(\mathcal{F}^{\text{o}}_q\) for \(\|y\|_{\infty} \leqslant r_q\).

A scalar model.

The structure to preserve is coercivity, and it is carried by the bare cube. In the scalar model the coordinate equation \(qy + \frac{\lambda}{3} y^3 = q \Phi_q\) has noise coefficient

large fields are more elliptic and less noisy. A good large-field extension keeps \(U\) untouched and freezes only the correction.

12.2Clipping

Let \(\tau : \mathbb{R} \to \mathbb{R}\) be smooth and odd, \(\tau (u) = u\) on \([- 1, 1]\), constant outside \([- 2, 2]\), with \(0 \leqslant \tau' \leqslant 1\), and set

The clip acts pointwise in space and maps every field into the ball, and the factor \(D \kappa_q\) is what keeps the clipped force a gradient. The cubic and linear parts are not clipped. By the chain rule, with factors \(| \kappa_q^{(b)} | \leqslant Cr_q^{1 - b}\), the clipped correction obeys the same bounds (11.2) globally:

for all \(y \in H_N\); the scale derivative includes the motion of the clip, \(\partial_q \kappa_q = O (r_q / q)\).

Idea. \(H^{- 1} = K_q + \lambda y^2 + M_q + D \tilde{\ell}_q\), and \(M_q + D \tilde{\ell}_q\) is small relative to \(K_q\): \(\|D \tilde{\ell}_q \| \leqslant Cg_q q\), and \(M_q\) is small after multiplication by \(C_q\) (the centred kernel is controlled by its first moment, Exercise 12.1). Write \(H = (I + G (M + D \tilde{\ell}))^{- 1} G\) with \(G = (K_q + \lambda y^2)^{- 1}\) and use the Neumann series and Lemma 10.1.\(\Box\)

So \(\bar{\mathcal{F}}_q\) is a globally coercive force, and Theorem 10.6 applies.

12.3The complete defect

On the set \(\{\|y\|_{\infty} \leqslant r_q \}\) the clip is the identity and the defect is \(\mathcal{D}^{\text{o}}\), of size (11.12). Outside it, the clipped force does not solve the force equation even approximately in relative terms. For \(q \geqslant q_{\ast}\) split the drift into its bounded-field and large-field parts,

and let \(\chi_q (y) = \mathbb{1}_{|y| > r_q}\) (pointwise) and \(t_q (y) = y \hspace{0.17em} \chi_q (y)\), the field restricted to the large-field set.

The complete proof is Lemma 21.2 of the Supplement. The first bound is (11.13) with \(n = 2\): on the ball \(|\mathcal{D}^{\text{o}} | \leqslant \textsf{S}_4 (\mathcal{D}^{\text{o}}) (2 r)^3 / 6\), and \(q^{- 1} \lambda^5 q^{- 3} r_q^3 = \lambda^{7 / 2} q^{- 5 / 2}\). The large-field part is as large as the bound \(\lambda q^{- 3 / 2} |y| = (g_q / q) |y|\) on the whole drift (no smallness beyond the factor \(g_q\) already present in \(b\)), but it is supported on the large-field set (up to nonlocal tails controlled by operator bounds). The last bound follows from the second since on \(\{|y| > r\}\), \(|t| + r^{- 1} y^2 \leqslant 2 r^{- 2} |t|^3\), and \((\lambda q^{- 3 / 2} r^{- 2})^2 = \lambda^4 q^{- 5}\).

Where the large-field error comes from.

Each group vanishes if \(\|y\|_{\infty} \leqslant r_q\), and each is bounded by \(g_q\) times \(|t|\) or \(r^{- 1} y^2 \chi\), in \(L^p\), by the same loop estimates (Section 21.3 of the Supplement; Exercise 12.2 for the one-site version). There is no cancellation to exploit: the clipping error is genuinely of the size \((g_q / q) |y|\) of the whole drift on the large-field set. What saves us is that this set is rarely visited, and its occupation is controlled by the coercivity of the actual dynamics. Lecture 6 controls it through the large-field action (13.7).

