Rational Ward identities for actual convergent products and iterates

Take an actual product $G=<f,Y1(w1,z1)Y2(w2,z2)w3>$ with |z1|>|z2|>0. Fix v in V. The module-current commutator with an intertwiner is $v_n\mathcal{Y}(w,z)-\mathcal{Y}(w,z)v_n={\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{n}{j}\right)z^{n-j}\mathcal{Y}(v_{j}w,z)\mathrm{.}$ To prove it, multiply its intertwiner Jacobi identity by the current variable to the integer power n, take its residue, and expand (z+x0)^n in nonnegative x0 powers. The x0 residue selects $v_{j}w$; $j\ge 0$ truncates at one bound depending on v,w, so the displayed sum is finite even for $n<0$. A term in the intermediate-weight sum containing the fixed current mode $v_n$ on the intermediate completed vector is thus equal to a term with the transposed current mode on the actual restricted-dual f minus finitely many terms with actual input $v_{j}w1$. The mode shifts intermediate generalized weight by the fixed integer wt(v)-n-1 and auxiliary degree by deg(v). Changing that summation index is bijective on the corresponding actual homogeneous components; absent components are zero. Taking absolute values bounds the entire intermediate-mode series by the finite sum of absolute binary product series for those actual modified exterior/input states. Those series converge by the given convergence hypothesis. This proves absolute legitimacy for every fixed current mode. For an iterate, a commutator on the completed inner value is not asserted to be a finite formula: even a fixed negative mode can involve an unbounded family of composite-state modes. The singular nonnegative-mode family needed for denominator clearing is instead treated by the explicit geometric resummation in part 4 of this proof. Current coefficients applied directly to the original exterior or an actual leaf are actual modified-state iterate pairings and are absolutely convergent by hypothesis. This argument is stronger than asserting cancellation in a merely formal completion: it supplies a finite absolute majorant before the cancellation. It does NOT yet interchange an infinite family of different current modes. The next step uses only finite polynomial operations before that issue arises.

Treat the current position T as a separate formal variable. Let $Q_{\mathrm{\infty }}(T)$ be the Laurent expansion of the outer-current expression $⟨f,Y_4(v,T){\mathcal{Y}}_1(w_1,z_1){\mathcal{Y}}_2(w_2,z_2)w_3⟩\mathrm{.}$ Every coefficient is an actual binary scalar pairing with a modified strong restricted-dual vector, as in part 1 of this proof. By the opposite-field formula on that dual, $Q_{\mathrm{\infty }}$ has a finite HIGHEST T power. This uses the finite state family $e^{xL1}v$ and lower truncation of the actual strong dual field; it does not assume the dual is C1-cofinite. Let $Q_{\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}}(T)$ be the expression with Y3(v,T) acting on w3 instead. It has a finite LOWEST T power by ordinary input truncation. Choose integers N1,N2 so that $v_j{w}_i=0$ for all $j\ge Ni$. The current commutator distribution, multiplied by $(T-z_i{)}^{Ni}$, is zero: its finitely many delta derivatives have orders below Ni, and multiplication by that factor kills them. Hence two successive applications of this weak commutativity give $P(T)Q_{\mathrm{\infty }}(T)=P(T)Q_0(T),P(T)=(T-z_1{)}^{N_1}(T-z_2{)}^{N_2}\mathrm{.}$ This equality is coefficientwise meaningful at the fixed z values. Multiplying by P takes only finitely many T coefficients and finite scalar combinations; part 1 of this proof justifies every intermediate-mode operation used in the two commutations. There is no infinite binomial inversion at this stage. The left side has finite highest T power, and the right side finite lowest T power. Their common value is therefore a Laurent POLYNOMIAL N(T). Consequently $Q(T)=\frac{N(T)}{(T-z_1{)}^{N_1}(T-z_2{)}^{N_2}}$ is a rational function of T, with possible finite poles only at 0,z1,z2. The two originally constructed coefficient series are its expansions in the outside and inside annuli, respectively. Indeed multiplication by P is invertible in either Laurent expansion ring: outside use powers $z_i\mathrm{/}T$, inside use $T\mathrm{/}{z}_i$. Their inverse coefficients are the ordinary integer-binomial series, and their absolute convergence on smaller annuli follows from ${\sum }_k(1+k{)}^M{\rho }^k<\mathrm{\infty }$ for $\rho <1$. N is finite, so no unproved sum over intermediate module weights is hidden in this inversion. For a current between the two intertwiners, the same polynomial locality identifies its product with P with N(T), coefficientwise by part 1 of this proof. It is thus the mixed expansion of the SAME rational Q on |z1|>|T|>|z2|. The inverse factors now use T/z1 and z2/T; both geometric bounds hold on that nonempty annulus. This specifies the formal expansions, not just an unspecified meromorphic continuation.

