The ledger of the OpenAI manuscript, version 1.1, October 1, 2026

Appendix B: Analytic profiles near the axis and their continuation (pp. 144 to 157)

MB.1: Symmetry-breaking axis velocity and the two-region split of the parameter interval

Node ns-mb-1-symmetry-breaking-axis-velocity-and-the-two-region-split, kind move, pp. 144.

• Statement: With Π0 from Lemma A.5 (real analytic near [−1, 1], even, Π0 ≤ −cP∗²f², ηΠ0η > 0 for η ≠ 0, f = (1 + η²)^{−1}) and h ≤ 10^{−2} fixed, choose 0 < j0 ≤ .05 and set U∗ = 4η + j0, H∗ = Dη + dU∗, W∗ = 1 − dU∗η − 2DηU∗, Z∗ = −A(1 − 2ηU∗)U∗ − H∗U∗η − dΠ0η + 4AηΠ0 (B.1). Then H∗ has exactly one zero η0 ∈ (−1, 0), |η0| ≍ j0; −W∗ > 2.8; Z∗(η0) ≥ cj0P∗² > 0. So on a compact I ⊃ [−1, 1] choose δ∗ > 0 with {|Z∗| ≤ δ∗} avoiding a neighborhood of η0, then σ∗ > 0 with χ = H∗²/(H∗² + σ∗²) > .99 wherever |Z∗| ≤ δ∗ (B.2): every η has χ > .99 or |Z∗| > δ∗.
• Obligation: The inner-edge inequality (B.19), which is vs > 2 + cex at X0 = 4/Λ (the viscous part of the admissible stress cone at the edge where the stress is born parallel to the shear; Proposition 4.10(ii), Theorem 4.6(iii)), must hold for every η. H∗ is the axis value of the coefficient Hc of the η-derivative in the transport terms, so the rotational mechanism dies at its zero; Z∗ is the axis value of the axial source Sn (so ns ≈ Z∗/L), so the axial-shear mechanism dies at its zeros. With j0 = 0 both vanish at η = 0 (U∗(0) = 0, H∗(0) = 0, and Π0η(0) = 0 by evenness), leaving the midplane uncovered; this is the upward-biased, mildly asymmetric axial profile motivated physically in Section 2.1. The constant −W∗ > 2.8 also supplies the order-one part of the angular source in (B.17), and the nonzero axis datum U∗ gives the axial velocity lower bound ‖uz(0)‖ ≍ τ^{−1/2−h} on the core (page 8).
• Mechanism: H∗/d = Dη/d + 4η + j0 increases strictly from −∞ to +∞ on (−1, 1) and equals j0 > 0 at η = 0, so its zero η0 is unique, negative, and about −j0/(D + 4). At η0 the H∗U∗η term of Z∗ drops out, 4Aη0Π0(η0) is positive (η0 < 0 and Π0 < 0) and at least cj0P∗², −dΠ0η(η0) ≥ 0 by the sign of ηΠ0η, and the remaining terms are O(j0), which the already large P∗ absorbs. So the zero set of Z∗ is separated from η0; δ∗ quantifies the separation, and σ∗ is then chosen so small that away from a neighborhood of η0 the regularized indicator χ of H∗ ≠ 0 exceeds .99. Every η then lies either in {χ > .99}, where the rotational branch of (B.19) will work, or in {|Z∗| > δ∗}, where the axial-shear branch will work. The formula for −W∗ is direct algebra from W∗ = 1 − 4d − 2Dη(4η + j0) with D = 1/2 − h.
• Antecedent: None cited. The physical motivation is Section 2.1 (pages 3 to 5).
• Cost: Parameters j0, δ∗, σ∗, the enlarged interval I and the complex domain Ω; the order j0 → δ∗, σ∗ → Λ in (B.40); dependence on the sign and size properties of the outer datum Π0 and on P∗ being already large. The offset also enters the final axial matching error (‖Gi − 4η‖ contains a Ckj0 term in the proof of Proposition 4.10), so j0 ≪ εm in the hierarchy of section 4.6.
• Backward question: At which values of η does each available amplification mechanism (axial transport of angular momentum versus radial shear of axial velocity) degenerate, and can a single small symmetry-breaking parameter keep those degeneracy sets disjoint?
• Checkable: With the actual Π0 of (A.21) (or, as a smoke test, the reference inner contribution −(5/2)P∗²f², which has the same parity and sign properties), tabulate H∗, W∗, Z∗ on I for fixed h and several j0 ≤ .05; root-find η0 and compare with −j0/(D + 4); confirm min(−W∗) > 2.8 (digest check: at h = .01, j0 = .05 the minimum over [−1, 1] is 2.871 and η0 = −0.01114); confirm Z∗(η0) > 0; then compute the largest admissible δ∗ and the largest σ∗ giving χ > .99 on {|Z∗| ≤ δ∗}.
• Depends on: MA.8 (the sign of Z*(η0) comes from Lemma A.5's datum: analytic, even, Π0 ≤ -(5/2)P*²f², and ηΠ0' > 0.); M4.4 (H*, W*, Z* are the axis values of the transport coefficient H_c, the factor W, and the axial source S_n of (4.8), (4.9).)
• Refs: p. 144, (B.1), (B.2); Lemma A.5, (A.21) to (A.22), p. 133; Proposition 4.10 proof, p. 37; Section 2.1, pp. 3 to 5; p. 8.
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.2: Steep, holomorphic swirl datum on the axis

Node ns-mb-2-steep-holomorphic-swirl-datum-on-the-axis, kind move, pp. 144-145.

• Statement: For Λ ≥ 1 define ζ∗ = −LH∗/(H∗² + σ∗²), ξ0 = Λζ∗, ϕ∗ = exp(Λ ∫0^η ζ∗(w) dw) (B.3). The integral is well defined on Ω, ϕ∗ > 0 on the real interval, and ϕ(0, η) = ϕ∗. Consequently ∂η log ϕ∗ = ξ0 and −H∗ξ0 = ΛLχ exactly (the manuscript writes H∗ξ0/(ΛL) = −χ). The amplitude C is chosen after Λ.
• Obligation: Makes the angular source Sq = −Wl − h(1 − 2ηU) − Hc(log E)η of (4.9) large and positive where H∗ is not small. This gives p1 = a > 0 on the stress-free region, the bound (B.17), and the rotational branch of (B.19), while keeping ϕ > 0 and analytic in η, as Theorem 4.6(i) and Lemma 5.1 require. It is the axis value ϕ(0, η) specified in the proof of Proposition 4.10, and its Λ fixes the radial scale Y = ΛX and hence Xa = 4/Λ.
• Mechanism: The term −Hc(log E)η in Sq is axial transport of angular momentum between η-layers. Prescribing (log ϕ)η ≈ −ΛL/H∗ would make it ≈ ΛL, a large positive source. The regularization H∗/(H∗² + σ∗²) in place of 1/H∗ keeps ϕ∗ holomorphic across the zero of H∗, at the price that the large source becomes ΛLχ and switches off near η0, which is exactly where the axial branch of MB.1 takes over. Physically, the swirl amplitude decreases in the direction of the axial characteristic speed H∗, so axial flow carries fluid from more rapidly rotating layers (Section 2.1). Because the leading source is proportional to Λ, balancing it against radial diffusion forces the radial scale X ~ 1/Λ, which is why the axis problem is posed in Y = ΛX.
• Antecedent: None cited.
• Cost: The large parameter Λ (Λ ≥ Λ0, chosen after σ∗). The manuscript notes that ϕ∗ can grow exponentially with Λ, which forces the amplitude condition C ≥ sup over Ω of |ϕ∗| in (B.16) and hence the Λ-dependent threshold C0(Λ). The stress-free region shrinks to X ≤ 4.1/Λ, X-derivative bounds carry factors Λ^{r−1}, and the O(Λ) slope of log ϕ∗ inflates ℓi = log(CE(Xi, ·)), which is why Tsh in (B.33) depends on Λ through Bk.
• Backward question: Can the sign of the angular-momentum source be forced by prescribing the η-dependence of the swirl on the axis alone, and how can the natural prescription (log ϕ)η ∝ −1/H∗, singular where H∗ vanishes, be made holomorphic without losing positivity?
• Checkable: Compute ζ∗ and ϕ∗ by quadrature on I for given j0, σ∗, Λ; verify the identity −H∗ξ0 = ΛLχ to rounding error; evaluate |ϕ∗| on a complex strip around I to size the threshold C0(Λ) ≥ sup|ϕ∗| and observe its growth in Λ.
• Depends on: MB.1 (uses H* and the regularization σ* chosen there, so χ = H*²/(H*² + σ*²) > .99 where |Z*| ≤ δ*, on the complex domain Ω.); M4.4 (the datum targets the term -H_c(log E)_η of the angular source S_q in (4.9), making it ≈ ΛLχ on the axis.)
• Refs: pp. 144 to 145, (B.3); p. 147, (B.16); p. 37 (proof of Proposition 4.10); (4.9), p. 26.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MB.3: Weighted analytic coefficient space Bρ and the radial inverses (Lemma B.1)

