Radial jets, calibrated labels and the free Maxwell comparison hierarchy · Part 4

16.10 Fixed-displacement evaluation and the triangular Wick calculation

Here are the three explicit evaluations used in (FS7)–(FS10). Put, and let for= 0. Its action on two-forms is the solution action is the unit antisymmetric plane tensor, the differentiated covariance (FM2) gives

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Kz(p, cab) = κz p(ρ(Hz)Eab),
κz = 4c4|z|−4 > 0.
(WE1)

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Indeed substitute the second derivatives of c4|z|−2 into the two curls: their radial terms combine to
the product of the two Householder matrix entries on the antisymmetric indices, with the displayed
common radial factor. This is an identity on the six-dimensional real space of antisymmetric order-
zero solution tensors.
Given X ∈S0, set W
= ρ(Hz)−1X.
Antisymmetry gives the plane expansion W
=
1
2
P
a,b cab(W)Eab. Define the one fixed functional

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X
wz,X = κ−1
z
1
2
a,b
cab(W)cab.

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Bilinearity and (WE1) now give

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X
Kz(p, wz,X) = 1
a,b
cab(W)p(ρ(Hz)Eab) = p(X).
(WE2)
2

Thus the evaluation witness used at displacement in (FS7) exists at that displacement itself. No change of the distant point or sum over different displacements is necessary.

Write) for the three-point Wick moment. At each point its two letters are Wick ordered, so a contraction cannot join two letters at that same point. In a pairing of the six letters, each point must therefore have one leg going to each of the other two points. For letters at the first point, two equal letters at the second, and at the third, there are two choices of which first-point letter goes to the second point. For either choice there are two choices of its partner among the equal’s and two choices of the partner among the equal’s. The third edge is then forced. Hence

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Tx,y,z([a, v], [p, p], [q, q]) = 4 dim g {Kx−y(a, p)Kx−z(v, q)
+ Kx−z(a, q)Kx−y(v, p)}Ky−z(p, q).
(WE3)

The color factor is exactly dim g: the three edges give Kronecker deltas requiring all three summed colors to coincide. All unequal-color summands are zero and each common color gives the same real scalar expression. Let= 0, with the third point= 0 fixed. The near covariance of two order-zero legs scales as, while that of orders one and zero scales as. Multiplication therefore gives

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4 dim g {tKu(a, p)Kx(t)−z(v, q) + Kx(t)−z(a, q)Ku(v, p)}Ky(t)−z(p, q).
(WE4)

Every covariance involving the distant point is continuous in this limit. The first term tends to zero; the second tends to 4 dim g). If the order-one letter is instead at the distant point, its two edges stay bounded and the only short edge is an order-zero covariance. Multiplication by leaves an extra factor, so that whole term tends to zero. Expand the panel in (FS6) in a basis as in (FS4), use (WE2) for the distant test, and apply these two limits to the three

silent terms. The common nonzero factor 4 dim g cancels; the odd sign of the order-one covariance under gives exactly (FS7). Finally take. Substitution in (FM2) gives, at every

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Kz(c01, c12) = 8c4z0z2(|z|2)−3.

Differentiation of the first leg in direction zero gives

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Kz(s001, c12) = 8c4{z2(|z|2)−3 −6z2
0z2(|z|2)−4}.

At, its squared norm is two and both displayed coordinates are one. The value = 0. The order-zero covariance is symmetric in its two plane tests; differentiating that symmetry gives the same value for). Thus (FS9) reads , proving

Source–readout formulation. A silent vector row is, in the terms of the method, a candidate vertical direction of the free readout: an operation on labels which no separated correlation with any spectators can see when it is applied to every insertion in turn (FS5). The two subsections above compute the fibre of such directions on degree-four fields and find that it is one dimensional, spanned by translation: by (FS10) a silent row acts there as the scalar2 times. The computation is a recovery of the Ward kernel from the hierarchy, not an assumption about it. Its inputs are the explicit kernel (FM2), the reflection characters, the Maxwell and Bianchi relations among the first jets, and one nonvanishing number: at the mixed covariance equals= 0, which is what forces the two constants and to agree through the symmetry (FS9). The fixed-displacement Wick evaluation (WE1)–(WE4) is triangular in the filtration, and it is this triangularity that allows the conclusion at weight four to be reached without knowledge of higher-weight rows. The exactness class is exact recovery within the free comparison sector.

