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Entry 14 · 2026-08-12 · marici.Nima · research

Direct Scalar Descent Theorem

Record

Date: 2026-08-12

Status: all-multiplicity theorem at generic formal massless tree kinematics. The scalar associated grade descends through the full Parke--Taylor relation ideal. Boundary resonance and the surface-function lift remain separate questions.

Verdict

The missing scalar-provenance step in entry 11 can be proved without identifying the answer as NLSM and without inserting the CHY representative.

Let

an(α)=gr⁡RAscalar,n(α)a_n(\alpha)=\operatorname{gr}_R A_{{\rm scalar},n}(\alpha)

for even (n), and set (a_n=0) for odd (n). The alternating scalar construction supplies the following data directly:

  1. cyclic symmetry;
  2. homogeneity of Mandelstam degree one;
  3. only simple allowed planar poles;
  4. scalar boundary factorization

Res⁡P2=0an=aLaR; \operatorname*{Res}_{P^2=0}a_n=a_La_R;

  1. an Adler zero when any external momentum is soft.

These properties imply the photon-decoupling identity and the fundamental BCJ relation at every even multiplicity. A general field-theory theorem then says that cyclic symmetry and the fundamental BCJ relation generate the full KK and general BCJ relations. Consequently,

an(ker⁡qn)=0,\boxed{a_n(\ker q_n)=0,}

where (q_n) sends cyclic words to Parke--Taylor twisted-cohomology classes. Thus the inverse pairing construction

Jn=(In♭)−1an\mathsf J_n=(I_n^\flat)^{-1}a_n

is representation independent at all multiplicities.

Combined with the period-separation theorem of entry 11, this gives

Jn=[(Pf⁡′An)2].\boxed{ \mathsf J_n=[(\operatorname{Pf}'A_n)^2]. }

Formal kinematic setting

Work first on the Gram-free massless Mandelstam vector space

Vn={sij=sji,sii=0,∑jsij=0}.V_n= \left\{ s_{ij}=s_{ji},\quad s_{ii}=0,\quad \sum_j s_{ij}=0 \right\}.

Its dimension is

dim⁡Vn=n(n−3)2.\dim V_n=\frac{n(n-3)}2.

The soft subspace for leg (i) is

Wi={s∈Vn:sij=0 for every j}≃Vn−1.W_i=\{s\in V_n:s_{ij}=0\text{ for every }j\} \simeq V_{n-1}.

An identity on this formal space specializes to any fixed spacetime dimension. Working here avoids accidental Gram relations during the contact-term argument.

For four distinct labels define the alternating four-cycle (C_{abcd}\in V_n) by

(ab,bc,cd,da)=(+1,−1,+1,−1)(ab,bc,cd,da)=(+1,-1,+1,-1)

and all other edge coordinates zero. One explicit basis of (V_n) is

{C123k,C12k3:4≤k≤n}∪{C12kℓ:4≤k<ℓ≤n}.\left\{C_{123k},C_{12k3}:4\leq k\leq n\right\} \cup \left\{C_{12k\ell}:4\leq k<\ell\leq n\right\}.

Its cardinality is

2(n−3)+(n−32)=n(n−3)2,2(n-3)+\binom{n-3}{2}=\frac{n(n-3)}2,

and independence follows directly in edge coordinates.

Two soft-contact lemmas

Linear lemma

Let (L\in V_n^*). If

L∣Wi=0for every i,L|_{W_i}=0 \qquad\text{for every }i,

then (L=0) for (n\geq5).

Every basis four-cycle omits at least one external label and therefore belongs to the corresponding (W_i). Hence (L) vanishes on a basis.

Quadratic lemma

Let (Q\in\operatorname{Sym}^2(V_n^*)). For even (n\geq6), if

Q∣Wi=0for every i,Q|_{W_i}=0 \qquad\text{for every }i,

then (Q=0).

Polarize (Q) to a symmetric bilinear form (b). The soft condition gives

b(Wi,Wi)=0.b(W_i,W_i)=0.

It is enough to test (b) on pairs of alternating four-cycles. If their supports omit a common label, both cycles lie in one (W_i), so their pairing vanishes.

For even (n\geq10), the union of two four-cycle supports contains at most eight labels and there is automatically a common omitted label. At (n=8), the only extra case has disjoint supports. Choosing a label (t) from the second support and using

Cabcd=Cabct+CatcdC_{abcd}=C_{abct}+C_{atcd}

reduces it to pairs whose support union has size seven.

At (n=6), support unions of size at most five are immediate. Union-six cases reduce by the same split unless the two labels unique to each cycle are adjacent in both cyclic orders. The remaining permutation orbit is represented by the following exact identity in Sym⁡2V6\operatorname{Sym}^2V_6:

C1342⊙C1562=C1243⊙C1235+C1243⊙C1263−C1243⊙C1236−C1253⊙C1245+C1245⊙C1256−C1345⊙C1356.\begin{aligned} C_{1342}\odot C_{1562} ={}&C_{1243}\odot C_{1235} +C_{1243}\odot C_{1263} -C_{1243}\odot C_{1236}\\ &-C_{1253}\odot C_{1245} +C_{1245}\odot C_{1256} -C_{1345}\odot C_{1356}. \end{aligned}

Every pair on the right omits a common label. Thus (b=0), and therefore (Q=0).