Node ns-mb-3-weighted-analytic-coefficient-space-b-and-the-radial, kind move, pp. 145-146.

• Statement: In Y = ΛX, for F = Σα Fα(η)Y^α let aαβ = 20^{−α}ρ^{−β}β! C(α + β, β)/((α + 1)²(β + 1)²) and ‖F‖ρ = sup over α, β ≥ 0, η ∈ I of |∂η^β Fα(η)|/aαβ (B.4); Bρ is complete. Lemma B.1: multiplication is bounded on Bρ; for ν = 1, 2 the inverse Jν of YGYY + νGY = F regular at 0 with G(0) = 0, (JνF)α+1 = Fα/((α + 1)(α + ν)) (B.5), is bounded, as are AX(F) = Y^{−1}IF, multiplication by Y, IF = ∫0^Y F dY' and ∂ηI; and ‖Jν[(∂ηF)(DXG)]‖ρ ≤ Cρ‖F‖ρ‖G‖ρ (B.6), also with either derivative omitted, with AX(F) for F, or with extra undifferentiated factors.
• Obligation: The stress-free equations (4.13) have a regular singular point at X = 0 and contain first-order η-derivatives (Hc∂η in the transport, ∂ηAX(U) inside W, η-derivatives of the pressure). An iteration in functions of Y alone loses one η-derivative per step. Without a space in which "increasing the radial degree compensates for a parameter derivative" (Section B.2, p. 145), the contraction for Proposition B.2 does not close, and neither analyticity in η nor regularity at the axis (a power series in X = r²/(2q)) would come out.
• Mechanism: The factor ρ^{−β}β! measures analyticity in η with radius about ρ, the factor 20^{−α} measures analyticity in Y with radius 20, and the binomial C(α + β, β) ties them: by (B.10), moving one unit from radial degree to η-derivative count costs (80/ρ)(i + 1), and the divisor i + 1 produced by one radial integration (or the divisor (α + 1)(α + ν) of Jν) pays for it. The quadratic denominators make Bρ a Banach algebra: in the Leibniz formula the derivative binomials cancel the factorials of the weights, (B.8) (a count of β-element subsets of a set split into two blocks) bounds the remaining binomial ratio by one, and (B.7), proved by splitting the sum at N/2, bounds the convolution of the (α + 1)^{−2} weights. In a mixed product (∂ηF)(DXG), the extra radial degree supplied by Jν is assigned to the factor carrying the η-derivative, while DX = Y∂Y multiplies a coefficient by its degree, which the (α + ν) divisor absorbs.
• Antecedent: Named ingredients only: the Leibniz formula, an elementary convolution estimate (B.7), and completeness via uniform convergence of coefficients and derivatives. No classical theorem is cited. (Digest's identification, not the manuscript's: a majorant-series norm of Cauchy-Kovalevskaya type adapted to a regular singular, Fuchsian-type, radial operator.)
• Cost: A fixed small η-radius ρ, later taken strictly below the distance from the working neighborhood to the boundary of Ω so that Cauchy's inequality absorbs the (β + 1)² factor; the fixed Y-radius 20, which must exceed 4.1; algebra constants Csq² and Cρ.
• Backward question: In which Banach space does inverting the regular singular radial operator gain exactly the one η-derivative that the transport terms cost, so that a fixed-point iteration closes without shrinking the domain?
• Checkable: Exact-arithmetic checks: (B.5) by substitution into YG'' + νG' = F; (B.8) by brute force over small indices; (B.9) and (B.10) as rational identities in the weights (digest check: both identities and bounds hold for all α, β < 40, and the supremum in (B.9) is 80, attained at α = β = 0); (B.7) numerically (digest check: the left side stays below 3.52 for N < 200, against the bound 8Σ i^{−2} = 4π²/3 ≈ 13.16).
• Depends on: M4.4 (J_1, J_2 invert the radial viscous operators of the zero-stress equations (4.13), regular at Y = 0, for U and for ϕ.); M4.3 (it bounds the radial average A_X of (4.6), which enters (4.13) through V0 and W, together with ∂_η I.)
• Refs: pp. 145 to 146, (B.4) to (B.10), Lemma B.1.
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.4: Explicit leading swirl profile f0 and its positivity window

Node ns-mb-4-explicit-leading-swirl-profile-f0-and-its-positivity, kind move, pp. 146.

