A Renormalisation Group Method. V. A Single Renormalisation Group Step

Abstract

This paper is the fifth in a series devoted to the development of a rigorous renormalisation group method applicable to lattice field theories containing boson and/or fermion fields, and comprises the core of the method. In the renormalisation group method, increasingly large scales are studied in a progressive manner, with an interaction parametrised by a field polynomial which evolves with the scale under the renormalisation group map. In our context, the progressive analysis is performed via a finite-range covariance decomposition. Perturbative calculations are used to track the flow of the coupling constants of the evolving polynomial, but on their own perturbative calculations are insufficient to control error terms and to obtain mathematically rigorous results. In this paper, we define an additional non-perturbative coordinate, which together with the flow of coupling constants defines the complete evolution of the renormalisation group map. We specify conditions under which the non-perturbative coordinate is contractive under a single renormalisation group step. Our framework is essentially combinatorial, but its implementation relies on analytic results developed earlier in the series of papers. The results of this paper are applied elsewhere to analyse the critical behaviour of the 4-dimensional continuous-time weakly self-avoiding walk and of the 4-dimensional $n$-component $|\varphi |^4$ model. In particular, the existence of a logarithmic correction to mean-field scaling for the susceptibility can be proved for both models, together with other facts about critical exponents and critical behaviour.

Appendices

Appendix A: Proof of Proposition 1.8

In this section we prove Proposition 1.8, which states that several normed spaces are complete.

We fix the scale $j$ and suppress it in the notation. Thus $\mathcal {C}(\mathbb {V})$ is the set of connected polymers at scale $j$. For $X\in \mathcal {C}(\mathbb {V})$, let $W (X,\phi )$ be a continuous positive function of $\phi$ in the normed space $\Phi (X^{\Box })$. This means that $W (X,\phi )$ is a function of $\phi$ in the space of fields in $\phi :\mathbb {V}\rightarrow \mathbb {C}$ but only depends on the restriction of $\phi$ to $X^{\Box }$. Let $\mathcal {S}(\mathbb {V})$ be the space of maps $F:\mathcal {C}(\mathbb {V}) \rightarrow \mathcal {N}(\mathbb {V})$ such that $F (X)$ is in $\mathcal {N}(X^{\Box })$ for $X$ in $\mathcal {C}(\mathbb {V})$. The following proposition provides the first step in the proof of Proposition 1.8.

Proposition A.1

For $\mathbb {V}=\varLambda$ or $\mathbb {V}={ {{\mathbb Z}}^d }$ the space $\mathcal {S}(\mathbb {V})$ is complete in the norm

$$\begin{aligned} \Vert F\Vert _{W} = \sup _{X \in \mathcal {C}(\mathbb {V}) , \phi \in \Phi (\mathbb {V})} \Vert F\Vert _{T_{\phi }}W^{-1} (X,\phi ) . \end{aligned}$$

(A.1)

Proof

We suppress the $\mathbb {V}$ argument. For $X\in \mathcal {C}$, $\phi \in \Phi$, $g\in \Phi$, define the linear functional

$$\begin{aligned} \lambda _{X,\phi ,g}: \mathcal {S}\rightarrow \mathbb {C}\quad \text {by}\quad F \mapsto \langle F (X),g \rangle _{\phi } W^{-1} (X,\phi ) , \end{aligned}$$

(A.2)

with the pairing on the right-hand side defined in [15, Definition 3.3]. Then

$$\begin{aligned} \Vert F\Vert _{\mathcal {S}} = \sup _{X \in \mathcal {C},\phi \in \Phi ,g \in B (\Phi )} \big |\lambda _{X,\phi ,g} (F)\big | . \end{aligned}$$

(A.3)

Therefore a sequence $F_{n}$ in $\mathcal {S}$ is Cauchy if and only if $\lambda _{X,\phi ,g} (F_{n})$ is Cauchy in $\mathbb {C}$, uniformly in the parameters $( X,\phi ,g)\in \mathcal {C}\times \Phi \times B (\Phi )$. Let $F_{n}$ be a Cauchy sequence in $\mathcal {S}$. By completeness of $\mathbb {C}$ the sequence $\lambda _{X,\phi ,g} (F_{n})$ has a limit $F_{X,\phi ,g}$ in $\mathbb {C}$. Since $F_{n}$ is uniformly Cauchy the convergence is uniform in the parameters. Therefore, to prove that $\mathcal {S}$ is complete, it suffices to prove that there exists $F$ in $\mathcal {S}$ such that $\lambda _{X,\phi ,g} (F) = F_{X,\phi ,g}$ for all values of the parameters. Thus we fix $X \in \mathcal {C}$, assume that $F_{n}$ is a sequence in $\mathcal {N}(X^{\Box })$, and it suffices to prove that there exists $F\in \mathcal {N}(X^{\Box })$ such that $\lambda _{X,g,\phi }(F_{n})\rightarrow \lambda _{X,g,\phi }(F)$.

It suffices to restrict the test function $g$ to a small class of test functions, as follows. Let $z$ be a sequence in $\varvec{\Lambda }^{*}$ and let $x$ and $y$ be the boson and fermion subsequences of $z$. Let the length $p (x)$ of the sequence $x$ be at most $p_{\mathcal {N}}+2$, where the $2$ allows for observables. We define a test function $\delta _z\in \Phi$ by setting $\delta _{z} (z')=1$ when $z' = z$ and $\delta _{z} (z')=0$ otherwise, for $z' \in \varvec{\Lambda }^{*}$. By the definition of $\Phi (X^{\Box })$ all elements of $\Phi (X^{\Box })$ are finite linear combinations of these special test functions. Thus it suffices to prove that $\lambda _{X,\phi ,\delta _z}(F_{n})\rightarrow \lambda _{X,\phi ,\delta _z}(F)$ since this gives the corresponding results for all $g$ in $\Phi (X^{\Box })$ and therefore also for all $g$ in $\Phi$.

Since $\delta _z$ can be normalised to be in $B (\Phi )$, $\lambda _{X,\phi ,\delta _z} (F_{n})$ is Cauchy in $\mathbb {C}$, uniformly in $\phi \in \Phi$. By the definition of the pairing,

$$\begin{aligned} p({x})!\langle F_{n},\delta _z \rangle _{\phi } = F_{n,x,y} = \left( \prod _{i=1}^{p (x)} \frac{\partial }{\partial \phi _{x_{i}}} \right) F_{n,y} (\phi ) \end{aligned}$$

(A.4)

is a partial derivative of $F_{n,y}$ with respect to $\phi$. By the definition of $\mathcal {S}$ this partial derivative is continuous in $\phi$ and the pairing is well defined on the equivalence classes $\phi \in \Phi (X^{\Box })$ and $g \in \Phi (X^{\Box })$. By hypothesis, $W (X,\phi )$ is bounded below uniformly on compact subsets of $\Phi (X^{\Box })$. Therefore the uniform convergence of $\lambda _{X,\phi ,\delta _z} (F_{n})$ implies that the partial derivative $F_{n,x,y}$ converges uniformly in $\phi$ for $\phi$ in compact subsets of $\Phi (X^{\Box })$. By the continuity of $F_{n,x,y}$ as a function of $\phi$ the limit of $F_{n,x,y}$ is continuous in $\phi$. By integration we find that the derivatives of the limit are the limits of the derivatives. Therefore there exists $F_{y} \in \mathcal {N}(X^{\Box })$ such that $F_{n,x,y} (\phi )$ converges to $F_{x,y} (\phi )$ for all $\phi$. Letting $F = \sum _y \frac{1}{y!}F_y(\phi )\psi ^y$ and noting that this sum over $y$ is a finite sum because $X$ is a finite set, we have $\lambda _{X,\phi ,\delta _z}(F_{n})\rightarrow \lambda _{X,\phi ,\delta _z}(F)$, and the proof is complete. $\square$

As in Sect. 1.7, we denote by $\mathcal {I}(\mathbb {V})$ the set of elements of $\mathcal {N}$ whose $T_{0}$ semi-norm is zero. Define $\mathcal {S}(T_0)$ to be the space of maps $F:\mathcal {C}(\mathbb {V}) \rightarrow \mathcal {N}(\mathbb {V})/\mathcal {I}(\mathbb {V})$ such that $F (X)\in \mathcal {N}(X^{\Box })/\mathcal {I}(\mathbb {V})$. Since we have factored out the ideal $\mathcal {I}(\mathbb {V})$, the $T_{0}$ semi-norm becomes a norm on this space.

Proposition A.2

For $\mathbb {V}=\varLambda$ or $\mathbb {V}={ {{\mathbb Z}}^d }$, the space $\mathcal {S}(T_{0})$ is complete.

Proof

Given $F\in \mathcal {N}$, we replace $\phi$ and $\psi$ by $t\phi$ and $t\psi$ and construct a polynomial $T \in \mathcal {N}$ of degree $p_{\mathcal {N}}$ by making a Taylor expansion in powers of $t$ to order $p_{\mathcal {N}}$ and setting $t=1$. Then derivatives of $T$ at $\phi =0$ match derivatives of $F$ up to and including order $p_{\mathcal {N}}$. Therefore $F-T\in \mathcal {I}(\mathbb {V})$, and the map $F\mapsto T$ identifies $\mathcal {N}(\mathbb {V})/ \mathcal {I}(\mathbb {V})$ with polynomials of degree $p_{\mathcal {N}}$. Then, for all $X$, $\mathcal {N}(X) /\mathcal {I}(\mathbb {V})$ is a finite dimensional space and therefore $\mathcal {S}(T_0)$ is complete in $T_{0}$ norm.

Proposition A.3

For either of the two choices $\mathbb {V}= { {{\mathbb Z}}^d }$ or $\mathbb {V}= \varLambda$, the spaces $\mathcal {F}(G)$, $\mathcal {F}(\tilde{G})$, are closed subspaces of $\mathcal {S}$ and are complete. Likewise, $\mathcal {F}(T_{0})$ is a closed subspace of $\mathcal {S}(T_{0})$ and is also complete.

Proof

The spaces $\mathcal {F}(G)$ and $\mathcal {F}(\tilde{G})$ are obtained when $W (X,\phi )$ is chosen as in (1.40). According to the definitions of the regulators in [17, (1.38), (1.41)], $W (X,\phi )$ is positive and continuous in $\phi$. Therefore, by Proposition A.1, with either choice of $W$, the space $\mathcal {S}$ is complete. Also, according to Proposition A.2, $\mathcal {S}(T_{0})$ is complete. Therefore it is sufficient to prove that $\mathcal {F}(G)$, $\mathcal {F}(\tilde{G})$ and $\mathcal {F}(T_{0})$ are closed subspaces. As discussed in Sect. 1.7.3, elements of $\mathcal {F}(G)$, $\mathcal {F}(\tilde{G})$ and $\mathcal {F}(T_{0})$ must satisfy the symmetry and field locality conditions of Definition 1.7. These conditions define closed subspaces.