These lemmas are the only uniqueness input below. No general NLSM uniqueness theorem is used.

Photon decoupling from scalar residues

Choose leg (1) and define

Dn=an(1,2,…,n)+∑i=2n−1an(2,…,i,1,i+1,…,n).D_n= a_n(1,2,\ldots,n) +\sum_{i=2}^{n-1} a_n(2,\ldots,i,1,i+1,\ldots,n).

The insertion-residue combinatorics is local on a cut polygon. On any allowed channel, retain only the insertion positions for which that channel is planar. Cutting the polygon maps these positions bijectively to all positions of leg (1) on one lower-point factor. The other factor is fixed. Therefore every residue has the form

Res⁡Dn=aLDm\operatorname*{Res}D_n=a_LD_m

or the same expression with left and right exchanged.

Assume lower even multiplicities vanish. Then (D_n) has no residues. Since every (a_n) has only simple physical poles, (D_n) is a local homogeneous polynomial of Mandelstam degree one. Every term has the scalar-derived Adler zero, so

Dn∣Wi=0D_n|_{W_i}=0

for every leg. The linear contact lemma gives (D_n=0).

The base case is

D4=s13+s23+s12=0.D_4=s_{13}+s_{23}+s_{12}=0.

Thus photon decoupling holds for all even (n).

Fundamental BCJ from scalar residues

For the same distinguished leg define

xi=∑j=2is1jx_i=\sum_{j=2}^{i}s_{1j}

and

Bn=∑i=2n−1xi an(2,…,i,1,i+1,…,n).B_n= \sum_{i=2}^{n-1} x_i\, a_n(2,\ldots,i,1,i+1,\ldots,n).

There are two channel types relative to the reference cut between (n) and (2).

Non-wrapping channel

For a consecutive block (P=[a,b]), collapse it to an internal label (I). The identity

s1I=∑j=abs1js_{1I}=\sum_{j=a}^{b}s_{1j}

turns the residue bracket exactly into the lower-point fundamental combination. Hence

Res⁡P2=0Bn=aLBm\operatorname*{Res}_{P^2=0}B_n=a_LB_m

up to exchanging the two factors.

Wrapping channel

Let

P=[a,…,n]∪[2,…,b].P=[a,\ldots,n]\cup[2,\ldots,b].

The allowed insertion slots are (i=b,\ldots,a-1). Write

zi=∑j=b+1is1j,xi=xb+zi.z_i=\sum_{j=b+1}^{i}s_{1j}, \qquad x_i=x_b+z_i.

The (z_i) terms form the lower fundamental BCJ combination, while the constant (x_b) multiplies the lower photon-decoupling sum. Thus

Res⁡P2=0Bn=aL(Bm+xbDm)\operatorname*{Res}_{P^2=0}B_n = a_L(B_m+x_bD_m)

up to exchanging factors.

Simultaneous induction on (D_m) and (B_m) kills every residue. Therefore (B_n) is a local homogeneous polynomial of Mandelstam degree two. It has an Adler zero for each leg: for legs other than (1), each amplitude term vanishes; when leg (1) is soft, every (x_i) vanishes. The quadratic contact lemma gives (B_n=0) for even (n\geq6).

At four points,

a4(2,1,3,4)=s23,a4(2,3,1,4)=s12,a_4(2,1,3,4)=s_{23}, \qquad a_4(2,3,1,4)=s_{12},

and (s_{12}+s_{13}=-s_{23}), so

B4=s12s23−s23s12=0.B_4=s_{12}s_{23}-s_{23}s_{12}=0.

Relabeling gives every fundamental BCJ relation.

Closure of the full Parke--Taylor ideal

The remaining step is algebraic rather than dynamical. At generic field-theory kinematics, cyclic symmetry together with the relabeled fundamental BCJ relations is a set of primary relations: it generates the general KK relations and all general BCJ relations. In particular, reflection and photon decoupling are included in the generated ideal.

Our scalar family has cyclic symmetry intrinsically: a cyclic step exchanges the two polygon colours and sends (delta\mapsto-\delta), while the extracted even grade is unchanged. The proof above supplies the other primary relation. Hence the scalar covector kills every relation among Parke--Taylor classes, and

an∈(Hn−)∗.a_n\in(H_n^-)^*.

This is exactly the descent criterion of entry 11.

Consequences and limits

Established now at generic genus-zero kinematics:

Not supplied by this proof:

Sources checked

Decision

Close the direct scalar-descent gap at tree level. The next Nima question is no longer whether (mathsf J) exists as a representation-independent normal symbol. It is whether the Jordan strictification acts intrinsically on that half-symbol. The surface/Cut-Equation naturality test remains with Frost, and the resonant six-point pairing block remains with YM.

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