• Statement: f0(z) = Σα≥0 (−z/2)^α/(α!(α + 1)!). For 0 ≤ z ≤ 4.1 and t = z/2: f0(z) ≥ 1 − t/2 + t²/12 − t³/144 ≥ 305719/1152000 > .265 and f0(z) ≤ 1 (B.11); the cubic is decreasing on [0, 2.05]. It is the unperturbed angular profile: Φ0 = (1 + T)^{−1}1 = f0(Yχ), the solution regular at Y = 0 with value 1 of 2(YΦYY + 2ΦY) = −χΦ. The companion quantity f0 + zf0' enters (B.19) through the bound f0 + zf0' ≤ 1 − t + t²/4 − t³/36 + t⁴/576.
• Obligation: Division by ϕ in the first equation of (B.15), and the use of log Φ in l and (log E)η, are legitimate only if Φ > 0 on the whole stress-free interval; (B.11) with (B.13) gives Φ ≥ c0 > 0 on 0 ≤ Y ≤ 4.1 for large Λ. f0 also fixes the rotational shear at the edge, a = −2YΦY/Φ ≈ −2zf0'(z)/f0(z) at Y = 4.
• Mechanism: When the large source ΛLχ dominates, the angular equation in Y becomes a linear regular singular ODE whose coefficient χ(η) does not depend on Y; its regular solution is a power series with alternating coefficients of decreasing magnitude on the relevant range, so partial sums bound it from both sides. The cubic partial sum equals 305719/1152000 at t = 2.05, and since 0 ≤ χ ≤ 1 the argument z = Yχ stays in [0, 4.1]. The edge Y = 4 lies beyond the point where −2zf0'/f0 first reaches 2 (digest computation: z ≈ 2.8916, the first zero of f0 + zf0') and well before f0 vanishes (digest computation: first zero at z ≈ 7.341).
• Antecedent: None cited; alternating-series bounds are used directly. (Digest's identification, not the manuscript's: f0(z) = J1(2√t)/√t and f0 + zf0' = J0(2√t) with t = z/2, Bessel functions of the first kind.)
• Cost: Caps the stress-free analytic region at Y ≤ 4.1 and places the stress activation at Y = 4 (X0 = 4/Λ, which is Xa of Proposition 4.10); forces t̄ with 4e^{2t̄} < 4.1 and tc with 4e^{tc} < 4.1 so that later cutoffs stay inside the analytic region.
• Backward question: When axial transport of angular momentum dominates the source, what linear equation does the swirl obey near the axis, and on what radial interval is its explicit solution still positive while its logarithmic slope already exceeds the viscous threshold 2?
• Checkable: Evaluate f0 by its series (or as J1(√(2z))/√(z/2)) on [0, 4.1] and confirm min > .265 (digest check: minimum .27111 at z = 4.1, maximum 1 at z = 0); confirm in exact arithmetic that the cubic partial sum at t = 41/20 equals 305719/1152000; confirm symbolically that f0 solves 2(zf'' + 2f') + f = 0 with f(0) = 1.
• Depends on: MB.2 (the large source ΛLχ from the steep axis datum reduces the angular equation to 2(YΦ_YY + 2Φ_Y) = -χΦ, solved by f0(Yχ).); M4.4 (that equation is the leading part of the angular zero-stress equation (4.13) in the variable Y = ΛX.); MB.3 (Φ0 = (1 + T)^{-1}1 is written with T = J_2χ/2, the radial inverse of Lemma B.1.)
• Refs: p. 146, (B.11); p. 148 (Φ0 = f0(Yχ)); pp. 149 to 150 (use at the endpoint).
• Verification: statement astra-spot-check-2026-10-01; hypotheses astra-spot-check-2026-10-01; computation checked False; Astra spot-check: correct

MB.5: Proposition B.2, the analytic stress-free axis profile by contraction

Node ns-mb-5-proposition-b-2-the-analytic-stress-free-axis-profile-by, kind move, pp. 146-148.

• Statement: With the axis data of Section B.1 there are Λ0 and, for each Λ ≥ Λ0, a threshold C0(Λ) such that every C ≥ C0(Λ) admits an analytic profile with vanishing leading residual stress on 0 ≤ Y = ΛX ≤ 4.1 of the form ϕ = ϕ∗Φ, U = U∗ + Λ^{−1}u, Π = Π0 + Λ^{−1}I(g²Φ²), g = ϕ∗/C (B.12), with Φ(0, η) = 1, u(0, η) = 0, analytic in Y and η on a common neighborhood of [0, 4.1] × I, and |∂Y^r ∂η^s (Φ − f0(Yχ), u + YZ∗/(2L))| ≤ Cr,s/Λ (B.13), uniformly in large Λ and C ≥ C0(Λ) (for X-derivatives the bound is Cr,sΛ^{r−1}).
• Obligation: This is the inner region of Theorem 4.6: (4.13) holds and T0 = 0 on 0 ≤ X ≤ Xa (Theorem 4.6(ii), Proposition 4.10(i)). F = ϕ/C, U, Π and V0/X are analytic in X (power series in r²/(2q)), hence smooth at the axis, and with E = √(2X)ϕ/C and (4.4) to (4.5) the Cartesian field is smooth across r = 0 (Definition 3.2, Theorem 4.6(i)). The profiles are analytic in η on one complex neighborhood (Theorem 4.6(i), used by Lemma 5.1), and ϕ > 0 gives E > 0 for X > 0.
• Mechanism: In Y the stress-free system becomes 2(YΦYY + 2ΦY) = −χΦ + Λ^{−1}R1 and 2(YuYY + uY) = −Z∗/L + Λ^{−1}R2, where R1, R2 are explicit polynomials in Φ, u, their DX and η derivatives, W = W∗ + Λ^{−1}B, Hc = H∗ + Λ^{−1}du, and p = I(g²Φ²). The only products with both an unintegrated η-derivative and a DX derivative are (∂ηAX(u))DXΦ and (∂ηAX(u))DXu, controlled by (B.6); ∂ηp is bounded because ∂ηI is, and DXp = Yg²Φ². So (Φ, u) ↦ JνRi is bounded and locally Lipschitz on balls of Bρ². The angular term −χΦ is of order one, not small, but T = J2χ/2 raises the minimal radial degree by one and divides by (b + 1)(b + 2), so ‖T^k‖ ≤ (40Mχ)^k/(k!(k + 1)!) and Σ(−T)^k inverts 1 + T whatever the size of χ. The map (Φ, u) ↦ (Φ0 + (1 + T)^{−1}J2R1/(2Λ), u0 + J1R2/(2Λ)), with (Φ0, u0) = (f0(Yχ), −YZ∗/(2L)), is then a strict contraction on a fixed ball for large Λ, with fixed point within C/Λ of the center. The pressure couples to ϕ∗, which can be exponentially large in Λ; (B.16) gives |g| ≤ 1 on the complex neighborhood, so Cauchy's inequality bounds all η-derivatives of g uniformly in both large parameters (the manuscript stresses that this bound must hold on a complex neighborhood). Finally Σα C(α + β, β)(R/20)^α = (1 − R/20)^{−β−1} turns the coefficient norm into analyticity on |Y| ≤ R for any 4.1 < R < 20 with η-radius below ρ(1 − R/20); Cauchy estimates give (B.13); real data and uniqueness of the fixed point give a real solution; (B.11) gives Φ > 0.
• Antecedent: As cited in the text: a strict contraction and its unique fixed point, Cauchy's inequality, and Cauchy estimates, in the norm of Lemma B.1. The inversion of 1 + T is a Neumann series, not named as such. (Digest's identification, not the manuscript's: the resolvent series has Volterra-type factorial decay.)
• Cost: Λ ≥ Λ0; C ≥ C0(Λ) ≥ sup over Ω of |ϕ∗|, so C is chosen after Λ and may be exponentially large in Λ; ρ strictly below the distance to the boundary of Ω; the solution exists only for Y ≤ 4.1; the smaller common complex neighborhood fixed here is the one Corollary B.6 later preserves.
• Backward question: Can the same large parameter that steepens the swirl in η also rescale the radius so that the nonlinear stress-free system becomes a 1/Λ perturbation of an explicitly solvable linear problem, and can the pressure's coupling to an exponentially large swirl datum be neutralized by the still-free amplitude C?
• Checkable: Run the fixed-point iteration numerically on truncated Taylor coefficients in Y (coefficient rule (B.5) for J1, J2), with η discretized spectrally (Chebyshev), for several moderately large Λ and C ≥ max over a complex strip of |ϕ∗|. Check that sup|Φ − f0(Yχ)| and sup|u + YZ∗/(2L)| on [0, 4.1] × [−1, 1] scale like 1/Λ as in (B.13), that the Taylor coefficients decay at least like 20^{−α}(α + 1)^{−2} as membership in Bρ requires, and that Φ > 0.
• Depends on: MB.3 (the fixed point is found in Lemma B.1's space B_ρ, whose radial inverses and product bound (B.6) absorb the η-derivatives of (4.13).); MB.4 (the contraction is centered at Φ0 = f0(Yχ), with 1 + T inverted by its factorially decaying series.); MB.2 (the axis data are ϕ(0, η) = ϕ*, and C ≥ sup|ϕ*| on Ω (B.16) tames the pressure coupling through g = ϕ*/C.); M4.4 (the system solved is the zero-stress equations (4.13), so T0 = 0 on the stress-free region.)
• Refs: pp. 146 to 148, Proposition B.2, (B.12) to (B.16); Theorem 4.6(i) to (ii), p. 33; Proposition 4.10(i), p. 36; (4.4) to (4.5), p. 25; Lemma 5.1, p. 47.
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.6: Proposition B.3, positive angular source and the two-branch endpoint inequality

Node ns-mb-6-proposition-b-3-positive-angular-source-and-the-two, kind move, pp. 148-150.