Therefore, it only remains to prove for the cases $\mathcal {F}(G)$, $\mathcal {F}(\tilde{G})$ that the condition of vanishing at weighted infinity defines a closed subspace of $\mathcal {S}$. For this let $F_{1},F_{2},\dots$ be a sequence of elements of $\mathcal {S}$ that vanish at weighted infinity and are such that the sequence converges in $\mathcal {S}$ to a limit $F$. We must prove $F$ vanishes at weighted infinity. Let $\varepsilon >0$ and let $X$ be a polymer. By definition, there exists $N$ such that $\Vert F (X) - F_{N} (X)\Vert _{T_{\phi }}W^{-1} (X,\phi ) < \varepsilon$ uniformly in $\phi$. Therefore

$$\begin{aligned} \Vert F (X)\Vert _{T_{\phi }}W^{-1} (X,\phi )&< \varepsilon + \Vert F_{N}\Vert _{T_{\phi }}W^{-1} (X,\phi ) . \end{aligned}$$

(A.5)

Since $F_{N} \in \mathcal {S}$, it follows that

$$\begin{aligned} \limsup _{ \Vert \phi \Vert _{\Phi (X)} \rightarrow \infty } \Vert F (X)\Vert _{T_{\phi }}W^{-1} (X,\phi ) < \varepsilon , \end{aligned}$$

(A.6)

and since this holds for all $\varepsilon$, $F$ vanishes at weighted infinity, as was to be proved. $\square$

Proof of Proposition 1.8

In view of Proposition A.3, it suffices to consider $\mathcal{W}({\mathbb V})$. A Cauchy sequence $F_{n}$ in $\mathcal {W}(\mathbb {V})$ is Cauchy in each of the $\mathcal {F}(G)$ and $\mathcal {F}(\tilde{G})$ norms. Therefore it has limits $F_{G}$ and $F_{\tilde{G}}$ in the $\mathcal {F}(G)$ norm and the $\mathcal {F}(\tilde{G})$ norm. Both norms imply convergence pointwise in $X,\phi$ so $F_{G} = F_{\tilde{G}}$ and therefore $F_{n}$ is convergent in $\mathcal {W}(\mathbb {V})$. $\square$

Appendix B: Two Properties of the Expectation

In this section, we prove that the expectation is continuous in the mass, and that the expectation preserves the property of vanishing at infinity. We begin with the continuity statement.

1.1 B.1 Mass Continuity of the Expectation

1.1.1 B.1.1 Statement of Continuity

We consider the continuity properties of the expectation as a function of the covariance, and of the mass which defines the covariance. There are two fixed scales, $j$ and $j+1$, and the scale advances in norms when the expectation is taken. We omit the scale when it is $j$ and write $+$ to indicate scale $j+1$. The covariance $C$ is always considered to be a test function with two arguments and, furthermore, is assumed to be in the unit ball $B_1(\Phi _+)$ of the space $\Phi _+$ of test functions. Let $X \in \mathcal {C}$ be a connected polymer $X \in \mathcal {C}$. Recall from (1.35) to (1.36) that two norm pairs $\Vert \cdot \Vert _j$, $\Vert \cdot \Vert _{j+1}$ are defined on $\mathcal {N}(X^\Box )$. We write $\mathcal {X}$ for either of the normed spaces defined by the two choices of $\Vert \cdot \Vert _j$ and for each of these choices let $\mathcal {X}_+$ be the normed spaces defined by the accompanying choice of $\Vert \cdot \Vert _{j+1}$. The main continuity result is the following proposition, whose proof is given in the remainder of Sect. B.1.

Proposition B.1

For $F\in \mathcal {X}$, the map $C \mapsto \mathbb {E}_{C}\theta F$ from $B_{1} (\Phi _{+})$ to $\mathcal {X}_{+}$ is continuous.

Now we choose the covariance $C$ to be one of the $m^2$-dependent covariances $C=C_{j+1}$ for $j<N(\mathbb {V})$, or $C=C_{N,N}$ for $j+1=N(\varLambda )$, which arise in the finite-range decomposition of the covariance $(-\varDelta +m^2)^{-1}$ described in [17, Section 1.1.1]. Proposition B.1 then implies the continuity of the expectation as a function of the mass $m^2$.

Proposition B.2

For $F\in \mathcal {X}$, the map $m^2 \mapsto \mathbb {E}_{C}\theta F$ from $\mathcal {X}$ to $\mathcal {X}_{+}$ is continuous.

Proof

By Proposition B.1, it suffices to show that $m^2 \mapsto C$ is a continuous function from $I_j$ to $B_1(\Phi _+)$. This is a consequence of [9, Proposition 6.1(b)]. (In fact, [9, Proposition 6.1] does not directly address mass continuity in the $\Phi _{j+1}$ norm, but it does when augmented with the estimate [4, (1.15)]). $\square$

1.1.2 B.1.2 Reduction to Dense Subset

The following lemma is a standard result in functional analysis. We omit the proof, which is an $\varepsilon /3$ argument.

Lemma B.3

Let $\mathcal {B}$ and $\mathcal {B}_+$ be Banach spaces. Suppose that the sequence of linear operators $T_{n}:\mathcal {B}\rightarrow \mathcal {B}_{+}$ is uniformly bounded in operator norm, and suppose that $T:\mathcal {B}\rightarrow \mathcal {B}_{+}$ is a bounded linear operator. If $T_{n} F \rightarrow T F$ for all $F$ in a dense subset of $\mathcal {B}$, then $T_{n} F \rightarrow T F$ for all $F\in \mathcal {B}$.

The dense subset we use is the subspace $\mathcal {X}_{0}$ of $\mathcal {X}$ whose elements are compactly supported in $\phi$, namely, given $X \in \mathcal {P}$,

$$\begin{aligned} \mathcal {X}_{0} = \{F\in \mathcal {X}: \exists R \;\text {such that}\; \Vert F\Vert _{T_{\phi }} =0 \;\text {if}\; \Vert \phi \Vert _{\Phi (X^\Box )} \ge R\}. \end{aligned}$$

(B.1)

Lemma B.4

The set $\mathcal {X}_{0}$ is dense in $\mathcal {X}$.

Proof

Let $\chi :{\mathbb C}\rightarrow {\mathbb R}$ be a smooth non-negative function of compact support that is bounded by 1, equals 1 on a neighbourhood of the unit disk and has support inside the disk of radius $2$. For $R \ge 1$, $x\in Y$, and $\phi \in \Phi$, let $\chi _{R,x} (\phi ) = \chi (\phi _{x}/R)$. Let $F \in \mathcal {X}$ and let $\varepsilon >0$. We will show that $R$ can be chosen so that $\Vert F-\chi _RF\Vert _\mathcal {X}< \varepsilon$, and this suffices since $\chi _RF\in \mathcal {X}_0$.

By the definition of the $T_{\phi }$ norm (see [15, Definition 3.3]), for $R$ large depending on $\mathfrak {h}$,

$$\begin{aligned} \Vert \chi _{R,x} \Vert _{T_{\phi }} \le \sum _{p=0}^{p_\mathcal {N}} \frac{1}{p!} \chi ^{(p)} (\phi _{x}/R ) (\mathfrak {h}/R)^p \le 1+O \left( \mathfrak {h}/R\right) \le 2 . \end{aligned}$$

(B.2)

Also, since $\Vert 1-\chi _{R,x}\Vert _{T_\phi }=0$ for $|\phi _{x}| < R$,

$$\begin{aligned} \Vert 1 - \chi _{R,x} \Vert _{T_{\phi }} = {1\!\!1}_{|\phi _{x}| \ge R}\, \Vert 1 - \chi _{R,x} \Vert _{T_{\phi }} \le 3{1\!\!1}_{\Vert \phi \Vert _{\Phi } \ge R} . \end{aligned}$$

(B.3)

Let $\chi _{R} (\phi ) = \prod _{x\in Y} \chi _{R,x} (\phi )$. Let $\succ$ be any total ordering of the points in $Y=X^\Box$. We apply (B.2) and (B.3) and the product property of the $T_{\phi }$ semi-norm to obtain

$$\begin{aligned} \Vert 1 - \chi _{R}\Vert _{T_{\phi }}&\le \sum _{y \in Y} \Vert 1 - \chi _{R,y} \Vert _{T_{\phi }} \prod _{x\in Y, x \succ y} \Vert \chi _{R,x} \Vert _{T_{\phi }} \le {1\!\!1}_{\Vert \phi \Vert _{\Phi (Y)}\ge R}\, 3|Y| 2^{|Y|} . \end{aligned}$$

(B.4)

By hypothesis, $\Vert F \Vert _{T_{\phi }}\mathcal {G}^{-1} (X,\phi ) \rightarrow 0$ as $\Vert \phi \Vert _{\Phi (Y)}\rightarrow \infty$. By the product property, for $R$ sufficiently large depending on $\varepsilon ,\mathfrak {h}, Y$, this gives

$$\begin{aligned} \Vert F - \chi _{R} F\Vert _{\mathcal {X}} \le 3|Y| 2^{|Y|} \sup _{\phi :\Vert \phi \Vert _{\Phi (Y)}\ge R} \Vert F \Vert _{T_{\phi }}\mathcal {G}^{-1} (X,\phi ) < \varepsilon . \end{aligned}$$

(B.5)

This completes the proof. $\square$

1.1.3 B.1.3 Continuity of Expectation in Covariance

Before proving Proposition B.1, we first prove the following lemma concerning a norm equivalence. We write $Y=X^\Box$ below, to simplify the notation. The normed space $\mathcal {X}$ is defined above Proposition B.1.

Lemma B.5

Let $S\subset \mathbb {C}^{Y}$ and let $F_{n}\in \mathcal {X}$ for $n\in {\mathbb N}$. Then $F_{n}$ is convergent in $T_{\phi }$ semi-norm uniformly in $\phi \in S$ if and only if $F_{n,y}$ and its derivatives up to order $p_{\mathcal {N}}$ converge uniformly in $\phi \in S$ for the finitely many possible sequences $y$ arising from $Y$. This is the same as $F_{n,y}$ being convergent in the $C^{p_{\mathcal {N}}} (S)$ topology for each such $y$.

Proof

The proof is closely related to that of Proposition A.1. Given $\phi \in S$ and $g\in \Phi$, we define the linear functional $\lambda _{\phi ,g}: \mathcal {X}\rightarrow \mathbb {C}$ by $F \mapsto \langle F,g \rangle _{\phi }$. Then

$$\begin{aligned} \sup _{\phi \in S}\Vert F\Vert _{T_{\phi }} = \sup _{\phi \in S,g \in B (\Phi )} \big |\lambda _{\phi ,g} (F)\big | . \end{aligned}$$

(B.6)

Therefore the sequence $F_{n}$ in $\mathcal {S}$ is convergent in $T_{\phi }$ uniformly in $\phi \in S$ if and only if $\lambda _{\phi ,g} (F_{n})$ is convergent in $\mathbb {C}$, uniformly in the parameters $(\phi ,g)\in S \times B (\Phi )$.

