Version 1 · earlier version; the current one is v2 · External review, round 1: minor issues

# Curvature of the derived background

- **Subproject:** 03-ilang-space
- **Package:** 19-curvature
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

**Model (question.md).** Probe $a$ with places $h$ labelled by $x_h\in U\subseteq\mathbb R^k$ ($U$ open). Medium $c$ with $d_c:=\dim\mathcal H_c$. Optional body $b$ at $\beta_0$ with $B:=B_{\beta_0}$ ($B=0$ without a body). All types are pairwise different, so (1.11) imposes nothing. The contract is $C=\sum_h\lvert h\rangle\langle h\rvert_a\otimes K(x_h)\ \bigl(+\sum_\beta\lvert\beta\rangle\langle\beta\rvert_b\otimes B_\beta\bigr)$ with $x\mapsto K(x)$ smooth and self-adjoint. The start is $\lvert\phi\rangle_a(\otimes\lvert\beta_0\rangle_b)\otimes\lvert\chi\rangle_c$ with $\phi_h\neq0$.

**Quoted inputs.** (3.5): $\cos\alpha(x,x')=\sqrt{W(x,x')}$. (5.6) for $B=0$ and (6.6) in general: $W(h,h';\lambda)=\lvert\langle\chi\vert e^{\mathrm i(K_{h'}+B)\lambda}e^{-\mathrm i(K_h+B)\lambda}\vert\chi\rangle\rvert^2$. (5.10): $\alpha(h,h';\lambda)/\lambda\to d_0(h,h')=\lVert u_h-u_{h'}\rVert$. (5.11): $d_0$ is the chord distance of the record set in $(\mathcal H_c,g)$. (6.9)–(6.11), with $D=K_h-K_{h'}$:
$$
W=1-\lambda^2\operatorname{Var}_\chi(D)+\lambda^3c_3^{(\beta_0)}+O(\lambda^4),\qquad
c_3^{(\beta_0)}=\underbrace{\operatorname{Im}\langle K_{h'}\tilde D^2\rangle}_{c_3^{(0)}}+\operatorname{Im}\langle B\tilde D^2\rangle .
$$

**Notation.** $\langle X\rangle$, $\tilde X$, $u(x)=\widetilde{K(x)}\chi$ and $g(v,w)=\operatorname{Re}\langle v\vert w\rangle$ are as in question.md. With $g$, the space $\mathcal H_c$ is a real Euclidean space $E\cong\mathbb R^{2d_c}$. In addition:
- $M(x):=K(x)+B$, so $\psi_x(\lambda)=e^{-\mathrm iM(x)\lambda}\chi$.
- Coordinate indices are $a,b,m,n,p,q\in\{1,\dots,k\}$, with summation convention; they are not object names. $K_a:=\partial_aK(x)$, $K_{ab}:=\partial_a\partial_bK(x)$, $\tilde K_a:=K_a-\langle K_a\rangle$.
- For self-adjoint $X,Y$: $\operatorname{Cov}_\chi(X,Y):=\tfrac12\langle\{X,Y\}\rangle-\langle X\rangle\langle Y\rangle=g(\tilde X\chi,\tilde Y\chi)$.
- Curvature convention: $R_{abmn}$ is normalized so that $R_{abab}/(g_{aa}g_{bb}-g_{ab}^2)$ is the sectional curvature of the plane $(\partial_a,\partial_b)$; the unit sphere has $+1$. For $k=2$, the Gaussian curvature is $\kappa=R_{1212}/\det g$. $\hat\kappa$ is the Gaussian curvature of $\hat g$, and $\kappa_g=\lambda^{-2}\hat\kappa$ is that of $g=\lambda^2\hat g$, because sectional curvatures scale inversely with a constant factor of the metric.

**Assumptions.** These are the setting of question.md and $\lambda\ge0$. Expansions in $\lambda$ are taken at fixed $x$ as $\lambda\to0^+$. They hold together with all $x$-derivatives and locally uniformly in $x$, because $\hat g$ is jointly smooth in $(x,\lambda)$ (Step 3). Curvatures are computed only where the metric is non-degenerate. The coordinates on $U$ are used only to state density and to differentiate (M5); all geometry comes from $\alpha$.

## Derivation

### Step 1. Dense family and the witness function on $U$

By (5.6) and (6.6), $W(h,h';\lambda)=W(x_h,x_{h'};\lambda)$, where $W(x,x';\lambda)=\lvert\langle\psi_{x'}(\lambda)\vert\psi_x(\lambda)\rangle\rvert^2$ is a single function on $U\times U$. It depends only on $K(\cdot)$, $B$ and $\chi$. It does not depend on $\phi$ or on which other places the description contains.