The coordinate correction.

using \(DU (y) = \frac{\lambda}{3} y^3\). The correction is a small contraction, \(\|D \varrho_q (y)\|_{p \to p} \leqslant \varepsilon_q := C (g_q + g_q^2)\), for all \(y\) and \(1 \leqslant p \leqslant \infty\), and \(\varrho_q (0) = 0\) by oddness. So, up to a small Lipschitz perturbation, the forecast is the solution of the cubic elliptic equation \(K_q (Y + \varrho) + \frac{\lambda}{3} Y^3 = K_q \Phi_q\) driven by the field at resolution \(q\). This is the equation whose coercivity Lecture 6 exploits.

12.4Why the Polchinski expansion is not enough

Lecture 4 gives two exact descriptions of the flow: the Burgers equation for the force \(F\) at the field \(\Phi\), and the force equation at the forecast \(\psi\). We use the second, here and in Lecture 6. The first looks more economical, and the following programme is natural. Keep the bare cube \(- \frac{\lambda}{3} \varphi^3\) unclipped, so that the approximate force stays coercive at all fields. Expand the rest of the force in perturbation theory and clip only these higher orders. Pay for the clipping in the source of the equation for the remainder \(R = F - f\), using the a priori estimates. No change of coordinates would be needed.

Here we test the programme where everything can be computed: on one site. There is no spatial structure and no renormalisation, so whatever goes wrong has nothing to do with divergences.

Conventions. As in Exercise 10.4: \(N = 1\), \(\Delta = 0\), \(K_q = q\), \(V (\varphi) = \frac{\lambda}{12} \varphi^4\) and \(r_q = (q / \lambda)^{1 / 2}\). We write \(S = C_q = 1 / q\) for the remaining covariance, and \(x\) for the value of the field at resolution \(q\).

The exact force is linear at large fields.

where \(\mathbb{E}_x\) is the expectation for the probability density proportional to \(\mathrm{e}^{- \frac{\lambda}{12} y^4 - \frac{q}{2} (y - x)^2}\). It solves the Burgers equation

Let \(m_q (x) =\mathbb{E}_x Y\). Integrating by parts in \(y\) gives

This is Proposition 10.4(b): the posterior mean is the forecast, \(\psi_q = m_q (\Phi_q)\). So \(F_q\) measures how far the forecast lies behind the field.

Proof. Integrating the derivative of \(y\) times the density gives \(\frac{\lambda}{3} \mathbb{E}_x Y^4 + q\mathbb{E}_x Y^2 - qx\mathbb{E}_x Y = 1\). With \(M = (\mathbb{E}_x Y^4)^{1 / 4}\) this gives \(\frac{\lambda}{3} M^4 \leqslant 1 + q |x| M\), and Young's inequality gives \(\frac{\lambda}{6} M^4 \leqslant 1 + (q|x|)^{4 / 3} (2 \lambda / 3)^{- 1 / 3}\). Finally \(|m_q | \leqslant M\).\(\Box\)

Therefore, by (12.7), the two regimes of the force are

The limits cannot be interchanged. The crossover is at \(|x| \sim r_q\), where the cube equals the Gaussian restoring force: \(\frac{\lambda}{3} r_q^3 \sim qr_q\). Beyond it the force is linear, because convolution with a Gaussian of variance \(S\) cannot produce a force steeper than \(x / S\).

So no estimate can make \(R\) a small fraction of the retained cube uniformly in the field. The cube in \(R\) is not a new counterterm, since there is nothing to renormalise here. It is the compensation for having frozen the leading term of a force that does not stay cubic.

The field equation does not see the cube.

But by Proposition 12.4, \(J_q R_q\) cancels the cube on the left side at large fields. Inserting (12.7) shows what is left:

There is still confinement, but it is linear, and it comes from the sublinear growth of \(m_q\).

Proof. The mean \(m\) has the sign of \(x\), the density being an even density tilted by \(\mathrm{e}^{qxy}\). Jensen's inequality in the identity of Lemma 12.3 gives \(\frac{\lambda}{3} m^4 + qm^2 - qxm \leqslant 1\). If \(x \geqslant r_q\) and \(m > \frac{9}{10} x\), then \(\frac{\lambda}{3} m^3 + q (m - x) > x \left( \frac{243}{1000} \lambda x^2 - \frac{q}{10} \right) \geqslant \frac{143}{1000} qx\), as \(\lambda x^2 \geqslant q\), so the left side is at least \(\frac{1287}{10000} qx^2 \geqslant \frac{1287}{10000} q^2 / \lambda > 1\), a contradiction. So \(| \eta_q (x) | \geqslant \frac{q}{10} |x|\) by (12.9), and we conclude with the sub-Gaussian bound.\(\Box\)

At \(R = r_q\) the exponent is at least \(q^2 / (200 \lambda)\). This estimate treats the combined restoring force through the posterior moment. It never uses the cube as a coercive term against a remainder regarded as a source.

