Phase decomposition and Wightman reconstruction · Part 3

Thus moment orders can be chosen before shrinking the spectral bands. Here is the interpolation that controls low energy. For distinct nodes, let be their Lagrange basis polynomials, 0. Polynomial interpolation gives

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1
x + y −
X
Y
Lj(x)
zj + y =
1
x + y
zj −x
zj + y ,
y ≥0,
x + y ̸= 0.
(RT4)
j<m
j<m

For completeness, multiply the left side by. It is a polynomial of degree at most, zero at every, and equal to one at. These + 1 values determine it and give the product on the right. Rescale. Multiplying by the even Laplace numerator 2 gives the exact response after subtracting the finite high-band interpolant. For moment-controlled profiles the subtracted moments are exactly those furnished by (RT4). The residual response is nonnegative for0, and the choice of the central profile gives

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Rm,a(E) ≥
cm
E + a,
0 ≤E ≤1,
0 < a ≪1,
(RT5)

with At zero this is the continuous value of the full response, including the profile’s zero-energy contribution; one does not divide a bare factor 2 by and discard an atom. The numerator factors in (RT4) stay of one sign because the high nodes lie outside the primary profile band. Their fixed ratios give; their denominators give the stated reciprocal bound.

The reason this positive response is small is a sign identity for the same tested position correlation. Suppose has reflection parities of opposite sign on axes. Write

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Pi(h; f) =
Z
Li(h)(pi)
Y
l∈Q\{i}
fl(pl) dνA
i .

Here are nonnegative even Schwartz frequency profiles and is the real even Laplace response of the inverse Fourier profile in direction. The two spectral formulas for that one tested correlation give. If some transverse axes are absent, insert profiles equal to one on larger and larger intervals; bounded dominated convergence gives the same identity for that subset. Split each active profile into its Lagrange interpolant and residual. Both residuals have nonnegative response. Consequently the primary residual on axis is bounded by the sum of the two cross terms containing an interpolant. The interpolants are finite sums of profiles in a high band.

For clarity the choice of orders in this repeated operation is important. Put, and scale each frequency profile as 0. A bound will be called a bound of exponent in this paragraph. If a high band has exponent 0, a transverse 2-th moment from (RT3) bounds its far remainder with exponent at most

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−2ka′ + K|Q \ {i, j}| + max(K, dD),
(RT6)

where and is the earlier chosen band exponent. Given any 0, choose so that the negative term exceeds the other fixed losses, and then choose 0 so small that At each switch the coordinate just treated is removed from the list still requiring a bound. If its parity agrees with the next required coordinate, insert an opposite-parity coordinate first. A finite induction on that list terminates; the terminal case has only nonshrinking transverse widths and is bounded by (RT3). Choose moment orders and admissible band exponents backwards through this finite list. This proves an exponent, for every 0, for the residual response. Equation (RT5) therefore gives

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Q
l∈Q\{i} f(N)
l
(pl)
pi + λq
dνA
i ≤Cηλη,
q < 0,
(RT7)
Z
0≤pi≤1

when the collapsed cluster has both parities on

A word on a coordinate line is collapsed on the other axes but can have a genuine time gap on . Subtract there a shifted positive-time profile. Its response is positive, dominates a fixed multiple 1, and its damped transverse moments cost at most the actual gap. The same switch pays this loss by increasing in (RT6). Thus (RT7) holds on , bound the additional-profile by its supremum. The resulting bound has exponent. This proves the line estimate even when the time-axis cluster is not collapsed.

For a diagonal stress reading a further estimate is necessary. Subtract its finite position Taylor jet with a cutoffequal to one at zero. The remainder vanishes to the required order: repeated integration along the segment from zero bounds its-th derivative by when derivatives 1 vanish. Tensor products of these subtractions preserve reflection parities. Conservation, followed by integration by parts on these tests, gives the three tested frequency expressions

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B = p2
i A + R1,
C = p2
i B + R2.
(RT8)

On profiles of width at most0, the remainders have respectively bounds) and ), with an arbitrarily small exponent loss. On profiles with the prescribed zero moments the remainder is exactly zero: it is a finite contact Taylor jet and each of its moments is zero.

