Free shell charges and separation of the Ward kernel

Free shell charges and separation of the Ward kernel

The free shell calculation is the comparison model for physical Ward charges. Compact cutoffs ensure that all shell integrals avoid the insertions, and the fixed exterior degree horizon controls the tests used to separate defects. The argument does not assume that a low-degree test alone detects an arbitrary high-degree label; the displacement and Wick calculations below provide the additional separating probes.

We give the polynomial calculations underlying the free charge separation used in the rotation and trace arguments. This also specifies the exterior tests needed when a charge acts on high-degree fields.

The compact stress shell identity

Let be a smooth compactly supported function equal to one near a distinguished point, whose support avoids all spectator points. For a current, define its outward shell functional by

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Z
(∂aχ)(y)Ja(y) d4y.
(FS1)
Eχ(J) = −
X
a

The integrand is supported in a compact annulus away from all insertions. All exchanges of finite Wick sums and this integral are consequently justified by smoothness there. A one-leg source at produces the Maxwell pulse obtained by applying its jet functional to the covariance. For a smooth source-free Maxwell field on a neighborhood of the support of, Green’s formula for the bilinear Maxwell stress current gives

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EχJv(pulsep,x, f) = −samplex(p, δvf).
(FS2)

Indeed the divergence vanishes offthe source by the Maxwell and Bianchi equations. Integrating the differentiated fundamental covariance against the cutofftransfers the derivatives to the smooth field. The fundamental-solution identity leaves precisely its jet at. The boundary term at infinity is absent because the cutoffis compact. The variation of a two-form is . For the three vector fields used here this variation preserves the Maxwell equations. The stress is symmetric and traceless, so its contraction with the symmetric derivative of a constant, skew-linear, or radial vector field contributes zero to the divergence.

Here is a regularized derivation of the fundamental-solution step in (FS2). With the covariance normalization), put

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gt(y) =
c4
t2 + |y|2 ,
ρt(y) =
8c4t2
(t2 + |y|2)3 ,
t > 0.
(FS2a)

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R ρ1 = 1: the sphere area is
Direct differentiation gives −∆gt = ρt.
The radial integral gives
R ∞
2π2 and
0 r3(1 + r2)−3dr = 1/4. Also t4ρt(ty) = ρ1(y). Hence for every continuous compactly
R ρt(y −x)h(y)dy →h(x), by the change of variable y = x + tz and the integrable
supported h,
majorant ρ1(z)∥h∥∞.

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For a plane source il, form the potential Ab = δib∂lgt−δlb∂igt, and take its exterior derivative. It
satisfies the Bianchi identity by commutation of derivatives, while its divergence is δlb∂iρt −δib∂lρt.
The divergence of its bilinear stress with the source-free field f is therefore the contraction of
this expression with f; the terms from the symmetric derivative of v cancel by tracelessness and
the other terms cancel by skew symmetry.
Integration by parts against χ transfers the source
derivatives to the smooth compact test made from χ, v, f. The preceding approximate-identity
limit evaluates their difference at x, giving −δvfil(x). On the shell the regulated exterior derivative
converges smoothly to the pulse, since that compact shell avoids x. Its flux thus converges to the
left side of (FS2). This proves the plane-source case with the stated sign.

Differentiate the source center in any direction while keeping the cutoffand vector field fixed. All differentiated shell integrands have a common bound on a fixed compact annulus, so differentiation passes through the integral. A center derivative raises the source leg by that direction; on the right it differentiates the sample of. Induction gives (FS2) for all derivative legs; linearity gives every solution-jet functional. The wave property of for the three motions ensures these samples respect the Maxwell jet relations.

For a constant vector, (FS2) gives minus the derivative-leg sample. For), with skew, it gives minus the infinitesimal tensor rotation. For, a homogeneous solution

jet of derivative order has Euler contribution, and the two form indices contribute two more. Thus the one-leg radial charge is + 2) times that jet. This specifies the engineering weight, including the two contributed by curvature rather than its derivatives.

When both pulses are centered at, use cutoffindependence and homogeneity instead of multiplying distributions at their common pole. Two cutoffs equal to one near have a difference supported away from; source-free integration by parts makes their fluxes equal. Replace the cutoffby its dilation by a factor two about. A jet-order- pulse scales by 2; the other scales ; the volume and cutoffderivative together contribute 2, and a motion homogeneous of degree = 0 or 1 contributes 2. Thus the common flux equals 2. That coefficient is strictly smaller than one, so = 0. When neither pulse is centered at, their fields are smooth throughout the support of the cutoff, and the source-free divergence identity gives zero. When precisely one pulse is centered there and the other is at a spectator point, (FS2) gives minus the covariance with the transformed distinguished leg.

Now insert the quadratic Wick stress in an arbitrary Wick monomial word. Double Wick extraction expresses its correlation as one half the sum over ordered choices of two occurrences, times the bilinear pulse current for those occurrences and the Wick pairing of all remaining occurrences. There are exactly the three cases just considered. The terms with both distinguished occurrences or with neither distinguished occurrence vanish. In each remaining term exactly one occurrence of the distinguished monomial is transformed. The two orders of the selected pair cancel the one half. Summing over its occurrences is the Leibniz rule for the derivation on that monomial. By multilinearity this proves, for arbitrary polynomial words,

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X
!
*
+
b
vb(y)⟨Tab(y)P(x)
Y
l
Pl(xl)⟩
(DvP)(x)
Y
l
Pl(xl)
Eχ
= −
.
(FS3)

Here is respectively translation, rotation, or the engineering-weight derivation. This proves the three shell laws, including their common minus sign, and includes every contraction term. In particular conservation of the free stress alone was not being used as a substitute for its charge normalization.