Let $z \in \varvec{\Lambda }^{*}$. We define a test function $\delta _z\in \Phi$ by setting $\delta _{z} (z')=1$ when $z' = z$ and $\delta _{z} (z')=0$ otherwise, for $z' \in \varvec{\Lambda }^{*}$. All test functions are finite linear combinations of these special test functions and they comprise a finite basis for the test functions in $\Phi (Y)$. Therefore $\lambda _{\phi ,g} (F_{n})$ is convergent uniformly in $(\phi ,g)$ if and only if $\lambda _{\phi ,g} (F_{n})$ is convergent uniformly in $\phi \in S$ when $g$ is a basis test function. Exactly as in (A.4), $\langle F_{n},\delta _{x,y} \rangle _{\phi }$ is a partial derivative of $F_{n,y}$ with respect to $\phi$. Therefore, uniform convergence of $\lambda _{\phi ,g} (F_{n})$ is equivalent to uniform convergence of partial derivatives, as claimed. $\square$

Proof of Proposition B.1

By Proposition 5.3, the linear map $T_C: F \mapsto \mathbb {E}_{C}\theta F$ from $\mathcal {X}$ to $\mathcal {X}_{+}$ is a bounded operator. The proof of Proposition 5.3 shows that $\Vert T_C\Vert$ is bounded uniformly in $C\in B_{1} (\Phi _{+})$. Let $C_n$ be a sequence of covariances in $B_1(\Phi )$ that converges to $C$, and let $T_{C_n}: F \mapsto \mathbb {E}_{C_n}\theta F$. By Lemmas B.3 and B.4, it suffices to show that $T_{C_n}F \rightarrow T_CF$ for all $F\in \mathcal {X}_0$, with $\mathcal {X}_0$ the dense subset of $\mathcal {X}$ defined by (B.1). An element $F\in \mathcal {X}_{0}$ has the form $F = \sum _{y}\frac{1}{y!}F_{y} \psi ^{y}$, and this is a finite sum because there are finitely many fermion fields with labels in $Y$. Therefore it suffices to show that, for each finite sequence $y$, $\mathbb {E}_{C_n}\theta F_{y}\psi ^{y}$ converges to $\mathbb {E}_{C}\theta F_{y}\psi ^{y}$. By [15, (2.39)], $\mathbb {E}_{C}\theta F = (\mathbb {E}_{C}\theta F_{y}) (\mathbb {E}_{C}\theta \psi ^{y})$, so it suffices to prove that $\mathbb {E}_{C_n}\theta F_{y} \rightarrow \mathbb {E}_{C}\theta F_{y}$ and $\mathbb {E}_{C_n}\theta \psi ^{y} \rightarrow \mathbb {E}_{C}\theta \psi ^{y}$.

Since $F_{y}\psi ^{y}\in \mathcal {N}(Y)$ we can regard $C$ as an element of the finite-dimensional vector space $\Phi _{+} (Y)$, on which all norms are equivalent. In particular a sequence $C_{n}$ of covariances converges in $\Phi _{+} (Y)$ if and only if the sequence converges in the sense of convergence of matrix elements of $Y \times Y$ matrices. By definition, $\mathbb {E}_{C}\theta \psi ^{y}$ is a polynomial in $\psi _{x}$ for $x \in Y$, with coefficients that are polynomials in matrix elements of $C$. The space of such polynomials is finite-dimensional, so the map $C \mapsto \mathbb {E}_{C}\theta \psi ^{y}$ is continuous.

It remains to prove that $C \mapsto \mathbb {E}_{C} \theta F_{y}$ is continuous as a map from a domain of fixed-size matrices to $\mathcal {X}_{+}$, for $F_{y}$ a compactly supported $p_{\mathcal {N}}$ times continuously differentiable function of $\phi \in \mathbb {C}^{Y}$. However, the map $\mathbb {E}_{C} \theta$ represents convolution by a Gaussian, and from this it can be seen that $\mathbb {E}_{C} \theta F_{y}$ is jointly continuous in $\phi$ and $C$. Derivatives commute with convolution so the same is true for derivatives. Therefore, $\mathbb {E}_{C} \theta F_{y}$ is continuous in $C$, as a map into $C^{p_{\mathcal {N}}} (\mathbb {C}^{Y})$. By Lemma B.5, it is therefore also continuous in the topology of convergence in $T_{\phi }$ norm uniformly in $\phi$. This is a stronger topology than the norm on $\mathcal {X}_{+}$, so the proof is complete. $\square$

1.2 B.2 Expectation and Vanishing at Weighted Infinity

We now prove that the property of vanishing at weighted infinity is preserved by the expectation. Since we only take expectations in finite volume we consider the vector space $\mathcal {K}(\varLambda )$ with the $\mathcal {F}(\mathcal {G})$ norm defined in (1.41) in terms of the weight

$$\begin{aligned} W (X,\phi ) = \rho ^{f (a ,X)}\mathcal {G}(X,\phi ) \end{aligned}$$

(B.7)

of (1.40), with $\rho$ given by (1.43). There is also the space $\mathcal {F}_+(\mathcal {G}_+)$ defined in terms of the weight

$$\begin{aligned} W_+ (X,\phi ) = \rho _+^{f_+ (a_+ ,X)}\mathcal {G}_{+}(X,\phi ), \end{aligned}$$

(B.8)

where we assume that $a<a_+$. Our norm pairs (1.35) and (1.36) are such that $\mathcal {G}=G_j$ is paired with $\mathcal {G}_+=T_{0,j+1}$, and $\mathcal {G}_j=\tilde{G}_j$ is paired with $\mathcal {G}_+=\tilde{G}_{j+1}^{\gamma }$. As a first step, we prove the following lemma.

Lemma B.6

Let $f_{R}={1\!\!1}_{\{\Vert \xi \Vert _{\Phi (X)}\le R\}}$. Then for $X\in \mathcal {P}$ and $\phi \in \mathbb {C}^{X^\Box }$,

$$\begin{aligned}&\displaystyle \mathbb {E}_{+} W (X,\phi + \xi ) \le W_{+}(X,\phi ), \end{aligned}$$

(B.9)

$$\begin{aligned}&\displaystyle \lim _{R\rightarrow \infty } \sup _{\phi \in \Phi } \frac{1}{W_{+}(X,\phi )} \mathbb {E}_{+} \left[ (1-f_{R}(\xi )) W (X,\phi + \xi ) \right] = 0 . \end{aligned}$$

(B.10)

Proof

By definition of the regulators, and by the inequality $\Vert \phi +\xi \Vert ^2 \le 2(\Vert \phi \Vert ^2+\Vert \xi \Vert ^2)$,

$$\begin{aligned} \mathcal {G}(X,\phi + \xi ) \le \mathcal {G}^2(X,\phi ) \mathcal {G}^2(X,\xi ) \le \mathcal {G}^2(X,\phi ) G^2(X,\xi ). \end{aligned}$$

(B.11)

Using [17, Lemma 1.2], we obtain

$$\begin{aligned} \mathcal {G}(X,\phi + \xi ) \le \mathcal {G}_{+} (X,\phi ) G^{2}(X,\xi ) . \end{aligned}$$

(B.12)

By [17, (1.74)],

$$\begin{aligned} \mathbb {E}_{+} G^{2}(X,\xi ) \le 2^{|X|_j} , \end{aligned}$$

(B.13)

and hence

$$\begin{aligned} \mathbb {E}_{+} \mathcal {G}(X,\phi + \xi ) \le 2^{|X|_{j}}\mathcal {G}_{+} (X,\phi ) . \end{aligned}$$

(B.14)

By (B.7), (B.14), and the fact that $\rho ^{f(a,X)} 2^{|X|_j} \le \rho _+^{f_+(a_+,X)}$ by definition (for small $\tilde{g}$),

$$\begin{aligned} \mathbb {E}_{+} W (X,\phi + \xi ) \le \rho ^{f (a,X)} 2^{|X|_{j}}\mathcal {G}_{+} (X,\phi ) \le W_{+} (X,\phi ) . \end{aligned}$$

(B.15)

This completes the proof of (B.9).

To prove (B.10), we repeat the steps in the proof of (B.9) but with the factor $1-f_{R}$ included. This factor then appears under the expectation in (B.13), and (B.10) then follows by dominated convergence. $\square$

A second ingredient we need is that for a function $f=f(\xi )$ of the fluctuation field $\xi$,

$$\begin{aligned} \Vert \mathbb {E}_{+}fF(X)\Vert _{T_{\phi ,+}(\mathfrak {h}_+)} \le \mathbb {E}_{+}\left[ |f(\xi )|\, \Vert F(X)\Vert _{T_{\phi +\xi }(\mathfrak {h})} \right] . \end{aligned}$$

(B.16)

This follows from a slight adaptation of [17, (7.2)–(7.3)], with the improved version of [15, Proposition 3.19] provided by [15, (3.68)] to include the function $h=f_R$. These give the inequality

$$\begin{aligned} \Vert \mathbb {E}_{+}fF(X)\Vert _{T_\phi (\mathfrak {h}/2)} \le \mathbb {E}_{+}\left[ |f(\xi )|\, \Vert F(X)\Vert _{T_{\phi +\xi }(\mathfrak {h})} \right] . \end{aligned}$$

(B.17)

With the monotonicity in $\mathfrak {h}$ of the $T_\phi$ norm provided by [17, Lemma 3.2], (B.16) then follows from $\mathfrak {h}_+ \le \frac{1}{2} \mathfrak {h}$. This application of [17, Lemma 3.2] requires that $F \in \mathcal {N}$ is gauge invariant and such that $\pi _{ab}F=0$ when $j<j_{ab}$, so we make this assumption throughout the rest of the section without further mention.

Proposition B.7

Suppose $F\in \mathcal {K}$ vanishes at $W$-weighted infinity. Then $\mathbb {E}_{+}\theta F$ vanishes at $W_+$-weighted infinity.

Proof

Let $C_{F} = \Vert F \Vert _{W}$ and let $X$ be a polymer. Given $R>0$, let $f_{R}={1\!\!1}_{\{\Vert \xi \Vert _{\Phi (X)}\le R\}}$. We write $I_{R} = \mathbb {E}_{+} f_{R}\theta F (X)$ and $I_{\not R} = \mathbb {E}_{+} (1-f_{R})\theta F (X)$ so that

$$\begin{aligned} \mathbb {E}_{+}\theta F (X) = I_{R} + I_{\not R}. \end{aligned}$$

(B.18)

By (B.10), we can choose $R$ large such that

$$\begin{aligned} \mathbb {E}_{+} \left[ |1-f_{R}(\xi )|\, W (X,\phi + \xi ) \right] \le C_{F}^{-1}\varepsilon W_{+}(X,\phi ) . \end{aligned}$$

(B.19)

Therefore, by (B.16),

$$\begin{aligned} \Vert I_{\not R}\Vert _{T_{\phi } (\mathfrak {h}_{+})}&\le \mathbb {E}_{+} |1-f_{R}(\xi )|\, \Vert F (X)\Vert _{T_{\phi +\xi }(\mathfrak {h})} \nonumber \\&\le \mathbb {E}_{+} |1-f_{R}(\xi )|\, C_{F} W (X,\phi +\xi ) \le \varepsilon W_{+}(X,\phi ) , \end{aligned}$$

(B.20)

and hence

$$\begin{aligned} \limsup _{\Vert \phi \Vert _\Phi \rightarrow \infty } \frac{1}{W_{+} (X,\phi )} \Vert I_{\not R}\Vert _{T_{\phi } (\mathfrak {h}_{+})} \le \varepsilon . \end{aligned}$$

(B.21)

Let

$$\begin{aligned} M (\phi ) = \sup _{\xi } |f_{R}(\xi )| \frac{1}{W (X,\phi +\xi )} \Vert F (X)\Vert _{T_{\phi +\xi } (\mathfrak {h})} . \end{aligned}$$

(B.22)

By (B.16) and (B.9),

$$\begin{aligned} \Vert I_{R}\Vert _{T_{\phi } (\mathfrak {h})} \le \mathbb {E}_{+} |f_{R}(\xi )|\, \Vert F (X)\Vert _{T_{\phi +\xi }(\mathfrak {h})} \le M(\phi ) \mathbb {E}_{+} W(\phi +\xi ) \le M(\phi ) W_{+} (X,\phi ). \end{aligned}$$

