03-ilang-space / 19-curvature
19curvatureverified
Summary The curvature of the derived background, defined from the witness angle, and how it changes with the state of the medium and with a body.
# External verification: 19-curvature
- **Subproject:** 03-ilang-space
- **Package:** 19-curvature
- **Verified version:** v1
- **External round:** 1 of 2
- **Date:** 2026-10-08T23:10:16+02:00
- **Focus points:** none
---
VERDICT: minor issues
## Summary
The exact metric formula, its small-\(\lambda\) expansion, the leading curvature formula, and the first-order body-induced curvature variation are correctly derived on patches where \(\hat g^{(0)}\) is non-degenerate. The three examples give the correct metric and curvature results for positive \(\lambda\). Two subsidiary statements need correction: one-dimensional flatness does not imply a vanishing second fundamental form, and the commuting example incorrectly includes \(\lambda=0\) in its non-degeneracy and curvature claims.
## Issues
### I1. One-dimensional flatness is incorrectly attributed to \(\mathrm{II}=0\)
- **Location:** Step 7, case (iii), following Eq. (19.11); Result, Item 2.
- **Severity:** minor
- **Problem:** The text says that the sufficient condition \(\mathrm{II}\equiv0\) holds when \(k=1\). This is false: an immersed curve can have a nonzero second fundamental form while its intrinsic Riemann curvature vanishes. For example, with \(\chi=\lvert0\rangle\) and \(K(t)=\cos t\,\sigma_x+\sin t\,\sigma_y\), the record map is \(u(t)=e^{it}\lvert1\rangle\). Its induced metric is non-degenerate, but \(u''=-u\) is normal to the curve, so \(\mathrm{II}_{11}\ne0\). The claimed intrinsic flatness is nevertheless correct, as the consistency checks establish by antisymmetry.
- **Suggested fix:** Remove \(k=1\) from the cases asserted to satisfy Eq. (19.11). State separately that every non-degenerate one-dimensional metric has zero intrinsic curvature, directly from Eq. (19.9).
### I2. The commuting example includes the degenerate point \(\lambda=0\)
- **Location:** Step 12, Eq. (19.18) and the following non-degeneracy statement; Result, Item 4(b).
- **Severity:** minor
- **Problem:** Although \(g=\lambda^2(\mathrm dx^2+\mathrm dy^2)\) is valid at every \(\lambda\ge0\), \(g\) is identically zero at \(\lambda=0\). Thus it is not non-degenerate there, and its Gaussian curvature \(\kappa_g\) is undefined. Equation (19.18) incorrectly states \(\kappa_g=0\) “for every \(\lambda\),” followed by an unrestricted claim of non-degeneracy. The smoothly extended rescaled metric \(\hat g\), by contrast, remains Euclidean at \(\lambda=0\).
- **Suggested fix:** Restrict the non-degeneracy of \(g\) and the assertion \(\kappa_g=0\) to \(\lambda>0\). State separately that \(g(x;0)=0\), while the smooth extension of \(\hat g\) has zero curvature also at \(\lambda=0\).
## Focus points
None given.