Low-degree vector charges

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Let Sj be the real space of homogeneous order-j Maxwell solution jets and Vj = S ∗
j .
Their
elements are the real one-leg test functionals. A variable at order j has engineering weight j + 2.
For p ∈Vj, write Bc(p) for its linear polynomial in color c, and set

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[p, q] =
X
c
Bc(p)Bc(q).

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This bracket denotes a color-contracted quadratic polynomial, not a Lie bracket. For a linear map
D : Sj →Vl, its polynomial is
C(D) =
X
ν
[eν, Deν],
(FS4)

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where (eν) is any basis and (eν) its dual. This is basis-independent, and C(p ⊗q) = [p, q]. If j ̸= l,
assigning independent jets in these two orders in one nonzero color proves that C is injective.
Consider a vector row Ti on the invariant polynomial filtration, killing the unit, increasing
filtration by at most one, real under conjugation, and signed-covariant. Suppose it is silent, meaning
X

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l
⟨P1(x1) · · · (TiPl)(xl) · · · Pn(xn)⟩= 0
(FS5)

for every allowed exterior word and separated configuration. At input weight four the invariant space is spanned by the unit and [] withV. At output weight at most five the only further possibility is a color-contracted pair of orders zero and one. Total inversion is even on weight four and odd on such a pair. The vector law anticommutes with total inversion, and therefore removes the weight-four part of the output. Reality permits real coefficients in (FS4). Consequently

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Ti[p, p] = C(Di(p)),
Di(p) : S0 →V1.
(FS6)

Let) denote the real single-color covariance between two test legs at displacement To extract the constraint on (FS6), put two quadratic insertions at and zero, and a third at a fixed nonzero point, and multiply (FS5) by. The covariance of jet orders zero and one is homogeneous of degree5. The triangle terms where the transformed order-one leg is on either near insertion have a finite limit; the term with that leg at the distant insertion tends to zero. The double contractions contribute 4 dim g. Choose the distant one-leg functional so that its covariance with a zero-order test is evaluation on a prescribedS. Cancelling the common nonzero factor gives (FS7)

The availability of these evaluations follows from covariance separation of solution jets. First suppose a test functional of order has zero covariance with every order-zero plane test at every nonzero displacement. Its pulse is then zero offits source, in every color component. Pair its radial stress shell with the homogeneous polynomial Maxwell solution having any prescribed order- solution jet. The shell integral is zero because the pulse is zero there; (FS2) makes it times the test evaluated on that arbitrary jet. The functional is zero. For partner order vanishing against all such partners makes the radial derivative of its covariance with any order- partner zero. That covariance is homogeneous of degree); Euler’s rule makes the radial derivative that nonzero scalar times the covariance. Thus it vanishes, and induction reduces to order zero. Finite-dimensional duality now makes these covariance evaluations span the required full dual. Thus no inverse of a covariance matrix at a single momentum is being assumed. This also proves that the solution-jet space is nonzero at every order: otherwise separation against that order would force the nonzero plane evaluation at order zero to vanish. If) = 0, (FS7) first makes) = 0 for= 0. Given) = 0, add a test with nonzero value at, apply that result to the sum, and subtract. Covariance separation gives = 0. Hence kerker). Choose with= 0; subtracting) from proves (FS8)

Set(0) = 0. For two nonzero linear functionals there is an on which both are nonzero: use a witness for one, a witness for the other, and their sum if needed. Substitution in (FS7) yields

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Kz(fi(p), q) = Kz(fi(q), p).
(FS9)

Pair the difference) with an arbitrary one-leg test, use (FS9), and cancel by linearity in its other slot. Separation makes the difference zero. The same argument with a scalar proves homogeneity. Thus VV is real linear. Signed covariance of, injectivity of (FS4), and cancellation of a nonzero test functional give simultaneous signed covariance of the family

We classify this family explicitly. Write for evaluation of a zero jet in the-plane and for the corresponding first-jet functional. Under reflection of axis, the latter has character equal to the product of the three signs belonging to Project onto a reflection character by the four commuting projections (1 The character of) leaves only

. Average the permutation interchanging axes two and three. The Maxwell relation gives, so this fixed subspace is the line spanned by. Similarly the character of) leaves only. Antisymmetrize the interchange of axes one and two; the Bianchi relation leaves the line spanned by. Hence these two values have scalar coefficients

Use (FS9) with, and The explicit differentiated covariance (FM2), evaluated below after (WE4), gives the same nonzero value for and Therefore. Permutations fixing zero give all remaining plane evaluations, which span V permutations moving zero give the other directions. We conclude

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fi = a∂i,
Ti[p, p] = a[p, ∂ip].
(FS10)

Since], the scalar of the translation derivation is2. Squares span the symmetric quadratic polynomials by polarization, and both rows kill the unit. This proves the entire degree-four assertion, with its factor two, rather than just a classification of invariant labels.

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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wz,X = κ−1
z
1
2
X
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

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