16.11 Detection with a fixed exterior degree bound

We first establish the collision extraction that will be used in the polynomial argument. Put . Say that is cleared at order when, initially defined offzero, extends to a real analytic function on all of. A harmonic pole of degree is analytic offzero, homogeneous of degree under positive dilations, harmonic there, and cleared at order1. The differentiated single-leg covariance of jet orders is such a pole with. A cleared homogeneous term of degree means a function homogeneous of degree cleared at order max( product of poles of positive degrees is a cleared homogeneous term of degree, because the product of their analytic numerators clears it at order (

The pole assertion for the differentiated kernels has a short direct proof. The Green function is homogeneous of degree2; multiplying it by leaves the constant. Away from zero its Laplacian vanishes, since the radial Laplacian in dimension four gives = 0 for . Suppose is a pole of degree2 with numerator. For a fixed direction , differentiation gives the numerator identity

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RrDvF = R1Dvg −(r −1)(DvR1)g.
(HC0)

Its right side is defined with the same regularity on all real space. Differentiating the dilation identity makes homogeneous of degree + 1), and commuting ordinary derivatives with the Laplacian shows it is harmonic offzero. Thus differentiation raises the pole degree by one and

retains the required clearing order. Finite linear combinations of equal-degree poles retain all these properties. The two curls in (FM2) are a finite combination of second derivatives of the Green function, so have degree four. The two jet orders add derivatives, with the factor ( for derivatives at the second endpoint. Color pairings and arbitrary probe coordinates are fixed linear maps. This proves exactly the degree 4 + and clearing hypothesis used above. It also gives a global bound: bound the continuous function on the unit sphere and use positive dilation homogeneity. All uses here are offthe real collision point; no contact value or complex cone division is involved.

For a harmonic homogeneous of degree, Euler’s identity, together with the product rule for the four-dimensional Laplacian, gives

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∆(Ra+1F) = 4(a + 1)(a + 2 −r)RaF.
(HC1)

For example ∆(; the general formula follows either by the same differentiation or induction. If2 and the pole is cleared already at order2, it is zero. Indeed, starting with its analytic numerator at that order, (HC1) repeatedly produces an analytic numerator one order lower. At each step3, so the displayed coefficient is nonzero. At the last step itself extends analytically. Its homogeneity then implies0. For clearing at order zero gives this conclusion immediately.

Here is the exact finite expansion needed; it includes terms more singular than the pole being extracted. For integers, write

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

F(tx) −
X
j<M
Wj(tx)
tr
−→K(x)
(x ̸= 0),
(HC2)

where each is a cleared homogeneous term of degree, and is a harmonic pole of degree ) = 0 for all sufficiently small positive, then = 0, for2. To prove this, peel the finite sum from its largest degree. A nonzero term of degree would diverge after scaling by multiplying the convergent expression by forces = 0. Repeat until only degrees at most remain. Degrees below tend to zero; the limit is the degree- term. Thus, or is zero if that degree is absent. In either case is cleared at order2, and (HC1) makes it zero. This finite peeling, rather than a choice of a large scaling exponent, removes the more singular double contractions.

To apply this statement, let contain jets through order, let a pair of test legs of orders with, be inserted at, and keep the other vertices separated. The contraction of an order- probe with the order- variables of is the differential operator

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j
X
Ba(p, z)P =
X
ν
Kz(p, ea,ν) ∂a,νP,
B(p, z)P =
a=0
Ba(p, z)P.
(HC3)

The index includes the jet basis and color. Wick’s rule applies the two operators to every ordered pair of vertices. Set and + 1. There are four cases.