(B.23)

When $\Vert \xi \Vert _{\Phi }\le R$, if $\Vert \phi \Vert _\Phi \rightarrow \infty$ then also $\Vert \phi +\xi \Vert _\Phi \rightarrow \infty$. Since $F$ vanishes at weighted infinity, it follows that $M (\phi ) \rightarrow 0$ as $\Vert \phi \Vert _\Phi \rightarrow \infty$, and hence

$$\begin{aligned} \lim _{\Vert \phi \Vert _\Phi \rightarrow \infty } \Vert I_{R}\Vert _{T_{\phi } (\mathfrak {h})} W_{+}^{-1} (X,\phi ) = 0 \end{aligned}$$

(B.24)

With (B.21) and (B.18), this concludes the proof. $\square$

Appendix C: Polymer Geometry

We now prove some geometric lemmas used in our analysis. They concern $f_{j} (z,a,X)$, which is defined for $z \ge 0$ and $a\in (0,2^{-d}]$ and $X \in \mathcal {P}_{j}$ by (5.21). We begin with the following elementary but useful observation. We claim that for $X \in \mathcal {P}$, $0 \le a' < a$, $C \ge 1$, and for $\varepsilon$ sufficiently small,

$$\begin{aligned} C^{|X|_{j}}\varepsilon ^{f (z,a , X)}&\le C^{2^d} \varepsilon ^{f (z,a',X)}. \end{aligned}$$

(C.1)

This follows from

$$\begin{aligned} C^{|X|_{j}} \varepsilon ^{f (z,a ,X)}&\le \varepsilon ^z C^{2^d} (C\varepsilon ^a)^{(|X|_{j}-2^d)_+} \le \varepsilon ^z C^{2^d} (\varepsilon ^{a'})^{(|X|_{j}-2^d)_+} = C^{2^d} \varepsilon ^{f (z,a',X)}, \end{aligned}$$

(C.2)

when $\varepsilon$ is small enough that $C\varepsilon ^a \le \varepsilon ^{a'}$.

The following is a subadditivity property of $f_j$. Fix any $a \in [0,2^{-d}z]$, and let $X = \bigcup _i X_{i}$ be a nonempty union of disjoint sets $X_{1},\cdots X_{n} \in \mathcal {P}_{j}$. Then

$$\begin{aligned} f_{j} (z, a ,X) \le \sum _{i} f_{j} (z, a ,X_{i}) . \end{aligned}$$

(C.3)

To prove this, we observe that for $|X|_{j} \le 2^{d}$ the inequality reduces to $z \le \sum _{i}z$, and otherwise the left-hand side equals

$$\begin{aligned} z - a 2^{d} + \sum _{i} a |X_{i}|_{j} \le \sum _{i} \left( z - a 2^{d} + a |X_{i}|_{j} \right) \le \sum _{i} f_{j} (z, a ,X_{i}) . \end{aligned}$$

(C.4)

For $F,G \in \mathcal {K}_{j}$ it is straightforward to check that $F\circ G$ is in $\mathcal {K}_{j}$. We use the following estimate for the circle product several times.

Lemma C.1

Fix $0 < a_{\mathrm {out}}< a_{\mathrm {in}}\in (0,2^{-d}]$ and let $\bar{\epsilon }$ be sufficiently small depending on $a_{\mathrm {out}},a_{\mathrm {in}}$. Let $\varepsilon \in (0,1)$ and $\delta =2^{-2^{d}}\varepsilon$. If $F,G \in B_{\mathrm {in}} (\delta )$ then $F \circ G \in B_{\mathrm {out}} (\varepsilon )$.

Proof

By the triangle inequality, the product property of the norm, the hypotheses and the subadditivity (C.3) of $f_{j}$, we have, for $Z$ connected and $\bar{\epsilon }\le 1$,

$$\begin{aligned} \Vert (F \circ G ) (Z)\Vert _{j} \le \sum _{X \in \mathcal {P}(Z)} \Vert F (X)\Vert _{j}\, \Vert G (Z\setminus X)\Vert _{j} \le \delta \sum _{X \in \mathcal {P}(Z)} \bar{\epsilon }^{f_{j} (a_{\mathrm {in}},X)} \bar{\epsilon }^{f_{j} (a_{\mathrm {in}},Z\setminus X)} \end{aligned}$$

(C.5)

$$\begin{aligned} \le \delta 2^{|Z|} \bar{\epsilon }^{f_{j} (a_{\mathrm {in}},Z)} \le \delta 2^{2^{d}} \bar{\epsilon }^{f_{j} (a_{\mathrm {out}},Z)} = \varepsilon \bar{\epsilon }^{f_{j} (a_{\mathrm {out}},Z)} . \end{aligned}$$

(C.6)

The last inequality is obtained from (C.1) and requires $\bar{\epsilon }$ to be small. This completes the proof. $\square$

Lemma C.2

For $z \ge 0$, $z_\mathrm{lead}\ge a \ge 0$, and $X, Y$ disjoint with $X \ne \varnothing$,

$$\begin{aligned} f (z, a,X ) + z_\mathrm{lead}|Y|_{j} \ge f (z,a,X \cup Y). \end{aligned}$$

(C.7)

Proof

Case $|X |_{j}\ge 2^{d}$. The left-hand side equals

$$\begin{aligned} z + a \big (|X |_{j}-2^{d}\big ) + z_\mathrm{lead}|Y|_{j} \ge z + a \big (|X \cup Y|_{j}-2^{d}\big ), \end{aligned}$$

(C.8)

which equals the desired right-hand side.

Case $|X |_{j} < 2^{d}, \, |X\cup Y|_{j} \ge 2^{d}$. Since $X$ is not empty the left-hand side equals

$$\begin{aligned} z + z_\mathrm{lead}|Y|_{j}&> z + a |Y|_{j} + a \big (|X|_{j}-2^{d}\big ) = z + a \big (|X\cup Y|_{j}-2^{d}\big ) , \end{aligned}$$

(C.9)

which equals the desired right-hand side.

Remaining case $|X\cup Y|_{j} < 2^{d}$. Since $X$ is not empty, the left-hand side equals

$$\begin{aligned} z + z_\mathrm{lead}|Y|_{j} \ge z = z+ a \big (|X\cup Y|_{j}-2^{d}\big )_{+} , \end{aligned}$$

(C.10)

which equals the desired right-hand side. $\square$

The following lemma is stated (but not proved) above [24, Lemma 2]. A consequence of the lemma is that if $X\in \mathcal {S}_j$ then $\overline{X} \in \mathcal {S}_{j+1}$. The important geometrical constant $\eta = \eta (d)>1$ used in Lemma 5.6 is introduced in Lemma C.3.

Lemma C.3

There is an $\eta = \eta (d)>1$ such that for all $L \ge L_0(d)=2^d+1$ and for all large sets $X \in \mathcal {C}_{j}$,

$$\begin{aligned} |X|_j \ge \eta |\overline{X}|_{j+1}. \end{aligned}$$

(C.11)

In addition, (C.11) holds with $\eta =1$ for all $X\in \mathcal {P}_{j}$ (not necessarily connected, and possibly small).

Proof

Fix $L \ge L_0(d)=2^d+1$ (this restriction enters only in the third paragraph of the proof). It is clear that for any $m \ge 1$ the closure of any set of $m$ $j$-blocks contains at most $m$ $(j+1)$-blocks, and hence (C.11) always holds with $\eta =1$.

Assume that $X$ is a large connected set. Let $\varDelta = \varDelta (d)$ denote the maximum possible number of blocks that touch a connected set of $2^d+1$ blocks. We will prove (C.11) by induction on $|\overline{X}|_{j+1}$, with $\eta = 1+1/(2^d+1+2^d\varDelta )$.

To begin the induction, we claim that if $|\overline{X}|_{j+1}=2^d+1$ then $|X|_j \ge 2^d+2$, and hence

$$\begin{aligned} \frac{|X|_j}{|\overline{X}|_{j+1}} \ge \frac{2^d+2}{2^d+1} = 1+ \frac{1}{2^d+1} \ge \eta . \end{aligned}$$

(C.12)

To prove the claim, we proceed as follows. The maximum possible value of $|\overline{X}|_{j+1}$ is $|X|_j$, so we only need to rule out the case $|X|_j = |\overline{X}|_{j+1} = 2^d +1$, which we now assume. Let $D(X)$ be the integer part of $L^{-j}\max _{x,y \in X}|x -y |_\infty$; this is a measure of the diameter of $X$ counted in number of $j$-blocks. Then $D(X) \le 2^d+1\le L$. Also, every $j$-block in $X$ lies in a different $(j+1)$-block in $\overline{X}$. However, any set of $2^d+1$ $(j+1)$-blocks contains a pair of blocks $B_1,B_2$ that do not touch. Therefore $D(b_1\cup b_2) >L$ for every pair of $j$-blocks $b_1\in B_1$ and $b_2\in B_2$, so that $b_1 \cup b_2 \subset X$ is not possible. This contradiction proves the claim.

To advance the induction, suppose that (C.11) holds when $2^d+1 \le |\overline{X}|_{j+1} \le n$, and suppose now that $|\overline{X}|_{j+1}=n+1$. We remove from $\overline{X}$ a connected subset of $2^d+1$ blocks. The complement of this connected subset consists of no more than $\varDelta$ connected components (since if there were more then one of these components is not adjacent to the removed subset nor to any of the at most $\varDelta$ components adjacent to the removed subset, and hence $X$ would be disconnected). We list these components as $\overline{X}_1, \ldots , \overline{X}_\varDelta$, and choose $k \in \{0,1,\ldots ,\varDelta \}$ such that $|\overline{X}_i|_{j+1} \ge 2^d+1$ for $i \le k$ and $|\overline{X}_i|_{j+1} \le 2^d$ for $i >k$ (some of the latter components may be empty). Let $M =\sum _{i=1}^{k}|\overline{X}_i|_{j+1}$ and $m= \sum _{i=k+1}^{\varDelta }|\overline{X}_i|_{j+1}$. By the induction hypothesis applied to $\overline{X}_i$ for $i \le k$, and by (C.11) with $\eta =1$ for $i > k$,

$$\begin{aligned} \frac{|X|_j}{|\overline{X}|_{j+1}}&\ge \frac{2^d+2+ \eta M + m}{2^d+1+ M + m} = 1 + \frac{1+ (\eta -1) M}{2^d+1+ M + m} \nonumber \\&\ge 1 + \frac{1+ (\eta -1) M}{2^d+1 + M + \varDelta 2^d } = 1 + \frac{1+ (\eta -1) M}{\frac{1}{\eta -1} + M} = \eta , \end{aligned}$$

(C.13)

where we used our specific choice for the value of $\eta$ in the penultimate step (note that the last equality is true no matter what the value of $M$). This advances the induction and completes the proof. $\square$

Lemma C.4

Suppose that either $X_K$ has at least two components, or $X_K$ has at least one component and $X_{\delta I} \ne \varnothing$. Let $n_{\delta I} =|X_{\delta I}|_j$ and write $X_{K_i}$ for the connected components of $X_K$. Let $z \ge z_0 >0$. Let $0<a \le 1$ and let $\tilde{a}\in (a ,\eta a)$. There exist positive $\delta ,v$, depending on $d,z_0,\tilde{a},a$, such that

$$\begin{aligned} n_{\delta I} + \sum _i f_{j} (z,a , X_{K_i}) \ge v + \delta |\overline{X_{\delta I} \cup X_K}|_{j+1} + f_{j+1} (z, \tilde{a}, \overline{X_{\delta I} \cup X_K}) . \end{aligned}$$

(C.14)

Proof

Suppose first that $\overline{X_{\delta I} \cup X_K} \in \mathcal {S}_{j+1}$. Then the right-hand side is at most $v + \delta 2^{d} + z$. In the two cases listed at the beginning of the statement of the lemma, the left-hand side is at least $2z$, $1+z$. There exist $v,\delta$ positive so that each of these is greater than $v +\delta 2^{d} + z$.