Apply (RT4) at widths to. After division by at levels 2, denote their six band readings by, and their residual differences by The positive spectral responses have signs0. Equations (RT8) and the two-band subtraction give small bounds for

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v −a,
V −4b,
r −v,
R −4V,
f −(a −b),
g −(v −V ),
h −(r −R).

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These imply a bound for f with the same exponent. Indeed if the sum of the seven absolute errors
is e, these equalities within error e imply a −b ≥−e, a −4b ≤3e, a −16b ≥5(−e); eliminating a, b
bounds f by 100e. In the exact case they force a = b = f = 0; the numerical factor safely bounds
the finite error elimination. The constant-moment term of each interpolant cancels between W and
2W; every other term has at least a factor W 2. Combining this bound with the positive residual
lower bound gives
Z

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0≤E≤1
EW 4
(E + W)4
Y
j̸=i
f(N)
j
(pj) dνJii
i
≤Cηλ4a+η.
(RT9)

This deduction uses positivity of spectral measures and of the indicated scalar responses. It does not differentiate a positive matrix or posit positivity of its derivatives.

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Now approximate the plane Killing field xaeb −xbea by compactly supported smooth vector
fields. Choose frequency widths λ−100 on axes a, b, λ−150 on the remaining line direction if present,
and λ−300 on the other axes. An odd primitive in direction a, multiplied by λ100, and its partner
with a, b interchanged give the two components. Their derivatives in the a, b pair agree exactly.
Since Jab = Jba, these two terms in P
i,j ∂iFjJij cancel before any integral is split. The remaining
derivatives are supported outside the insertion neighborhoods, where no contact term is present.

Split each such tested spectral integral into1 with transverse frequencies inside the box widened by, and its complement. On the complement, the Schwartz bound of order gives an arbitrary factor in a transverse direction. For 1 the actual exterior time gap gives for every fixed this is at most, eventually. The finite damped moments from (RT3) then pay the whole complement. Inside the box use (RT2) with the two reciprocal weights from (RT7) for shear components, and (RT7),(RT9) for a diagonal component. Split the latter once more at. Below that threshold (including zero), the large gain in the line estimate pays a

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crude stress bound; above it (RT9) applies. With the stated widths these estimates, including the
primitive amplitude λ100, leave a bound Cλ−1. Thus the exterior flux tends to zero.
Here are explicit definitions of the responses and contact expressions in this argument. For a
real Schwartz frequency profile f, write Dsf(p) = s−1f(p/s), mq(f) =
R pqf(p) dp, and bf(t) =
R e−iptf(p) dp, with the Fourier constants used in (RT2)–(RT4).
Choose an even nonnegative
smooth u of integral one supported in 1 ≤|v| ≤2, and set uk(v) = v−2ku(v), which is smooth
because its support misses zero. Write Lj(X) = P
k ljkXk for the node polynomial in (RT4). The
actual finite interpolant is

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Bm,sf =
X
X
k
ljkm2k(f)s−2kDsDj+2uk,
Rm,sf = f −Bm,sf.
(RT11)
j<m

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Integrating against a continuous scalar function and changing variables in each term turns its kernel
into
R u(v) P
j Lj((p/(sv))2)g(sv(j + 2)) dv. Putting g(p) = (p2 + E2)−1 and using (RT4) gives

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Rm,s,f(E) = 2E
Z
u(v)
Z
f(p)
p2 + E2
Y
(sv(j + 2))2 −p2
(sv(j + 2))2 + E2 dp dv
(E > 0).
(RT12)
j<m

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It is the real part of
R e−|t|E \
Rm,sf(t) dt, not a rational expression assigned independently at E = 0.
The latter integral gives its continuous value there, namely pref f(0), since the interpolant vanishes
near zero. The odd-parity response uses
R sgn(t)e−|t|E bf(t) dt = −2i
R pf(p)/(p2 + E2) dp for E > 0.
Its subtraction is obtained by applying the even interpolant to pf(p) and dividing the result by p;
division is on support away from zero. Even and odd residuals have the required first 2m moments
zero.
For f ≥0 supported in [−s, s], every product factor in (RT12) lies between zero and one,
because |v| ≥1 and j + 2 ≥2. Each is at most (2(j + 2))2(s/E)2; if s ≥1, it is at least 1
2(1 + E)−2.
The lower estimate follows from (sv(j + 2))2 ≥4s2 and p2 ≤s2. For the central profile f = Dwχ,
where χ = 1 on [−1, 1], integration just over 0 < p < wE/(E + w) gives

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2E
Z Dwχ(p)
p2 + E2 dp ≥(E + w)−1.

