证明
Lemma 32 ([Table flags and binary prescriptions]). The following statements…Lemma 27 ([Label support]). Every w∈W of rank at most R_* has a representation…Lemma 28. Put L_0^ lab =R_*+14 . At an endpoint, the union of the nonzero label…观点
Equal keys imply U_iv_ iz =U_zv_ zi , and the role sums in supply the witness parity. It remains to construct the gradients. We distinguish abstract Gram entries from the random event that the actual entries agree with them. A baseline of bounded primal rank. At each component and sign choose a splitting of the table space as J=D⊕ J' , append the key space, and append an additional primal complement. The independence in Lemma 32 ([Table flags and binary prescriptions]) makes these splittings possible. Retain the small table on the product of the two table spaces and the actual frozen entries in any pin or key row or column. These prescriptions agree on their overlaps by admissibility. Set the free blocks from an additional primal complement to the opposite J' to zero, in both orientations, and set the additional-primal by additional-primal block to zero. This defines baseline cross bilinear forms. Fix an opposite endpoint z . Let F^0_ z,e :_ O,e ⟶^h, G^0_ z,e :_ O,e ⟶^h be evaluation against its Y_z,X_z channels. On additional primal inputs, evaluation of an opposite channel factors through its projection from J onto D ; hence the rank there is at most D . The remaining primal inputs are in bounded-dimensional projected pin and key spaces. It follows that the ranks of both baseline maps on all primal inputs are bounded independently of r_0,h,n . Arbitrary channel-channel entries of the small table do not affect this bound. A uniform bound for either map on the combined primal inputs of the two endpoints is B_ lin =3K+28: the opposite pin projection costs at most K dimensions, the two individual projected primal spaces cost at most 2K , and the two endpoints have at most twenty-eight key rays. Let u_z^0U_i denote the resulting contractions. They give the table values on C_i . On an attack atom toward z they are zero, since its tensor is the equal-key tensor at z and u_zU_z=0 . On an attack toward z' they equal p_ z' of its matched atom, since that tensor is the corresponding tensor at z' . These are precisely the values in. Allowed changes. Let A_ O,e ^ ± be the span of all key directions on the indicated side of the unit, and put F_ O,e ^ ± =D_ O,e ^ ± + A_ O,e ^ ± . Write bars for the quotients _ O,e =_ O,e /( F_ O,e ^ + +_iH_ i,1 ^ + ), _ O,e =_ O,e /( F_ O,e ^ - +_iH_ i,1 ^ - ). We may change F_z^0,G_z^0 by any maps δF_ z,e :_ O,e ⟶(H_ z,0 ^-)^⊥, δG_ z,e :_ O,e ⟶(H_ z,0 ^+)^⊥. Here each map is extended to a bilinear change supported on its opposite channel block and zero on the opposite primal blocks. It vanishes on all frozen entries: the domain quotient kills frozen directions on one side, while on the other side a pin's channel projection lies in H_ z,0 and is annihilated by the allowed values. A reciprocal-role change in the same Gram block contributes nothing on this map's primal inputs, because it is supported on a channel argument there. On private-by-private blocks both changes are zero. They may add further channel-channel values, which are irrelevant to the primal contractions. Thus all choices below may be made independently for opposite endpoints and for the reciprocal unit roles as far as primal-input values are concerned. The unrestricted linear response. Fix z and solve simultaneously for the two i∈O . Consider the linear space of responses formed by F_z^0·δG_z+δF_z·G_z^0 and all bilinear forms on _ O,e ×_ O,e , with their contractions summed over components on each individual cut profile. We claim that the pair of target forms (r(v_ iz )-u_z^0U_i)_ i∈O belongs to this response space. Let (x_i)_ i∈O ∈^O annihilate all responses. Annihilating the arbitrary pure bilinear forms says that at each component the sum of the two nominal tensors has zero image in the product of barred quotients. Project further onto the individual primal spaces modulo S_ i,e ^ ± and the individual key spans. This projection is well defined on the barred spaces, since every pin projects into S_ i,e ^ ± . It shows that x_ i,e vanishes after both such quotients. A matrix killed after quotienting its two factors by spaces of dimensions d_+,d_- has rank at most d_++d_- : in bases extending those spaces all entries outside their rows and columns vanish. Consequently x_ i,e ≤2K+28≤R_*. Apply Lemma 27 ([Label support]) to every component. For an attack label at i , compare its label summand with all key directions in that component. Their combined label set has at most R_*+14 elements, whose direction spaces p_s⊗^ 1+d_ b form a direct sum. Any vector of S_ i,e ^ ± in this sum has zero attack-label component, by Lemma 28. Thus the attack summand projects to zero modulo just its individual key ray if the atom uses this component, and is zero if there is no such ray. In the former case, writing its affine base vector as