• Statement: After increasing Λ and then C0(Λ): Φ ≥ c0 > 0 on [0, 4.1] × [−1, 1]; Sq ≥ 2.5 + .95ΛLχ (B.17); with p1 = ps,1 = XQs/L, p2 = ps,2 = XNs/(LE), ns = Ns/L, one has p1/X ≥ c1 > 0, 0 < p1 ≤ C1, ns = −2UX = Z∗/L + O(Λ^{−1}) on 0 < X ≤ 4.1/Λ (B.18); at X0 = 4/Λ, p1 + p2²/p1 > 2 + cex (B.19), for example with cex = .2; Φ, log Φ, u, p1, ns have η-derivative bounds of every fixed order uniform in C ≥ C0(Λ).
• Obligation: In the stress-free region ps = s = (a, −bs), so vs = a + bs²/a = p1 + p2²/p1, and (B.19) is exactly vs > 2 + cex at the inner annulus edge. Since the stress is born parallel to the shear there (Theorem 4.6(iii)), this is the viscous inequality of the admissible stress cone at the edge, stated in Proposition 4.10(ii) as a(Xa) > 0 and vs(Xa) > 2 + cex. The source bound gives p1 = a > 0 (so ts and vs are defined) and the lower bounds that Lemma B.4 propagates.
• Mechanism: For the source, (B.20) follows from Φ ≈ f0(Yχ) and |H∗χ'| ≤ 2‖H∗'‖∞χ; then −Hc(log ϕ)η ≈ ΛLχ with errors absorbed by Young's inequality (allocating .05ΛLχ), and with l = 1 + DX log Φ, −W∗ > 2.8 and |h(1 − 2ηU)| ≤ 10h ≤ .1 one gets (B.17) without dividing by χ near the zero of H∗. Qs > 0 follows from the positive representation Qs = ∫0^X X'Φ(ΛX')Sq dX'/(X²Φ(ΛX)). Integrating the stress-free equations from the axis gives p1 = a = −2YΦY/Φ and ns = −2uY. At Y = 4 there are two branches. If χ > .99, then t = 2χ ∈ (1.98, 2] and f0 + zf0' ≤ 1 − t + t²/4 − t³/36 + t⁴/576 < −.18 (the quartic equals −75535511/400000000 < −.188 at t = 1.98 and decreases on [1.98, 2]), so a = −2zf0'/f0 = 2 − 2(f0 + zf0')/f0 > 2.36 and p1 > 2.3 for large Λ: the rotational shear alone clears the threshold. If χ ≤ .99, then |Z∗| > δ∗ by (B.2), so |ns| ≥ δ∗/2, and at the endpoint |p2| = C(2/Λ)^{1/2}|ns|/(ϕ∗Φ) grows linearly in C for fixed Λ while p1 ≤ C1; enlarging C0(Λ) gives p2²/p1 > 2.3: a small swirl amplitude 1/C lets the axial shear dominate. This is the analytic form of Section 2.1's statement that the radial shear of axial velocity supplies the amplification near the middle plane while the rotational mechanism suffices farther away.
• Antecedent: Young's inequality and alternating-series remainder bounds, used directly; nothing classical cited.
• Cost: A further increase of Λ and of C0(Λ) (C must beat a Λ-dependent bound on ϕ∗Φ at the endpoint); the constants c0, c1, C1; the margin cex = .2, which reappears as vs > 2 + cex in Proposition 4.10(ii).
• Backward question: At the radius where the stress will be switched on, can vs > 2 be guaranteed for every η, and which free parameter enlarges the rotational part a, and which the axial part bs²/a, of vs?
• Checkable: Exact rational check that the quartic partial sum at t = 99/50 equals −75535511/400000000; evaluate a(z) = −2zf0'(z)/f0(z) on [3.96, 4] (digest check: 3.326 to 3.389, well above the proved 2.36); symbolic check that p2 = XNs/(LE) with E = √(2X)ϕ∗Φ/C equals C(X/2)^{1/2}ns/(ϕ∗Φ), which is C(2/Λ)^{1/2}ns/(ϕ∗Φ) at X = 4/Λ; with the numerical profile of MB.5, evaluate p1 + p2²/p1 at Y = 4 over η ∈ [−1, 1] for increasing C.
• Depends on: MB.5 (reads all bounds off Proposition B.2's stress-free profile and its closeness (B.13) to f0(Yχ) and -YZ*/(2L).); MB.1 (the two branches at Y = 4 are the split χ > .99 (rotational) or |Z*| > δ* (axial shear), with -W* > 2.8 in the source.); MB.4 (on χ > .99 the bound on f0 + zf0' gives a = -2zf0'/f0 > 2.36 at Y = 4, and (B.11) gives Φ ≥ c0 > 0.); M4.7 (with p_s = s in the stress-free region, p1 + p2²/p1 is v_s of (4.20), so (B.19) is v_s > 2 + c_ex at the edge.)
• Refs: pp. 148 to 150, Proposition B.3, (B.17) to (B.21); Proposition 4.10(ii), p. 37; Theorem 4.6(iii), p. 33; Section 2.1, p. 5.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MB.7: Lemma B.4, reference continuation with frozen logarithmic slopes

Node ns-mb-7-lemma-b-4-reference-continuation-with-frozen-logarithmic, kind move, pp. 150-151.