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Consequently
Rm,s,Dwχ(E) ≥2−m(E + w)−1(1 + E)−2m.
(RT13)

Continuity proves the same inequality at zero; on 01, (RT5) follows with. The fixed subtraction band1 and the shrinking primary width are distinct. At high energy the product upper bound gives, for, a bound . These are the precise positive majorants used in the parity switch. For an arbitrary profile of either parity, a nonnegative even compact majorant and the odd division construction bound its subtracted response by a constant times an even response of this form.

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To specify the contacts in (RT8), let Rab,cd denote the compensated product-test distribu-
tion (RT0b), let Tab,cd be its value after Fourier transforming all four test slots, and let Qbcd =
i Fourier(P
j ∂jRjb,cd). Writing Pj for multiplication of frequency slot j by pj, the source identity
is Qbcd = P
j PjTjb,cd. Thus the quantities called A, B, C in (RT8) are

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A = Tii,ii,
B =
X
j,a̸=i
PjPaTai,ji,
C =
X
j,a,b,c̸=i
PjPaPbPcTcj,ab.
(RT14)

Their contact remainders are exactly

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R1 =
X
j̸=i
PjQiji −PiQiii,
R2 =
X
j,a,b̸=i
PjPaPbQjab −Pi
X
j,a̸=i
PjPaQaji.
(RT15)

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Expand Q = P PT, use the exchange symmetries of the two tensor pairs, and commute the scalar
multipliers Pj; this gives B = P 2
i A + R1 and C = P 2
i B + R2 exactly. The compensated Q has no
contact jet of total order below three by (RT0b). On widths ≤λa this costs 3a with arbitrarily
small exponent slack. Multiplication by one frequency costs a and by three costs 3a; hence (RT15)
costs 4a and 6a, respectively. This proves the powers asserted after (RT8). Vanishing on profiles
with the required moments follows directly by evaluating the finite contact polynomial on those
moments.
The exterior-flux powers can also be made explicit. Put E = pd, HN(p) = Q
j̸=d Dλ1+wj χ(pj),
and

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Dq(p) = 1E≤1
HN(p)
E + λq ,
Zz(p) = 1E≤1
Eλ4z
(E + λz)4 HN(p).
(RT16)

The exterior-word integral of has exponent, and otherwise, by the line estimate. For a shear stress, the stress integral of has exponent one. The squared response is at most). Weighted Cauchy–Schwarz therefore gives exponent 3 +, then = 1 and giving at most, then300 and150, giving at most5. These exhaust the shear case; both bounds are stronger than exponent If, split instead at. On the upper part use. Its stress integral has exponent 1 + 4(395, by (RT9) with slack one, and the squared response is at most . The resulting exponent is at most 3+(150101. On the lower part use the crude stress weight, whose integral has exponent at most 3300 = 900. The squared response is at most, giving exponent at most 3101. These pointwise estimates are valid on the nonnegative spectrum and include = 0. The complement of the low box has arbitrary negative exponent by the damped moments and Schwartz tails discussed after (RT9). Thus the spectral reading has exponent101. Multiplying by the angular primitive’s leaves exponent1, as claimed in (RT0c). Decomposition into the finitely many reflection parities preserves the bound; parity-mismatched terms integrate to zero by reflecting their odd axis.

The compact cutoffflux identity is therefore (RT1) equal to zero for every separated tuple on a coordinate line. Translation along that line changesby the fixed vector and leaves the correlation unchanged. Subtracting the two angular identities gives the corresponding sum of the translation shell maps equal to zero. Empty tuples give empty sums throughout.

19.17 The order of choices in the spectral sign walk

We record the quantitative induction that supplies the small mixed-band bound in (RT6). Let be the finite set of axes being tested, with two signs assigned so that each axis has an opposite-sign partner. Let be the axes whose width exponent is still allowed to be negative, and let be the active axis. A finite route alternates signs and removes the currently visited axis from. Such a route exists by induction on: if the active axis is in, remove it and move to an opposite partner. If it is not, choose an axis still in; reach it directly if its sign is opposite, and otherwise take one opposite-sign detour. This reduces the nonempty set after at most two such moves.