Remark 12.6. (This does not transfer to the lattice) With a linear term \(ay\) in the force, Lemma 12.3 acquires the term \((3 (q + a)_- / \lambda)^{1 / 2}\). On the lattice the bare mass makes \(a\) divergent and negative, so this elementary moment bound is not uniform until the renormalised cancellations have been built in. The scalar tail bound is therefore not an independent input for the field theory.

What clipping does to the Burgers equation.

with \(f_1 = - \frac{\lambda}{3} (x^3 + 3 Sx)\), the cube and its Wick correction, and \(f_2 = \frac{\lambda^2}{3} Sx^5 + \ldots\), the first tree. Keep \(f_1\) unclipped and clip the correction: \(f^c = f_1 +\mathcal{C}f_2\) with \(\mathcal{C}f_2 = \kappa_q \prime \cdot f_2 \circ \kappa_q\), the clip \(\kappa_q\) of (12.1). Then

Without the clip only the last two terms, of higher order, would be present. The clip reintroduces a defect at exactly the order that \(f_2\) was built to cancel. For \(|x| \geqslant 2 r_q\) we have \(\mathcal{C}f_2 = 0\), so \(f^c = f_1\), and the Burgers defect is exactly

This is the price of the clipping. At the edge of the clip it has size \((3 q / \lambda)^{1 / 2}\), which is not integrable in \(q\), so it can only be paid through the small probability of reaching the clip. Along the exact process \(\hspace{0.17em} \mathrm{d} \Phi_q = S^2 F_q (\Phi_q) \hspace{0.17em} \mathrm{d} q + S \hspace{0.17em} \mathrm{d} W_q\), Itô's formula gives for \(R_q = F_q (\Phi_q) - f_q^c (\Phi_q)\) the equation

The transport term has a favourable sign at large fields: \((f^c)' = - \lambda (x^2 + S)\) for \(|x| \geqslant 2 r_q\), which is the dissipation one hopes to use (near the clip the sign fails: on the ball the quintic takes over beyond \(|x| \approx 0.77 r_q\)). But at large fields the bracket is \(\frac{\lambda^2}{3} S^2 x^5 - \lambda S^2 x^2 R\), and for the true \(R \sim \frac{\lambda}{3} x^3\) the two terms cancel at leading order. The source and the dissipation are the same quantity, so estimating them separately gives nothing. To close, one would need, for instance, a signed estimate that retains this cancellation, or a tail bound for \(\Phi\) obtained independently of \(R\). On one site Proposition 12.5 is such a bound: the probability of reaching the clip is at most of order \(\exp (- cq^2 / \lambda)\), so in expectation the source is exponentially small, and the frozen cube is not ruled out there. But that bound comes from the posterior moment, and by the remark above it does not transfer. The field equation alone cannot replace it, since its restoring force is the cube and \(R\) together, which is what we are trying to estimate.

Going further in perturbation theory does not help: any finite depth of expansion leaves the same structure (Exercise 12.5), and we shall see in a moment that the series of trees stops converging inside \(r_q\).

Integrating the cube.

The characteristics give the solution in closed form.

Proof. Differentiate \(x = y + \frac{\lambda}{3} Sy^3\) with \(d = 1 + \lambda y^2 / q\): \(y' = 1 / d\), \(\partial_S y = - \frac{\lambda}{3} y^3 / d\), \(f' = - \lambda y^2 / d\) and \(f'' = - 2 \lambda y / d^3\). Hence \(\partial_S f = \frac{\lambda^2}{3} y^5 / d = ff'\), which is (12.13) as \(\partial_q = - S^2 \partial_S\). For (a), the stationarity condition is \(\frac{\lambda}{3} y^3 + q (y - x) = 0\). For (b), use Lagrange inversion on \(y = x - \frac{\lambda}{3} Sy^3\); the nearest singularity is where \(d = 0\), that is at \(y = \pm \mathrm{i} r_q\), \(x = \pm \frac{2}{3} \mathrm{i} r_q\). Finally (c) and (d) are read off from the formulas, since \(d = 1 / (qH_q (y))\) with \(H_q (y) = (q + \lambda y^2)^{- 1}\).\(\Box\)

Item (b) explains why the depth of the expansion is not the issue. The tree series has the same complex singularity \(y = \pm \mathrm{i} r_q\) as the one-loop seed of Section 12.1, and its truncations at even \(k\) point outward at large fields.