**Limit.** Take a sequence of finite descriptions $\mathcal D_\nu$ (A1) with the same medium, $\chi$, $K(\cdot)$ and body. Let their label sets $X_\nu=\{x_h\}$ become dense: for every $x\in U$ there are $x_\nu\in X_\nu$ with $x_\nu\to x$. The witness weights of every $\mathcal D_\nu$ are restrictions of the same smooth function $W$ ($K$ is smooth and the exponential is entire). Hence $W(x,x')=\lim_\nu W(x_\nu,x'_\nu)$, and with it $\alpha=\arccos\sqrt W$, is fixed on all of $U\times U$. No further limit enters. On $U$, $\alpha$ is a pseudometric: points with $W=1$ are identified in $X_V$, as in 03.

### Step 2. The metric is the infinitesimal witness angle (item 1a)

Fix $\lambda$, $x$ and $v$, and let $\psi(\varepsilon):=\psi_{x+\varepsilon v}(\lambda)$. Write $\dot\psi=v^a\partial_a\psi$ and $\ddot\psi$ for the derivatives at $\varepsilon=0$. Differentiating $\lVert\psi(\varepsilon)\rVert^2=1$ once and twice gives $\operatorname{Re}\langle\dot\psi\vert\psi\rangle=0$ and $\operatorname{Re}\langle\ddot\psi\vert\psi\rangle=-\lVert\dot\psi\rVert^2$. Then
$$
W(x,x+\varepsilon v)=\bigl\lvert1+\varepsilon\langle\dot\psi\vert\psi\rangle+\tfrac{\varepsilon^2}{2}\langle\ddot\psi\vert\psi\rangle\bigr\rvert^2+O(\varepsilon^3)
=1-\varepsilon^2\bigl(\lVert\dot\psi\rVert^2-\lvert\langle\psi\vert\dot\psi\rangle\rvert^2\bigr)+O(\varepsilon^3).
$$
The bracket equals $g_{ab}v^av^b$ for two reasons. First, $\lVert\dot\psi\rVert^2=v^av^b\operatorname{Re}\langle\partial_a\psi\vert\partial_b\psi\rangle$. Second, $\langle\psi\vert\partial_a\psi\rangle$ is purely imaginary, so $\lvert\langle\psi\vert\dot\psi\rangle\rvert^2=v^av^b\langle\partial_a\psi\vert\psi\rangle\langle\psi\vert\partial_b\psi\rangle$, and this product is real. Since $\sin^2\alpha=1-W$ and $\alpha\to0$:
$$
\alpha(x,x+\varepsilon v;\lambda)^2=\varepsilon^2g_{ab}(x;\lambda)v^av^b+O(\varepsilon^3),\qquad
\lim_{\varepsilon\to0^+}\frac{\alpha(x,x+\varepsilon v;\lambda)}{\varepsilon}=\sqrt{g_{ab}(x;\lambda)v^av^b}.
\tag{19.1}
$$
By polarization, $\alpha$ determines $g$: the metric $g$ is the infinitesimal form of the witness angle. Because $W$ is invariant under $\psi_x\to e^{\mathrm if(x)}\psi_x$, so is $g$ (M6). Since $\psi_x(0)=\chi$ for every $x$, we have $g(x;0)=0$, which is why the rescaled metric $\hat g$ is needed.

### Step 3. Exact form of $\hat g$

Since $\partial_aM=K_a$, Duhamel's formula gives $\partial_ae^{-\mathrm iM\lambda}=-\mathrm i\int_0^\lambda e^{-\mathrm iM(\lambda-s)}K_ae^{-\mathrm iMs}\,\mathrm ds$. Hence $e^{\mathrm iM\lambda}\partial_a\psi_x=-\mathrm i\lambda\bar K_a\chi$, with the self-adjoint operator
$$
\bar K_a(x;\lambda):=\frac1\lambda\int_0^\lambda e^{\mathrm iM(x)s}K_a(x)\,e^{-\mathrm iM(x)s}\,\mathrm ds=\sum_{j\ge0}\frac{(\mathrm i\lambda)^j}{(j+1)!}\,\mathrm{ad}_M^{\,j}K_a,\qquad \mathrm{ad}_MX:=[M,X].
$$
Insert this into the definition of $g$. The unitary $e^{\mathrm iM\lambda}$ is taken at the same $x$ in both factors, and $\langle\psi_x\vert\partial_b\psi_x\rangle=-\mathrm i\lambda\langle\bar K_b\rangle$. The result is
$$
\hat g_{ab}(x;\lambda)=\operatorname{Cov}_\chi\bigl(\bar K_a,\bar K_b\bigr)=g\bigl(\widetilde{\bar K_a}\chi,\widetilde{\bar K_b}\chi\bigr).
\tag{19.2}
$$
This is exact for $\lambda>0$, entire in $\lambda$ and jointly smooth in $(x,\lambda)$. **Commuting case:** if $[M(x),K_a(x)]=0$ for all $a$, then $\bar K_a=K_a$ and
$$
\hat g_{ab}(x;\lambda)=\operatorname{Cov}_\chi(K_a,K_b)\qquad\text{for every }\lambda .
\tag{19.3}
$$