The induction statement is uniform in a width vector with, and outside, for every 0. It asserts a residual budget 0 after sufficiently high moment order, for nonnegative even compact profiles and all fixed support factors. If is empty, the direct spectral moment bound of order zero has a finite price exponent. Choose 0 with The nonnegative widths then pay its residual by that budget.

At an induction step from axis to an opposite-sign axis, first obtain the future bandwidth 0 from the induction hypothesis for. Put and = max(). Choose

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an integer
k > (p + z)/a′.
(RSW1)

Only now purchase the moment estimate of order, with its finite price exponent, and choose with. The positive profile majorant constructed by the finite interpolant bounds the baked profile and keeps the remainder nonnegative on the spectral half-line. The far part of the two-axis comparison has exponent at most

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−2ka′ + p + max(−wi, dD) ≤−2ka′ + p + z < 0.
(RSW2)

The near part is paid by the future hypothesis after replacing the active profile by that majorant and its exponent by the nonnegative. The two residuals have nonnegative spectral integrands. The signed comparison inequality from (RT5) therefore bounds their sum by the two paid positive quantities; no spectrum is transferred from one axis to another. Moment orders and thresholds can be enlarged to their maximum. Induction along the finite route gives the bound for arbitrary initial nonpositive widths, with one sufficiently small positive. This explains why the future bandwidth is chosen before the present moment order and why the contact-price exponent is chosen before the present final bandwidth. It supplies the precise quantifiers behind the mixed-band estimate used to derive the collinear Ward identities.

19.18 Normalized two-band diagonal estimate

For clarity, the six normalized readings used in (RT8)–(RT9) have the following common normalization. Let0, and let denote the order- profile from (RT4) at width with all other profiles fixed. Put

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a0 = Re A(q(W)
0
),
b0 = Re A(q(2W)
0
),

Re 1 Re 1 Re 2 Re 2 (DB1) The normalization on the second band is still by, not by 2. Multiplication of the order- profile by the frequency square gives ( times its order- predecessor. Consequently (RT8) gives small errors for . The first contact error has unnormalized order , the second order; their respective normalizations leave order, including the arbitrary small exponent slack. The zero-moment contact subtraction has the same bound: its constant interpolation coefficient cancels between the two bands, and every nonconstant even coefficient has at least a factor Let be the real spectral gap readings at these three orders, normalized by 1 Their signs are0. The difference between a gap profile and its two-band difference is exactly the finite interpolation debris just bounded. Hence the additional errors ) have the same small budget. Let be the sum of the absolute values of these seven errors. From0 we get; from0 we get. Therefore , and (DB2)

Finite sums and fixed scalar multiples preserve the budget exponent, so (DB2) proves the small diagonal gap bound. This is the algebraic step in (RT9); it needs positivity of the three response signs and the contact bounds, and makes no positivity claim for the six individual band readings.

19.19 Ward kernel transfer and the rotation action

We explain how these identities extend to general tuples and to a group action. Differentiating a shell integral is allowed because its fixed compact annulus remains separated under small movements of the distinguished point. Differentiation acts both on the correlation and on the centered angular coefficient. The latter derivative is the relevant constant translation coefficient. Hence every position derivative of a line Ward test is again a finite sum of line Ward tests for successor-degree labels. The all-order scaling comparison (WS7) transports these tests to the free quotient, where the translation and angular actions are explicit derivatives of polynomials. Finitely many

separated reading maps suffice at each finite degree, by the detector argument in (WS4)–(WS5). Their matrices converge to injective free matrices. This transfers the free rank statements for the spaces of possible internal defects and forces any such defect to vanish. Thus the translation shell map is one positive scalar times the already constructed derivative, and the angular rows, divided by, satisfy the Ward identity at every separated tuple.