So suppose now that $\overline{X_{\delta I} \cup X_K} \not \in \mathcal {S}_{j+1}$. For non-empty $X_{K}$ we let $\sum _i$ denote the sum over components $X_{K_{i}}$. We reduce $v,\delta$, if necessary, so that $\tilde{a}+ \delta \le \eta a$ and $v - \tilde{a}2^{d} \le - a 2^{d}$. By Lemma C.3, using $a \le 1$ and (C.3), we have

$$\begin{aligned} v + \delta |\overline{X_{\delta I} \cup X_K}|_{j+1} + f_{j+1} (z, \tilde{a}, \overline{X_{\delta I} \cup X_K})&\le v + z - \tilde{a}2^{d} +\eta a |\overline{X_{\delta I} \cup X_K}|_{j+1} \nonumber \\&\le v + z - \tilde{a}2^{d} + a n_{\delta I} + a |X_K|_{j} \nonumber \\&\le z - a 2^{d} + n_{\delta I} + a |X_K|_{j} \nonumber \\&\le n_{\delta I} + f_{j} (z, a , X_K) \nonumber \\&\le n_{\delta I} + \sum _{i} f_{j} (z, a , X_{K_{i}}), \end{aligned}$$

(C.15)

as required.

Lemma C.5

Let $0<z<2z'$. Recall the definition of $\mathcal {Y}_0(W)$ below (D.17). There exists $c = c (d)$ such that for $a_{\mathrm {in}}\in (0,c)$, $a_{\mathrm {out}}\in [0,a_{\mathrm {in}}]$, and for $(X, \{(U_B,B)\}, U_{M}) \in \mathcal {Y}_0(W)$,

$$\begin{aligned} z'|X|_j + \sum _i f_j (z,a_{\mathrm {in}},U_{M,i}) \ge (a_{\mathrm {in}}- a_{\mathrm {out}}) |W|_j + f_j (z,a_{\mathrm {out}},W) . \end{aligned}$$

(C.16)

Proof

Let $U_{M,i},\, i = 1,\cdots ,n_{M}$, be the components of $U_{M}$. Let $S$ denote the number of small sets $U$ that can contain a given block $B$. Then $|X^{\Box }|_j\le 2^d S|X|_j$, and hence, since $W=X^{\Box }\cup U_M$,

$$\begin{aligned} |W|_j \le 2^d S|X|_j + \sum _i |U_{M,i}|_j . \end{aligned}$$

(C.17)

Letting $u = a_{\mathrm {in}}- a_{\mathrm {out}}$ we rewrite this as

$$\begin{aligned} u|W|_j + a_{\mathrm {out}}|W|_j + z-a_{\mathrm {out}}2^{d} \le a_{\mathrm {in}}2^d S|X|_j + \sum _i a_{\mathrm {in}}|U_{M,i}|_j + z-a_{\mathrm {out}}2^{d} . \end{aligned}$$

(C.18)

The definition of $\mathcal {Y}_{0}$ excludes the case $X = \varnothing$ so we assume $X \not = \varnothing$ and we can also assume $W \not \in \mathcal {S}$, because $W=X^{\Box }\cup U_M$ and $X^{\Box } \not \in \mathcal {S}$. Then $f_j (z,a_{\mathrm {out}},W) = z-a_{\mathrm {out}}2^{d} + a_{\mathrm {out}}|W|_j$. Therefore the left-hand side is $u|W|_j + f_j (z,a_{\mathrm {out}},W)$. Let $v = z-a_{\mathrm {in}}2^{d}$ so that $v + a_{\mathrm {in}}|U|_j \le f_j (z,a_{\mathrm {in}},U)$. Then we can rewrite the inequality as

$$\begin{aligned} u|W|_j + f_j (z,a_{\mathrm {out}},W)&\le a_{\mathrm {in}}2^d S|X|_j + \sum _i a_{\mathrm {in}}|U_{M,i}|_j + v + u 2^{d} \nonumber \\&= a_{\mathrm {in}}2^d S|X|_j + \sum _i \left( v + a_{\mathrm {in}}|U_{M,i}|_j \right) + (1 - n_{M}) v + u 2^{d} \nonumber \\&\le a_{\mathrm {in}}2^d S|X|_j + \sum _i f_j (z,a_{\mathrm {in}},U_{M,i}) + (1 - n_{M}) v + u 2^{d} . \end{aligned}$$

(C.19)

We choose $a_{\mathrm {in}}> 0$ sufficiently small that $v = z - a_{\mathrm {in}}2^{d}\ge 0$. Decreasing $a_{\mathrm {in}}$ if necessary we have $a_{\mathrm {in}}2^dS + u 2^{d} \le z'$. If $n_{M} \ge 1$ then we use $a_{\mathrm {in}}2^dS|X|_j + u 2^{d} \le z' |X|_j$ to obtain the desired result.

Now we consider the case $n_{M}=0$, which is the same as $U_{M} = \varnothing$. Decreasing $a_{\mathrm {in}}$ if necessary, and using $z<2z'$, we have $a_{\mathrm {in}}2^d S + \frac{1}{2} z + u 2^{d} \le z'$. The definition of $\mathcal {Y}_{0}$ requires $|X|_j\ge 2$ when $U_{M} = \varnothing$ so

$$\begin{aligned} a_{\mathrm {in}}2^d S|X|_j + (1-n_{M})v + u 2^{d}&= a_{\mathrm {in}}2^d S|X|_j + v + u 2^{d} \nonumber \\&\le a_{\mathrm {in}}2^d S|X|_j + z + u 2^{d} \nonumber \\&\le \big ( a_{\mathrm {in}}2^d S + {\textstyle {\frac{1}{2}}}z + u 2^{d} \big )|X|_j \nonumber \\&\le z' |X|_j . \end{aligned}$$

(C.20)

This completes the proof. $\square$

Appendix D: Change of Variables

In this section, we prove Proposition 4.1, which for convenience we restate here as Proposition D.1. For further discussion of this proposition, see [13, Section 5]. This section applies for any norm $\Vert \cdot \Vert$ on $\mathcal {N}$ which obeys the product property [17, (1.44)]. We make use of (4.1)–(4.5), and in particular recall that $M$ is defined in (4.5), for $K_{\mathrm {in}}\in \mathcal {K}$ and $U \in \mathcal {C}$ and with $\bar{J} (U,B)=I_{\mathrm {in}}^UJ(U,B)$, by

$$\begin{aligned} M(U) = K_{\mathrm {in}}(U) - \sum _{B \in \mathcal {B}(U)} \bar{J} (U,B) . \end{aligned}$$

(D.1)

Proposition D.1

Let $a_{\mathrm {in}}$ be small as specified in Lemma C.5. Let $a_{\mathrm {out}}<a_{\mathrm {in}}$ and $z' > \frac{1}{2} z$. Let $\rho$ be sufficiently small depending on the difference $a_{\mathrm {out}}-a_{\mathrm {in}}$. Let $\varepsilon \in (0,1)$. Let $J,I_{\mathrm {in}}$ be as in (4.1)–(4.4). Suppose that $K_{\mathrm {in}}\in \mathcal {K}$ and $J$ satisfy

$$\begin{aligned}&\sup _{{\mathcal {D}}(J)} \Vert I_{\mathrm {in}}^{U} J (U,B) \Vert \le \varepsilon \rho ^{z'}, \end{aligned}$$

(D.2)

$$\begin{aligned}&M\in B_{\mathcal {F}_\mathrm{in}} (\varepsilon \rho ^{z}) . \end{aligned}$$

(D.3)

Then there exists $K_{\mathrm {out}}\in \mathcal {K}$ such that

$$\begin{aligned}&(I_{\mathrm {in}}\circ K_{\mathrm {out}}) (\varLambda ) = (I_{\mathrm {in}}\circ K_{\mathrm {in}})(\varLambda ), \end{aligned}$$

(D.4)

$$\begin{aligned}&K_{\mathrm {out}}\,\text {is polynomial in}\,I_{\mathrm {in}},\bar{J},K_{\mathrm {in}}, \end{aligned}$$

(D.5)

$$\begin{aligned}&K_{\mathrm {out}}= M+E\,\text {with}\,E\in B_{\mathcal {F}_\mathrm{out}} (\varepsilon \rho ^{z+(a_{\mathrm {in}}-a_{\mathrm {out}})/2}). \end{aligned}$$

(D.6)

If $K_{\mathrm {in}}=0$ and $J=0$, then $K_{\mathrm {out}}=0$.

Proof

For $U \in \mathcal {C}$ and $B \in \mathcal {B}$, with $\bar{J} (U,B)=I_{\mathrm {in}}^UJ(U,B)$, let

$$\begin{aligned} \bar{J}(U) = \sum _{B \in \mathcal {B}(U)} \bar{J}(U,B) . \end{aligned}$$

(D.7)

For $U_J\in \mathcal {P}$, let

$$\begin{aligned} \bar{J}(U_J) = \prod _{U \in \mathrm{Comp}(U_J)} \bar{J}(U). \end{aligned}$$

(D.8)

By the definition of $M$ and the component factorisation property of $K_{\mathrm {in}}$,

$$\begin{aligned} (I_{\mathrm {in}}\circ K_{\mathrm {in}})(\varLambda )&= \sum _{U_{\mathrm {in}}\in \mathcal {P}} I_{\mathrm {in}}^{\varLambda \setminus U_{\mathrm {in}}} K_{\mathrm {in}}(U_{\mathrm {in}}) \nonumber \\&= \sum _{U_{\mathrm {in}}\in \mathcal {P}} I_{\mathrm {in}}^{\varLambda \setminus U_{\mathrm {in}}} \prod _{U \in \mathrm{Comp}(U_{\mathrm {in}})}\big (\bar{J} (U) + M(U)\big ) \nonumber \\&= \sum _{U_{\mathrm {in}}\in \mathcal {P}} I_{\mathrm {in}}^{\varLambda \setminus U_{\mathrm {in}}} \sum _{\hat{U}_M \subset \mathrm{Comp}(U_{\mathrm {in}})} \bar{J} (U_{\mathrm {in}}\setminus U_M) M(U_M) , \end{aligned}$$

(D.9)

where $U_{M}$ is the union of components in $\hat{U}_{M}$.