• Statement: Lemma B.4. Let X0 = 4/Λ, Xi = 110, y = log(X/X0), 0 < t1 ≤ t̄ with 4e^{2t̄} < 4.1. The reference (ϕr, Ur) is the analytic profile for y ≤ t1, has its slopes ∂y log ϕr, ∂yUr cut to zero by the step σ of (A.5) on t1 < y < 2t1 (B.22), and is constant in log X after; pressure, Qs, Ns are integrated from the axis with datum Π0. For Λ, then C, large and then t1 small, on [X0, Xi] × [−1, 1]: Sq,r ≥ .94LΛχ + 2.4, lr ≤ 1, vr = p1,r + p2,r²/p1,r > 2 + c (B.23), p1,r > 0, p1,r > 3 on [100, 110]; η-norms of CEr, 1/(CEr), p1,r, ns,r are bounded uniformly in C, t1; ‖Πr − Π0‖ ≤ CkC^{−2} (B.24).
• Obligation: The analytic solution exists only for Y ≤ 4.1, but the profile must reach X ≈ XRe^{−5}, a radius growing like C^{10}. The shear reduction of Proposition B.5 needs a comparison profile on the whole interval whose integrated coefficients ps,r are uniformly bounded and signed, with vr > 2 + c, so that the stress created by lowering the shear lies in the admissible cone.
• Mechanism: Cutting the logarithmic slopes of ϕ and U to zero and then holding the fields constant in log X means nothing grows, however long the interval: Ur stays within O(Λ^{−1}) + o(1) of U∗ (radial averaging is a contraction in sup norm, so AX(Ur) does too, independently of the interval length), log ϕr stays a bounded distance from log ϕ∗, and the pressure moves only by O(C^{−2}) because E ∝ 1/C. The source keeps its large gradient part LΛχ (replacing H∗ by Hc,r costs Cσ∗√χ + o(1), absorbed) and its constant part from −W∗ > 2.8. Because the analytic profile has p1 = a > 0, its ϕ slope is nonpositive, so freezing it gives lr ≤ 1. The equations (B.25) are linear with regular initial values; with lr ≤ 1 and either Sq,r ≥ .94LΛχ or Sq,r ≥ 2.4, integrating factors give the exponential and the linear lower bounds for p1,r, keeping p1,r above 2.3 on χ > .99 and above 3 from X = 100. On χ ≤ .99 the axial source keeps |ns,r| bounded below and p2,r = Xns,r/Er grows with C, so vr > 2 + c everywhere.
• Antecedent: None cited; the smooth step (A.5) is the manuscript's own.
• Cost: The width t1 (0 < t1 ≤ t̄) and t̄ with 4e^{2t̄} < 4.1; the fixed local radii X0, 100, 110; the order Λ, then C, then t1; the constant c in vr > 2 + c.
• Backward question: How can a profile known only on a short analytic interval be carried out to arbitrarily large radius with every quantity in the cone test bounded uniformly in the length of the interval and in the amplitude?
• Checkable: From the numerical axis profile of MB.5, integrate (B.22) and then (B.25) in y up to log(110/X0); check Sq,r ≥ .94LΛχ + 2.4, lr ≤ 1, vr > 2 + c, and p1,r > 3 on [100, 110]; check the two comparison inequalities pointwise.
• Depends on: MB.6 (propagates Proposition B.3's source bound, p1 > 0, and the endpoint inequality v > 2 + c from X0 out to X_i = 110.); MB.5 (the reference equals the analytic profile for y ≤ t1 and then freezes its logarithmic slopes.); M4.4 (Q_s, N_s are integrated from the axis by (4.9), giving the linear equations (B.25) for p1,r and n_s,r.)
• Refs: pp. 150 to 151, Lemma B.4, (B.22) to (B.25); (A.5), p. 129.
• Verification: statement completeness audit 2026-10-01: FRAGMENT; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.8: Proposition B.5, first part: flat activation of the stress by shear reduction

Node ns-mb-8-proposition-b-5-first-part-flat-activation-of-the-stress, kind move, pp. 151-153.

• Statement: Proposition B.5, first part. Fix C large and t∗ ≤ t̄ with Lemma B.4 uniform for 0 < t1 ≤ t∗; y = log(X/X0). For 0 < κ0 < 1/2 set on 0 < y < t1: ea = (1 − κ0)σ(y/t1), κ = 1 − ea, a = κp1,r, DXU = −κXns,r/2, ∂y log ϕ = −κp1,r/2 (B.26). Then ps − ps,r, Er/E − 1 are O(yea) (B.28); vs = κvr + O(yea), Pc = vr + O(yea), Jc = O(yea) (B.29); so Pc − vs ≥ cea, Pc > 2 + c, and the admissible cone holds where vs > 2, the strict relaxed cone elsewhere. Also T0 = eaB0, B0 smooth, B0(0, η) = F(X0)ps,r(X0) ≠ 0 (B.30), and |T0| ≥ ce^{−t1²/y²}, |∂^I T0| ≤ CIe^{−t1²/y²}y^{−NI} (B.31).
• Obligation: Theorem 4.6 at the inner edge Xa = X0: T0 = 0 up to Xa and nonzero just after (part (ii)); T0 flat at Xa, so its extension by zero is smooth; the unit direction n extends smoothly to Xa and is parallel to (a, −bs) there, with the directional margin (4.26) on the inner collar (part (iii)); the weighted bounds (4.27) with ζ comparable to e^{−t1²/ya²} (part (iv)); and the factorization (4.33) of Proposition 4.10(ii). Proposition C.3 derives these conclusions directly from (B.30) and (B.31).
• Mechanism: In the stress-free region the integrated inviscid vector equals the shear, ps = s, so T0 = F(ps − s) = 0. The stress is switched on by lowering the shear, not by changing the sources: the actual shear is prescribed as κ(p1,r, p2,rEr/E), a fraction κ = 1 − ea of the reference's integrated vector (which equals the reference shear while the reference is still stress-free). The vector ps depends only on profile values and cumulative radial integrals (the identities (4.16)), and the profile values differ from the reference by integrals of ea, which are O(yea); hence ps stays at ps,r up to O(yea) while s drops by the factor κ, and T0 = F·ea·ps,r + O(yea). The stress is therefore born along the old shear direction, where Jc = 0, the most interior direction for the quadratic cone test: the ratio of (vs − 2)+Jc² to (Pc − vs)² is at most Cy². The common factor κ cancels from ts = −bs/a, so no inverse power of κ0 enters the error constants. The flat step makes ea vanish to infinite order at X0, and Lemma A.9 (the substitution u = δ/(1 + δ²v)^{1/2}) shows that ea-weighted primitives are ea times y³ times smooth functions; all field differences lie in eay³C∞ (moments and pressure in eay⁶C∞), a class closed under products, smooth compositions and reciprocals of nonvanishing fields, so division by ea is smooth.
• Antecedent: Lemma A.9 (pp. 141 to 142), the quadratic cone test of Lemma 4.5, and the moment formulas (4.16) of Lemma 4.3; nothing classical cited.
• Cost: κ0 ∈ (0, 1/2) and the activation width t1, chosen in the order C, then κ0 (small relative to Vmax and the family constants), then t1; the same t1 becomes the inner exponent of the global weight ζ = exp(−t1²/ya² − 4/yb²) in (C.18); radial derivatives of the narrow cutoffs can be large (only η-derivatives are controlled uniformly).
• Backward question: How can the leading stress be switched on from zero smoothly, flatly, and already strictly inside the admissible cone, without disturbing the profile values and cumulative integrals on which the outer fields depend?
• Checkable: With numerical reference data at X0 (from MB.5 and MB.7), integrate (B.26) and (B.27) on a fine y-grid in (0, t1) using σ from (A.5); compute ps via (4.16), s via (4.11), and ts, vs, Pc, Jc via (4.20); verify T0/ea → F(X0)ps,r(X0), Pc − vs ≥ cea, and |Jc|/(yea) bounded. Verify Lemma A.9's factorization numerically: y^{−3}(1/ea(y))∫0^y ea(u)b(u)du → b(0)/(2t1²) as y → 0, a consequence of B(0, η) = b(0, η)/c in (A.47) (digest check with t1 = .3, κ0 = .25, b = 1 + 2u: 5.657 at y = .01 and 5.609 at y = .005, approaching 5.556 at rate O(y)); and σ(y/t1)e^{t1²/y²} → e as y → 0, so ga(0) = (1 − κ0)e.
• Depends on: MB.7 (the shear is prescribed as the fraction κ = 1 - e_a of Lemma B.4's reference, whose p_s,r is bounded with v_r > 2 + c.); M4.5 (p_s depends only on values and cumulative integrals (4.16), so O(ye_a) value changes keep p_s = p_s,r + O(ye_a) while s drops by κ.); MA.13 (Lemma A.9 with c = t1², j = 0 makes e_a-weighted primitives e_a y³ times smooth factors, giving (B.30) and (B.31).); M4.7 (the stress is born along the shear, with J_c = O(ye_a) and P_c - v_s ≥ ce_a, inside Lemma 4.5's quadratic cone test.)
• Refs: pp. 151 to 153, Proposition B.5, (B.26) to (B.31); Lemma A.9, (A.47), pp. 141 to 142; Proposition 4.10(ii), (4.33), p. 37; Theorem 4.6(ii) to (iv), p. 33; Proposition C.3, (C.18) to (C.19), pp. 163 to 164.
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.9: Proposition B.5, second part: small-shear continuation to Xi with the barrier p1 > 2

Node ns-mb-9-proposition-b-5-second-part-small-shear-continuation-to, kind move, pp. 153.