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The finite-dimensional rank arguments just used have two different domains. For translation
rows, the domain at cap d is the real subspace of Q4
i=1 HomC(Fd, Fd+1) consisting of rows that kill
the unit, preserve each lower filtered image, commute with conjugation, and transform as a vector
under signed permutations. A reading is a position derivative at a fixed separated collinear tuple of
the Ward sum P
lS(v1, . . . , Tivl, . . . , vn). The free readings have common kernel precisely the real
line spanned by the derivative row. Select finitely many such readings by minimizing the dimension
of their common kernel. Adjoining a further reading cannot lower that minimal dimension, so this
finite kernel already equals the full kernel. Pullback by Uk turns the physical readings into matrices
on this one fixed real space; their convergence is (WS7). Eventual rank and the two known rows
in their kernels then give the translation normalization stated above.
For the internal angular defect, the domain is instead the real subspace

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Id ⊂HomR(so(4), HomR(so(4), EndC Fd)) .
(RT17)

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Its elements T(K, L) are alternating in K, L, kill the unit, commute with conjugation, preserve every
lower filtered image, and intertwine simultaneous conjugation of K, L by orientation-preserving
signed permutations with their action on labels.
The reading for a separated tuple, labels vl,
and K, L, is P
lS(v1, . . . , T(K, L)vl, . . . , vn). The free internal Ward separation says the common
kernel of these readings is zero at the required sufficiently large caps; lower caps follow by inclusion.
Select finitely many readings by the same decreasing-kernel-dimension argument. Their joint map
is injective. Transport by T 7→UkTU −1
k , retaining both Lie algebra slots, identifies the actual and
free domains; reality, signed covariance, unit preservation and lower-image preservation follow from

(WS1). On fixed free basis inputs the actual reading matrices converge to the free ones by order-zero joint comparison. Linear independence of finitely many matrix columns is open, so they are eventually injective. This permits testing the unknown, scale-dependent pulled-back defect without assuming it is bounded. Its readings vanish by the Ward identity; injectivity makes the defect itself zero.

The free internal separation used here has a concrete angular argument. Identify fields with their Wick polynomials as in (FM3)–(FM7). For a fixed pair, the assumed zero Ward sums say that the corresponding polynomial endomorphism is silent in every moment; in particular its two-point pairings obey (RT18)

It preserves the degree-four subspace by the lower-image condition. The explicit curvature contraction (FM2), expanded on its degree-four invariant polynomials, gives at equal radii= 0 the

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identity
⟨p(x), q(0)⟩= ⟨(ρ(HyHx)p)(y), q(0)⟩,
(RT19)

where is reflection in the hyperplane normal to, and acts on all tensor legs. This follows by substituting the orthogonal factor in the degree-four Wick pairings; the radial factors agree, and contractions use the invariant color inner product. The explicit low-degree evaluations and the extension to every filtered cap are (FS6)–(FS14), with (WE1)–(WE4) and (HC1)–(HC4) supplying their concrete evaluation and collision steps. Apply (RT18) at and , use (RT19) in both terms, and subtract. Positive definiteness of the reflected polynomial pairing at forces to commute on degree four with). Taking, with the matching radius, makes this product a signed quarter-turn in the-plane.

Signed covariance of the two Lie algebra slots now makes, for each degree-four invariant under simultaneous quarter-turn conjugation of. An alternating bilinear map with this invariance is zero. To check this without representation theory, use the plane basis of so(4). Equal arguments vanish by alternation. For two planes sharing one axis, the square of a quarter-turn in the remaining two axes preserves one plane and negates the other. For disjoint planes, two quarter-turn substitutions exchange the arguments with the sign which, combined with alternation, negates the value. Every coefficient is therefore its own negative. This proves that the internal endomorphism kills the whole degree-four subspace, including the free stress.

The local angular Ward identity for the free stress then propagates this to higher degrees: insert the stress in the silent moment sum, apply the compact angular shell integral, and use that its-term is zero. The remaining identity is the moment pairing of the commutator of with the free angular generator. Nondegeneracy of the Wick polynomial tests, proved by the reflected Gram argument following (FM7), makes this commutator zero. Finite-dimensional ODE uniqueness exponentiates that commutation to each quarter-turn at the higher cap. Signed covariance and the same plane-basis calculation again force = 0. This proves the separation of for every used in (RT17), rather than assuming an additional internal symmetry classification.