Given $X \in \mathcal {P}$, let $B_1,\ldots , B_n$ be a list of the blocks in $\mathcal {B}(X)$, and let

$$\begin{aligned} \mathcal {U}(X) =&\{ \{(U_{B_1},B_1), \ldots , (U_{B_n},B_n)\} : \nonumber \\&U_{B_i} \in \mathcal {P}, \; U_{B_i} \supset B_i,\; U_{B_i}\,\,\text {does not touch}\,\, U_{B_j} \text {for}\,i \ne j\} . \end{aligned}$$

(D.10)

(In particular, $\mathcal {U}(X)$ is empty if any two blocks of $X$ touch each other.) Given an element of $\mathcal {U}(X)$, we write $U_J = \cup _{B \in \mathcal {B}(X)}U_B$, and write $\mathcal {P}'(\varLambda \setminus U_J)$ for the set of polymers that do not touch $U_J$. By interchanging the sums over blocks $B$ in (D.7) and polymers $U_B$, we obtain

$$\begin{aligned} (I_{\mathrm {in}}\circ K_{\mathrm {in}})(\varLambda )&= \sum _{X\in \mathcal {P}(\varLambda )} \sum _{\{(U_B,B)\}\in \mathcal {U}(X)} \left( \prod _{B \in \mathcal {B}(X)} \bar{J} (U_B,B) \right) \nonumber \\&\quad \quad \quad \times \sum _{U_{M} \in \mathcal {P}'(\varLambda \setminus U_J)} M(U_{M}) I_{\mathrm {in}}^{\varLambda \setminus (U_{M} \cup U_J)}. \end{aligned}$$

(D.11)

Recall the definition of the small set neighbourhood of $X$ in Definition 1.6. For $W=X^{\Box } \cup U_M$, we write

$$\begin{aligned} I_{\mathrm {in}}^{\varLambda \setminus (U_{M} \cup U_J)} = I_{\mathrm {in}}^{\varLambda \setminus W} I_{\mathrm {in}}^{W \setminus (U_{M} \cup U_J)}. \end{aligned}$$

(D.12)

With this notation the claim (D.4) holds with $K_{\mathrm {out}}$ given by

$$\begin{aligned} K_{\mathrm {out}}(W) = \sum _{(X, \{(U_B,B)\}, U_{M}) \in \mathcal {Y}( W)} \left( \prod _{B \in \mathcal {B}(X)} \bar{J} (U_B,B) \right) M(U_{M}) I_{\mathrm {in}}^{W \setminus (U_{M} \cup U_J)} \quad (W \in \mathcal {P}) .\nonumber \\ \end{aligned}$$

(D.13)

Here $\mathcal {Y}(W)$ denotes the set of triples $(X, \{(U_B,B)\}, U_{M})$, with $X \in \mathcal {P}(W)$, $\{(U_B,B)\} \in \mathcal {U}(X)$, $U_{M} \in \mathcal {P}'(\varLambda \setminus U_J)$, and $X^{\Box } \cup U_{M} = W$ (see Fig. 3). Note that the small set neighbourhood $X^\Box$ contains all possible unions $U_J$ of small sets in the summation over the $U_B$.

Fig. 3

This figure illustrates an element of $\mathcal {Y}(W)$. The white squares are the blocks of $X$, and are centred in two larger squares whose union is $X^\Box$. Each white square $B$ is contained in a small set $U_B$, which is itself contained in $B^{\Box }$. The components of $U_M$ are the two shaded components without white squares, and $W$ is the total area

Note that the above formula implies that if $J = 0$ then $K_{\mathrm {out}}=K_{\mathrm {in}}$. In particular, as claimed, $K_{\mathrm {out}}=0$ when $K_{\mathrm {in}}=0$ and $J=0$. Also, it shows that $K_{\mathrm {out}}$ is polynomial in $I_{\mathrm {in}},\bar{J},K_{\mathrm {in}}$ as claimed in (D.5). As an explicit example, for the case where $W$ is a single block $B$, (D.13) gives

$$\begin{aligned} K_{\mathrm {out}}(B)&= M(B) + \sum _{U : (U,B) \in {\mathcal {D}}(J)} \bar{J}(U,B) . \end{aligned}$$

(D.14)

The properties that define $\mathcal {Y}(W)$, together with the hypothesis that $K_{\mathrm {in}}\in \mathcal {K}$ and that $J$ obeys (4.3), can be used to verify the claim that $K_{\mathrm {out}}\in \mathcal {K}$. In particular, $K_{\mathrm {out}}$ obeys the factorisation property of Definition 1.7 by construction, and the field locality property holds because we have constructed $K_{\mathrm {out}}(W)$ as a polynomial in the local objects $\bar{J}$, $M$, $I_{\mathrm {in}}$ evaluated on sets contained in $W$.

For the bound claimed in (D.6), we first show that the contribution to (D.13) due to triples with $|X|=1$ and $U_{M} = \varnothing$ vanishes. This feature is a crucial ingredient. In this case, $X$ is a single block $B$, $K_{\mathrm {out}}(W)=0$ unless $W=X^{\Box }=B^{\Box }$ and thus $B$ is uniquely determined by $W$, and by (4.2), the contribution to (D.13) is

$$\begin{aligned} \sum _{U\in \mathcal {S}: U \supset B} J (U,B) I_{\mathrm {in}}^{B^{\Box }} = 0 . \end{aligned}$$

(D.15)

Let $W\in \mathcal {C}$. As in (5.21), we write

$$\begin{aligned} f_{j} (z, a,X) = {\left\{ \begin{array}{ll} z + f_{j} (a, X) &{}X\not = \varnothing \\ 0&{}X = \varnothing , \end{array}\right. } \end{aligned}$$

(D.16)

for $z \ge 0$, $a\in (0,2^{-d}]$ and $X \in \mathcal {P}_{j}$. We apply the triangle inequality, product property [17, (1.44)], and the hypotheses to (D.13), to obtain

$$\begin{aligned} \Vert K_{\mathrm {out}}(W) - M(W)\Vert \le \sum _{(X, \vec {U}, U_{M}) \in \mathcal {Y}_0(W)} \varepsilon \; \rho ^{z'|X|} \prod _i \left( \rho ^{f (z,a_{\mathrm {in}},U_{M,i})}\right) \alpha _{I}^{|W\setminus (U_J\cup U_{M})|} ,\nonumber \\ \end{aligned}$$

(D.17)

where $\mathcal {Y}_0(W)$ imposes the constraints on $(X,\{(U_B,B)\}, U_{M})$ required in (D.13) with the additional constraint that the terms with $X=\varnothing$, $U_{M} = W$, and with $|X|=1$, $U_{M} = \varnothing$ are omitted (the first omission is because $M$ is subtracted in (D.17) and the second is due to the cancellation in (D.15)). Since $\alpha _{I}\ge 1$, and since $W\setminus (U_J \cup U_{M}) \subset X^{\Box }$ and $|X^{\Box }| \le 2^dS|X|$, the power of $\alpha _{I}$ above can be replaced by $\mathrm{const}^{|W|}$. By Lemma C.5, for $(X, \{(U_B,B)\}, U_{M}) \in \mathcal {Y}_0(W)$, and with $2u = a_{\mathrm {in}}- a_{\mathrm {out}}$, we have

$$\begin{aligned} z'|X| + \sum _i f (z,a_{\mathrm {in}}, U_{M,i}) \ge 2u|W| + f (z,a_{\mathrm {out}},W). \end{aligned}$$

(D.18)

Therefore,

$$\begin{aligned} \Vert K_{\mathrm {out}}(W) - M(W)\Vert \le \varepsilon \rho ^{2u|W| + f (z,a_{\mathrm {out}},W)} \mathrm{const}^{|W|} |\mathcal {Y}_0(W)|, \end{aligned}$$

(D.19)

where $|\mathcal {Y}_0(W)|$ denotes the cardinality of $\mathcal {Y}_0(W)$. Let $S$ denote the number of small sets that can contain a given block $B$. For fixed $X$, there are at most $S^{|X|}$ possible choices of the small sets $U_B$ specified in the definition of $\mathcal {Y}_0(W)$. We use this, and also relax the summation to disjoint $X$ and $U_M$ in $W$. Since there are at most $3^{|W|}$ ways to partition $W$ into $X,U_{M},W\setminus (X\cup U_{M})$, we can absorb $|\mathcal {Y}_0(W)|$ into $\mathrm{const}^{|W|}$. Finally, we choose $\rho$ sufficiently small depending on $u$ so that $\mathrm{const}^{|W|} \le \frac{1}{2} \rho ^{-u|W|}$. Then

$$\begin{aligned} \Vert K_{\mathrm {out}}(W) - M(W)\Vert \le {\textstyle {\frac{1}{2}}} \varepsilon \, \rho ^{u|W| + f (z,a_{\mathrm {out}},W)} \le {\textstyle {\frac{1}{2}}} \varepsilon \, \rho ^{u + f (z,a_{\mathrm {out}},W)}, \end{aligned}$$

(D.20)

which implies (D.6). $\square$

Appendix E: Approximate Isometry Between Finite and Infinite Volume

The proof of Theorem 2.5 uses Lemmas E.4–E.6, which are given below. Lemmas E.1–E.3 are used in the proof of Lemmas E.4 and E.5. In this section $k$ is any scale, in particular it can be $j$ or $j+1$. Let $X\in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$. A coordinate map $\iota$ from $X$ to a torus $\varLambda$ exists for all $\varLambda$ with $\mathrm{diam}(\varLambda ) \ge 2 \mathrm{diam}(X)$. Given a coordinate map $\iota :X\rightarrow \varLambda$ and a test function $g\in \Phi (\iota X)$, we define a test function $g_{\iota } \in \Phi (X)$ by $(g_{\iota })_{z} = g_{\iota z}$, where $\iota z$ is defined by letting $\iota$ act on the sequence $z$ componentwise.

For the first lemma, recall the pairing in the definition of the $T_\phi$ semi-norm in [15, Definition 3.3].

Lemma E.1

Let $X \in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$. For a coordinate map $\iota :X \rightarrow \varLambda$, $F \in \mathcal {N}(X)$, $g \in \Phi (\varLambda )$, and $\phi \in \mathbb {C}^{\varLambda }$,

$$\begin{aligned} \langle \iota F, g \rangle _\phi = \langle F , g_{\iota } \rangle _{\phi _{\iota }} \end{aligned}$$

(E.1)

Proof

Both sides are linear in $F$. By (1.81), it suffices to consider $F=F_y \psi ^y$. Then $\iota F = \iota (F_y) \psi _{\iota }^y$. According to (1.82),

$$\begin{aligned} \langle \iota F, g \rangle _\phi = \sum _{x \in (\iota X)^{*}}\frac{1}{x!} \Big (F_{y}(\phi _{\iota })\Big )_{x}g_{x,\iota y} = \sum _{x \in X^{*}}\frac{1}{x!} F_{x,y}(\phi _{\iota }) g_{\iota x,\iota y} = \langle F , g_{\iota } \rangle _{\phi _{\iota }} . \end{aligned}$$

(E.2)

$\square$

For real $t>0$ and a nonempty polymer $X\in \mathcal {P}_k( { {{\mathbb Z}}^d })$, let $X_{t}\subset { {{\mathbb Z}}^d }$ be the smallest subset that contains $X$ and all points in ${ {{\mathbb Z}}^d }$ that are within distance $tL^{k}$ of $X$. The next lemma expresses a sense in which coordinate maps are approximately isometries as maps between spaces of test functions. Our norms on test functions (see [15, Example 3.2] and [15, (3.37)]) depend on a parameter $\mathfrak {h}$. For the next lemma, we exhibit this dependence by writing $\Phi _{k} (\iota X,\mathfrak {h})$, etc.

