A two-dimensional area law from a global spectral gap

Proof. On-site and nearest-neighbor interactions have graph range at most one. Each inner boundary vertex is incident with between one and four crossing edges, so |\partial_{\rm in}A|\le|\partial_\Lambda A| \le4|\partial_{\rm in}A|. Counting both endpoints gives |Z_A|\le2|\partial_\Lambda A|\le4|Z_A|. These inequalities include empty boundaries.□

Proof. We include the mixture argument for the bound in [41, Lemma 2]. If \tau=\sum_j p_j\tau_j, adding a classical register that records j shows

For the lower bound, condition on that register and use strong subadditivity. For the upper bound, use the ordinary mixing upper bound on S(DE) and concavity of S(E). The ordinary bound follows by recording a duplicate of the classical register: its conditional entropy given DE is nonnegative after this further conditioning.

If \delta>0, write the Jordan decomposition \rho-\sigma=\delta(\tau_+-\tau_-), where \tau_\pm are states. Then

Apply the mixture bounds to both expressions and use |S(D|E)_{\tau_+}-S(D|E)_{\tau_-}|\le2\log d_D. This proves the first inequality. The second follows from the elementary bound h(u)\le C\sqrt u. The case \delta=0 is immediate. Finally, I(D:E)=S(D)-S(D|E), so the two displayed entropy bounds apply.□

Proof. Such a set A is a union of graph components. Every support of finite graph diameter lies in one component, so H is the tensor sum of component Hamiltonians. Its ground eigenspace is the tensor product of their ground eigenspaces. If the former is one-dimensional, so is each latter space. The ground vector factors over the components and therefore over the cut.□

Proof. Put M(u)=\sum_jp_j^{1-u} and L(u)=\log M(u). Let P be the support projection of \rho. On P\mathcal H_B define F=\rho^{-u/2}, and extend F by the identity on the kernel of \rho. The normalized tilted vector and its marginal are

where the last power is zero on the kernel. Since F is invertible on the whole space,

An interaction supported in B^c is unchanged by this conjugation. An interaction supported in B has unchanged expectation, because

by [F,\sigma]=0 and cyclicity. Thus only crossing interactions contribute to the excitation energy of \Theta.

Fix such an interaction. For every complex z, use the support convention

This is an entire scalar function. Although \sigma^0=P, the vector \Theta lies in its support, so f_i(0)=\langle\Theta,h_i\Theta\rangle. At

the two scalar normalization factors cancel, giving

Indeed, on the support, \sigma^{z_u}=M(u)^{-z_u}F and \sigma^{-z_u}=M(u)^{z_u}F^{-1}. The endpoint support projections in the expectation remove any matrix elements entering the kernel.

On the imaginary axis the powers are unitary on the support, and hence |f_i(\mathrm i y)|\le c_0. Decompose h_i across the cut by matrix units on its portion in B:

Here each factor acts as the identity outside the appropriate portion of D_i. If s_j are the positive Schmidt probabilities of \Theta, a product contributes terms with scalar coefficient s_j^{1/2+z}s_k^{1/2-z}. For |\Re z|\le1/4,

Moreover, in the two Schmidt bases, Cauchy–Schwarz bounds every row and column sum of |(c_\alpha)_{jk}(d_\alpha)_{jk}| by \norm{c_\alpha}\norm{d_\alpha}. Compression to the Schmidt supports does not increase either norm. Summing first over a row or a column therefore gives

These bounds are uniform in \Im z. Apply the three-lines theorem on each of the two strips between real parts 0 and \pm1/4. Writing \ell_i=\log(\mathrm e d_i), we obtain

because \log(2d_i^2)\le2\ell_i. The case c_0=0 is immediate; otherwise this application may be made to f_i/c_0.

Hermiticity gives f_i(-\overline z)=\overline{f_i(z)}. The even Taylor coefficients at zero are therefore real, and the odd coefficients are purely imaginary. Cauchy’s coefficient estimate on the disk of radius \delta_i, followed by summation of the even terms, yields, for real |x|\le\delta_i/2,

In particular, it is the real linear term that vanishes; we do not assert that f_i'(0)=0.

For u in the interval of the lemma, |u|\le1/32 and |z_u|\le|u|. Since \ell_i\le\sqrt{\mathcal B}, the estimate (3.6) applies to every crossing interaction. Using (3.3) and (3.4), we conclude that

The full-system gap inequality \mathsf H-E_0\ge g_0(\id-\ket{\Psi}\bra{\Psi}) now gives

The last subtracted quantity is at most \mathrm e\vartheta/[8(1+\vartheta)]<1/2. Taking logarithms and using -\log(1-v)\le2v for 0\le v\le1/2 gives

Iterate this inequality. Every successive halving remains in the permitted interval, for either sign of u, and 2^jL(u/2^j)\to uS because L(0)=0 and L'(0)=S. The sum of the remaining geometric series is at most two, proving (3.1). Finally apply the exponential Markov inequality with u=\pm[32\sqrt{(1+\vartheta)\mathcal B}]^{-1}. The quadratic term in (3.1) is at most \mathrm e/2. Adding the two one-sided estimates proves (3.2).□

Proof. If X is empty, the conclusion is immediate. Suppose henceforth that X\ne\varnothing. We will compare two estimates for the largest norm obtained by applying positive filters on nested regions around X. The filters need not commute. Stationarity and the gap give a comparison involving the entropies of those regions. For the opposite comparison, we pin a Schmidt vector of X and then average over its index. This removes X from each region and contributes a single entropy cost S(X). The difference of the two comparisons will therefore control a weighted sum of conditional entropies of X.

Choose an integer D>2R+2. Temporarily leave the integer C_{\mathrm{pad}}>D+1 variable, and put

All these rectangles lie in the safe parent Q. Any interaction meeting an X_j lies wholly in A: otherwise a graph path of length at most R from a support site in X_j to a support site of the opposite color would contain a point of Z within ambient distance R of Q. This contradicts safety. The same observation rules out any nearest-neighbor color-crossing edge incident with X_j. Consequently the edge boundary of X_j has at most 4(r+2d_j) edges, all crossing the ambient rectangle’s perimeter. Missing sites in \Lambda only reduce this count.

Every designated interaction support splits at most one of the cuts X_j\mid X_j^c. More precisely, a support split by X_j is disjoint from every X_i with i<j and contained in every X_i with i>j. Indeed, successive rectangles have an ambient margin D>R, whereas the ambient diameter of the support is at most its graph diameter R. A term that does not split a contour is either contained in it or disjoint from it. The support budget and (3.8) therefore give, with constants independent of C_{\mathrm{pad}},

where \mathcal B_j is the parameter in Lemma 3.1 for the marginal on X_j.

Assign a positive weight a_j to each contour. The energy estimate below charges contour j by a constant times (r+d_j)a_j^2. We choose weights of total mass two: in the eventual conditional-entropy comparison this will give the factor two needed for (3.9). At this fixed mass, Cauchy–Schwarz gives

Equality holds for weights proportional to 1/(r+d_j). Accordingly, put

Their sum is two, and their quadratic cost attains this lower bound. Integral comparison gives the uniform lower bound

For example, compare the sum with the integral of (2r+Dx)^{-1} over 1\le x\le m+1, use 2r+D(m+1)>(C_{\mathrm{pad}}+1)r, and use r\ge1 in the denominator. Thus we may make \mathcal E as small as needed, uniformly in r. We take C_{\mathrm{pad}} large enough that a_j\le1/2 for all j. We shall increase it once more after the energy estimate below; all constants up to that choice are independent of C_{\mathrm{pad}}.

Stationarity of the optimized filters. Fix a sufficiently small \epsilon>0, and maximize

over positive operators L_j on X_j satisfying

The trace is over \mathcal H_{X_j}. Choose \epsilon so small that (\dim\mathcal H_{X_j})\epsilon^{2/a_j}<1 for every j. The feasible sets are nonempty and compact. A maximum exists and is positive, since every feasible product is invertible. At an optimum write

We claim that

Prove this in descending order. For an anti-Hermitian operator B on X_j, vary L_j to \mathrm e^{tB}L_j\mathrm e^{-tB}. This preserves both constraints in (3.13). With K_{>j}=L_m\cdots L_{j+1}, the derivative of the logarithmic norm is

Assume that (3.14) is already known for indices greater than j. In this expectation the outer factor L_m can be removed by cyclicity of the trace on X_m, because its marginal commutes with L_m and the operator inside is supported on X_m. Repeat this argument down to L_{j+1}. Stationarity, together with the fact that the expectation of B is purely imaginary, gives

The matrix L_j^{-1}\rho_{\phi,X_j}L_j is therefore Hermitian. The equality to its adjoint, multiplied on both sides by L_j, shows that \rho_{\phi,X_j} commutes with L_j^2, hence with L_j. This proves the induction. Only expectations have been simplified; no commutation between distinct filters is asserted.

In a common eigenbasis let l_i be the eigenvalues of L_j, let p_i be those of \rho_{\phi,X_j}, and set x_i=l_i^{2/a_j}. A diagonal variation and the same descending trace argument give

The ratios p_i/x_i are equal to a common value \lambda on free coordinates and are at most \lambda on coordinates at the floor. There is a free coordinate by strict dimensional slack. Moreover \lambda>0: if all free coordinates had zero probability, transfer from one of them to a positive-probability floored coordinate would increase the norm. Consequently

The last inequality follows because taking a maximum with a common positive floor contracts differences of logarithms.

The optimized vector has small excitation energy. We have the exact identity KHK^{-1}\phi=E_0\phi for the original full Hamiltonian. Consider one interaction in its expectation. All outer filters on regions containing its support can be removed by (3.14) and cyclicity, as above. All remaining inner filters whose regions are disjoint from its support commute with the interaction. The single-split property leaves at most one conjugation L_jh_iL_j^{-1}.

In Schmidt bases for X_j\mid X_j^c, Hermitian pairing replaces the real part of this conjugation by coefficients involving

For positive p_i,p_k, (3.15) and a_j\le1/2 bound this expression by C a_j^2(p_i+p_k). To verify the uniform constant, put z=\tfrac12\log(p_i/p_k) and use

Pairs containing a zero probability have zero coefficient. A finite-range interaction has a product decomposition across this cut with a uniformly bounded sum of norm products. The row and column Cauchy–Schwarz estimate used in Lemma 3.1 now bounds its contribution by C a_j^2. Summing (3.10) gives

The constant depends only on q,R,J, not on the number of filters or their supports’ dimensions.

The full-system gap converts (3.16) into 1-|\langle\Omega,\phi\rangle|^2\le C\mathcal E/\Delta. Choose C_{\mathrm{pad}} large enough that this is at most 1/4. At the same time arrange that

Both choices are uniform in r: by (3.10), the right side is bounded below by a parameter-dependent constant times (r+d_j)^{-1/2}, whereas the left side is at most 4/[Z_r(r+d_j)]. Equation (3.12) completes the choice of C_{\mathrm{pad}}, which is now fixed.

The first norm comparison. Let N_j be the optimum for the prefix of filters 1,\ldots,j at the same floor, and set N_0=1. The preceding stationarity and energy arguments apply separately to each prefix: its summed cost is at most \mathcal E. Thus, for a parameter-dependent constant C, its normalized output has overlap at least \sqrt{1-C\mathcal E} with \Omega, where C\mathcal E\le1/4. For an optimizing prefix, put K_{<j}=L_{j-1}\cdots L_1. Cauchy–Schwarz gives

For any feasible L_j, Schatten Hölder gives

Apply Lemma 3.1 with u=-a_j/(1-a_j), permitted by (3.17). We obtain

After summing logarithms over the prefixes and using \log(1-C\mathcal E)\ge-C'\mathcal E, this gives

The constants in this estimate are independent of the floor.

The opposite comparison by Schmidt pinning. Let the floor decrease to zero at this fixed finite instance. The optima converge to the optimum over nonnegative filters with the same trace constraints. Indeed, these latter feasible sets are compact, and every feasible nonnegative operator is a limit of strictly positive, trace-normalized operators. Taking their minimal eigenvalues as admissible floors proves the assertion. We continue to denote the limiting optimum by N; (3.18) remains valid.

Write the Schmidt decomposition at X as

and let T_j=X_j\setminus X. For a fixed index i use trial filters

They are feasible in the zero-floor limit and give N\ge\sqrt{p_i}\,N_i, where

under the displayed constraints. We claim that

For fixed invertible choices M_j, consider F_i(z)=M_m^z\cdots M_1^z w_i on 0\le\Re z\le1. It is bounded and analytic there. Its norm on the imaginary axis is one. On the other boundary, put U_j=M_j^{\mathrm i y} and V_j=U_j\cdots U_1, with V_0=\id. Moving these unitaries to the left gives the exact identity

Each positive factor is conjugated only by unitaries supported on smaller nested regions. The remaining factors are therefore still positive, supported on their respective T_j, and have the same trace-power normalizations. The leftmost product of unitaries does not change the norm, so \norm{F_i(1+\mathrm i y)}\le N_i. The vector form of the three-lines theorem, obtained by testing against unit vectors and applying the scalar lemma [38, Lemma 5.10], gives \log\norm{F_i(x)}\le x\log N_i for 0\le x\le1. Differentiating from the right at zero yields

For 0<\delta<1, take

These are strictly positive and satisfy \Tr M_j^{2/a_j}=1. As \delta\downarrow0, their expectations of \log M_j tend to -a_jS_{w_i}(T_j)/2, since the marginal gives zero weight to its kernel. This proves (3.19).