At the point zi, let $q_i(t)$ be the scalar series with module current $Y_i(v,t)$ acting on the actual input wi, all other insertions fixed. Each coefficient is an actual convergent binary product with a modified wi, so it is defined and absolutely convergent; input truncation gives one finite lowest t power. The binary intertwiner Jacobi identity at that vertex identifies $q_i(t)$ with the expansion of Q(zi+t) in powers of t. Here are the coefficient operations making this usual scalar-duality implication precise. First multiply Jacobi by the finite polynomial P(T) and take its current-variable coefficients. The two current-order expressions, after part 2 of this proof, become the appropriate outside/inside expansions of the one Laurent polynomial N(T). Its delta substitution sends T to zi+t. The resulting identity says $P(z_i+t)q_i(t)=N(z_i+t)\mathrm{.}$ The power $t^{Ni}$ in P(zi+t) is explicit; the other factors have nonzero constant term because the insertion points are distinct. Their formal inverses are ordinary binomial power series. Dividing in the local Laurent ring is legitimate coefficientwise: $q_i$ has finite lowest power and an inverse unit with nonnegative powers, so every coefficient is a finite sum. This yields $q_i(t)={\iota }_{T=zi}Q(T)$, INCLUDING every nonnegative t coefficient, not only the principal part. The delta substitution just used can be checked without an analytic duality theorem: verify it for each Laurent monomial $T^m(T-zi{)}^n$ by the integer binomial expansion in the three prescribed regions; then use linearity, clear the finite powers of T and T-zi, and divide by the local units. All of those steps involve finite Laurent-polynomial coefficients before local unit inversion, whose coefficient finiteness was just verified. The same argument handles the input at zero and the current at infinity. Therefore for any rational test f(T) with poles at the marked points, ${\sum }_{i=1}^3{\mathrm{Res}}_{t=0}f(z_i+t)G(\dots ,Y_i(v,t)w_i,\dots )={\mathrm{Res}}_T{\iota }_{\mathrm{\infty }}f(T)Q(T),z_3=0.$ This is the rational Ward identity (W). To prove the scalar residue rule, partial-fraction-expand the rational function fQ. A pole term $c(T-zi{)}^{-k}$ has the same total finite residue and coefficient of $T^{-1}$ at infinity (both zero for $k>1$, both c for $k=1$); a polynomial contributes zero to both. Add the finitely many terms. Thus the residue equality uses no contour deformation or unproved interchange. Each local residue on the left is FINITE on the actual input: f has a finite principal part and the input field has a finite singular part. The right residue is the finite output-dual rational-current action by strong dual truncation. This recovers precisely the rational action and sign convention in the local formal-action supplier.

For the iterate $G=<f,Y^1(Y^2(w1,a)w2,b)w3>$ use absolute convergence on |b|>|a|>0 and global leaf positions $c=a+b,b,0$. Fix the first input w1 for the $P^{(1)}(b)$ passage. Here the current on the completed inner value needs one extra estimate; it cannot be treated as a uniformly truncated algebraic state. For $k\ge 0$ the actual inner commutator is $v_k{\mathcal{Y}}^2(w_1,a)w_2={\mathcal{Y}}^2(w_1,a)v_k{w}_2+{\sum }_{j=0}^k\left(\genfrac{}{}{0px}{}{k}{j}\right)a^{k-j}{\mathcal{Y}}^2(v_j{w}_1,a)w_2\mathrm{.}$ The first term truncates at one k bound on w2, and j in the second term truncates at one j bound on w1. In the outer current OPE, this is multiplied by (T-b)^(-k-1). For |T-b|>|a|, each surviving j has the exact absolutely convergent sum ${\sum }_{k\ge j}\left(\genfrac{}{}{0px}{}{k}{j}\right)a^{k-j}(T-b{)}^{-k-1}=(T-b-a{)}^{-j-1}\mathrm{.}$ Its absolute majorant is $\mathrm{\mid }T-b{\mathrm{\mid }}^{-j-1}{\sum }_{r\ge 0}\left(\genfrac{}{}{0px}{}{\{}{r}\right)+j\}\{j\}$ |a/(T-b)|^r. The binomial is a fixed-degree polynomial in r, so the series is finite uniformly on smaller overlap regions. Multiplying this geometric-polynomial bound by the ABSOLUTE iterate projection sum for $Y^1(Y^2(v_{j}w1,a)w2,b)w3$ bounds the full joint k/intermediate-weight sum. There are only finitely many j. Hence the current on the composite inner value has precisely the finite principal parts at $T=c$ and $T=b$, rather than a falsely asserted uniform truncation in T-b. Multiply those rational principal parts by (T-c)^N1(T-b)^N2 before collecting current coefficients. Their denominators cancel. Each coefficient is now a finite combination of actual iterates with modified leaf/exterior vectors; its projection sums are absolutely convergent by hypothesis. The full inner Jacobi gives the local delta identities after this same finite cancellation. Thus the same two binary Jacobi identities give polynomial current locality at the THREE leaf positions: first apply it inside $Y^2$ for w1,w2, and then at $Y^1$ for its output and w3. Each fixed mode operation is absolutely legitimate by part 1 of this proof. On the selected nonempty overlap |c|>|b|>|a|, outside and intermediate annuli are available. Equivalently, the finite-clearing step can be checked directly at the outer Jacobi residue. In its local variable $y=T-b$, the cleared inner current is $(y-a{)}^{N_1}{y}^{N_2}{Y}_M(v,y)I^2(w_1,w_2)\mathrm{.}$ The factor (y-a)^N1 allows inner weak locality to move the current through $I^2$. The factor $y^{N2}$ then kills all negative y powers of the resulting current on w2. The whole cleared local term is regular in y. Its residue against the outer commutator delta, expanded in nonnegative y, is zero. This proves $P{Q}_{\mathrm{\infty }}=P{Q}_{\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}}$ without invoking associativity or a uniform truncation bound for $I^2$'s completed value. The geometric estimate above supplies absolute legitimacy for the singular-family resummation. The same denominator clearing proves a rational Q(T) with possible poles at c,b,0, and the same local Jacobi substitution identifies ALL its local coefficients with the currents on the corresponding leaf. The resulting Ward equality is therefore the actual iterate Ward identity, not an assumed consequence of a product-to-iterate factorization. Subtracting the finite first-input residue leaves the pair at b,0, exactly the reverse partial identity in the formal supplier §7. For the deformation c→uc with b fixed, take $a_s=uc-b$ and restrict to the common geometric neighborhood; the same proof applies for each s. All field-truncation bounds involve fixed input states or their finite nilpotent families, so one common finite set of denominators and state cutoffs works there.