Lemma E.2

Let $X \in \mathcal {P}_{k}({ {{\mathbb Z}}^d })$, let $s>0$, and let $\iota$ be a coordinate map from a polymer containing $X_{s}$ into a torus $\varLambda$. There exists $\mathfrak {h}_s>0$ and $c>0$ (independent of $X,k,\iota ,\varLambda$), with $1 \le \mathfrak {h}/\mathfrak {h}_s \le 1+cs^{-1}$, such that for $g \in \Phi _{k} (\iota X,\mathfrak {h})$,

$$\begin{aligned} \Vert g_{\iota }\Vert _{\Phi _{k}(X,\mathfrak {h})} \le \Vert g\Vert _{\Phi _{k}(\iota X,\mathfrak {h})} \le \Vert g_{\iota }\Vert _{\Phi _{k}(X,\mathfrak {h}_{s})}, \end{aligned}$$

(E.3)

and likewise for the $\tilde{\Phi }_{k}$ semi-norm.

Proof

We give the proof for the case $\varvec{\Lambda }= \varLambda$, because the general case is merely an elaboration of notation. We write $\Phi = \Phi _{k}$. By the definition of the norm on test functions, we see that it is sufficient to fix an integer $p\ge 1$ and prove the lemma for the case where the test function $g\in \Phi (\iota X,\mathfrak {h})$ is zero except on sequences of length $p$. The domain $\iota X$ is contained in a torus $\varLambda$. By thinking of $\varLambda$ as a hypercube in a lattice of hypercubes paving ${ {{\mathbb Z}}^d }$, we identify a test function on $\varLambda$ with a function on $({ {{\mathbb Z}}^d })^{p}$ which is periodic in each component. The $\Phi (\iota X,\mathfrak {h})$ norm of $g$ is the infimum of $\Vert g'\Vert _{\Phi (\varLambda )}$ over extensions $g'$ of $g$ to $\varLambda ^{p}$; by the identification and the definition of $g_{\iota }$ this is the infimum of $\Vert g'_{\iota }\Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})}$ over extensions $g'_{\iota }$ of $g_{\iota }$ to functions of $({ {{\mathbb Z}}^d })^{p}$ that are periodic in each component. The norm $\Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h})}$ is the same but the extensions are not constrained to be periodic. Therefore

$$\begin{aligned} \Vert g\Vert _{\Phi (\iota X,\mathfrak {h})} \ge \Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h})} \end{aligned}$$

(E.4)

which is the lower bound claimed in (E.3).

Let $r=\frac{s}{3}$. By the definition of $\Phi (X,\mathfrak {h})$, there exists an extension $\tilde{g}_{\iota } \in \Phi ({ {{\mathbb Z}}^d },\mathfrak {h})$ of $g_{\iota }$ such that

$$\begin{aligned} \Vert \tilde{g}_{\iota }\Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})} \le (1+ r^{-1}) \Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h})}. \end{aligned}$$

(E.5)

By [16, Lemma 3.3], there exists a function $\chi =\chi _r$, which is equal to $1$ on $X^{p}$ and $0$ on ${ {{\mathbb Z}}^d }^{p}\setminus X_{2r}^{p}$, and a constant $c_0>0$ (independent of $p$, $X$, and $L^j$), such that

$$\begin{aligned} \Vert \tilde{g}_{\iota }\chi \Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})} \le \big (1 + c_0 r^{-1} \big )^p\Vert \tilde{g}_{\iota }\Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})}. \end{aligned}$$

(E.6)

In combination with (E.5), this gives the existence of $\mathfrak {h}_s$ obeying the desired bound, such that

$$\begin{aligned} \Vert \tilde{g}_{\iota }\chi \Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})} \le \Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h}_{s})} . \end{aligned}$$

(E.7)

By hypothesis, the domain of $\iota$ strictly contains $X_{2r}$, so we can invert $\iota$ on $\iota X_{2r}$. Therefore $(\tilde{g}_{\iota } \chi )_{\iota ^{-1}}$ is an extension of $g|_{\iota X}$ to the subset $\iota X_{2r}$ of $\varLambda$. Provided $L$ is large enough so that $rL^{k}\ge p_{\Phi }$ for all $k$, the derivatives up to order $p_{\Phi }$ in each argument of this extension are zero near the inner boundary of $\iota X_{2r}$ so we can further extend by zero to all of $\varLambda$. Call this extension $G$. Then, by definition of $\Phi (\iota X,\mathfrak {h})$,

$$\begin{aligned} \Vert g\Vert _{\Phi (\iota X,\mathfrak {h})} \le \Vert G\Vert _{\Phi (\varLambda ,\mathfrak {h})} = \Vert \tilde{g}_{\iota }\chi \Vert _{\Phi ({ {{\mathbb Z}}^d },\mathfrak {h})} \le \Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h}_{s})} \end{aligned}$$

(E.8)

and this proves the upper bound of (E.3). $\square$

The next three lemmas express senses in which coordinate maps are isometries, provided a small change is made in the parameter $\mathfrak {h}$.

Lemma E.3

Let $X \in \mathcal {P}_{k}({ {{\mathbb Z}}^d })$, let $\iota : \tilde{X} \rightarrow \varLambda$ be a coordinate map with $X_{N/4} \subset \tilde{X}\in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$, and let $\phi \in \mathbb {C}^\varLambda$. The induced map $\iota : \mathcal {N}(X)\rightarrow \mathcal {N}(\iota X)$ is defined in (1.82). This map is linear, and there exists $\mathfrak {h}^-=\mathfrak {h}^-(N) \le \mathfrak {h}$, with $\mathfrak {h}^- \rightarrow \mathfrak {h}$ as $N=N (\varLambda )\rightarrow \infty$, such that $\Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h}^-)} \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \le \Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h})}$ for all $F \in \mathcal {N}(X)$.

Proof

The linearity of the map is clear. We write $\Phi =\Phi _{k}$. Let $F\in \mathcal {N}(X)$ and $g\in \Phi (\iota X,\mathfrak {h})$. By Lemma E.1, the definition of the $T_{\phi }(\mathfrak {h})$ norm, and Lemma E.2,

$$\begin{aligned} |\langle \iota F,g \rangle _{\phi }| = |\langle F,g_{\iota } \rangle _{\phi _{\iota }}| \le \Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h})} \Vert g_{\iota }\Vert _{\Phi (X,\mathfrak {h})} \le \Vert F\Vert _{T_{\phi _{\iota }(\mathfrak {h})}} \Vert g\Vert _{\Phi (\iota X,\mathfrak {h})} . \end{aligned}$$

(E.9)

Taking the supremum over $g$ with unit norm, we have $\Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \le \Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h})}$ which is one of the desired inequalities. For the reverse estimate, we consider $N(\varLambda ) \rightarrow \infty$, assume $\mathrm{{diam}}{X} < \frac{1}{4} \mathrm{{diam}}{\varLambda }$, let $s= \frac{1}{4} N(\varLambda )$ and write $h^-$ for $h_s$ of Lemma E.2. Note that $\mathfrak {h}^- \uparrow \mathfrak {h}$ as desired. By the second bound in Lemma E.2, for a test function $f\in \Phi (X,\mathfrak {h})$,

$$\begin{aligned} |\langle F,f \rangle _{\phi _{\iota }}| = |\langle \iota F,f_{\iota ^{-1}} \rangle _{\phi }| \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \Vert f_{\iota ^{-1}}\Vert _{\Phi (\iota X,\mathfrak {h})} \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \Vert f\Vert _{\Phi (X,\mathfrak {h}^-)} . \end{aligned}$$

(E.10)

Taking the supremum over $f$ with $\Vert f\Vert _{\Phi (X,\mathfrak {h}^-)}=1$, we have $\Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h}^-)} \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})}$, which completes the proof. $\square$

Lemma E.4

Let $X\in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$, $F\in \mathcal {N}(X^\Box )$, and let $\iota : \tilde{X} \rightarrow \varLambda$ be a coordinate map with $X_{N/4} \subset \tilde{X}\in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$. The map $F \mapsto \iota F$ is a linear map from $\mathcal {N}(X^\Box )$ to $\mathcal {N}(\iota X^\Box )$, and obeys

$$\begin{aligned} \Vert \iota F \Vert _{\mathcal {G}} \le \Vert F \Vert _{\mathcal {G}} \end{aligned}$$

(E.11)

for either choice of the regulators $\mathcal {G}=G$ or $\mathcal {G}= \tilde{G}$ (recall (1.35) and (1.36)).

Proof

The linearity of $F \mapsto \iota F$ is clear. By the definition of the norm, followed by Lemma E.3 and then Lemma E.2 (to bound the norm in the regulator),

$$\begin{aligned} \Vert \iota F \Vert _{\mathcal {G}} = \sup _{\phi \in \mathbb {C}^{\varLambda }} \Vert \iota F\Vert _{T_{\phi }} \mathcal {G}^{-1} (\iota X,\phi ) \le \sup _{\phi \in \mathbb {C}^{\varLambda }} \Vert F\Vert _{T_{\phi _{\iota }}} \mathcal {G}^{-1} (\iota X,\phi ) \nonumber \\ \le \sup _{\phi \in \mathbb {C}^{\varLambda }} \Vert F\Vert _{T_{\phi _{\iota }}} \mathcal {G}^{-1} (X,\phi _{\iota }) \le \Vert F \Vert _{\mathcal {G}} , \end{aligned}$$

(E.12)

and the proof is complete. $\square$

Lemma E.5

Let $\mathfrak {h}^-$ be as in Lemma E.3, and let $\gamma ^+= \gamma (\mathfrak {h}/\mathfrak {h}^-)^2$. For $X\in \mathcal {P}_{k}({ {{\mathbb Z}}^d })$, $F\in \mathcal {N}(X^{\Box })$, $\gamma \in (0,1]$, and for a coordinate map $\iota : \tilde{X} \rightarrow \varLambda$ with $X_{N/4} \subset \tilde{X}\in \mathcal {P}_{k} ({ {{\mathbb Z}}^d })$,

$$\begin{aligned}&\Vert F\Vert _{T_{0}(\mathfrak {h}^-)} \le \Vert \iota F\Vert _{T_{0}(\mathfrak {h})} \end{aligned}$$

(E.13)

$$\begin{aligned}&\Vert F\Vert _{\tilde{G}^{\gamma ^+}(\mathfrak {h}^-)} \le \Vert \iota F\Vert _{\tilde{G}^{\gamma }(\mathfrak {h})}. \end{aligned}$$

(E.14)

Proof

We only prove (E.14), because (E.13) is a specialisation of the same method to $\phi =0$. Let $\phi \in \mathbb {C}^{\varLambda }$. By Lemma E.2,

$$\begin{aligned} \Vert \phi \Vert _{\Phi (\iota X,\mathfrak {h})} \le \Vert \phi _\iota \Vert _{\Phi (X,\mathfrak {h}^-)} = (\mathfrak {h}/\mathfrak {h}^-) \Vert \phi _\iota \Vert _{\Phi (X,\mathfrak {h})}, \end{aligned}$$

(E.15)

and hence, by definition of the regulator, $\tilde{G}^\gamma (\iota X,\phi ) \le \tilde{G}^{\gamma ^+}(X,\phi _\iota )$. Therefore, by Lemma E.3,

$$\begin{aligned} \Vert F\Vert _{T_{\phi _{\iota }}(\mathfrak {h}^-)} \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \le \Vert \iota F\Vert _{T_{\phi }(\mathfrak {h})} \tilde{G}^{-\gamma }(\iota X,\phi ) \tilde{G}^{\gamma ^+}(X,\phi _{\iota }) . \end{aligned}$$

(E.16)

We divide by $\tilde{G}^{\gamma ^+}$ and take the the supremum over $\phi$ to complete the proof. $\square$

Let $\mathcal {X}\subset \mathcal {C}_{k} (\mathbb {V})$, and let $F:\mathcal {X}\rightarrow \mathcal {N}$ have the properties listed in Definition 1.7 except that Euclidean covariance is replaced by the restricted version that if $X,Y \in \mathcal {X}$ and $E$ is a Euclidean automorphism such that $Y=EX$ then $E (F (X)) = F (EX)$. Let $W:\mathcal {C}_{k} (\mathbb {V})\rightarrow {\mathbb R}_{+}$ be a Euclidean invariant function such that $\Vert F (X)\Vert _{k} \le W (X)$ for $X \in \mathcal {C}(U)$. The following lemma, whose proof does not depend on the other lemmas in this appendix, shows that $F$ has an extension to an element of $\mathcal {K}(\mathbb {V})$.