For each Schmidt index the trial norm now gives

Average with probabilities p_i and use concavity of each T_j entropy. Their averaged states are the original marginals, so

The two norm estimates have now served their purpose. Combining (3.18) and (3.20) gives

Since the T_j are nested and \sum_j a_j=2, strong subadditivity bounds the left side below by 2S(X\mid T_m). Finally T_m\subseteq T, so conditioning further decreases the conditional entropy. Because m\le C_{\mathrm{pad}}r/D and all parameters have now been fixed, the resulting error is at most C_{\mathrm{buf}}r. This proves (3.9).□

Proof. Choose an integer M below, and consider a safe parent rectangle of size at most Mr. If its size is at most r, its entropy is already bounded by F_{\mathrm{box}}(r). Otherwise partition each axis into chunks of r sites and, if needed, one final shorter chunk. Their Cartesian products give at most M^2 children Q_i, each safe and of size at most r. Put X_i=A\cap Q_i.

Call a child interior if its C_{\mathrm{pad}}r padding lies in the parent. There are at most

noninterior children: in either coordinate only the first or last C_{\mathrm{pad}}+2 chunk positions can violate this containment. This estimate remains valid when the parent is thin and all of its children are noninterior.

For an interior child its padded rectangle meets at most (2C_{\mathrm{pad}}+5)^2-1 other children. Order all children by a uniform random permutation. The probability that this child comes after every one of those other children is at least

On that event, its preceding physical sites contain the entire buffer in Lemma 3.2. The corresponding term of the entropy chain rule is therefore at most S(X_i)/2+C_{\mathrm{buf}}r. In all other cases it is at most S(X_i). The latter bound also holds for noninterior children. Taking the expected entropy chain rule consequently gives

For completeness, the probability of the buffer event may vary among children. Its lower bound p_0 controls the negative term -p_0S(X_i)/2, since S(X_i)\ge0; its positive error can always be bounded by C_{\mathrm{buf}}r. Thus the displayed common bound holds without assuming equal event probabilities. Parents of size at most r satisfy the same inequality once its coefficient is at least one.

Choose M\ge16(C_{\mathrm{pad}}+2) and M\ge2. Then the coefficient in (3.22) is at most

Iterate the recursion at r=1,M,M^2,\ldots. The inhomogeneous terms are bounded by a geometric series with ratio M/\lambda<1, yielding F_{\mathrm{box}}(M^j)\le C\lambda^j for all j\ge0. Set

For a general r\ge1, comparison with the next power of M and monotonicity of F_{\mathrm{box}} prove (3.21) after a fixed change of constant.□

Proof. Fix B_y and define

For t\ge0, fix one such A, put f(t)=[\tau_t(A),B_y], and let Q_X(t)=\sum_{j:X_j\cap X\ne\varnothing}\tau_t(h_j). Since the terms disjoint from X commute with A, differentiating and using the Jacobi identity gives the exact equation

The propagator of the homogeneous equation is unitary conjugation. Variation of constants, followed by the supremum over A, consequently gives

The same argument with -H handles negative times.

Starting from X=X_i, iterate (4.5). At each step the sum of the available interaction norms is at most \kappa:=v_Rb_0, by (4.2). A chain X_i,X_{j_1},\ldots,X_{j_m} with consecutive intersections and with y\in X_{j_m} has d_\Lambda(a_i,y)\le(m+1)R. The remainder after M iterations is at most 2\|B_y\|(2\kappa t)^M/M!, since F(X,u)\le2\|B_y\|; it tends to zero. For every \mu>0, the resulting chain sum satisfies

This proves (4.3). If the distance is infinite, no finite chain contributes and the vanishing remainder proves exact zero.

For completeness, write the expectation in (4.1) as a product of on-site twirls T_y. These are commuting contractions, and telescoping that product shows, for any B,

where U_y ranges over on-site unitaries. Sites in another component contribute zero. The number of sites at graph distance d is at most C(d+1)^2, because the graph ball lies in an ambient lattice ball. Applying (4.3) and summing \sum_{d>l}(d+1)^2e^{-cd}\le C'e^{-c'l} proves the exponential bound in (4.4). Contractivity gives its other bound.□

Proof. We give the derivative estimate responsible for the exponent. For real u\in(-1,1) set w=1-u^2. Choose \eta_p>0 sufficiently small. On the complex disk |z-u|\le\eta_pw one has

The last inequality follows by continuity at \zeta=0, with p fixed. The function e\exp[-(1-z^2)^{-p}] is analytic on the disk, and Cauchy’s estimate gives

For each fixed m this tends to zero at both endpoints. Extension by zero is therefore smooth. To make the estimate uniform in u, substitute s=w^{-p} and maximize s^{m/p}e^{-c_ps}. Its maximum is bounded by C^{m+1}(m!)^{1/p}, using m!\ge(m/e)^m; the case m=0 is absorbed in C. Scaling from u to \omega yields

Integration by parts m times has no boundary terms. Consequently, for t\ne0,

For sufficiently large |t|, choose m=\lfloor\eta |t|^{p/(p+1)}\rfloor with a fixed, sufficiently small \eta>0. The expression in parentheses is at most 1/2, proving the decay in (4.6); bounded t is covered by increasing C. In particular f\in L^1(\mathbb R). Fourier inversion gives the first identity in (4.6). It can also be obtained directly by inserting e^{-\varepsilon t^2}, interchanging the two integrals, and letting the resulting Gaussian approximate identity converge to the continuous function \chi. Reality and evenness follow from those properties of \chi.□

Proof. Use Lemma 4.2 with \delta=\Delta/2, and define

The integral converges in norm and M_i is Hermitian. In an eigenbasis of H, its uncentered ground-to-energy-E matrix element equals the corresponding matrix element of h_i multiplied by \chi(E-E_0). All excited energies satisfy E-E_0\ge\Delta, where \chi vanishes. The ground component is unchanged because \chi(0)=1, and the scalar in (4.10) removes it. Hence M_i\Omega=0. Also, since \tau_t(H)=H and \int f=1,

Choose c_*:=\max\{1,J(\|f\|_1+1)\} and set k_i=|M_i|/c_*. For a Hermitian operator, |M_i|\ge M_i and \ker|M_i|=\ker M_i. These observations and (4.11) prove (4.8).

It remains to check locality after taking the absolute value. The scalar in (4.10) is fixed by E_{i,l}. For |t|\le\varepsilon l, choose \varepsilon>0 small enough that (4.4) is at most Ce^{-c'l} for A=h_i. For the remaining times use its bound 2J. Thus

The last tail estimate follows by substituting u=|t|^\alpha and absorbing the resulting fixed power of u into the exponential. The case l=0 follows from the uniform norm bound.

We use the dimension-independent inequality

Here is an order proof. The integral formula

follows from scalar integration and the spectral theorem. It shows that the square root is operator monotone: inversion reverses the order of positive invertible operators, and T(T+sI)^{-1}=I-s(T+sI)^{-1}. With \epsilon=\|U-V\|, monotonicity and scalar functional calculus give U^{1/2}\le(V+\epsilon I)^{1/2}\le V^{1/2}+\sqrt\epsilon I. Interchanging U,V proves (4.13).

Apply this with U=M_i^2 and V=(E_{i,l}M_i)^2. Both factors in the latter square are Hermitian, and their norms are at most c_*. Therefore

The operator |E_{i,l}M_i| is supported in N_l(a_i). Since E_{i,l} fixes it and is contractive,

After division by c_* this proves (4.9); its order bounds follow from positivity and unitality of E_{i,l}.

Finally, every original support lies in one component. The Hamiltonian is a tensor sum over components, so \tau_t(h_i) is supported in the component of a_i. The integral, scalar centering, and functional calculus in the above construction all stay in that component’s operator algebra. This proves the exact component assertion.□

Proof. Apply (4.13) to k_i,k_{i,l} and to I-k_i,I-k_{i,l}, using (4.9). The order and ground-vector assertions follow from functional calculus. Each displayed channel is a sum of two maps of the form B\mapsto A^*BA; this proves complete positivity, while the sum of the squared Kraus operators is I. More explicitly, the map v\mapsto G_iv\oplus K_iv is an isometry and \mathcal E_i(B) is the compression of B\oplus B by that isometry. This proves contractivity, and the same proof works for the local maps and after adjoining an identity tensor factor. Finally,

also on every auxiliary extension. Applying this to both summands proves (4.15), where the completely bounded norm is the supremum of these extended operator norms. An operator commuting with the ball algebra commutes with both local square roots, so its image is itself times G_{i,l}^2+K_{i,l}^2=I.□

Proof. We first prove the summation estimate for any fixed \alpha\in(0,1) for which (4.9) holds, and any integer r_0\ge1. At each finite distance d from S_0 there are at most C\mu_R|S_0|(d+1)^2 labels: every anchor at that distance belongs to N_d(x) for some x\in S_0, and anchor multiplicity is bounded by \mu_R. Labels at infinite distance have no error by definition. The triangle inequality therefore gives

To justify the final inequality, the terms d\le2r_0+2 have sum at most C(r_0+1)^3e^{-cr_0^\alpha}. For d>2r_0+2 use \lfloor d/2\rfloor\ge d/3; comparison with the integral of (x+1)^2e^{-c(x/3)^\alpha}, followed by u=x^\alpha, bounds the remaining tail by a fixed power of r_0+1 times e^{-c''r_0^\alpha}. All such powers are absorbed by decreasing the positive exponential constant. Thus (4.21) has no factor involving |\Lambda|.

Now take \alpha=1/2, |S_0|\le C_0n^2, and r_0=\lceil C_1(\log n)^2\rceil. The right-hand side of (4.21) is at most C'C_0n^{2-c'\sqrt{C_1}}. Taking C_1 sufficiently large proves (4.17), including its g/4 bound for all n\ge2.

Positivity gives \widetilde E_0\ge0, and testing \widetilde H on \Omega gives \widetilde E_0\le\epsilon_n. The min–max principle and (4.17) put the second eigenvalue of \widetilde H at least g-\epsilon_n. The difference between its first two eigenvalues is consequently at least g-2\epsilon_n\ge g/2, which also proves uniqueness. Moreover,

The trace distance of two pure states equals the square root of one minus their squared overlap, giving the second estimate in (4.18). Choose the overlap real and nonnegative, and use 1-s\le1-s^2 for 0\le s\le1 to obtain the first estimate.

Finally, if a designated support meets X\subseteq S_0, then its anchor has finite distance from S_0. If r_i>r_0, formula (4.16) gives r_i=\lfloor d_i^0/2\rfloor<d_i^0, so that its ball cannot meet S_0. Hence every label in \mathcal C_X has r_i=r_0. If its anchor lies in X, take a shortest path from the anchor to a point of the ball outside X; otherwise take one to a point inside X. This path stays in the ball and contains an edge of \partial_\Lambda X. Thus the anchor is within r_0 of an endpoint of a crossing edge. There are at most 2b_X endpoints, at most C(r_0+1)^2 sites in each such ball, and at most \mu_R labels per anchor. This proves the first bound in (4.19). The ball volume bound also gives \log d_i=|\widetilde X_i|\log q\le C(r_0+1)^2. Summing its squared version proves (4.20). Since r_0+1\le C(\log n)^2 for n\ge2, its stated consequence follows.□

Proof. Purity and strong subadditivity give the displayed identities and \eta\ge0. It suffices initially to take \sigma,\tau faithful. Write d_U=S(x|U)_\theta and d_P=S(x|P)_\theta. Consider

The midpoint is the vector in (5.4). All imaginary powers here are unitary. In particular G is bounded on this strip, with bounds allowed at this intermediate step to depend on its fixed invertible factors.