For the translation detector there are two further points behind its free one-dimensional kernel. A separated target tuple can be joined to a tuple on any prescribed coordinate line by a segment staying in configuration space: order the line coordinates by the target coordinates on that axis, break ties by distinct arbitrarily small increments, and interpolate. Strict inequalities prevent a collision along the segment, and tied target coordinates remain separated until the endpoint. All position derivatives of the Ward sum vanish at the line tuple when all the scene readings vanish. Analyticity of the explicit Wick kernels gives zero in its neighborhood and then along this segment. Thus scene vanishing is the full silent-moment identity. The color-vector, reality and filtration restrictions of (RT17)’s translation analogue then give precisely the derivative line: the free Wick coefficient calculation identifies its degree-four charge as a real scalar translation charge; subtract that scalar derivative. The remaining silent row kills the free stress. The radial shell identity makes its commutator with the engineering-weight derivation a ghost. The fixed exterior-cap separation (FS13) kills that commutator, and the inversion-parity calculation (FS14) kills the row on every higher cap. The complete charge and polynomial-separation calculations are (FS1)–(FS14), built on the hierarchy (FM1)–(FM7) and the invariant spaces (FC1)–(FC15). The zero scalar case is included.

Apply two orbital derivatives to that identity and subtract in opposite orders. The vector fields have bracket with the opposite sign to the matrix bracket, while the substitutions in a multilinear correlation have the same opposite sign. The resulting defect is [ ]), tested in the internal defect space just eliminated. It is zero. This proves the Lie relations on the actual quotient.

There is a useful additional detail in fitting this representation to the free one: the deformed matrices need not be normal. Put , and work at one finite degree in the space of linear maps between the two quotients. Its relative generators are). Their sum of squares commutes with every: expand the commutator, use the plane matrix brackets, and pair the two opposite ordered-plane terms. On the free side this sum is negative semidefinite in the invariant inner product. A real polynomial in it projects onto the zero eigenspace, by interpolation on its finite real spectrum. Evaluate that same polynomial on the nearby deformed

sum of squares. The resulting matrices converge to the free projection, commute with all the deformed generators, and obey 0; they are not yet projections. Correct them by iteration of, direct polynomial multiplication gives (RT10)

For bounded and sufficiently small initial defect, the defect decreases at least geometrically by 2, and the increments are bounded by a fixed constant times that defect. The iterates are Cauchy and converge to an exact idempotent, with0. Polynomial iteration preserves every commutation and every closed real symmetry constraint.

It remains to show that the relative generators vanish on its range. Each plane generator is the bracket of two other plane generators. Since commutes with them and is idempotent, the restricted matrices satisfy those same perfect bracket identities. They tend to zero because the free generators kill is the largest of their finitely many norms, the bracket identities give . Eventually2, hence = 0. Apply to the comparison map corresponding to the identity: its limit is the identity, so the result is invertible for large and is an exact intertwiner. Reality and signed permutations preserve it, because signed conjugation permutes the plane generators with signs whose squares are one. This proves the required compatibility without assuming a spectral theorem for the deformed matrices.

Transport the free orthogonal representation by this intertwiner. Its exponential derivative is ). Alongand the simultaneously moved labels, the ordinary product rule and the Ward identity show that the derivative of the correlation is zero. The product rule is valid jointly in positions and labels: expansion in fixed finite bases is a finite sum of products of smooth position functions and polynomial coefficient functions. Thus the correlation is constant along every plane rotation. Plane rotations align any unit vector with a coordinate axis; every hyperplane reflection is a conjugate of a coordinate reflection by such rotations. Reflections generate the orthogonal group, so covariance holds for all of

All actions preserve the inclusions. This is first true for their generators by the quotient derivative and Ward identities; the two sides of an inclusion intertwining equation then solve the same finite-dimensional linear differential equation along every exponential, with the same initial value. Uniqueness proves it on the exponentials and the group generators. Finally average the normalized stress, with its two tensor indices moved as well, over normalized Haar measure on compact (4). Conservation, reality and each shell identity are finite-dimensional linear identities, so they pass through the integral. This produces a symmetric conserved covariant stress with the normalized translation and angular charges on the same hierarchy. The next subsection uses precisely this Euclidean action.

Source–readout formulation. The four subsections above establish a symmetry of the recovered hierarchy that is not a symmetry of the realized lattice source. Three statements should be kept distinct.