Lemma E.6

Any $F:\mathcal {X}\rightarrow \mathcal {N}$ as above has an extension to an element $\hat{F} \in \mathcal {K}(\mathbb {V})$ such that $\Vert \hat{F} (X)\Vert _{k} \le W (X)$ for $X \in \mathcal {C}(\mathbb {V})$. The map $F \mapsto \hat{F}$ is linear, and if $F (X)$ satisfies (1.37) for $X$ in $\mathcal {X}$, then the same is true for $\hat{F} (X)$ for all polymers $X$.

Proof

For $X\in \mathcal {C}_{k} (\mathbb {V})$ such that $X = EY$ for some automorphism $E$ of $\mathbb {V}$ and some $Y\in \mathcal {X}$, define $\hat{F} (X)= EF (Y)$. If there exists another $Y' \in \mathcal {X}$ and an automorphism $E'$ such that $X=E'Y'$, then $A=E^{-1}E'$ is a Euclidean automorphism such that $AY'=Y$. By hypothesis, $E'F (Y') = EAF (Y') = EF (AY') = EF (Y)$, so this definition of $\hat{F}$ is not dependent on choices. If there is no pair $Y,E$ such that $X=EY$ then define $\hat{F} (X)=0$. By construction $\hat{F}$ has the properties listed in Definition 1.7, the extension is bounded by $W$, linear in $F$, and preserves the property (1.37). $\square$

Appendix F: Aspects of Symmetry

We now prove properties of the polynomial $Q$ of (1.70), and prove in particular that it lies in $\mathcal {Q}$ as claimed below (1.70). In addition, we prove that Gaussian expectation preserves defining properties of the space $\mathcal {K}$ in Definition 1.7; this is used in the proof of Proposition 5.1.

We draw attention to a notational clash in this appendix: $Q$ denotes the polynomial (1.70) in Lemma F.2, whereas $Q$ denotes the supersymmetry generator (see [9, Section 5.2.1] or [21, Section 6]) in Lemma F.3.

For the following lemma, we write $F|_0$ for the constant part of $F\in \mathcal {N}$, which results from setting $\phi =0$ and $\psi =0$ in $F$.

Lemma F.1

For $F \in \mathcal {N}$ and $X \subset \varLambda$, the constant monomial of $\mathrm{Loc} _X F$ is $F|_0$.

Proof

It is the defining property of $\mathrm{Loc} _X F$ in [16, Definition 1.6] that $\langle F,g \rangle _0 = \langle \mathrm{Loc} _X F,g \rangle _0$ for all test functions $g$ in the space $\varPi$ of polynomial test functions. One such test function is $g_\varnothing = 1$ (a test function with no arguments). By setting $g=g_\varnothing$ in the pairing, we obtain $F|_0=(\mathrm{Loc} _X F)|_0$. Since $(\mathrm{Loc} _X F)|_0$ is the constant monomial of $\mathrm{Loc} _X F$, the proof is complete. $\square$

Lemma F.2

The formula $Q (B) = \sum _{Y \in \mathcal {S}(\varLambda ) : Y \supset B} \mathrm{Loc} _{Y,B} I^{-Y} K (Y)$ of (1.70) defines an element $Q\in \mathcal {Q}$.

Proof

The operator $\mathrm{Loc}$ preserves Euclidean covariance, gauge invariance, and supersymmetry, according to [16, Proposition 1.9] and [16, Proposition 1.14]. Since $V$ and $K$ have these properties, therefore $Q$ also has them. It is then a consequence of [9, Lemma 5.3] that $Q$ lies in $\mathcal {Q}$, once we prove that $Q$ cannot have a constant term. But by Lemma F.1, the constant monomial in $\mathrm{Loc} _Y I^{-Y}K(Y)$ equals the constant part of $I^{-Y}K(Y)$, and this is zero by the assumption that $K \in \mathcal {K}$ and $V \in \mathcal {Q}$. Therefore, the constant monomial in $\mathrm{Loc} _{Y,B} I^{-Y} K (Y)$ is also zero, and hence so is the constant monomial in $Q(B)$, as desired.

To understand the role of the block $B$ in more detail, we first note that any choice of $B$ determines $\pi _{\varnothing }Q$, because by the Euclidean invariance of $\pi _{\varnothing }K$ specified in Definition 1.7, (1.70) assigns the same value to $\pi _{\varnothing }Q$ for all choices of $B$. For the observable terms, because (1.9) contains indicator functions, (1.70) does not determine the coupling constants in $\pi _{\alpha }Q$ ($\alpha = a,b,ab$) unless $B$ contains $a$ or $b$. Taking all choices of $B$, (1.70) consistently determines a unique element $Q$ in $\mathcal {Q}$. $\square$

Lemma F.3

Let $F \in \mathcal {N}$ and suppose that $\mathbb {E}_{j+1} \theta F$ exists.

• (i) If $F$ is gauge invariant or Euclidean covariant, then so is $\mathbb {E}_{j+1}\theta F$.

• (ii) The supersymmetry generator $Q$ commutes with $\mathbb {E}_{j+1}\theta$, i.e., $Q\mathbb {E}_{j+1}\theta = \mathbb {E}_{j+1}\theta Q$. In particular, if $F$ is supersymmetric then so is $\mathbb {E}_{j+1}\theta F$.

• (iii) If $F$ is supersymmetric, then $\mathbb {E}_{j+1} F = F|_0$. In particular, if $F$ has zero constant part, then so does $\mathbb {E}_{j+1} \theta F$.

Proof

Throughout the proof, we write simply $\mathbb {E}$ for $\mathbb {E}_{j+1}$, and we omit some details. All forms in the proof have even degree.

• (i) Let $A_{j+1}=C_{j+1}^{-1}$. By definition,

  $$\begin{aligned} (\mathbb {E}\theta F)(\sigma ,\bar{\sigma },\phi ,\bar{\phi },\psi ,\bar{\psi }) = \int e^{-S_{A_{j+1}}(\xi ,\bar{\xi },\eta ,\bar{\eta })} F(\sigma ,\bar{\sigma },\phi +\xi ,\bar{\phi }+\bar{\xi },\psi +\eta ,\bar{\psi }+\bar{\eta }),\nonumber \\ \end{aligned}$$

  (F.1)

  where the action $S_{A_{j+1}}$ is Euclidean and gauge invariant. The claim can be seen to follow from this.

• (ii) From [9, (5.13)–(5.14)], we know that $\hat{Q} =(2\pi i)^{-1/2}Q$ commutes with $\mathcal {L}$ and hence also with $e^{\mathcal {L}}$. Since the action of $\mathbb {E}\theta$ on polynomials is the same as the action of $e^\mathcal {L}$ by [15, Lemma 4.2], this implies that $\mathbb {E}\theta QP = Q\mathbb {E}\theta P$ for polynomials $P \in \mathcal {N}$. A proof for general integrable elements of $\mathcal {N}$ can be based on the argument of [14, Lemma A.2], and we provide a sketch. By definition, $\theta F$ is a function of fluctuation and other fields, and the expectation integrates out the fluctuation fields leaving dependence on the others. We denote integration with respect to the fluctuation fields by $\int _1$, with respect to the other fields by $\int _2$, and with respect to all fields by $\int _{21}$. Then for a form $K_{12}$ depending on both fields, and a form $K_2$ depending on the other fields, since the integral of any $Q$-exact form is zero (see [21, p. 58]), we have $\int _{21}Q(K_2K_{12})=0$. Therefore, since $Q$ is an antiderivation and $K_2$ has even degree,

  $$\begin{aligned} \int _2K_2\int _1QK_{12}=\int _{21}K_2(QK_{12})= - \int _{21}(QK_2)K_{12} = - \int _{2}(QK_2)\int _{1}K_{12} .\qquad \end{aligned}$$

  (F.2)

  Similarly, $\int _{2} Q(K_2\int _1K_{12})=0$, and hence $\int _{2} (QK_2)\int _1K_{12}= -\int _{2} K_2Q\int _1K_{12}$. Thus we have shown that

  $$\begin{aligned} \int _2K_2\int _1QK_{12}= \int _{2} K_2Q\int _1K_{12}. \end{aligned}$$

  (F.3)

  Since $K_2$ is arbitrary, this implies that $Q\int _1K_{12}= \int _1QK_{12}$. We set $K_{12}=e^{-S_A} \theta F$ and use $Q e^{-S_{A}}=0$ (for $A=C_{j+1}^{-1}$) to conclude that

  $$\begin{aligned} Q\mathbb {E}\theta F= Q \int e^{-S_A}\theta F = \int Q(e^{-S_A} \theta F) = \int e^{-S_A} Q\theta F = \mathbb {E}Q\theta F. \end{aligned}$$

  (F.4)

  In particular, $Q\mathbb {E}\theta F = \mathbb {E}Q\theta F$. It suffices finally to show that $Q\theta F = \theta QF$. Since $Q$ is an antiderivation and $\theta$ is a homomorphism, it is enough to verify that $Q\theta F =\theta Q F$ for $F = f (\phi ,\bar{\phi })$, $F = \psi _{x}$, and $F=\bar{\psi }_x$. These are readily verified using $Q = d +\iota _{X}$ (see [21, (6.4)]).

• (iii) Suppose that $F$ is supersymmetric. Let $\tilde{F} = F - F|_0$, which is supersymmetric and has zero constant part. For $m \ge 0$, let $\tilde{F}(m) = e^{-m\sum _{x\in \varLambda } \tau _x}\tilde{F}$. We claim that $\mathbb {E}\tilde{F}(m)$ is independent of $m$. Indeed, let $v_{x} = \phi _x\psi _x$. Then $\tau _x = \hat{Q}v_x$ and since $\hat{Q}\tilde{F}=0$,

  $$\begin{aligned} \frac{\partial }{\partial m} \mathbb {E}\tilde{F}(m) = \sum _{x\in \varLambda } \mathbb {E}(\tau _x \tilde{F}(m)) = \sum _{x\in \varLambda } \mathbb {E}(\hat{Q}(v_{x} \tilde{F}(m))). \end{aligned}$$

  (F.5)

  The right-hand side of (F.5) is zero, since the integral of any $Q$-exact form vanishes (see [21, p. 58]). It follows that $\mathbb {E}\tilde{F} = \lim _{m\rightarrow \infty } \mathbb {E}\tilde{F}(m)$, and this limit vanishes since $\tilde{F}$ has zero constant part. Therefore, $\mathbb {E}F = F|_0$. In particular, since $(\mathbb {E}\theta F)|_0=\mathbb {E}F$, if $F$ has zero constant part then so does $\mathbb {E}\theta F$.$\square$