### Step 4. Expansion in $\lambda$ (item 1b)

From the series, $\bar K_a=K_a+\lambda\Delta_a+O(\lambda^2)$ with $\Delta_a:=\tfrac{\mathrm i}{2}[M,K_a]=\tfrac{\mathrm i}{2}[M,\tilde K_a]$, which is self-adjoint. Expand (19.2) using the bilinearity of $\operatorname{Cov}_\chi$.

**Order 0.** Since $\partial_au=\partial_a\bigl(K\chi-\langle K\rangle\chi\bigr)=\tilde K_a\chi$,
$$
\hat g^{(0)}_{ab}(x)=\operatorname{Cov}_\chi(K_a,K_b)=g(\partial_au,\partial_bu),\qquad\text{i.e. }\ \hat g^{(0)}=u^*g .
\tag{19.4}
$$
Thus $\hat g^{(0)}$ is the pullback by the record map of the flat metric of $E$. The quadratic form is $\hat g^{(0)}(v,v)=\lVert v^a\partial_au\rVert^2=\operatorname{Var}_\chi(v^aK_a)$, which vanishes iff $\chi$ is an eigenvector of $v^aK_a$. So $\hat g^{(0)}(x)$ is non-degenerate iff no nonzero real combination $v^aK_a(x)$ has $\chi$ as an eigenvector, i.e. iff $u$ is an immersion at $x$. Since $\langle\chi\vert\tilde X\chi\rangle=0$, the map $u$ and all its derivatives take values in $\chi^\perp$, a real subspace $E_\perp\subset E$ of dimension $2(d_c-1)$.

**Order 1.** $\hat g^{(1)}_{ab}=\operatorname{Cov}_\chi(K_a,\Delta_b)+\operatorname{Cov}_\chi(\Delta_a,K_b)$. Let $X,Y,M$ be self-adjoint with $\langle X\rangle=\langle Y\rangle=0$. Then $\operatorname{Cov}_\chi(X,Z)=\tfrac12\langle\{X,Z\}\rangle$, and the identity $\{X,[M,Y]\}+\{[M,X],Y\}=[M,\{X,Y\}]$ holds because the terms $XMY$ and $YMX$ cancel. Use it with $X=\tilde K_a$ and $Y=\tilde K_b$, together with $\langle[M,Z]\rangle=2\mathrm i\operatorname{Im}\langle MZ\rangle$ for self-adjoint $Z$, and split $M=K+B$. This gives $\hat g^{(1)}=\hat g^{(1,0)}+\hat g^{(1,B)}$ with
$$
\hat g^{(1,0)}_{ab}=\frac{\mathrm i}{4}\bigl\langle[K,\{\tilde K_a,\tilde K_b\}]\bigr\rangle=-\frac12\operatorname{Im}\bigl\langle K\{\tilde K_a,\tilde K_b\}\bigr\rangle ,
\tag{19.5}
$$
$$
\hat g^{(1,B)}_{ab}=\frac{\mathrm i}{4}\bigl\langle[B,\{\tilde K_a,\tilde K_b\}]\bigr\rangle=-\frac12\operatorname{Im}\bigl\langle B\{\tilde K_a,\tilde K_b\}\bigr\rangle ,
\tag{19.6}
$$
where every operator is taken at $x$. The body-free part (19.5) is all of $\hat g^{(1)}$ when $B=0$. The body part (19.6) is linear in $B$.