v=(1,y) , a symmetric matrix killed in the two quotients by v has the form Z=cvv+vd+dv. Indeed choose v as the first basis vector: the surviving matrix can be supported only on its first row and column. The moment identities in the original affine coordinates now give Z_ ii =cv_i, Z_ 0i =cv_i+d_i+d_0v_i, so d_i=d_0v_i and Z=cvv . The cut relation on any triangle of tags separates label by label: the union of the three supports has at most 3R_* labels, within the interpolation bound. At this attack label the nonzero components lie on its tag star; the triangle relation forces their scalar coefficients to be equal. Subtract that scalar multiple of the attack atom from x_i . Do this for every attack label. Each subtracted atom annihilates the response space, because both its factor directions are keys. The remaining profiles still annihilate the response and have only nonattack label support. Removing these separated label blocks does not increase their component ranks or label support sizes. At one component contract a remaining x_ i,e on the minus side by a primal functional φ at i annihilating S_ i,e ^- and all individual key directions. Extend φ by zero on other endpoint and channel blocks; it is a functional on _ O,e . Pure annihilation shows that the resulting individual plus vector d is a sum of a pin, key directions, and private channel directions. In this representation all key coefficients vanish. To see this, project onto each individual primal block. At the other endpoint the projected pin would be a sparse sum of attack rays. At endpoint i it is the difference of the nonattack vector d , supported on at most R_* labels, and a sum of at most fourteen attack rays. In both cases the label exclusion prevents any attack coefficient. The channel components of the pin lie in the protected spaces H_0 , whereas the remaining channel term lies in their chosen complements H_1 . They must both vanish. Therefore d∈D_ O,e ^+∩_i. Test the derivative response using δG_ z,e =φ⊗c for arbitrary c∈(H_ z,0 ^+)^⊥ . It gives F_ z,e ^0(d)·c=0 . Hence the table evaluation of d is zero modulo H_ z,0 ^+ , and injection in yields d=0 . The reciprocal contraction proves the same assertion in the minus mode. It follows that the remaining component tensors lie in the products of the individual projected pin plus key spaces. Their nonattack support removes the key directions by the same sparse-label argument. Thus the remaining profiles belong to C_i . We have proved that every annihilator is a sum of effective-space profiles and attack atoms. The target vanishes on the former by and on the latter by and the baseline values. In a finite-dimensional vector space the image of a linear map is the annihilator of the kernel of its dual: this follows by extending a basis of the image to a basis of the target space. It proves the claimed linear solvability. A low-rank representation of the target. For one i , the form r(v_ iz ) has a matrix representation on the components of rank at most 15r_0+14J|| per component. Here is an explicit routing for its tester part. Let w=(w_t) be the representation given by its atom list. The list condition gives χ_*(w)=∑q=1 . On an allowed O_ d,t block at any tag the coefficient in a+T^0(v_ iz ,·) is 1+a(v_ iz )+b_t(v_ iz ) =1+χ_*(w)+η_S(w_t)=η_S(w_t). There are at most seven tags t with nonzero η_S(w_t) . Route this tester from each allowed tag l∈I_d to d through the component l,d . It contributes nothing at d , where O_ d,t is disallowed. At a component this uses at most fourteen ordinary testers, allowing either endpoint as center. For the shared tester, route a star centered at t with coefficient a_t(v_ iz ) . Its contribution at the center is (g-1)a_t(v_ iz )=0 , and at any other tag it gives the desired term in a_t(v_ iz )b_t . Consolidating gives at most one shared tester per component. Each ordered tester matrix has rank r_0 , proving the 15r_0 bound. For T^1(v_ iz ,·) , each of the two orientations of an indexed term supplies a matrix of rank at most the component rank of v_ iz . This profile is a sum of at most seven atoms, so its total component rank is at most 7(g-1) . The total contribution per component has rank at most 14J(g-1)≤14J|| . This proves . Removing the bounded-dimensional obstructions. Let M_ i,e be the preceding representing matrices. On each individual primal factor choose a projection A_ i,e ^ ± whose kernel is the sum of its projected pin and key spaces. The matrix (A_ i,e ^+) M_ i,e A_ i,e ^- , extended by zero outside that individual block, is a permitted pure quotient form. Its sum over i has rank at most twice . Moreover M_ i,e -(A_ i,e ^+) M_ i,e A_ i,e ^- , = (I-A_ i,e ^+) M_ i,e +(A_ i,e ^+) M_ i,e (I-A_ i,e ^-),; (M_ i,e -(A_ i,e ^+) M_ i,e A_ i,e ^-) , ≤(I-A_ i,e ^+)+(I-A_ i,e ^-) ≤2K+28. Subtract these pure forms from the linear task. Its residual, including the baseline contraction, has a per-individual matrix representation of bounded rank independent of r_0,h,n , and remains linearly soluble. In any