On the left boundary, discard the leftmost unitary factors and move the remaining imaginary powers past the commuting real powers. With \rho=\rho_{xU}, Q=I_x\otimes\widehat\rho_U, and \tau_u=Q^{-iau}\tau Q^{iau}, this gives

Hölder’s inequality and (5.3) imply

We retain the slack in this bound: a large slack already improves the left boundary, while a small slack will control the distance from the right-boundary input to \theta. Choose a universal C_0 larger than half the constant in this last bound and set

The normalized Hölder overlap then satisfies

We need to relate this overlap to \rho, independently of the smallest eigenvalues of either marginal. Taking \rho in place of \tau_u for this comparison gives

The trace is real and nonnegative. Thus 1-\Tr\rho^{2a}\xi^{1-2a}\le C a^2\ell^2. The function b\mapsto1-\Tr\rho^{[b]}\xi^{[1-b]} is nonnegative and concave on [0,1]: expand in their two eigenbases for concavity, and use Hölder for nonnegativity. For 2a\le1/2, concavity and its nonnegative value at zero give

Consequently

We next show how the slack in (5.5) controls the state arriving at the opposite boundary. Define

This is a unit vector with the same P marginal as \theta. We claim

Here and in the verification all kernel phases are extended by one. First,

For the remaining comparison of \tau_u and \rho insert the phase of \xi. For any densities A,B and real b,

To check this, expand in their two spectral bases. For positive eigenvalues with ratio v, use |1-v^{ib}|\le C(1+|b|)|1-\sqrt v|. This follows from the mean-value theorem when 1/4\le v\le4, and the bound two outside that interval. At a zero eigenvalue of A the same assertion follows from the bound two and |1-\sqrt0|=1; zero eigenvalues of B contribute nothing to the left side.

Replacing \sqrt\rho by \sqrt\xi in the \tau_u,\xi phase difference costs at most 2H. On \sqrt\xi, give a pair of eigenvalues p of \xi and q of \tau_u the probability p|\langle p,q\rangle|^2 and put v=\log(q/p). Faithfulness of \tau_u makes this finite, and

by (5.6). On v\le0,

and

The last step uses e^{2av}-1\le2a(e^v-1) for v>0. On v>0 the squared phase difference is at most C a^2u^2 e^v. These estimates, followed by (5.9) for \xi,\rho and (5.7), prove (5.8).

The phase comparison now lets us recover the other conditional entropy. On the right boundary, the remaining P phases merely conjugate \sigma. Apply (5.3) and Hölder on xP to the input \theta_u, whose P marginal is unchanged. It follows that

The continuity bound for conditional entropy in Lemma 2.1, using only \dim x, and (5.8) now give

On the left boundary the bound is exactly -ad_U+C_0a^2\ell^2-\lambda(u).

For clarity, the logarithmic interpolation used here assigns density

to each boundary line when evaluated at the midpoint. Each line has total mass 1/2. This follows by mapping the strip to the upper half-plane by z\mapsto e^{\pi iz} and using harmonic measure at i. Applied to the logarithm of a scalar analytic function it follows from the submean inequality, first on bounded domains and then by a limit. Pairing G with a unit vector in its midpoint direction gives the vector-norm assertion. All boundary norms above are bounded above and away from zero for the fixed faithful factors, so this application and the limit are legitimate.

Pair the two boundary estimates at the same u. Young’s inequality gives

After integration, this term, the quadratic errors, and the remaining a^{5/4}\ell^{3/2} term are all at most C a^{5/4}\ell^2. The entropy contribution is -a(d_P+d_U)/2, proving (5.4) for faithful \sigma,\tau. Approximation by faithful densities removes that restriction by continuity of positive powers. Finally normalizing subdensities only increases the left side; the zero subdensity case is immediate.□

Proof. The proof writes \eta as the integral of a nonnegative resolvent defect. Two moment bounds, involving only d, control the ends of the integral; the entropy controls the part between them. Initially let \rho_f=\rho_{xUF} be faithful, and put

On their Hilbert–Schmidt matrix spaces define positive definite operators

where L and R denote left and right multiplication. Define the isometry

It is an isometry by taking the partial trace of \rho_f. Taking matrix elements in the same way with \sigma_f shows V^*AV=B: both sides have the form \Tr X^*\sigma_s Y. Put b_0=\sqrt{\rho_s} and a_0=Vb_0=\sqrt{\rho_f}. For v>0, let

Expansion of its weighted squared norm, using V^*(A+vI)V=B+vI, gives the resolvent compression identity [14, Lemma 2.1]:

The logarithmic resolvent formula gives

where D(\rho\|\sigma)=\Tr\rho(\log\rho-\log\sigma). Indeed -\langle\sqrt\rho,\log(L_\sigma R_{\rho^{-1}}) \sqrt\rho\rangle=D(\rho\|\sigma), and the identity terms in the log integral cancel. The last equality in (5.12) follows by expanding the four entropies.

The local domination proved in (5.1) now reads \rho_f\le d^2\sigma_f. Therefore

The first resolvent quadratic form in (5.11) is therefore at most d^2. For another bound use (A+vI)^{-1}\le v^{-1}I and (B+vI)^{-1}\ge v^{-1}I-v^{-2}B. We obtain

For -1<\Re z<0, scalar integration and spectral calculus give

One may approach z=-iu from -1/4\le\Re z<0 under this integral: (5.13) gives an integrable majorant d\,v^{-3/4} near zero and v^{-3/2} near infinity. Hence

For \eta>0, split the integral at y/d^2 and 1/y. The two ends cost at most C\sqrt y by (5.13); Cauchy–Schwarz and (5.12) bound the middle by

If \eta=0, nonnegativity and continuity imply h(v)=0 for all v>0, so the discrepancy is zero. To identify the resulting vectors, use the definition of V:

After cancelling the common scalar phase d^{iu} and reversing the sign of the difference, its Hilbert–Schmidt norm is

For any operator T on xUF one has \|T\theta\|=\|T\sqrt{\rho_f}\|_2. This proves (5.10) in the faithful case.

For a possibly singular \rho_f, replace it by (1-\epsilon)\rho_f+\epsilon I/\dim(xUF) and use its reductions in the preceding proof. Conditional mutual information converges in finite dimension. The phase products on the square root converge in Hilbert–Schmidt norm in the required order. For example, the innermost lifted phase of \rho_{xU,\epsilon} acts on \sqrt{\rho_{f,\epsilon}}. Its limit lies in \supp\rho_{xU}\otimes F, because \supp\rho_f\subseteq\supp\rho_{xU}\otimes F and the square-root mass outside this space tends to zero. On this space the inner phase converges, since the mollified marginal commutes with its limit. Its output lies in x\otimes\supp\rho_U\otimes F, so the outer \rho_{U,\epsilon} phase also converges there. The same argument applies to the xUF,UF pair. All phases have norm one, which controls the vanishing complementary pieces. Thus the limit is the product of support phases, equivalently the unitary extensions appearing in (5.10). No operator-norm convergence of phases on a limiting kernel is being used. The faithful proof was on Hilbert–Schmidt spaces, so it also applies when a faithful approximation would require a larger purifying system than P. This establishes (5.10) in full generality.□

Proof. Put p=\langle\theta,h\theta\rangle\in[0,1], so f(0)=p. We first bound f on a strip using the support dimension of h, then compare its imaginary-axis values with p, and finally estimate its real-axis deviation from [0,\infty). Expand h=\sum c\otimes d' across P_0,x using matrix units on P_0. There are at most d_h^2 terms, with \|c\|\|d'\|\le1 for each. For 0\le\Re z\le1/4 the contribution is the inner product of

Each norm is bounded by the norm of its local operator: for a marginal \rho and 0\le\Re z\le1/2,

by Schatten Hölder. This is also valid at the endpoints using support powers. Reflection f(-\bar z)=\overline{f(z)} treats negative real parts. Thus |f(z)|\le d_h^2 on |\Re z|\le1/4. On the imaginary axis it is an expectation in a unit vector and has absolute value at most one. Three-lines on the two half-strips gives

for a sufficiently small universal c_0>0.

On the imaginary axis Schmidt mirroring across P,YF gives

Compare this with w_u^0=\widehat\rho_U^{iu}\widehat\rho_{UF}^{-iu}\theta. The latter is a unitary acting only on UF, so \langle w_u^0,hw_u^0\rangle=p. Multiplication of both vectors first by \widehat\rho_P^{iu} and then by \widehat\rho_U^{-iu}, and a second use of Schmidt mirroring, show that

Lemma 5.2 therefore gives \|w_u-w_u^0\|\le C|\sinh(\pi u)|\mathcal R_\eta, where \mathcal R_\eta is defined there with d=\dim x.

It remains to pass from phases to the real-power skew. Positivity of h and the identity \langle w_u^0,hw_u^0\rangle=p imply

In particular, differentiating at zero and using (5.10) gives

The imaginary character also follows from f(-\bar z)=\overline{f(z)}.

Set m=\min(1,\mathcal R_\eta). The function F(z)=(f(z)-p)e^{z^2} has absolute value at most Cm on the imaginary axis: use (5.10), the Gaussian factor e^{-u^2}, and also |f(iu)-p|\le2. On the two outer lines \Re z=\pm b it is bounded by a universal constant by (5.15). Three-lines on each half-strip therefore bounds |F| by Cm^{1/2} on |\Re z|\le b/2. If m=0, the identity theorem already gives f\equiv p. Otherwise Cauchy’s estimates on a circle of radius b/2 centered at zero, applied also to the bounded factor e^{-z^2} on this circle, give

Write f'(0)=ib_1 with b_1\in\R. If p>0, then

if p=0, (5.17) makes b_1=0. Adding the remainder changes |f|-\Re f by at most twice its absolute value. Thus

Since d=\dim x\le d_h, the quantity supplied by Lemma 5.2 satisfies, for 0<\eta\le1, \mathcal R_\eta\le C\sqrt{\eta\log(ed_h/\eta)}; for \eta\ge1, the bound \eta\le2\log d\le2\ell_h gives \mathcal R_\eta\le C\ell_h. These two cases imply

absorbing the logarithm of 1/\eta by the smaller power of \eta. Taking the universal c in the lemma small enough to ensure t=a/2\le b/8 proves (5.14).□

Proof. The first formula in (6.3) is the value of the Weyl character at the identity. To derive the second, multiply the character decomposition of (x_1+\cdots+x_q)^k by \det[x_i^{q-j}] and extract the coefficient of x_1^{l_1}\cdots x_q^{l_q}. The strict decrease of the l_i selects only label \lambda and the identity term in its numerator alternant. On the other side the coefficient is

Here a reciprocal negative factorial is zero and (u)_m is the falling factorial. The last determinant is a Vandermonde because its columns are monic polynomials of the indicated degrees. The estimate \log n!=n\log n-n+O(\log(n+2)) gives the entropy formula, including zero parts, with 0\log0=0. The fixed shifts from \lambda_i to l_i and the Vandermonde contribute only O_q(\log(k+2)). There are at most (k+1)^q labels, and the dimension formula bounds each \dim V^{(q)}_\lambda by (k+q)^{q(q-1)/2}.

Each label projector is a central character idempotent in the permutation algebra:

This follows from character orthogonality, or directly from Schur’s lemma applied to the central sum and its trace on each irreducible. Conjugation by U_{QE}(\pi) sends a central function of the Q permutations to itself. This proves nested commutation; disjoint commutation is immediate. On \mathcal S_k, U_Q(\pi)z=U_{Q^c}(\pi^{-1})z. A permutation is conjugate to its inverse, so \chi_\lambda(\pi)=\chi_\lambda(\pi^{-1}) and the two central-projector sums agree there. This also shows that all Specht modules are self-dual: their dual characters are the same.

For disjoint subsystems, [\nu] is a constituent of [\lambda]\otimes[\mu] under diagonal permutations. This proves the first inequality of (6.4). The identity

and self-duality prove the second. Conditional on \lambda,\mu, a product of invariant states is the identity divided by d_\lambda d_\mu on the two Specht factors. If g_{\lambda\mu\nu} is the multiplicity of [\nu] in their tensor product, its conditional label probability is exactly g_{\lambda\mu\nu}d_\nu/(d_\lambda d_\mu). The ambient decomposition of (Q\otimes E)^{\otimes k} implies

so g_{\lambda\mu\nu} is polynomially bounded. With R=d_\lambda d_\mu/d_\nu\ge1, the conditional expectation in (6.5) is \sum_\nu g_{\lambda\mu\nu}R^{b-1}\le\sum_\nu g_{\lambda\mu\nu}, again polynomial.

For the copy-group assertion, choose a copy of [\alpha]\otimes[\beta] inside the restriction of [\lambda] to S_{k-r}\times S_r. Its dimension gives the lower bound. The translates of this subspace span [\lambda], by irreducibility. Choosing one representative of each subgroup coset bounds the span dimension by \binom{k}{r}d_\alpha d_\beta. Central elements of S_k commute with its subgroup algebra, proving the commutation claim.

An invariant density has blocks \rho=\bigoplus_\lambda I_{[\lambda]}\otimes M_\lambda. Every positive eigenvalue u of M_\lambda is repeated d_\lambda times, so u d_\lambda\le1. This gives F_Q\le L_\rho and

where the inner sum counts eigenvalues with multiplicity in M_\lambda.