The defect is now bounded for all fields, and it gives a global estimate.

Proof. Use \(S = 1 / q\) as time, starting from \(S = 0\). Subtracting, \(\hat{R} = F - f\) solves the linear problem

The zeroth-order coefficient has the good sign, as \(- f' \geqslant 0\). With \(z = (\lambda S / 3)^{1 / 2} y\), the source obeys \(\frac{1}{2} |f'' | \leqslant C_{\ast} (\frac{\lambda}{3 S})^{1 / 2}\), which is integrable at \(S = 0\), so \(\beta (S) = 2 C_{\ast} (\lambda S / 3)^{1 / 2}\) satisfies \(\mathfrak{L} \beta \geqslant \frac{1}{2} |f'' |\). To apply the maximum principle on the line, take \(\chi = \mathrm{e}^{3 S} (1 + x^4)\). Then \(\mathfrak{L} \chi \geqslant 0\), since \(6 x^2 \leqslant 3 (1 + x^4)\) and \(xF (x) \leqslant 0\); the last holds because \(F = - \frac{\lambda}{3} \mathbb{E}_x Y^3\), and \(\mathbb{E}_x Y^3\) has the sign of \(x\), the density being an even density tilted by \(\mathrm{e}^{qxy}\). Both \(F\) and \(f\) grow at most cubically, uniformly for bounded \(S\): \(|f| \leqslant \frac{\lambda}{3} |x|^3\) since \(|y| \leqslant |x|\), and \(\frac{\lambda}{3} \mathbb{E}_x Y^6 + q\mathbb{E}_x Y^4 - qx\mathbb{E}_x Y^3 = 3\mathbb{E}_x Y^2\) gives \(\mathbb{E}_x Y^4 \leqslant (|x| + (3 S)^{1 / 2})^4\), up to constants, hence \(|F| \leqslant C \lambda (|x|^3 + S^{3 / 2})\). Also \(\hat{R}\) is continuous at \(S = 0\), since the posterior concentrates at \(x\). So \(\pm \hat{R} - \beta - \varepsilon \chi\) is negative on the boundary of a large cylinder, hence inside it. Let the cylinder grow and then \(\varepsilon \to 0\).\(\Box\)

Figure 12.1. The force on one site, for \(\lambda = 3\), \(q = 4\). Left: the frozen cube leaves the exact force at \(x \sim r_q\), while the integrated cube follows it at all fields: the two curves coincide at this scale, their difference being at most \(0.06\) in these units. Right, on a logarithmic scale: the remainder of the frozen cube grows like the cube itself, while the remainder of the integrated cube stays below the bound of Theorem 12.8.

The integrated cube is the forecast coordinate. The equation \(qy + \frac{\lambda}{3} y^3 = qx\), at \(x = \Phi_q\), is the coordinate equation of Exercise 10.4, so \(y_q (\Phi_q) = Y_q\). Equations (12.7) and (12.14) then say that \(F_q - f_q = q (m_q - y_q)\): the remainder is \(K_q Z_q\), and the error \(Z\) of (10.11) is the posterior mean minus the posterior mode. By (12.16), on one site \(|Z_q | \leqslant 2 C_{\ast} (\lambda / 3)^{1 / 2} q^{- 3 / 2}\) surely, for all fields.

Why the forecast coordinate has no such problem. In the variable \(y\) the bare cube does not flow at all: all the trees of (b) are contained in the map \(x \mapsto y\). The second term of (10.9) is a loop, and the defect \(\mathcal{D}_q (y) = \lambda yH_q (y)^2\) of \(\bar{\mathcal{F}}_q = \frac{\lambda}{3} y^3\) is bounded by \(\lambda^{1 / 2} q^{- 3 / 2}\) for all \(y\). This is why Section 12.2 can leave the cube untouched and clip only the corrections: what is clipped consists of loops, not of trees whose large-field sum is forced to change the growth of the force. This removes the frozen-cube problem specifically: the clipping error of Section 12.3 is still there, and it is what Lecture 6 pays for.