unrestricted solution, replace the derivative output maps by bounded-rank maps in the same allowed value spaces, preserving all their dot products with the opposite baseline values on primal inputs. Such replacement is possible as follows. The map from the allowed value space to the dual of the span of those baseline values has bounded rank. Choose a linear section of its image and compose with that map. This has bounded-dimensional image and preserves every observed pairing. Postcomposing the original derivative map with this value projection leaves its primal derivative response unchanged and preserves its domain quotient. The two new derivative maps have bounded rank. Their extra product δF_z·δG_z is also a bounded-rank pure form, which we include with opposite sign in the remaining pure task. That task still has an unrestricted pure solution. Its per-individual target matrices have rank at most R_ res =2K+28+4B_ lin =14K+140. Here the first term bounds the projection difference, one B_ lin bounds the baseline contraction, two bound the derivative terms, and one bounds their extra product. Compression of the remaining pure task. We show that the latter task has a pure solution of bounded rank. At endpoint i , expand all projected pin and key vectors, and all row and column forms in its bounded-rank target representation, along selector coordinates and base blocks. Their counts depend on earlier construction parameters but not on r_0,h,n . On each ordinary base block let U be the span of the vectors to fix and K_0 the common kernel of the forms to preserve. Choose a complement in K_0 to U∩K_0 , and project along that complement onto a subspace containing U . The resulting endomorphism fixes U , preserves all the forms, and has rank at most U+ K_0 . There are at most 2||(K+14) vectors to fix and at most 2||R_ res forms to preserve at one endpoint before selector expansion. Thus an explicit bound for the rank on an ordinary block is D_ blk =2p_*||(K+14+R_ res ). Use these endomorphisms blockwise, fix the constant base coordinate, and act trivially on the selector factor. Use the same resulting map at every component and in both modes at this endpoint. It sends each allowed point to an allowed point and therefore preserves all W_l and the cut domain. It fixes the individual projected pin and key vectors. Taking identity on channels consequently fixes every mixed pin itself and preserves private channel subspaces. The map descends to the barred spaces and has bounded rank there: only the bounded protected channel dimensions survive their quotients. On either barred space a uniform bound is D_ quo =2p_*(1+(g^2+3)D_ blk )+2K. The first term counts the two compressed primal blocks and the second bounds their surviving channel projections. Precompose an unrestricted pure solution on its two arguments with these maps. Its rank per component becomes bounded. On individual cut profiles its value is unchanged, because the compressed profile remains in that cut domain and every form in the target representation is preserved. It is therefore the required bounded-rank pure solution. Factoring the pure forms through channels. Restore the earlier projected pure forms. Their sum with the compressed correction has rank at most 30r_0+28J||+D_ quo per component for the fixed opposite endpoint. In particular the additive constant is independent of r_0,h,n . In the two allowed channel value spaces impose, in addition, orthogonality to the opposite baseline and bounded derivative values on primal inputs. These are only boundedly many linear restrictions. Each remaining value space has codimension at most K+2B_ lin , so their dot pairing has rank at least h-2K-4B_ lin : restricting a nondegenerate pairing by codimensions c_1,c_2 loses at most c_1+c_2 in rank. With h=1000r_0 , the sufficient requirement is 970r_0>28J||+D_ quo +2K+4B_ lin . Every quantity on its right is fixed before r_0 , so this is one of the permitted requirements on that choice. Factor a rank- r desired pure bilinear form as ∑_ a=1 ^rℓ_a⊗m_a . Choose vectors f_a,g_a in these remaining channel spaces with f_a·g_b equal to the Kronecker delta; they exist by the pairing rank. The changes ∑_aℓ_af_a and ∑_am_ag_a realize that form as their dot product. Their cross terms with the baseline and the previous small derivative changes vanish on primal inputs by the additional orthogonality conditions. They obey and hence preserve every frozen entry. This realizes the target exactly for both endpoints i∈O . Repeat for both opposite endpoints and both unit roles. The independence of primal-input changes already proved ensures that the resulting two cross Gram blocks realize all equations simultaneously. Only primal-channel entries have been used. In each of the two blocks there are at most 8h such entries per component, proving the asserted count. Actual agreement with them gives all gradient equations; equal keys and the role sums give the other witness conditions. All four cross pairs are therefore holes.