It remains to identify the marked-copy branches. Multiplying a Weyl numerator alternant by \sum_i x_i gives the sum of alternants with one exponent increased by one. Terms with equal exponents vanish; the remaining terms correspond exactly to adding a legal box to a partition. Thus V^{(q)}_\nu\otimes\C^q=\bigoplus_{\lambda=\nu+e_i} V^{(q)}_\lambda, with multiplicity one. Comparing the two Schur–Weyl decompositions of the first k-1 copies and the last copy shows that the \nu branch in the \lambda block has rank d_\nu on [\lambda], and acts identically on V^{(q)}_\lambda. The invariant state is uniform on [\lambda], giving the probability in (6.8). The dimension formula gives the displayed quotient. In its Vandermonde ratio, a factor involving an earlier row is at most two, and a factor involving a later row is at most one. Positivity holds on a legal removable branch.

For the star eigenvalue, let E_{ab}^{\rm tot} be the sum over copies of the matrix unit E_{ab} on \C^q. Direct multiplication gives

The right side is scalar on a Schur block. The lexicographically leading weight in its Weyl character is \lambda; a vector of that weight is annihilated by E_{ab}^{\rm tot} for a<b, since those operators would increase the weight. On this vector the diagonal terms contribute \sum_i\lambda_i^2, and the two terms for a<b contribute \lambda_a-\lambda_b, using [E_{ab}^{\rm tot},E_{ba}^{\rm tot}]=E_{aa}^{\rm tot}-E_{bb}^{\rm tot}. The scalar is therefore \sum_i\lambda_i(\lambda_i+q+1-2i). Subtract the formula for k-1 copies with label \lambda-e_i. The difference of the transposition sums is kJ_{Q,k}, and its value is \lambda_i-i.□

Proof. Begin in dimension d. Write \Delta(s)=\prod_{i<j}(s_i-s_j) and choose s on the simplex s_i\ge0, \sum_i s_i=1, with density proportional to \Delta(s)\Delta(s^t). This is a nonnegative, integrable density, positive off the collision hyperplanes in the simplex interior. It is the simplex normalization of the Laguerre Muttalib–Borodin weight; the gamma integral below is the corresponding form of [20, Section 2.3, equation (26)]. We include the calculation before verifying the common-subsystem and ratio properties. For d=1 use the point mass at s_1=1. Choose an independent Haar-distributed unitary eigenbasis and let \sigma have eigenvalues s. The integral of (\sigma^t)^{\otimes k} commutes both with permutations and with the diagonal unitary group; hence it is central in the Schur decomposition. Its eigenvalue on \lambda is the mean of \chi_\lambda^{(d)}(s^t)/\dim V^{(d)}_\lambda.

Put l_i=\lambda_i+d-i. The Weyl denominator cancels \Delta(s^t) in the density. Expanding the two remaining determinants and using the simplex integral

gives, before dividing by the Weyl dimension and the density’s normalization,

The simplex identity follows, for example, by changing variables x_i=rs_i in the product of the defining gamma integrals; the Jacobian contributes r^{d-1}. Factor \Gamma(1+t l_i) from each row. The remaining entries are rising-factorial monic polynomials in t l_i of degrees d-j, so their determinant is t^{d(d-1)/2}\Delta(l). The Weyl dimension cancels \Delta(l). Normalizing at k=0 gives exactly (6.10).

For d_Q\le d, choose an isometry E:\mathcal H_Q\to\C^d. Compression of a permutation on (\C^d)^{\otimes k} by E^{\otimes k} is the same permutation on Q^{\otimes k}. Compressing its central label function therefore gives the identical function w_k(\lambda) on all labels that occur in dimension d_Q. On each sample, write B=E^\dagger\sigma^t E and \tau=B^{1/t}. The min–max principle bounds the decreasing eigenvalues of B by the corresponding eigenvalues of \sigma^t. Consequently \Tr\tau\le\Tr\sigma=1, while B>0 almost surely. Thus the compressed integral has the form (6.12), with the distribution of \tau as \zeta_Q. This proves the asserted representation without changing the common label function. Exact complement equality follows from Lemma 6.1.

The gamma estimate \log\Gamma(1+u)=u\log u-u+O(\log(u+2)), with 0\log0=0, applied to the fixed number of factors in (6.10), gives \log w_k(\lambda)=-tkH(\lambda/k)+O_{d,t}(\log(k+2)). The terms containing \log t cancel. Formula (6.3) proves (6.11).

For the marked ratio, now put l_i=\lambda_i+d-i using the full dimension. On a removable branch its exact eigenvalue is

Gamma ratios give r_{\lambda,i}\asymp_{d,t}((l_i+1)/k)^t. When l_i/k is bounded away from zero, the same ratios give uniform convergence to (l_i/k)^t. When l_i/k is small, both r_{\lambda,i} and \max((l_i-d)/k,0)^t are uniformly small. First separate at l_i/k=\varepsilon, then let k\to\infty and \varepsilon\downarrow0. The star eigenvalue formula proves (6.13) and a uniform bound for \norm{R_{Q,k}}.

The Q marginal of a symmetric vector is permutation invariant. Its conditional branch probabilities are bounded by C_d l_i/k: Lemma 6.1 first uses \lambda_i+d_Q-i, which is no larger than l_i. Hence, conditional on any \lambda,

This proves the inverse vector bound. On the spectral part J_{Q,k}\ge\delta, the inverse approximation follows in operator norm from (6.13). On the remaining part, l_i/k\le\delta+d/k. The true inverse’s squared norm there is at most C_{d,t}(\delta+d/k)^{1-2t}; the clipped inverse’s squared norm is at most C_d(\delta+d/k)\delta^{-2t}. Taking the limit superior and the square root proves (6.14). All bounds were conditional on an arbitrary label, so they are uniform in the original symmetric vector.□

Proof. For a positive definite C and unit vector z, Cauchy–Schwarz gives

It therefore suffices to bound the inverse quadratic form of C_{x,k} at \theta^{\otimes k}. Since C_{x,k}^{-1}=A_k^{1/2}B_k^{-1}A_k^{1/2},

To obtain the second line, cancel the disjoint F factors and commute the remaining nested or disjoint central factors. This is an identity of the displayed norms; no commutation of A_k with B_k is asserted.

Let \rho_R be the marginal of \theta for R=P,Y_0, and set

with the positive power zero on the kernel. The invariant density \rho_R^{\otimes k} has every eigenvalue in its \lambda block at most d_\lambda^{-1} by eigenvalue repetition, as in the proof of (6.7). Thus (6.11) implies

uniformly in \theta, including marginals of deficient rank. Furthermore T_P commutes with both numerator factors in (6.20), and so does T_{Y_0}. For the overlapping numerator this follows because W_{R,k} is central on the nested subsystem and (\rho_R^t)^{\otimes k} commutes with every simultaneous permutation on xR; the other numerator is disjoint from R.

Extend \rho_R by eigenvalue one on its kernel and denote this extension by \widetilde\rho_R. The product of the two marginal support projectors fixes \theta, since P and Y_0 are disjoint. We can therefore insert T_P,T_{Y_0} and move them to the left in (6.20), obtaining the upper bound

Here and below the polynomial can be enlarged to absorb its square. Expand the two numerator integrals (6.12) in their stated order. The triangle inequality and the tensor-power norm identity bound this by

where \sigma,\tau are positive definite subdensities on the indicated subsystems. Normalizing a subdensity multiplies its t-power by a scalar at least one, so it suffices to take density matrices. Since P is disjoint from xY_0, its inverse power commutes with \tau_{xY_0}^t. The one-copy word is therefore

exactly the word in Lemma 5.1. That lemma bounds its log norm by -a\eta_\theta/2+C a^{5/4}\log^C(e\dim x). Substituting in (6.22), and then using the inverse Cauchy–Schwarz inequality at the start of the proof, gives (6.19). Every polynomial bound used in this argument was uniform in \theta.□

Proof. We first replace the marked ratios by bounded polynomial words, then prove a product calculus for those words, and finally pass to absolute values. The inverse approximation must be applied on symmetric input vectors; no full-space bound for R_{Q,k}^{-1} is needed.

Bounded approximation on symmetric inputs. Between symmetric vectors, h^{(k)} may replace \bar h. The F factors commute with this term and cancel. Since W_{Q,k-1} commutes with W_{Q,k} and is disjoint from h^{(k)}, the remaining form is the compression of

Write h as a finite sum of products c_P\otimes e_Y. Such a decomposition follows by expanding matrix units on the two parts of its support; its length and norm bounds are fixed here. For one product and a symmetric unit vector z, the quadratic form of (6.28) equals

Both inverses act directly on z. The uniform forward operator bound and inverse vector bound from Lemma 6.2 bound these two vectors uniformly and therefore bound the numerical radius of O_k. The elementary inequality \norm T\le2\sup_{\norm z=1}|\langle z,Tz\rangle|, obtained by polarizing the quadratic form, proves uniform boundedness of O_k.

In (6.29), replace each inverse by \max(J_{Q,k},\delta)^{-t} using (6.14); then replace each forward ratio by (J_{Q,k})_+^t using (6.13). The first replacement has limiting error at most C_{\mathcal V,t,h}\delta^{(1-2t)/2}, uniformly on symmetric unit vectors. At fixed \delta the second has error tending to zero. Approximate the continuous functions u\mapsto u_+^t and u\mapsto\max(u,\delta)^{-t} uniformly on [-1,1] by polynomials. All star spectra lie in this interval. Polarization, applied after summing the finitely many product components, gives compressed operators B_k, each a fixed polynomial word in marked operators and stars, such that

Here any \varepsilon>0 is achieved by first choosing \delta, then the polynomials, and then taking the limit in k. The approximants may be chosen with a common bound on \limsup_k\norm{B_k}; an approximation-dependent threshold for k does not affect any of the limits below.

The coherent measure on a fixed number of copies. We next establish the uniform coherent-measure calculation for a fixed number of copies. If G acts on \mathcal V^{\otimes m}, let \mathcal T_{k,m}(G) be its average over all ordered injections of these m copies into k copies. It preserves \mathcal S_k. For k\ge m, symmetry gives \Tr\sigma\mathcal T_{k,m}(G)=\Tr\sigma^{(m)}G. For the integral of its product-vector expectation, (6.23) gives the finite-copy identity [15, equation (5)]

In the permutation expansion of \Pi_{k+m}, the fraction mapping every one of the last m indices into the first k is (k)_m/(k+m)_m=1-O_m(k^{-1}). Any such permutation is equivalent, by a permutation of the first k indices on each side, to m disjoint swaps matching last indices with distinct first indices. Indeed reorder the old images of the last indices on the left and their old preimages on the right, then eliminate the remaining permutation among old indices. Both old permutations are absorbed because U(\pi)\sigma=\sigma U(\pi)=\sigma. The matched swaps contribute exactly \Tr\sigma^{(m)}G by partial trace. Every other term has absolute value at most \norm{\sigma\otimes G}_1\le d^m\norm G. Also D_k/D_{k+m}=1+O_{d,m}(k^{-1}). Consequently

uniformly in \sigma.

Products after compression. The remaining issue is multiplication after compression. For any ambient operator B, define its full-copy permutation average

It commutes with all U(\pi), hence preserves symmetry, and

Products of compressed words are therefore restrictions of products of their full permutation averages. The multiplication of these averages uses disjoint placements of fixed copy blocks, as in the mean-field symmetrization argument of [35, Lemma IV.1]. For fixed operators G,H on m,n copies, respectively,

for sufficiently large k. Indeed two independently chosen injections have intersecting images with probability O_{m,n}(k^{-1}). Conditional on disjoint images, their union is a uniform ordered injection of m+n copies. The intersecting terms and the change in normalization each cost at most that probability times \|G\|\|H\|.

Expand a fixed star word as sums over its donor indices, with one marked center. Repeated donor indices have total coefficient weight O(k^{-1}), so their contribution has norm O(k^{-1}) for that fixed word. After averaging each word separately, both its center and its donors occupy a uniformly relocated block of indices. For a monomial containing r stars, its distinct-donor terms give an injection average with coefficient (k-1)_r/k^r=1+O_r(k^{-1}). Thus each averaged word is, up to operator-norm error O(k^{-1}), a finite sum of injection averages of fixed operators. Iterating (6.33) puts a product of words on disjoint blocks, with a separate center for each word.

On a product vector, a donor appearing in one partial swap contracts to \rho_Q on the center. In matrix units this is the identity

Tracing distinct donors successively preserves the order of the operators on each center. The separate centers give multiplication of scalar expectations, since

It follows that a polynomial approximant has the symbol obtained by replacing its stars by the corresponding marginal densities, its adjoint has the complex conjugate symbol, and products multiply these scalar symbols. Applying (6.32) proves these assertions uniformly in the symmetric density matrix, for each fixed polynomial word.

Removing clipping and taking absolute values. We have proved the product assertion with fixed clipping. To remove clipping on the one-copy side, use

for the eigenvalues of any marginal density. This bounds the squared norm of its inverse-power tail on \theta, uniformly in \theta. The clipped one-copy vectors corresponding to (6.29) therefore converge uniformly to their support-power limits, and their inner product converges uniformly to the zero-kernel-power expression f_\theta. In particular this symbol is bounded and continuous, as a uniform limit of the clipped continuous symbols. Telescoping fixed products in (6.30), using the common limiting norm bound, now proves (6.25).