• Statement: Proposition B.5, second part. After the activation keep (B.26) with κ = κ0; for κ0 small (with κ0 sup p1,r < .8), then t1 small, Pc > 2 + c/2 and vs < 1, so the strict relaxed cone holds. At X = 100 cut the axial prescription for DXU by a smooth β from 1 to 0 (Pc > 2, vs < 1), then interpolate a convexly to .8 with DXU = 0. On the final interval a = .8, l = .6, Sq > 1, so at a downward crossing p1 = 2 one would get DXp1 = XSq/L − .6p1 > X − 1.2 > 0, a contradiction. The continuation reaches Xi = 110 with E > 0, a = .8, DXU = 0, p1 > 2, and is unchanged for X ≤ X0.
• Obligation: Proposition 4.10(ii) requires a > 0, Pc > 2 and vs < U(Pc, Jc) on all of (Xa, Xh]. The state at Xi (a = .8, so l = .6, the angular-momentum slope of the reference power law E ∝ X^{1/10}, and U constant in X) is the endpoint of the first continuation from Proposition B.5 as listed in the proof of Proposition 4.10, and it is the starting state from which (B.34) can reach the outer reference profile.
• Mechanism: With the shear held small (vs < 1), the relaxed cone (4.21) collapses to the single scalar inequality Pc > 2, because U(Pc, Jc) > 2 whenever Pc > 2. Pc stays large because it is essentially the reference's vr and the integrated angular coefficient p1 keeps growing under a positive source. The axial shear is then turned off (bs = 0, so ts = 0 and Pc = p1), and a is steered to .8 = 2 − 2(.6), where .6 = 1/2 + 1/10 is the slope l of the reference power law. The barrier uses the scalar equation DXp1 = XSq/L − lp1, which follows from (4.9) and p1 = XQs/L.
• Antecedent: A barrier argument for a scalar ODE (the manuscript calls the same device a barrier on p. 155); nothing classical cited.
• Cost: The condition κ0 sup p1,r < .8; the two final transitions, whose total logarithmic width ωfin enters the matching error through (B.32); the fixed radii 100 and 110.
• Backward question: Once the stress exists, what is the weakest condition that can be maintained over a long radial interval, and to what normalized state must the profile be steered so that it can be glued to the outer power law?
• Checkable: Continue the numerical integration of MB.8 through the κ = κ0 stage, the β cutoff at X = 100, and the interpolation of a to .8, up to X = 110, recomputing ps and the cone coordinates from (4.16), (4.11), (4.20); confirm Pc > 2 and vs < 1 throughout, and p1 > 2, a = .8, DXU = 0 at X = 110. Arithmetic: .6 × 2.8 = 1.68, and X − 1.2 > 0 on [100, 110].
• Depends on: MB.8 (continues the prescription (B.26) with κ = κ0, whose estimates (B.29) give P_c > 2 + c/2 and v_s < 1.); MB.7 (Lemma B.4's reference gives v_r ≤ V_max and p1,r > 3 on 100 ≤ X ≤ 110 for the final transitions.); M4.7 (with v_s < 1 and P_c > 2 the relaxed cone (4.21) holds whatever J_c, since U(P_c, J_c) > 2 when P_c > 2.); M4.4 (the barrier p1 > 2 uses the scalar equation D_X p1 = XS_q/L - lp1 obtained from (4.9).)
• Refs: p. 153 (proof of Proposition B.5); Proposition 4.10(ii) and its proof, p. 37.
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.10: Corollary B.6, a quarantined analytic collar

Node ns-mb-10-corollary-b-6-a-quarantined-analytic-collar, kind move, pp. 153.

• Statement: There is tc > 0 with tc < t1 and 4e^{tc} < 4.1 such that E/√(2X), U, V0/X, Π are smooth in (X, η) through X = 0 on [0, X0e^{tc}] × [−1, 1] and analytic in η on one complex neighborhood, every fixed radial derivative being analytic on the same neighborhood and bounded on a smaller one; on X0 < X ≤ X0e^{tc} the admissible stress cone and (B.30) hold; all later profile modifications are supported strictly to the right of X0e^{tc}.
• Obligation: Theorem 4.6(i) requires a fixed Xan ∈ (Xa, Xb) with analyticity in η on [0, Xan] × [−1, 1], and an inner collar [Xa, Xan] carrying the directional margin; Proposition C.3 obtains this from Corollary B.6, and Appendix C chooses Xan inside this rectangle and modifies the profile only beyond it. Lemma 5.1 needs this analytic rectangle to solve the positive-order systems, which contain ∂η terms, on a fixed radial interval.
• Mechanism: On the first collar every ingredient (the reference's Qs, Ns and the prescriptions (B.26)) is built from coefficients analytic in η, their η-derivatives, and forward radial integrals, multiplied by cutoffs that depend on y alone, and cutoffs in y do not affect η-analyticity. Denominators stay nonzero on a small complex neighborhood (Φ has no zeros there), and ϕ is an exponential, hence nonvanishing. Because pressure and the cumulative moments are integrated forward from the axis, no edit supported at larger X can change anything on the rectangle.
• Antecedent: None cited.
• Cost: The width tc, strictly inside the first collar; neither tc nor the radial derivative bounds are uniform as C grows and the widths shrink (they are fixed once the finite choices are made).
• Backward question: Which later edits could destroy analyticity in η near the axis, and can a rectangle be quarantined that no later edit or forward integral can reach?
• Checkable: Structural bookkeeping only: in an implementation, assert that every later edit's support begins at X > X0e^{tc} and that recomputed forward integrals and fields on [0, X0e^{tc}] are unchanged. Otherwise none: pure argument.
• Depends on: MB.5 (on [0, X0] the profile is Proposition B.2's solution, analytic in Y and η on one common neighborhood.); MB.8 (on the first collar the activation (B.26) uses analytic coefficients times cutoffs in y alone and yields the admissible cone and (B.30).); M4.5 (pressure and the cumulative integrals are integrated forward from the axis, so edits farther out cannot reach the rectangle.)
• Refs: p. 153, Corollary B.6; Theorem 4.6(i), p. 33; Appendix C opening, p. 158; Proposition C.3, p. 164; Lemma 5.1, p. 47.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MB.11: Lemma B.7 and the amplitude-independent shape transition (B.33) to (B.34)

Node ns-mb-11-lemma-b-7-and-the-amplitude-independent-shape-transition, kind move, pp. 154-155.