In the coordinate \(y\) the cube is genuine: \(qy + \frac{\lambda}{3} y^3 = q \Phi_q = (q - q_0) \psi_q + \eta_q\), i.e. \(q_0 y + \frac{\lambda}{3} y^3 = \eta_q + (q - q_0) (\psi_q - y)\). This is the forecast equation of Lecture 4 with the explicit force in place of the exact one, the equation (13.3) of Lecture 6 on one site.

What the model points to. The scalar model points to the integration of the cube. The force equation in the forecast coordinate and the FBSDE of Theorem 10.6 are one way to organise it; another would keep the original field, with the force \(f_q\) given by the variational formula of (a), and work with the defect \(\frac{1}{2} \text{Ctr}_{C_q^2} D^2 f_q\). On the lattice the variational formula still defines this force, with a nonlocal response, but its defect contains the loop contractions, starting with the tadpole, and must be renormalised. A complete proof along that route is not given in these lectures.

The scalar model proves less than a no-go theorem. It shows that a global estimate treating the remainder of a frozen cube as a small cubic perturbation is false. It does not show that every construction based on the Polchinski equation fails, nor does it give the asymptotics of the force on the lattice. There is also a limit to the integrated cube itself. With a linear term in the force, the map \(y \mapsto y + S (\frac{\lambda}{3} y^3 + ry)\) has derivative \(1 + S (r + \lambda y^2)\), which vanishes somewhere when \(1 + Sr < 0\) (Exercise 12.6). A divergent negative mass cannot simply be placed inside this monotone map, and this is one reason why the mass correction \(M_q\) of (12.1) is kept apart from the cube and is required to be small relative to \(K_q\) (Lemma 12.1).

12.5The force below the auxiliary scale

The force above is built for \(q \geqslant q_{\ast}\), because its expansion needs \(g_q\) small. Nothing below \(q_{\ast}\) requires a construction: we freeze it. For \(0 < q < q_{\ast}\) set

Then \(K_q + D \bar{\mathcal{F}}_q = K_{q_{\ast}} + D \bar{\mathcal{F}}_{q_{\ast}}\), so the coordinate equation \(K_q y + \bar{\mathcal{F}}_q (y) = K_q x\), at \(x = \Phi_q\), reads \(K_{q_{\ast}} y + \bar{\mathcal{F}}_{q_{\ast}} (y) = K_q x\). Thus for \(q < q_{\ast}\),

The Itô computation of Theorem 10.6 holds for any such family. So the system (10.12) holds on \([q_0, \infty)\), with \(Y_{q_0} = 0\) for \(\nu_N\): the initial value is deterministic, and no large mass is used. The force is continuous in \(q\) and piecewise \(C^1\), which is all that Itô's formula needs. The defect for \(s < q_{\ast}\) is explicit:

Here \(\mathcal{U}\) is the loop part of \(\mathcal{D}_{q_{\ast}}\); it obeys \(\|\mathcal{U}(y)\|_p \leqslant Cg_{q_{\ast}} \|y\|_p\), which is part of the proof of the global linear bound of Proposition 12.2 (Lemma 21.2 of the Supplement). Uniformly in \(N\), for \(s < q_{\ast}\),

with \(b = - H\mathcal{D}= H (y -\mathcal{U})\). Here we used \(\|H\|_{p \to p} \leqslant C / q_{\ast}\). These are all the properties of \(s < q_{\ast}\) we use. Note that \(\mathcal{D}\) jumps at \(q_{\ast}\); the jump is harmless, and the measure does not depend on \(q_{\ast}\). More precisely \(\bar{\mathcal{F}}, y, \mathcal{D}, b, Y, Z\) depend on \(q_{\ast}\), while, with \(\mu_N\) fixed (\(b_N\) then depends on \(q_{\ast}\), Theorem 11.2), \(\nu\), \(\psi\), \(\text{Cov}\) and the law of \(\Phi\) do not.

Exercises

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