In particular, both O_k^\dagger O_k and O_kO_k^\dagger have symbol |f_\theta|^2. Choose an interval [0,M^2] containing all their spectra and all values |f_\theta|^2. Uniform polynomial approximation of the square root on that interval gives symbols |f_\theta| for |O_k| and |O_k^\dagger|, with arbitrarily small uniform expectation errors. This argument applies whether or not O_k is normal. Combining with the linear terms proves (6.26). Finally apply Lemma 5.3 pointwise to f_\theta to obtain (6.27). Purity gives the two expressions for \eta_\theta; enlarging x to the designated supporting subsystem is allowed by that lemma. The proof is unchanged after exchanging P and F.□

Proof. We first prove the assertions for one binary vertex. For an invertible S, put \widehat A=S^*AS and U=A^{1/2}S\widehat A^{-1/2}. Then U is unitary and

Functional calculus proves (S^*AS)\#_p(S^*BS)=S^*(A\#_pB)S. After congruence to A=I, the identity A\#_pB=B\#_{1-p}A reduces to C^p=C^{1/2}(C^{-1})^{1-p}C^{1/2}. For 0<p<1, the scalar formula

follows by substituting \lambda=xt in the beta integral. The inverse reverses the order of positive definite operators, so C(C+\lambda I)^{-1}=I-\lambda(C+\lambda I)^{-1} is increasing in C. Thus C\mapsto C^p is operator monotone. This proves monotonicity in B, and symmetry proves it in A. The endpoint cases are immediate. Scalar homogeneity at a vertex is (cA)\#_p(dB)=c^{1-p}d^p(A\#_pB).

For the vector inequality let r=\langle z,Az\rangle and y=A^{1/2}z/\sqrt r. Scalar Jensen applied to the spectral measure of C=A^{-1/2}BA^{-1/2} in the unit vector y gives

Induction proves the congruence, monotonicity, homogeneity, and vector inequality for the full tree.

To prove (7.4), fix 0<\varepsilon<1. For every leaf,

Choose a common \delta>0 so small that (1-\varepsilon)c_j\delta\leq\varepsilon\lambda_{\min}(A_j) for all leaves. Monotonicity and homogeneity applied to the strictly positive inputs (1-\varepsilon)c_j(P+\delta I) give

Let \varepsilon decrease to zero. This argument needs no definition of the mean on singular inputs.

For the last assertion, cross-band commutation implies

Indeed the inverse square root of the first product factors, its whitened second input is the product of the commuting positive operators D_g^{-1/2}E_gD_g^{-1/2}, and the pth power factors. Induction over the tree completes the proof.□

Proof. Differentiating (7.5) gives

The integral converges in operator norm: its integrand is O(\lambda^p) at zero and O(\lambda^{p-2}) at infinity. Each integrand is a completely positive sandwich map. Formula (7.8), together with the congruences in (7.1), proves complete positivity for the second child. Exchange symmetry proves it for the first child. Scalar homogeneity gives \partial_D M[D]=wM, proving unitality. The chain rule, with the square-root normalizations cancelling at successive vertices, proves (7.7). Composition preserves both properties; the map along an empty path is the identity. In the derivative integral all sandwich factors are block diagonal when the inputs are block diagonal; their first blocks depend only on the first blocks of the inputs. The same is true after exchanging the children and composing the maps.□

Proof. For fixed real z\ne0, integrate e^{iuz}m_s(u) around a rectangle between the lines \R and \R+i. The vertical integrals tend to zero because the denominator grows exponentially with |\Re u|. The two poles in the strip are i(1/2-s) and i(1/2+s), with residues of m_s equal to 1/(2\pi i) and -1/(2\pi i), respectively. Since m_s(u+i)=m_s(u), the residue theorem gives

Division gives (7.10); dominated convergence gives its value at zero. Positivity follows from the displayed denominator.

The algebraic identities in (7.11) follow by factoring e^{\pm sz}-1. The nearest poles of m_{s/2} have imaginary parts of absolute value 1/2-s/2, larger than s/2. Shifting its Fourier contour by \pm is/2 therefore crosses no pole and gives the densities in (7.12). To check the bound without losing a constant, write c=\cosh(2\pi u). At s=1/4,

Consequently

□

Proof. We first identify the derivative states and prove the entropy estimate. We then construct an auxiliary block tree for each physical term and derive (7.22) by a matrix Schwarz inequality.

The derivative at an interpolating vertex. At the vertex M_h=A_h\#_pB_h, put V_h=C_h^{p/2}A_h^{1/2}M_h^{-1/2}. Since V_h^*V_h=I, this matrix is unitary, and direct differentiation gives

This is also the normalized old-child derivative applied to \log C_h. To verify that assertion, perturb the old input to A_h^{1/2}(I+\delta\log C_h)A_h^{1/2} while keeping B_h fixed. After congruence the two inputs commute, so their mean is

Its derivative at zero is (1-p)C_h^p\log C_h. The normalization in (7.6) divides by 1-p and proves the assertion. Composing with the ancestor maps therefore yields

The coefficient here is w_h, whereas the weight of the same terminal leaf in the full tree is (1-p)w_h.

To differentiate the norm, diagonalize \mathcal M with eigenvalues x_r>0 and substitute \mathrm{pre}=N\mathcal M^s v. The matrix coefficient, between the components \overline v_r and v_t, multiplying the (r,t) entry of \mathsf H is

with the divided difference interpreted as a derivative when the eigenvalues agree. The common value in that case is 2s. Consequently

Insert (7.24) and the Fourier formula (7.10). Lemma 7.2 implies that (7.19) is positive and has trace one. Moving each map to its trace adjoint proves (7.20).

Transfer of the relative metric lower bound. By Lemma 7.1, the old metrics, the conditional roots, the interpolated root \mathcal M=\prod_g\mathcal M_g, and the relative metrics factor over bands. Here \mathcal M_g is the full tree evaluated on the inputs from band g alone. More precisely, if B_{h,g} is the conditional tree of new metrics in band g, then

Congruence covariance identifies C_{h,g} as the conditional choice tree of the single-move relative metrics A_{h,g}^{-1/2}A_{h,c,g}A_{h,g}^{-1/2}. Lemma 6.3 gives, for every unit \theta,

where b_k>0 and -\log b_k=O_{\mathrm{fixed}}(\log(k+1)). The same b_k and constant can be used for the finitely many moves. For a move of the empty subsystem the relative metric is the identity, and the assertion holds after decreasing b_k to at most one. Projection transfer, (7.4), gives

The common polynomial factor occurs once, because the conditional weights sum to one.

Here is the logarithmic passage from this pointwise rank-one inequality to an estimate valid in every state. Define the unital positive map

The identity \mathcal Q_k(1)=I follows from the invariance of the integral and irreducibility of \Sym^k\mathcal V under the full unitary group; its trace fixes the scalar. For a positive bounded function f bounded away from zero, integration of the positive scalar block matrix with entries f,1;1,f^{-1} gives

Apply this also to f+\lambda and integrate the scalar resolvent formula

It proves

The same inverse-order argument proves operator monotonicity of the logarithm. All these integrals converge in norm on compact positive spectral intervals.

Integrating (7.27) against Haar probability gives

Thus operator log monotonicity and (7.28) turn the factor D_k into an additive logarithmic loss:

The entropies are bounded in the fixed one-copy dimension, so the exponential symbol used here is positive and bounded away from zero for each fixed k. Also D_k=\binom{k+\dim\mathcal V-1}{\dim\mathcal V-1} is polynomial in k. Taking the expectation in the trace-one state \sigma_{h,u}, summing (7.26), and using \sum_hw_h=1 and \int m_s=1/2 proves (). In particular the loss is logarithmic in k, without an additional factor D_k after taking expectations.

We have obtained the entropy gain against the old-leaf states. To control the energy in these same states, the remaining argument represents the commutator with each \bar h_i by a block matrix and transports its square through the same tree.

An auxiliary positive metric for each energy term. Fix i. Let T:\mathcal S_k\longrightarrow\mathcal D be h_i^{1/2} on the last copy, restricted to \mathcal S_k, with codomain its image. Then T^*T=\bar h_i: matrix elements of any marked copy coincide between symmetric vectors, and their average is the restriction in (7.16). If T=0, the term contributes nothing and can be omitted. Otherwise T is surjective onto \mathcal D.

We will add a second diagonal block to each leaf metric. This block is chosen so that the squared off-diagonal commutator recovers the absolute-value expression of Lemma 6.4. The first block remains the original metric, so Lemma 7.2 will preserve its derivative maps when the enlarged tree is evaluated. For every terminal leaf with input A_j, we construct a positive definite operator B_{i,j} on \mathcal D. When [A_j,\bar h_i]=0, use the polar decomposition T=U|T|. Here U is unitary from \supp|T| onto \mathcal D, and this support is invariant under A_j. Set B_{i,j}=U(A_j|_{\supp|T|})U^*. Commutation with |T| gives

For a split leaf put

For any operator O, the notation |O| means (O^*O)^{1/2}. The range of O_{i,j} is the range of S_j^*, namely A_j^{-1/2}\operatorname{ran}T^*. Since S_j is surjective, S_jS_j^* is positive definite on \mathcal D. Define

The operator Q_j is strictly positive on \operatorname{ran}S_j^*. Thus (7.31) is positive definite on \mathcal D and S_j^*B_{i,j}S_j=Q_j; the latter follows because S_j^*(S_jS_j^*)^{-1}S_j is the projection onto \operatorname{ran}S_j^*. Define Y_{i,j},Z_{i,j} by the formulas in (7.30). They now satisfy

The map Y_{i,j} is surjective. Consequently Y_{i,j}(Y_{i,j}^*Y_{i,j})^+Y_{i,j}^*=I_{\mathcal D}, where the plus denotes the inverse on the support and zero on the kernel. Multiplying on the two sides by Z_{i,j}^* and Z_{i,j} shows

The last equality follows, for example, from a singular value decomposition of O_{i,j}. Hence

For an unsplit leaf use \mathsf D_{i,j}=0; it agrees with the same formula because O_{i,j}=\bar h_i. The construction has therefore included singular \bar h_i without ever inverting it on its kernel.

Transporting the square with both Fourier kernels. Evaluate the full terminal tree on the block inputs

Its root is \mathbb M_i=\operatorname{diag}(\mathcal M,\mathcal B_i). Denote its normalized leaf maps by \widehat\Phi_{i,j}. Their first diagonal-block restrictions are exactly \Phi_j, by Lemma 7.2, although the second blocks depend on i. Let

Rotating every input by e^{itL_i} rotates the root by the same unitary, by congruence covariance. Differentiate at zero, normalize at the root, and cancel the common factor -i. The chain rule gives

The lower-left block of K_{i,j} is Y_{i,j}-Z_{i,j}, its upper-right block is the negative adjoint, and the upper-left block of K_{i,j}^*K_{i,j} is \mathsf D_{i,j}. In particular K_{i,j}=0 for j\notin\mathcal J_i.

Apply h_\pm(\operatorname{ad}_{\log\mathbb M_i}) to (7.33). By (7.11), the lower-left block of the left side is

The Fourier formula for the right side and preservation of blocks give an expression for (0,E_{i,\pm}v) as a sum over split leaves of

The output phase is unitary. Cauchy–Schwarz with respect to the positive measure assigning mass \pi_j|q_\pm(u)|\,du to (j,u) therefore bounds the squared norm by

For a unital completely positive map \Psi,

Indeed apply its two-by-two amplification to the positive block matrix \left(\begin{smallmatrix}K^*K&K^*\\K&I\end{smallmatrix}\right) and take the Schur complement of the identity. By the first-block restriction of \widehat\Phi_{i,j}, this bounds the norm squared in the preceding display by \Tr(\sigma_{j,u}\mathsf D_{i,j}). Finally (7.12) and \int m_s=1/2 give the exact estimate

This proves the bound for both signs. The first factor W_i comes from Cauchy–Schwarz, and the sum contains a second total split weight. For example, a uniform bound on its trace integrals gives a bound proportional to W_i^2, rather than W_i.

From the transported square to physical energy. At a split leaf only one band fails to commute with \bar h_i. The other factors cancel in its similarity transform O_{i,j}. At such a leaf j\in\mathcal J_i, the assumed bound \log(e\dim D_i)\leq\ell applies. Our designated-support assumptions put this two-part split within the hypotheses of Lemma 6.4; Lemma 5.3 then gives, uniformly over density matrices \sigma on \mathcal S_k,

There are finitely many split term–leaf pairs, so one common r_k\geq0 can be chosen for all of them; take r_k=0 if there are none. Unsplit pairs have \mathsf D_{i,j}=0 exactly and incur no symbol remainder or support-dimension factor. These terminal metrics do not depend on the interpolation parameter. Uniformity over \sigma makes (7.37) applicable to every \sigma_{j,u}, regardless of p, u, or the normalized vector.