1. Both legs contract at. The resulting finite sum consists of second partial derivatives of times products of two kernels. Their degrees are 8 +, with. They are the cleared terms in (HC2), with harmonic part zero.

2. The order- leg contracts at, and the order- leg at another vertex. The distant contraction is analytic at = 0. Multiplication by selects only); the terms have smaller pole degree and vanish. The distant contraction tends to its value at

3. The order- leg contracts at, and the other leg is distant. Its largest pole degree is 4+ , so the scaled limit is zero.

4. Both vertices are distant. Every coefficient is regular at = 0, so the scaled limit is zero.

These conclusions require only continuity of the distant Wick coefficients for the scaled limit; the coefficients are in fact analytic. Summing the finitely many placements proves (HC2). In case 2, summing the distant placements is Wick’s first-leg identity at. Summing colors in the invariant two-leg tester therefore gives harmonic part

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K(z) = ⟨(Cj(p, q; z)P)(x0)
Y
i
Vi(xi)⟩,
Cj(p, q; z)P =
X
c
B(q, c) Bj(p ⊗c, z)P.
(HC4)

Here) is the order-, color- jet variable. Orthogonality of the color action makes the sum invariant. The operator replaces a variable of weight + 2 by one of weight + 2, so it strictly lowers engineering weight whenever If is a ghost at exterior cap, and 4 +, the inserted invariant pair is an allowed spectator. For every fixed nonzero direction, all the inserted points stay separated for sufficiently small positive. Permuting the inserted pair past, ghost vanishing makes the left side before subtraction in (HC2) zero. Finite peeling and (HC1) now give) = 0 for every= 0. Since the spectator word was arbitrary, is itself a ghost at the same cap. This is the cluster extraction used below; it retains arbitrary spectator arity while keeping each spectator’s degree bounded.

The passage to arbitrary input weight needs a uniform form of polynomial separation. An exterior bound limits the weight of each spectator field, while allowing arbitrarily many spectators. Call a polynomial a ghost at bound if all its separated correlations with all such spectator words vanish. We prove that one finite, depending on the group, detects polynomials of every input weight. This is stronger than a Gram statement at a fixed input cap.

First, separated single-leg covariances with fixed partner jet order span the dual of each homogeneous solution-jet space. Their annihilator is zero by the differentiated covariance separation used in (FS7), and finite-dimensional duality proves the span assertion. Consequently if every single contraction of a polynomial with those tests is zero, all its partial derivatives in the corresponding jet variables are zero. Indeed a single contraction is the sum of those partial derivatives multiplied by the covariance coordinates; linear combinations of the spanning covectors isolate each derivative.

Next we must separate colors using invariant tests. Put = dim g. For every color index 1, choose a nonzero functionalV and a solution jet with) = 1. Such jets exist at every order. Form the polynomial matrix

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Mcb = Bb(pc).
(FS11)

Its determinant is not the zero polynomial: assign the order- jet to color with value, and all other color components at that order to zero. This evaluation makes the identity. The argument takes place in the polynomial ring, which is an integral domain. It does not assert that is invertible at every field configuration.

Suppose a polynomial has no jet variables above order +1. Extract a mixed cluster consisting of one highest-order contracted leg and one of the low-order color tests just chosen. Vanishing of these extracted correlations says = 0, where is the single contraction in color with the highest-order variables. Multiply by the adjugate: det( = 0. Since the determinant is a nonzero polynomial, = 0. The preceding dual-span argument makes every highest-order partial derivative zero. Over characteristic zero a variable occurring with a positive power has a nonzero

partial derivative, so the polynomial contains no variable at that order. This removes the highest jet order one layer at a time.