• Statement: Lemma B.7. Fix the data, Λ and an order k. With ℓi = log(CE(Xi, η)), Gi = U(Xi, η), ωfin the total width of the final transitions, ‖ℓi‖ ≤ Bk and ‖Gi − U∗‖ ≤ Ck^nat/Λ + Ck^join(Λ)(t1 + κ0 + ωfin) in C^k in η (B.32), constants independent of large C. Take Tsh ≥ 20‖σ'‖∞(B0 + ‖log f‖∞) (B.33), B0 = Bk at k = 0; with yi = log(X/Xi) set log E = −log C + yi/10 + (1 − σ(yi/Tsh))ℓi + σ(yi/Tsh) log f, U = Gi (B.34). Then l ∈ [.55, .65], a ∈ [.7, .9], bs = 0, Sq > 1, and the barrier DXp1 > 0 at p1 = 2 keeps p1 > 2: the strict relaxed cone holds through the transition and after, up to restoration.
• Obligation: The outer reference profile (A.7) has η-shape f = (1 + η²)^{−1}, while the axis profile's η-shape at Xi is exp(ℓi), which carries the steep factor ϕ∗. The shape must be changed over a logarithmic length Tsh that does not depend on the amplitude C chosen later; otherwise enlarging C to shrink the inner moment discrepancies (MB.12, MB.13) would lengthen the transition and undo the gain. Proposition 4.10 lists Tsh among the parameters fixed before C.
• Mechanism: ℓi = log(√(2Xi)ϕ(Xi)) involves only the logarithmic slopes accumulated from the axis, and these are bounded independently of C by Lemma B.4; only integrals of bounded cutoff values enter, so no inverse powers of the transition widths appear. A slow interpolation in log X between ℓi and log f, superposed on the fixed power 1/10, perturbs l = DX log H by at most ‖σ'‖∞(B0 + ‖log f‖∞)/Tsh ≤ .05, so the shear stays in a ∈ [.7, .9] with bs = 0. Positivity of Sq is checked term by term: the old gradient gives −Hcℓi' ≥ LΛχ minus absorbable errors (via (B.20)); the new gradient gives −Hc(log f)' = 2ηHc/(1 + η²) ≥ −Cj0² − C/Λ − o(1), after writing ηH∗ = (D + 4d)η² + dj0η and completing the square; the two are combined convexly; and −W ≈ −W∗ > 2.8 with l ≥ .55.
• Antecedent: None cited.
• Cost: Tsh (large, fixed before C, and dependent on Λ through Bk); the bounds Bk; the error term Ck^join(Λ)(t1 + κ0 + ωfin), which forces the order j0, Λ, then C, then κ0, t1, ωfin; the radius Xsep = Xie^{Tsh}.
• Backward question: Can the length of the shape-changing transition be bounded before the amplitude is chosen, so that the amplitude remains a free knob afterward?
• Checkable: Compute ‖σ'‖∞ from (A.5) (digest check: ‖σ'‖∞ = 8, attained at y = 1/2) and use ‖log f‖∞ = log 2 on [−1, 1], so (B.33) reads Tsh ≥ 160(B0 + log 2); given a numerical ℓi from MB.9, evaluate l along (B.34) and confirm l ∈ [.55, .65]; evaluate Sq along the transition from (4.9).
• Depends on: MB.9 (the transition starts from the state at X_i = 110 (a = .8, D_XU = 0, p1 > 2) reached by Proposition B.5.); MB.7 (ℓ_i involves only logarithmic slopes from the axis, bounded independently of C by Lemma B.4, which gives B_k in (B.32).); MA.4 (the target is the η-shape f = (1 + η²)^{-1} of the outer reference branch (A.7), interpolated with the step σ of (A.5).); M4.4 (along (B.34), l stays in [.55, .65], and the barrier D_X p1 = XS_q/L - lp1 with S_q > 1 keeps p1 > 2.)
• Refs: pp. 154 to 155, Lemma B.7, (B.32) to (B.34); Proposition 4.10 and its proof, pp. 36 to 38.
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

MB.12: Amplitude-radius exchange, XR = Xi(CP∗)^{10}

Node ns-mb-12-amplitude-radius-exchange-xr-xi-cp-10, kind move, pp. 155-156.

• Statement: Set XR = Xi(CP∗)^{10} (Proposition 4.10 writes XR = 110(CP∗)^{10}), x = X/XR, and Xsep = Xie^{Tsh}. Beyond Xsep, (B.34) is exactly the ideal angular profile, because C^{−1}f(X/Xi)^{1/10} = P∗f x^{1/10}. In normalized coordinates the end of the transition sits at xsep = Xsep/XR = e^{Tsh}/(CP∗)^{10} (B.38), which tends to 0 as C → ∞ with Tsh fixed.
• Obligation: The inner profile's contribution to the five cumulative integrals, measured in units of the outer scale, must fall below a tolerance fixed by the outer data (B.36), so that a fixed-size bump correction can cancel it. The matching radius must also exceed R∗ (Lemma 4.8) and be large enough that Pc = XRxG/L > 2 on the correction patch (B.37). Proposition 4.10(iii) draws the consequence: XR can be pushed past any later lower bound while Λ and Tsh stay as already chosen.
• Mechanism: The reference power law E ∝ X^{1/10} is scale covariant. The inner construction lives at swirl amplitude 1/C (E = √(2X)ϕ/C), and after the transition the profile follows C^{−1}f(X/Xi)^{1/10}, which reaches the outer amplitude P∗f x^{1/10} only after the radius is rescaled by (CP∗)^{10}. So the amplitude C, the last free parameter of the inner construction, is converted into radial separation: the inner structure, fixed in X up to Xsep before C is chosen, occupies [0, xsep] with xsep ∝ C^{−10} in outer units, and its moment contributions vanish as powers of xsep and of 1/C, as in (B.39).
• Antecedent: None cited.
• Cost: XR grows like C^{10}; C must be chosen after Tsh and Λ; section 4.6 imposes C ≥ max{C0, (R∗/110)^{1/10}/P∗} so that XR ≥ R∗.
• Backward question: Is there an exact scaling under which the inner region, fixed in X, looks arbitrarily small from the viewpoint of the outer profile, without redoing the inner construction?
• Checkable: Symbolic check of C^{−1}(X/Xi)^{1/10} = P∗(X/XR)^{1/10} for XR = Xi(CP∗)^{10}; tabulate xsep = e^{Tsh}/(CP∗)^{10} against C for given Tsh and P∗ and find the least C with xsep < e^{−8}; check that the reference pressure increment at xsep is (5/2)P∗²f²xsep^{1/5} = (5/2)f²e^{Tsh/5}C^{−2}.
• Depends on: MB.11 (beyond X_sep = X_i e^{T_sh} the transition (B.34) has reached C^{-1}f(X/X_i)^{1/10}, with T_sh fixed before C.); MA.4 (the target is the outer reference branch E = P*f x^{1/10} of (A.7), reached exactly when X_R = X_i(CP*)^{10}.)
• Refs: pp. 155 to 156, (B.38); Proposition 4.10, pp. 36 to 37; section 4.6 Step 1, pp. 39 to 40.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MB.13: Proposition B.8, exact matching of the five cumulative radial integrals

Node ns-mb-13-proposition-b-8-exact-matching-of-the-five-cumulative, kind move, pp. 155-157.