The two opposite powers of \mathcal B_i in (7.34) cancel in the pairing:

For the last inequality, expand the left pairing, bound each linear cross term by \frac14\norm{Tv}^2+\norm{E_{i,\pm}v}^2, and bound the remaining mixed term by half the sum of the two error squares. Summing the equality in () over i gives E_0: one has \mathcal M^s v=\mathrm{pre}/N, and \langle\mathcal M^{-s}v,\mathcal M^s v\rangle=1. Thus

Insert (7.36) and (7.37) to obtain (7.22). The total remainder is bounded by a fixed constant times r_k\sum_iW_i^2 and hence tends to zero uniformly in p and v.

Finally, all tree inputs are positive definite, so for fixed k the roots and N are smooth up to p=0,1. The normalized maps are used only where their branch weights are positive; their being unital ensures bounds uniform as p approaches an endpoint. The entropies are bounded in the fixed one-copy space, and all probability weights sum to one. Thus the entropy integrand and its error have the uniform bounds required to integrate () and pass to the endpoints. A finite sequence of rounds permits the same choices of polynomial losses and remainders by taking their maxima.□

Proof. Deterministic geometry and nesting. A truncated support meeting S_0 has radius r_0: when its anchor is at distance d>2r_0, the alternative radius \lfloor d/2\rfloor cannot reach S_0. Every splitting support meets a compact part contained in S_0, so this observation applies. If an r_0-ball meeting S_0 also contained a site of \Lambda\setminus A, a graph path inside the ball would contain a cut edge with an endpoint at ambient distance at most L+2r_0 from T. This contradicts (9.2).

At a charge the largest newly assigned oriented depth is at most j+D+r_0; subsequent movement of the nominal front only improves this bound. Each side receives at most \lceil nm/2\rceil fills, so advances through at most m/2+O(1) deterministic rows. Its charged lead is D+O(r_0)=o(m). The two fronts in a window remain separated by order m, as do the active regions of successive windows. This proves both that an opposite side cannot have consumed a scheduled slot and that a support cannot split two fronts. All nonblank slots strictly behind a nominal front have been assigned to that side.

A near set contains its completed deterministic prefix and is contained in the prefix ending at its maximal allowed lead. The difference has at most CnD sites. For UY, take the complement of the assigned far suffix within the original shell prefix: it is contained in the deterministic prefix ending at the nominal far front and differs from it only in O(D) rows. These deterministic comparison prefixes are all prefixes of one fixed ordered list of positive-depth slots and are consequently nested, even across histories and bands.

A full depth prefix has O(n) boundary edges. For its outer boundary at positive depth, use the last included row, since depth changes by at most one along an edge. For the boundary at depth zero, use the first positive row. Partial rows cost at most 4n more edges. There are no compact-to-opposite-color edges, by the clearance. Changing at most CnD sites then gives the claimed O(nD) bound for all status cuts and their differences. Split anchors themselves lie in O(D) rows around the nominal fronts, containing O(nD) anchors per band. Summing gives CKnD.

For two bands g'<g'', the entire near side together with the middle of the first band is contained in the near side of the second band, for every pair of histories. After adjoining C, this says P_{g'}Y_{g'}\subset P_{g''}. Mirror the central function on F_{g'} to its complement P_{g'}Y_{g'}, using Lemma 6.1. The resulting functions from the first band commute with those of the second: they are disjoint from its Y,F parts and invariant under the joint permutations whose center acts on its P part. This proves the asserted cross-band commutation.

Dilution by the initial offsets. For a fixed anchor to split, its oriented depth must lie between j-O(r_0) and j+D+O(r_0). Both formulas in (9.3) are affine, with coefficient 1 or -1, in the initial uniform offset. Thus only O(D) of the m offsets can permit this event, and only O(1) bands can reach the fixed anchor. This proves CD/m separately for the old and new leaf distributions, and hence also for their common-p mixture. It does not require conditional independence of the split event and the later charge choices.

Charge ancestry and availability. We finally bound the probability that charged sites get too far ahead of a nominal front. This will ensure that every old split lies in the charge list. Suppose a side has an assigned site of A leading its nominal front by more than D/2. Trace the site’s first assignment backwards through the charge ball that assigned it, choosing a previously assigned same-side site in that ball, until an initial or deterministic assignment is reached. An ancestry with j' charges changes oriented depth by at most 2r_0j'. If its first charge is at time t_1 and the history is tested at time t, then

The fixed fill schedule gives j(t)-j(t_1)\ge (t-t_1)/(4n)-O(1). Substitution in the preceding inequality yields both

Thus all ancestry times lie in this last interval of rounds. All sites in the ancestry are in A, including for the far side, because the charging balls are wholly in A.

For a fixed endpoint, test time, and length j', the time choices and spatial choices have at most

possibilities. The spatial bound permits both an anchor and a predecessor within O(r_0) graph distance at each step, and includes the bounded anchor multiplicity. Conditional on earlier choices, a prescribed hit has probability at most (cnD)^{-1}. The bound \binom{v}{j'}\le(ev/j')^{j'} therefore makes the total probability at most

The endpoints being counted are charged sites in S_0, of which there are O(n^2); the numbers of test times and bands are also polynomial in n. Thus no far-domain volume enters this factor. Since r_0 is polylogarithmic and D=n^{\kappa+o(1)}, the bound is smaller than every fixed inverse power of n.

On a good old history, a split anchor has oriented depth at least j-r_0, because its ball meets an unassigned site, and at most j+D/2+r_0, because it meets the side. For large n this interval lies in the sampling interval [j-r_0,j+D]. Choosing the relevant side and its labelled slot has probability 1/(2M_n)\ge c/(nD), and the resulting move is exactly the Y-part of that designated support.□

Proof. The transfer from \Omega to \widetilde\Omega costs O(n^{-248}) in any entropy in (9.4): the trace distance is O(n^{-500}), and the compact systems have O(n^2) sites, so Lemma 2.1 applies. A partial last depth row adds at most n\log q. Lemma 9.1 then gives the matched-prefix and mismatch bounds required by the comparators. It also gives the O(nD) status boundaries. A crossing support has Hilbert dimension at most q^{C(1+r_0)^2}, and its anchor lies within distance r_0 of a cut edge. Consequently the marginal-tail parameter of Lemma 3.1 is at most CnD(\log n)^C. At the target itself it is at most Cn(\log n)^C. The latter bound shows that the width w=n^{3/5} has failure mass at most n^{-100} for all sufficiently large n. In fact its tail is bounded by C\exp[-c n^{1/10}/(\log n)^C].

At a charge round write h for an old joint history, of probability w_h, and \sigma_{h,u} for its transported state in Proposition 7.4. Let \mu_\sigma denote the coherent probability measure of Lemma 6.4. The entropy derivative uses the old-history weights w_h; the same transported states occur in the energy estimate with terminal weights (1-p)w_h. We therefore measure the entropy cost of the old splits by \mathcal Q(p), defined as

Each split term in this sum has a unique splitting band, and its partition P,Y,F is understood to be that band’s partition. If term i splits on the near side, its moved set is x=Y\cap\widetilde X_i, the unassigned part of its designated r_0-ball, and \eta_{i,h}=S_\theta(x\mid P)+S_\theta(x\mid Y\setminus x); on the far side exchange P,F. These quantities are nonnegative. They are also bounded by C(\log n)^C, because the moved sites lie in an r_0-ball. The measure m_s(u)\,du has fixed finite mass 2s.

On every good old history each split term is sampled with probability at least c/(nD), and its sampled move has exactly the corresponding \eta_{i,h}. Equation () therefore gives, at a charge round,

At fill rounds retain just the same negative error; the entropy gains are nonnegative. Here and below “fixed” means that the complete single-copy instance and the scanner have been fixed. No such constant survives the replica limit.

Integrate over every full round, including fills. Adjacent endpoint metrics agree by construction, so the left sides telescope. Initially -\log N^2\ge-O_{\rm fixed}(\log(k+2)): the floor (6.18) passes to the initial mean tree by Lemma 7.1, and \|\mathrm{pre}\|\le1. At the last endpoint the rough upper comparison (8.8) gives

For the last inequality use \kappa\le e, L\le n, and \log d_*=S_{\widetilde\Omega}(X)+O(w+|\log z|). Restrict the positive charge integrals in (9.8) to [\epsilon/2,\epsilon], but pay the negative errors on all 2nm full rounds. Since aK=W, the integrated inequality, normalized by the number of charge rounds, is

The interval has length \epsilon/2, and m\asymp n^\mu. Averaging therefore gives a charge round and a p_k in this interval such that

The fixed factor two in the interval length is absorbed into constants. All integrands are bounded at fixed k, and continuous in the interior interpolation parameter, so ordinary averaging suffices for this selection.

Apply the energy estimate (7.22) at this same point. For each term its split weight is at most CD/m. There are at most CKnD split term occurrences in each joint leaf. On old good leaves the transported states are exactly those in (9.7); their full terminal weight differs only by the factor 1-p_k\le1. Hölder’s inequality, applied to the positive measure over histories, terms, u, and \theta, bounds their sum of \eta^{1/8} by

On bad old histories use the uniform local dimension bound and Lemma 9.1 with an arbitrarily large fixed tail power. On all new leaves use the same dimension bound and their total terminal weight p_k\le\epsilon. After increasing the fixed logarithmic power, the energy estimate is therefore

The factors in this expression have distinct origins: D/m is the split probability of one fixed term, KnD counts split occurrences, and a^2 is the local skew cost. Their product simplifies to

Subtracting \widetilde E_0 and dividing by \widetilde g\ge g/2 proves (9.5).

Only splitting terms contribute to this energy error; all others have zero skew exactly. Thus its coefficient contains no total-volume factor. The symbol remainders can be summed over all terms before taking k\to\infty, because that complete term set is finite and fixed. Their uniformity in replica states and in the interpolation parameter also justifies the possibly k-dependent selection in (9.9).□

Proof. Consider first one sampled piece. On an integer horizontal row its sites form an integer interval. Its left endpoint is a maximum of affine integer-valued functions with slopes in \{-1,0,1\}, and its right endpoint is a minimum of such functions; horizontal sides impose an interval constraint on the row index. To obtain these formulas, round the intercepts of the defining line inequalities. Because the slopes are integers, this rounding is independent of the integer row index. The difference of the lower and upper bounds is convex, so the nonempty rows are consecutive. On consecutive nonempty rows each endpoint moves by at most one. Consecutive row intervals overlap or are adjacent as integer intervals.

After integer sup-norm dilation by j, the interval on a row is the union of the original intervals in the window of rows at distance at most j, each extended by j on both ends. This union is still an integer interval. Its endpoints are the sliding minimum and maximum of the old endpoints, minus or plus j. They again move by at most one between consecutive nonempty rows: the nonempty index windows at two consecutive rows have Hausdorff distance at most one, and the old endpoint functions are 1-Lipschitz. One more dilation changes each endpoint on an old row by at most two and adds at most two rows, each of length O(s_0+j+1). This proves a bound C(s_0+j+1) for both the new layer and the edge boundary of one piece. Summing over the pieces proves (9.10), with the layer constant absorbed in C_{\rm tpl}. The area bound follows from |P_a\cap\Z^2|\le C(s_0+1)^2.

We also need a count at intermediate dyadic scales. In the usual dyadic partition of the lattice into integer squares, the number of side-u squares containing both a site of a dilated piece and a site outside it is

Indeed, in a block of u consecutive nonempty rows the two endpoints vary by at most u. Only constantly many square columns near each endpoint can be mixed. At the first and last blocks count instead the whole width, which costs O((s_0+j+1)/u+1) in total. A square mixed for a union, or for a difference of two unions, must be mixed for at least one of the constituent sets. Thus the corresponding count for T, T_j, or T_j\setminus T, with j\le s_0, is at most C(N_0+n/u).

Cover T_j\setminus T by maximal contained dyadic squares, with side capped at the largest power of two not exceeding L. These squares partition the set: a single site is an admissible square, and the dyadic nesting makes distinct maximal squares disjoint. At the cap there are at most Cn/L squares because |T_j\setminus T|\le nj\le nL. Below the cap, the parent of each maximal square is mixed. Equation (9.12) therefore bounds the number of squares of side u by C(N_0+n/u)\le Cn/u, since u\le L\le s_0 and N_0s_0\le n. All these squares are safe: their sites lie in T_j, their sides are at most s_0, and

Subadditivity and the assumed box bound now give

Covering T in the same way, with cap comparable to s_0, uses O(n/s_0) squares at the cap and O(n/u) below it, and gives the first bound in (9.11).