We spell out the bounded-degree and induction aspects of the cluster extraction. Set +1. The polynomial ring on jets through order has finitely many variables and is Noetherian. Form its ideal generated by all positive polynomial-length homogeneous invariant polynomials. Choose finitely many such generators of this ideal, and let bound their engineering weights. They generate every invariant polynomial as an algebra: for an invariant polynomial of length 0, express it in the ideal, take its length- component, and average the coefficients over the compact group. Averaging is a projection onto invariant polynomials and satisfies

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for invariant g.
For a generator of positive length l, its coefficient after this projection is the
averaged length-k −l part of the old coefficient; it has length strictly smaller than k. Induction on
length expresses it as products of the same generators. Constants are empty products. Thus every
invariant polynomial homogeneous of length k on this shelf lies in the span of products Q
i Vi, with
each Vi invariant of engineering weight at most r, homogeneous of length ki, and P
i ki = k. This
is the finite hosting bound; the number of factors is unrestricted.

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Here is the precise coalescence argument that uses it. Suppose a ghost P has all length compo-
nents below k zero. Fix x ̸= y and a hosted product Q = Qn−1
i=0 Vi of total length k. Place P at x
and Vi at y + tie0, for t > 0 sufficiently small. These points are distinct and avoid x. Expand all
polynomials into monomials and apply Wick’s rule. A monomial of P of length l > k has too many
letters to pair with the k exterior letters: it has no internal pair at x, so its moment is exactly zero.
If l = k, every exterior letter must pair with a letter at x, leaving no exterior-to-exterior pair. All
the resulting kernel arguments approach the nonzero vector x −y; their finite products therefore

converge. Terms are absent by hypothesis. The limit is exactly the two-point Wick pairing B), where is the length- component. Ghost vanishing at every positive separated makes this limit zero. Linearity extends it from products to their span. No estimate of kernels between coalescing exterior points is needed, since the pairing count removes those kernels before taking the limit.

Induct on. Let be the first possibly nonzero length component and reflect its complex . The shelf, invariance, engineering cap and conjugate across the time hyperplane, giving length are preserved by this operation. The hosting result puts in the allowed span. At points its reflected Wick pairing with is strictly positive if= 0, by the polynomial Gram calculation above. Coalescence has made the same pairing zero, a contradiction. Thus Every polynomial has only finitely many length components, so this induction proves that a ghost lying on this fixed shelf is zero. Define

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d = max{r, 4 + j + L}.
(FS12)

Induct on the engineering weight of the candidate ghost. If its highest jet order is, cluster a pair of jet tests of orders and. Its weight is 4 +, and. Apply (HC2)–(HC4) with highest shelf order: the finite cutoffis 8 + + 1, and the extracted pole degree is 4 +. The extracted mixed contraction is a ghost of strictly smaller engineering weight, hence is zero by induction. The color-matrix argument then removes the highest jet layer. Repeating leaves a polynomial on the shelf through, where the hosting tests separate it. The weight-zero case is the constant, detected by the empty word. This proves

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∃d ∀w ∀P of weight at most w,
P a ghost at bound d =⇒P = 0.
(FS13)

The same finite bound is used before the finite-dimensional Ward readers are selected. In particular one takes their cap at least max( 4); lower-cap conclusions then follow by the filtered

inclusions. Arbitrarily high output weight in (FS13) is permitted, but arbitrarily high individual spectator degree has not been silently introduced.

Source–readout formulation. Detection is the statement that the free readout is faithful on the physical label space when the exterior probes are restricted to a fixed finite degree. The quantifier order is the content: one degree bound serves all weights.

Lemma 16.6 (Finite peeling). Suppose

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tr
F(tx) −
X
j<M
Wj(tx)

−→K(x)
(x ̸= 0),

with cleared homogeneous of degree and a harmonic pole of degree for all small then

Proof. Peel from the largest degree: a nonzero with would diverge, and multiplying the convergent expression by forces = 0. Then degrees below tend to zero and = 0), cleared at order2, hence zero by (HC1).