• Statement: For C sufficiently large and then the activation transitions sufficiently small, the profile of Proposition B.5 continued by (B.34) continues to log x = −5 with the strict relaxed cone preserved, and there its fields, pressure Π, and all five integrals M, I, J, S, Cp agree exactly with those of the reference inner profile (A.7); it then follows the outer radial profile without changing Π0 or the later Qs, Ns.
• Obligation: Lemma 4.4(i): if two profiles agree beyond a radius and their five integrals agree there, with the same axis pressure datum, then pressure, V0, Qs, Ns, ps, a, bs and T0 agree at all larger radii. This lets the regular axis profile replace the temporary inner branch (A.7) of the outer profile, which is not regular at the axis, without changing the exterior: the pressure normalization (4.25), the moment identities (4.28), the heat exterior (4.29), and hence Lemma 4.9 and Lemma A.8 (T0 = 0 for X ≥ Xb), whose hypotheses require a regular axis and exact moments. It is Proposition 4.10(iii) and the joining in Step 1 of the proof of Theorem 4.6.
• Mechanism: The stress at a radius depends on the profile inside that radius only through five cumulative integrals (Lemma 4.3), so a gluing must match those five functions of η as well as the fields. Normalizing them by powers of XR removes XR from (B.35), so the Lipschitz constants of the map from moments to (Qs, Ns) and the inverse bounds of the correction depend only on outer data: one fixed tolerance serves every large XR, and enlarging XR only helps, since Pc ∝ XR. The discrepancy entering the correction is at most Ck‖Gi − 4η‖ in C^{k+1} plus a term that is o(1) as C grows, because the inner region is tiny in x (MB.12) and Gi is close to 4η once j0, then Λ, then the transition widths are small ((B.32) and the proof of Proposition 4.10). After the row operations the linearization is block triangular: the U-bumps move M and J − 4ηI, and the E-bumps move I, S − 8ηM and Cp; each block is a matrix of distinct powers integrated against bumps on ordered disjoint intervals, invertible by the Rolle argument of Lemma A.1; the remainder is exactly quadratic, so the contraction of Lemma A.2 gives smooth coefficients with the prescribed finite set of η-derivative bounds.
• Antecedent: Lemma A.1 (Rolle's theorem: a nonzero combination of m distinct powers has at most m − 1 positive zeros, plus multilinearity of the determinant), Lemma A.2 (a contraction plus the pointwise implicit function theorem), Corollary A.3 (the five-moment blocks with α = 1/10), and Lemmas 4.3 and 4.4.
• Cost: The restoration patch (−8, −7) and the correction patch (−6, −5) in log x; five bump amplitudes; the tolerance εm fixed by outer data; one extra η-derivative on the inputs, because (B.35) contains η-derivatives of the moments; smallness only for a prescribed finite set of η-derivative orders.
• Backward question: Which finite set of integral invariants carries everything the outer fields need from the inner profile, and in what units does correcting them have bounds independent of the huge matching radius?
• Checkable: Assemble the 5 × 5 Jacobian of the normalized moment map at the ideal profile on (−6, −5) in log x, using bumps that are rescaled copies of σ' and the weights (1, f x^{3/5}) and (x^{1/2}, f x^{1/10}, f x^{−9/10}); compute its determinant and condition number for η ∈ [−1, 1]; run c ↦ B^{−1}(d − Q(c, c)) on synthetic discrepancies and confirm convergence when 8β0²κ0d0 ≤ 1 (Lemma A.2). Arithmetic: .9 + .01/.7 = .914 < 1, and 1 − (.1/(1 + w))w/.7 ≥ 6/7 for all w ≥ 0. Recompute the reference integrals (4.34) to confirm the target values.
• Depends on: MB.12 (X_R = X_i(CP*)^{10} puts the inner structure at x ≤ x_sep → 0, so its moment contributions (B.39) fall below the tolerance.); MA.3 (Corollary A.3 at α = 1/10 makes the two U bumps and three E bumps on -6 < log x < -5 an invertible five-moment map.); M4.6 (once fields and all five integrals agree at log x = -5 with the same Π0, Lemma 4.4(i) makes every outer field agree beyond it.); MA.2 (the exactly quadratic five-moment system is solved by Lemma A.2's contraction, with bounds on finitely many η-derivatives.)
• Refs: pp. 155 to 157, Proposition B.8, (B.35) to (B.39); Lemma A.1, Lemma A.2, Corollary A.3, pp. 126 to 128; (A.7), p. 129; (A.25), p. 134; Lemmas 4.3 and 4.4, pp. 28 to 29; (4.34) and Proposition 4.10(iii), p. 37.
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MB.14: Remark B.9 and Corollary B.10, the order of choices and the completed connection

Node ns-mb-14-remark-b-9-and-corollary-b-10-the-order-of-choices-and, kind move, pp. 157.

• Statement: Remark B.9: for any finite list of η-derivative orders, choose in the order Md, Td, P∗, λ, h → moment tolerance, j0 → δ∗, σ∗, Λ → (Bk), Tsh → C, XR → κ0, t1, final widths (B.40), the radial frequency of Appendix C last. Corollary B.10: the analytic axis profile of Proposition B.2 has a smooth continuation, E > 0 for X > 0, matching the reference outer radial profile at log(X/XR) = −5 with exact fields, pressure and radial moments: stress-free through X0, then an inner collar, analytic in η, with the admissible cone and flat factor (B.30), then the strict relaxed cone to the matching point.
• Obligation: Guarantees the construction is not circular: each smallness condition refers only to earlier choices. In particular C0(Λ) may be exponentially large in Λ, the moment tolerance depends only on outer data, and Tsh is fixed before C. Corollary B.10 is the deliverable consumed by Proposition 4.10 (hence by Step 1 of the proof of Theorem 4.6) and by Appendix C as its fixed input profile.
• Mechanism: Every bound is arranged to be uniform in everything chosen later: the tolerance is uniform in XR by the normalization (B.35) to (B.37); the transition length is uniform in C by (B.32) to (B.33); the axis solution's η-bounds are uniform in C ≥ C0(Λ) by (B.13); the activation comparisons hold uniformly as κ0 and t1 decrease; the final widths enter only through small errors whose constants are fixed by earlier choices. Section 4.6 restates the same order as the hierarchy 0 < C^{−1} ≪ Tsh^{−1} ≪ Λ^{−1} ≪ σ∗ ≪ δ∗ ≪ j0 ≪ εm ≪ h ≪ λ ≪ P∗^{−1} ≪ Md^{−1} ≪ 1 and 0 < ωfin ≪ t1 ≪ κ0 ≪ C^{−1}, followed by N^{−1} ≪ ωfin.
• Antecedent: The smallness convention of Definition 3.3 (in a chain of constants the rightmost is fixed first); nothing classical cited.
• Cost: Constants may depend on all earlier choices; radial derivative bounds of cutoffs contain inverse powers of the final widths; only a prescribed finite set of η-derivative orders is made small (every other fixed order is merely finite).
• Backward question: Is there a single linear order of all parameter choices in which every smallness requirement refers only to quantities already fixed?
• Checkable: Encode the dependency graph of the constants (each condition's list of previously fixed quantities, read from (B.40), the proof of Proposition 4.10, and section 4.6) and check that it is acyclic and consistent with both stated orderings; the estimates themselves are pure estimates with unspecified constants.
• Depends on: MB.13 (Corollary B.10's matching at log(X/X_R) = -5 with exact fields, pressure, and moment functions is Proposition B.8.); MB.10 (the inner collar, analytic in η, admissible, and carrying (B.30), is Corollary B.6.); MB.9 (the strict relaxed cone from the collar out to X_i is the small-shear continuation of Proposition B.5.); MA.4 (the order (B.40) begins with the outer parameters M_d, T_d, P*, λ, h of (A.6), fixed before any axis parameter.)
• Refs: p. 157, Remark B.9, (B.40), Corollary B.10; section 4.6, pp. 39 to 40; Definition 3.3, p. 18.
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