Adding an arbitrary part of one depth row changes entropy by at most n\log q. Finally, near these sets there is no edge joining A to its complement: such an edge would have an endpoint in Z within distance s_0+1 of T. Thus their physical boundaries are bounded by the ambient template and shell boundaries just counted. This argument does not count missing lattice edges as physical boundary edges.□

Proof. Let e_0\in(0,1) be supplied by Proposition 3.3. If e_0\le e_*, the assertion already follows. Otherwise fix, once for the entire bootstrap,

These parameters will not change when the current entropy exponent improves.

Suppose F_{\rm box}(r)\le C_e r^{1+e}, where e_*<e\le e_0. Take any safe integer rectangle T of actual maximal side length s_0, and set n=\lceil C_2s_0\rceil for a sufficiently large fixed C_2. If A\cap T is empty, its entropy is zero, so assume otherwise. For large s_0, L=o(s_0), so (9.2) holds. In particular the weaker safe-box clearance D_0s_0, rather than the stronger template clearance of Definition 9.3, is enough for this application.

Here is the shell entropy input explicitly. The shell T_j\setminus T, j\le L, has area O(nL) and consists of a bounded number of rectangular strips. Its mixed dyadic squares of side u number O(n/u+1), by the boundary-line count. The maximal-contained-square argument of Lemma 9.4, capped at side comparable to L, therefore gives S_\Omega(Q_j)\le C_e nL^e. These covering squares are safe: their sides are at most L, and their distance from Z is at least D_0s_0-L>D_0L once s_0 is large. This also covers thin rectangles. The target entropy is at most C_e s_0^{1+e}. Adding a partial row costs O(n), so all the hypotheses of Proposition 9.2 hold.

We check its quantitative output. Since K\asymp n^{g_0-\ell}, a\asymp n^{-g_0+\ell}. In (9.6) the two powers before logarithms are at most

Thus \delta_n\le n^{-0.24g_0} for sufficiently large n, after constants and fixed logarithmic powers have been absorbed. The energy prefactor has power \ell+2\kappa\le0.00003g_0. Its slower decaying input is \epsilon=n^{-0.001g_0}, rather than \delta_n^{1/8}\le n^{-0.03g_0}. Hence

All scale conditions and the smallness of ar_0^2 hold at large n. No sharp marginal-moment condition has been used.

Put \tau=n^{-\nu/4}. Apply the rough upper and lower comparisons (8.8) and (8.9) at the selected interpolation. Write S_X=S_{\widetilde\Omega}(X). The upper pin cost has leading coefficient 2S_X; the lower coefficient is 4sWS_X=4S_X. Replacing \log d_* and S' by S_X costs O(w+|\log z|). The remaining relative loss in this coefficient is at most C(\tau+E_{{\rm def},k}/\tau), which is small after taking k\to\infty and then n sufficiently large. The two comparisons consequently imply

To justify the order of limits in this absorption, keep n fixed in both norm comparisons and use (9.14). The resulting scalar inequality has the same S_X for every k; all other remainders are uniform at the selected points and vanish. Only then choose the large-scale threshold making the coefficient loss smaller than the fixed margin between four and two.

For 0<\tau<1/2, h(\tau)\le C\tau\log(e/\tau). Therefore

because g_0<1/2. Also w=n^{3/5}\le n, and \kappa\le e_*/4<(1-\ell)e while e>e_*. Equation (9.15), followed by the negligible transfer back to \Omega, gives

Bounded s_0 are included by increasing the constant, using the volume bound. Taking the supremum over safe boxes thus replaces the current exponent e by (1-\ell)e.

Starting from e_0, finitely many repetitions reach an exponent at most e_*, which can be loosened to e_*. Every iteration uses only the current safe-box estimate on the shell covering squares. The parameters in (9.13), and the finite number of iterations, depend only on the fixed Hamiltonian parameters. Thus the final constant is uniform over the domain and cut.□

Proof. If X is empty there is nothing to prove. Otherwise Proposition 9.5 and Lemma 9.4 supply (9.4) with e=e_*=2\cdot10^{-6}. Since s_0\le n, the core entropy is at most Cn^{1+e}. The template clearance and L\le s_0 imply (9.2) at large n; the logarithmic truncation radius is negligible compared with L.

Use the following fixed parameter choices:

Then a\asymp n^{-0.74994}. The two terms in (9.6) have powers -0.248998 and -0.186485, before the fixed logarithmic factors. Hence \delta_n\le n^{-0.18} at sufficiently large scales. The power of the energy prefactor is

The \epsilon=n^{-0.001} term is again the limiting one, and Proposition 9.2 gives

Set \tau=n^{-0.00025}. All fixed multiples of a lie in the marginal-moment interval required for the sharp comparisons, since

At each leaf write, as before, the physical partition as X,U,Y,V. All entropies in the next calculation refer to \widetilde\Omega. Let

These are the quantities in the sharp norm comparisons. Their upper bounds are uniform in the status. In particular S(XU)\le S(X)+S(U)\le Cn^{1+e}, and by purity S(V)=S(XUY)\le S(X)+S(UY)\le Cn^{1+e}. Here the matched-prefix error CnD is absorbed because \kappa<e. Thus 0\le g_{\rm pre}\le Cn^{1+e}.

Denote the full terminal path and uniform band average by \langle\cdot\rangle_{\rm leaf}. Subtracting (8.10) from (8.11), and then dividing by 2sW, bounds 2S(X)+\langle g_{\rm pre}-g_{\rm post}\rangle_{\rm leaf} by a constant times

This calculation uses S'=S(X)+O(w+|\log z|) and the analogous estimate for \log d_*. It also restores the lost \tau\langle g_{\rm pre}\rangle_{\rm leaf} from the lower comparison using the preceding uniform bound.

For each leaf separately, purity gives the exact identity

Both V and YV contain the fixed subsystem \Lambda\setminus S_0. Monotonicity of mutual information therefore makes the left side of (9.20) at least I_{\widetilde\Omega}(X:\Lambda\setminus S_0), independently of the history. This remains true after the leaf average, even though the selected round and its weights can vary with k.

For completeness, after k\to\infty the terms in (9.19) have the following upper bounds on their powers of n, apart from fixed logarithms.

All are strictly below 1-\varepsilon=0.99999. Equation (9.18) controls the mass-loss term uniformly before the limit. This proves I_{\widetilde\Omega}(X:\Lambda\setminus S_0) \le Cn^{1-\varepsilon}.

Finally write mutual information as S(X)-S(X\mid\Lambda\setminus S_0) and use Lemma 2.1. Only \log\dim\mathcal H_X=O(n^2) enters, even if the far subsystem is very large. The trace distance O(n^{-500}) between the original and truncated ground states therefore changes this mutual information by O(n^{-248}). This proves (9.16). The final size and boundary assertions were established in Lemma 9.4.□

Proof. We first prove an operator estimate for unital channels. Define

where K_{i,l}=(E_{i,l}k_i)^{1/2}. When k_i\Omega=0, functional calculus gives G_i\Omega=\Omega and K_i\Omega=0, so

for every operator C and spectator vector \xi. Products of these channels therefore recover the contraction products when evaluated on the ground vector. We localize the channels first: their unitality makes each local channel fix the algebra outside its supporting ball. They are also completely positive, hence contractions in operator norm, including after adjoining spectators. Lemma 4.4 gives

The completely bounded norm here is the supremum of the operator norms after adjoining identity maps on finite spectator systems. In this case the estimate follows directly from the norm bounds for the two square roots, so it has no dimension factor.

Put \mathcal E_{i,-1}=\id and \Delta_{i,l}=\mathcal E_{i,l}-\mathcal E_{i,l-1} for l\ge0. By increasing C to include l=0,

Both channels in this difference fix operators commuting with the algebra on N_l(a_i). Thus the difference vanishes on that commutant. For an operator B define its oscillation at a physical site y by

Average over the on-site unitaries in a set Q, one site at a time. The total norm change is at most \sum_{z\in Q}d_z(B), and the result commutes with the algebra on Q. Consequently

This argument also applies to operators acting on spectators.

To bound the oscillation after one event, suppose first that m=d_\Lambda(a_i,y)<\infty. The channel \mathcal E_{i,m-1} does not act at y and commutes with conjugation by U_y; hence it cannot increase d_y. Telescope the shells from radius m onward and use d_y(C)\le2\|C\| to obtain

For m=0 the initial channel is the identity. If m=\infty, \mathcal E_i acts in a different component and cannot increase d_y. Exact component support follows either from the construction or by letting l exceed the component diameter in the approximation assumption.

Sum the increment in (10.9) over the retained anchor labels. Define the nonnegative matrix

with the constant from (10.9). It has a uniformly bounded weighted row sum:

for some a>0. Indeed, at radius l there are at most C(1+l)^2 anchor labels whose balls contain y, and at most C(1+l)^2 sites in each ball. Their distances from y are at most 2l. The left side is therefore bounded by

which is finite for sufficiently small a. There are no kernel entries between different components.

Let B_u be the result of the retained channel events up to time u, starting from B. The Poisson jump formula and (10.9) imply, in integral form, the differential upper bound

The site and anchor sums are finite for a fixed domain, and the radius sums converge absolutely with the uniform bounds in (10.10). Initially d_y(B)\le2\|B\|1_S(y). On components meeting S, weight the last inequality by e^{a d_\Lambda(y,S)^\alpha}. The triangle inequality and 0<\alpha<1 give

The weighted supremum thus grows at rate at most v. Gronwall’s inequality yields

Oscillations on components disjoint from S remain zero.

We can now remove the distant events. For an omitted anchor at finite distance d=d_\Lambda(a_i,S)>r, sum (10.8) over all l\ge0 and apply (10.11). When l\le d/2, every z\in N_l(a_i) has d_\Lambda(z,S)\ge d/2. The sum over these radii is bounded by C\|B\|e^{vu-a(d/2)^\alpha}. For l>d/2 use the shell decay and the ball cardinality instead. Absorbing its polynomial factor gives, in both cases,

At infinite distance the left side is zero.

Couple full and retained processes by using the same clocks. The difference of their products telescopes with earlier retained channels and later full channels. All the later channels contract the operator norm. The earlier retained process is independent of the omitted clocks, so their expected contributions are obtained by integrating the rate-one bound (10.12). There are at most C|S|(d+1)^2 anchor labels at distance d from S. Summing the stretched-exponential tail and integrating time therefore prove the coupled expected-norm estimate

In particular this is a bound on the expectation of the norm, not merely on the norm of a difference of averaged operators.

If k_i\Omega=0, evaluating both channel products on \Omega\otimes\xi and iterating (10.5) converts (10.13) into (10.4). Unitality is essential in the channel argument: the single map C\mapsto G_iCG_i need not fix operators on a disjoint system.

Finally there are at most C|S|(1+r)^2 retained labels, and hence their expected number of events is at most Ct|S|(1+r)^2. Each event’s contraction changes by at most Ce^{-c r^\alpha} under radius-r approximation. Telescoping the contraction products proves (10.3). A retained anchor is within r of S, so every approximating factor is supported on N_{2r}(S). This completes the proof.□

Proof. The case X=\varnothing is immediate; take r=1. Suppose that X is nonempty. We keep the entire physical system fixed while constructing the auxiliary operators; the estimates below are uniform in its volume.

Set w=n^{3/5}. Lemma 3.1, applied in the original ground state to the three cuts, provides spectral projections P_X,T_Y,Q_{S_0} onto their positive eigenvalues with surprisal in the intervals

respectively. The original interactions have uniformly bounded support dimensions and at most Cn terms cross each cut. Thus each failed projection has probability at most C\exp(-cn^{1/10}). For any fixed integer K\ge1, increasing the scale threshold makes these three probabilities at most

For a trivial subsystem its projection is the identity. All inverse eigenvalues used below are restricted to the selected positive spectrum, so no invertibility assumption is imposed on a marginal.

Choose a full eigenbasis \{|x\rangle\} for \rho_X, with eigenvalues p_x\ge0, and write a Schmidt expansion

Append two systems s',e', each a copy of the Hilbert space on X, with fixed unit blank vectors denoted |0\rangle. Consider the unit vectors

In \phi, the original target has moved to e', so its joint reduction with a remote physical system is unchanged after relabeling. In \eta, e' is purified by s' and is independent of all physical systems. The unitary we construct will leave e' untouched and act only on s' and a graph neighborhood of S_0.

For readability write Q=Q_{S_0} and T_0=T_Y. On s',X,Y define the square operator

It acts trivially on e' and Z'. The localization lemma applies to vectors of the form B(\Omega\otimes\xi) with a local operator B. To put R_0\phi in this form, define an operator supported on S_0,s',e' by

The summation over y includes zero-probability eigenvectors. Direct substitution gives

We establish three estimates for these operators before introducing the constraint product. The first is a norm bound controlled by \mathcal I. Let

Orthogonality of the output vectors in (10.22) gives R_0^\dagger R_0=|0\rangle\langle0|_{s'}\otimes |0\rangle\langle0|_X\otimes B_Y. Similarly, B_0^\dagger B_0=|00\rangle\langle00|_{s'e'}\otimes I_X\otimes B_Y. The typical spectral inequalities are

They imply

because purity gives S(X)+S(S_0)-S(Y)=I(X:Z')=\mathcal I.