Theorem 16.7 (Faithful readout with bounded probes).

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∃d ∀w ∀P of weight at most w :
P a ghost at bound d =⇒P = 0.

Proof. Induct on the engineering weight. If the highest jet order of is, cluster a pair of tests of orders and: its weight is 4 + and. By (HC2)–(HC4) with shelf order (cutoff8 + + 1, pole degree 4 +) the mixed contraction is a ghost of smaller weight, hence zero, and the color-matrix argument removes the highest layer. Repeating leaves a polynomial on the shelf, where the hosting tests separate it. Weight zero is the constant, detected by the empty word.

In source–readout terms the theorem says that the fibre of the readout “all separated free correlations against exterior words of degree at most” over zero is trivial on the quotient by the equation ideal. A ghost would be a nonzero vertical direction of this restricted readout; the peeling lemma removes it by extracting, at the leading scaled order, a harmonic pole that must vanish. The degree bound matters for the later transfer: only finitely many probes per weight are needed, so the rank conditions transferred to the interacting hierarchy involve finite matrices.

16.12 Elimination of higher-degree vector and angular defects

Subtract from a silent vector row the scalar translation obtained in (FS10), and denote the residual row by. It is silent, kills every degree-four field, and in particular kills the free stress. Insert that stress as an additional spectator in (FS5), apply its radial shell identity (FS3), and subtract the silent identity with the engineering-weight derivative applied to the distinguished field. The terms where the row acts on a spectator cancel in pairs. The stress term is zero. What remains says that

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[E, Qi]P
is a ghost,
E =
X
ν
wνXν∂Xν,
wν = jet order of Xν + 2.
(FS14)

All additional stress fields have degree four and the spectator cap is the fixed one in (FS13). Thus (FS13) gives [ Every polynomial is a finite sum of engineering-weight components, and map commuting with preserves each such eigenspace: projecting its image onto weight

gives () times that projection equal to zero. Total inversion acts on a weight- monomial , because a derivative-order- curvature variable has sign ( But the vector covariance law gives. On a preserved weight component these two rules give simultaneously and. Hence = 0, and on the whole cap. This proves the full one-dimensional free translation-charge kernel used after (RT17).

For the internal angular row the degree-four quarter-turn argument in (RT18)–(RT19) already makes it zero on the free stress. Apply the angular version of (FS3) in the same silent identity. It makes the commutator with each angular generator a ghost, and (FS13) makes that commutator zero. On every finite filtered space, exponentiating a finite matrix gives the unique solution of its linear differential equation. Commutation with the generator therefore gives commutation with its quarter-turn. Signed covariance of the two alternating Lie-algebra slots and the plane-basis calculation following (RT19) then make every component zero at that cap. This supplies the higher-degree angular separation with the required bounded exterior tests.

Finally the analytic segment argument following (RT19) converts the collinear scene readings and all their position derivatives into the full silent identity (FS5). Taking the cap in (FS12)– (FS13), the preceding arguments identify its kernel. Finite-dimensional decreasing-kernel selection gives a finite list with exactly that kernel. Its nonzero minors remain nonzero for the convergent interacting reading matrices. This proves the rank transfer used to normalize the actual translation charge and eliminate the actual internal angular defect.

Source–readout formulation. In the free sector the Ward readout (all silent identities) has kernel exactly the one-dimensional translation line and no angular component; the charge is therefore recovered up to one real scalar, which the shell law (FS3) fixes. This is an exact statement about the imported comparison realization. Its transfer to the interacting hierarchy is a rank transfer: a finite reader list with nonzero minors in the free sector keeps nonzero minors along convergent interacting reading matrices. The lift records the free statement as exact, and the interacting statement as conditional on those convergences, which are supplied by the comparison estimates of the trace chapter. Under the frame realization all objects of this section are functions of gauge-invariant labels and are unchanged.