The second estimate is an adjoint identity with no factor M_0. Let \mathcal R_X be the isometry that relabels X as e' and inserts the blank vectors on X,s'; thus \mathcal R_X\Omega=\phi. For every physical operator L' the coefficients \sqrt{p_x} in (10.21) cancel the selected inverse square roots, giving exactly

The three projections are contractions, whether or not they commute. Telescoping them on \Omega and using (10.19) therefore gives

Third, although R_0\phi may have large norm, its physical ground component has norm at most one. Define

Since Q commutes with \rho_{S_0}, one has 0\le\rho_{S_0}Q\le\rho_{S_0}. Cyclicity of the partial trace for operators on Y allows a second T_0 on the right. Hence

The last inequality follows by adding the analogous positive term with I-T_0; the two cross terms have zero partial trace. In particular M_X vanishes on \ker\rho_X. On its support put A_X=\rho_X^{-1/2}M_X\rho_X^{-1/2}, so 0\le A_X\le I. With \rho_X^+ denoting the inverse on the support and zero on the kernel,

The ancillary coefficient of |x\rangle_{s'}|y\rangle_{e'} in P_\Omega R_0\phi is (M_X)_{yx}/\sqrt{p_x} for selected x. It follows that

We have now constructed a restoring operator whose norm is e^{O(\mathcal I+w)}, whose adjoint is accurate on the desired state, and whose ground component is bounded. Positive constraints will suppress the remaining component while preserving the last two properties.

Run the clocks of Lemma 10.1 for time

where C and then C' are chosen sufficiently large. Use S=S_0, and put \zeta=R_0\phi. The full, retained, and truncated products are denoted D,D_{\rm ret},L' respectively. Equations (10.14) and (10.25) imply

Apply Lemma 10.1 with B=B_0, \xi=|00\rangle_{s'e'}, and S=S_0. The column representation (10.24) then gives

All of these estimates are on the same clock space.

Here is explicit control of the scale choices. First choose C so that gt_*\ge\mathcal I+3w+6K\log n. The first expectation is then at most n^{-6K}. By (10.15), both t_* and r are bounded by fixed powers of n, with constants depending on K. Their polynomial prefactors and |S_0| therefore cost only O_K(\log n) in the logarithm. Since \log M_0=(\mathcal I+3w)/2\le Ct_*, choosing C' larger makes each of the last two expectations at most n^{-3K} for large n. Markov’s inequality and a union bound consequently provide a realization satisfying

Indeed the three failure probabilities sum to at most n^{-4K}+2n^{-2K}<1 at sufficiently large scales.

Every exact factor fixes the ground vector in both directions. Thus D preserves the ground component in (10.28), and D_{\rm ret}^\dagger\Omega=\Omega. Since M_0\ge1, (10.30) gives

Combining the latter bound with (10.27), and writing R'=L'R_0, we obtain

Because \phi,\eta are unit vectors, the real overlap satisfies \Re\langle\eta,R'\phi\rangle\ge1-4n^{-K}. Expanding the square using (10.31) gives

The square operator R' is supported on s' and N_{2r}(S_0), and acts trivially on e'. Its polar decomposition extends on this finite local space to R'=U'H_+, where U' is unitary and H_+=(R'^\dagger R')^{1/2}. Such a unitary extension exists even when R' is singular: its two polar support spaces have equal dimension, as do their orthogonal complements. Tensor it with the identity elsewhere. For \xi=(U')^\dagger\eta we have

Subtract these two errors in the identity

Since \|(I+H_+)^{-1}\|\le1, this proves

In particular no bound on \|H_+\| is needed.

Let J\subseteq\Lambda\setminus N_{2r}(S_0) be arbitrary. The unitary U' changes neither e' nor J, so their joint reduction in U'\phi is exactly the original XJ reduction, with X relabeled as e'. In \eta that joint reduction is a product. Apply Lemma 2.1 to I(e':J)=S(e')-S(e'|J) and (10.33). Only the dimension of e' appears. It has \log\dim e'=|X|\log q\le Cn^2, and therefore

Choose the fixed integer K\ge4(M+3). This proves (10.17), uniformly for every such J. Discarding part of X proves the assertion for its subsets.

Finally (10.29) and (10.15) give r\le C_M n^{(1-\varepsilon)/\alpha}. The choices in (10.1) imply (1-\varepsilon)/\alpha<\beta and 1-\varepsilon<\beta, which prove (10.16). If a graph path of length at most 2r from S_0 reached \Lambda\setminus A, it would contain an endpoint of the original cut. A starting point of this path has ambient distance at most L from T, and each graph step changes ambient distance by at most one. That endpoint would be within L+2r of T, contradicting (10.18) for large n. Thus the graph neighborhood lies in A, as claimed.□

Proof. Write A_f=\bigcup_{i\in\mathcal I_f}X_i and E_f=\sum_{i\in\mathcal I_f}\varepsilon_i. The entropy chain rule and purity give

On the other hand, purity and subadditivity give

Adding the inequalities for f=0 and f=1 cancels every individual tile entropy and proves the claim.□

Proof. Consider a belt cell of side t=t_k. In its 10t-neighborhood only its own and adjacent layers can occur, by (11.6) and t_k/2^k\to0. All fine-cell sides there are t/2,t, or 2t. Cell corners, fan centers, and side midpoints therefore lie on the affine mesh o+(t/4)\Z^2. All active lines have one of the four allowed slopes and intercepts on this mesh. Marks have local scales comparable to t. Distinct marks have distance at least t/4, and a supporting line that does not contain a given mark is at sup distance at least t/8 from it. The same bounds apply to the nearest other endpoint of an incident segment. Choosing a_0 small gives both assertions locally. Apply this argument at the larger scale for a pair of nearby marks; pairs outside the local neighborhood already satisfy the asserted separation.

The preceding coloring shows that two sectors separated by an active ray have opposite labels. Coincident boundary segments or inactive seams are counted only once, so this describes the actual tile regions near the mark.□

Proof. Suppose the assertions hold at one generation. By (11.10), the squares of radius 6d(S) are contained in their parent stars. They are also disjoint from all other current removal squares. Indeed for parent marks (v,S),(w,T),

Thus each modification sees only its parent’s radial boundaries.

Across a new perimeter segment away from ray hits, the new interior label was chosen opposite to the exterior label. Across an interior ray, the two labels remain opposite because the two exterior labels were opposite. All other positive-length interfaces are unchanged. Contacts not lying in the relative interior of one of these segments are among the center, corners, or side midpoints of the new square. This proves the coloring and contact assertions.

Normalize one removal square to [-1,1]^2. Its post-replacement boundaries are subsets of the original eight radial rays together with the four lines supporting this square. The possible new marks are the nine points of \{-1,0,1\}^2. Every nonincident supporting line stays at a fixed positive distance from a mark, and every other endpoint stays at distance at least one. Thus a fixed sufficiently small a_0 gives an isolated radial star at each child, on the child’s scale d(S). This is the same finite line configuration at every generation, so no iteration of a shrinking constant is involved.

Two children of the same parent are at least d(S) apart. For children of different parents their separation is at least

which is larger than a_0\max(d(S),d(T)) by (11.10). Their isolated-star squares therefore satisfy the original separation invariant as well.

The child stars lie inside the parent star. Consequently they avoid all terminal squares from earlier generations. A terminal square created at another parent in the current generation is excluded by the same cross-parent distance estimate. For a direct statement covering all later generations, each successive child displacement is at most the preceding removal radius, and subsequent radii shrink by the fixed factor at most a_0/1000. Their sum, including the radius of the final descendant birth square, is less than 2d(S). Subtrees of distinct parents are therefore separated by the preceding inequalities, even if one subtree has already terminated. This proves the asserted terminal-square avoidance and descendant bound.

Finally, as long as a branch continues, its successive scales satisfy S_{j+1}=d(S_j)\le S_j^{\rho_0} with \rho_0<1. After finitely many steps the computed next halfside is below S_*, and that removal is terminal. This completes the induction.□

Proof. At the birth of an initial belt tile, compare its closure with the primary and belt closures at that time. At the birth of a replacement, compare it with the old remnants after that generation’s removal squares have been cut out, and with the peer replacements. Each final actual region is a subset of the corresponding compared region. In particular an ancestor is compared after its square has been removed; its original birth region need not be disjoint from the new one.

Restrict attention to a closed box obtained by enlarging the tile’s cell or removal square by 2u in each direction. Points outside this box are at distance at least 2u from the birth region. Translate by its cell corner, or by its removal center, and rescale lengths by u. Only finitely many local pairs of closed regions can arise:

• •

  For belts, the nearby active lines and endpoints lie on the affine quarter mesh, with their four prescribed slopes. In a bounded box there are only finitely many such lines and endpoints. The local closures are unions of faces of their arrangement; allowing all possible identifier groupings still leaves only finitely many pairs of unions. We classify only the restrictions to this box: disconnected pieces of an identifier outside it do not change the local pattern.

• •

  For a replacement, the enlarged box lies in the isolated star of its parent, by (11.10). It sees only the old radial layout and the new square and sectors. These lie on a fixed finite arrangement in the rescaled coordinates. Other parents’ simultaneous modifications do not enter this box.

Thus there is a common positive scale for the isolation neighborhoods used in all these comparisons.

The two compared closures of the same label have no positive-length interface. Their intersections, if any, are contact marks by the belt construction or Lemma 11.4. At a belt contact the mark scale is comparable with u (neighboring belt scales differ by at most two); at a replacement contact it is u. Choose disjoint open contact squares of radius \lambda u inside the isolated stars, for a fixed sufficiently small \lambda>0.

First consider pairs of points (x,y) in the two compared closures that do not both belong to the same open contact square. This condition specifies a closed subset of the compact product of the two local closures. On it the distance \norm{x-y}_\infty has a positive minimum: if the minimum were zero, x=y would be a common contact, and both points would belong to its open contact square. Since the list of rescaled patterns is finite, the minimum is uniformly at least c_2u whenever this subset is nonempty.

It remains to consider x,y in one contact star centered at v. Inside that star each closure is a union of closed radial sectors. Their direction sets are disjoint. A shared direction would give a shared ray segment, contrary to the absence of positive-length same-label contacts. Their boundary directions are among the eight axis and diagonal directions, so the angle between their direction sets is at least \pi/4. Elementary Euclidean geometry, followed by comparison of Euclidean and sup norms, gives

For example, the squared Euclidean distance between radial vectors of lengths a,b and angle \theta\ge\pi/4 is (a-b)^2+2ab(1-\cos\theta), which bounds a fixed multiple of \max(a,b)^2.

At the immediately following trim, the compared old or peer identifier loses a square about v of radius at least c_4u^{\rho_0}, by (11.9) and comparability of mark scales. This removal also occurs if the square is terminal. That identifier only shrinks afterward. Thus a point y in its final actual region has \norm{y-v}_\infty\ge c_4u^{\rho_0}, and (11.11) supplies the required lower bound for pairs inside the contact star. Together with the preceding compactness estimate and the exterior-of-box estimate this proves the lemma.□

Proof of Theorem 1.1. The empty domain and the empty or full region have zero entropy. Put b=|\partial_\Lambda A|. If b=0, Lemma 2.2 gives S_\Omega(A)=0.

Assume b>0. Choose n_* large enough for Proposition 9.6 and Proposition 10.2 with M=100 to apply to the templates of Proposition 11.2, uniformly with the stated template bounds. Increase it so that every radius r supplied by amplification at scale n\ge n_* satisfies L+2r<n^\beta. Let D,X_i,T_i,n_i,f_i be the resulting geometric decomposition, and put

Here [-L_i,L_i]^2_{\Z} denotes the integer square. For each i, let r_i be the radius supplied by amplification, using the size, boundary, and information bounds from Proposition 9.6. The resulting mutual-information estimate applies to A\cap T_i against any sites outside the graph neighborhood N_{2r_i}^{\Lambda}(S_i).

We verify this condition for the whole set J_i, rather than for its individual pieces. A graph path of length at most 2r_i from S_i stays within ambient distance L_i+2r_i<n_i^\beta of T_i. Equation (11.4) excludes every earlier same-family X_j jointly. Such a path also cannot reach \Lambda\setminus A: its first crossing of the cut would put an endpoint of a cut edge within that ambient distance of T_i, contradicting (11.3). Thus

Proposition 10.2 and monotonicity under discarding part of its first system yield

These estimates all concern the same original state \Omega, although the auxiliary construction used to prove each estimate may differ.

Apply Lemma 11.1 to the two labels. Since each site has dimension q,

Every choice of threshold and every constant in this proof depends only on the fixed local dimension, interaction range and norm bound, and spectral gap. None depends on the size, shape, connectedness, or holes of \Lambda, or on the choice of A. This proves the theorem.□