### Step 5. Relation to (5.10) and (6.9)–(6.11) (item 1c)

Put $x_h=x$ and $x_{h'}=x+\varepsilon v$. Then $D=-\varepsilon v^aK_a+O(\varepsilon^2)$, $\tilde D^2=\varepsilon^2(v^a\tilde K_a)^2+O(\varepsilon^3)$, $K_{h'}=K(x)+O(\varepsilon)$, and $v^av^b\{\tilde K_a,\tilde K_b\}=2(v^a\tilde K_a)^2$. Inserting these into the quoted (6.9)–(6.11) and comparing with (19.4)–(19.6):
$$
\begin{aligned}
\operatorname{Var}_\chi(D)&=\varepsilon^2\,\hat g^{(0)}_{ab}v^av^b+O(\varepsilon^3),\\
c_3^{(0)}(h,h')&=-\varepsilon^2\,\hat g^{(1,0)}_{ab}v^av^b+O(\varepsilon^3),\\
\operatorname{Im}\langle B\tilde D^2\rangle&=-\varepsilon^2\,\hat g^{(1,B)}_{ab}v^av^b+O(\varepsilon^3).
\end{aligned}
\tag{19.7}
$$
These relations can also be reached by an independent route. By (19.1) and Step 3, $1-W=\varepsilon^2\lambda^2\bigl[\hat g^{(0)}+\lambda\hat g^{(1)}+O(\lambda^2)\bigr](v,v)+O(\varepsilon^3)$, while (6.9) gives $1-W=\lambda^2\operatorname{Var}_\chi(D)-\lambda^3c_3^{(\beta_0)}+O(\lambda^4)$. Because $W$ is jointly smooth in $(\varepsilon,\lambda)$, the coefficients of $\varepsilon^2\lambda^2$ and $\varepsilon^2\lambda^3$ must agree. They do: Step 4 and the inputs from 06 agree, and the body-free and body parts match separately. In words:
- $\hat g^{(0)}$ is the infinitesimal form of $d_0$ in (5.10). Taking $\lambda\to0$ and $\varepsilon\to0$ in either order, $\alpha/(\lambda\varepsilon)\to\sqrt{\hat g^{(0)}(v,v)}$.
- By (5.11), $\hat g^{(0)}$ is the metric that the record set $u(U)$ inherits from $E$. It is intrinsic, whereas $d_0$ is the chord distance.
- $-\hat g^{(1)}$ is the infinitesimal form of the $\lambda^3$ coefficient $c_3^{(\beta_0)}$.

### Step 6. Curvature of $\hat g^{(0)}$ (item 2a)

By (19.4), wherever $\hat g^{(0)}$ is non-degenerate, $(U,\hat g^{(0)})$ is the immersed submanifold $u(U)$ of the flat space $E$, with the induced metric. Split the second derivatives into tangent and normal parts:
$$
\partial_a\partial_bu=\Gamma^p_{ab}\,\partial_pu+\mathrm{II}_{ab},\qquad g(\mathrm{II}_{ab},\partial_mu)=0 .
\tag{19.8}
$$
Differentiating (19.4) gives $\Gamma_{ab|m}:=g(\partial_a\partial_bu,\partial_mu)=\tfrac12(\partial_a\hat g^{(0)}_{bm}+\partial_b\hat g^{(0)}_{am}-\partial_m\hat g^{(0)}_{ab})$. So $\Gamma^p_{ab}=\hat g^{(0)pq}\Gamma_{ab|q}$ are the Levi-Civita symbols of $\hat g^{(0)}$, and $\mathrm{II}_{ab}=\partial_a\partial_bu-\hat g^{(0)pq}\Gamma_{ab|p}\partial_qu$ is the second fundamental form. The Gauss equation for a flat ambient space (standard) then gives
$$
\begin{aligned}
\hat R^{(0)}_{abmn}&=g(\mathrm{II}_{am},\mathrm{II}_{bn})-g(\mathrm{II}_{an},\mathrm{II}_{bm})\\
&=g(\partial_a\partial_mu,\partial_b\partial_nu)-g(\partial_a\partial_nu,\partial_b\partial_mu)-\hat g^{(0)pq}\bigl(\Gamma_{am|p}\Gamma_{bn|q}-\Gamma_{an|p}\Gamma_{bm|q}\bigr),
\end{aligned}
\tag{19.9}
$$
and for $k=2$
$$
\kappa^{(0)}=\frac{g(\mathrm{II}_{11},\mathrm{II}_{22})-g(\mathrm{II}_{12},\mathrm{II}_{12})}{\hat g^{(0)}_{11}\hat g^{(0)}_{22}-(\hat g^{(0)}_{12})^2}.
\tag{19.10}
$$

### Step 7. Flatness (item 2b)

By (19.9), a sufficient condition for flatness is
$$
\partial_a\partial_bu(x)\in\operatorname{span}_{\mathbb R}\{\partial_1u(x),\dots,\partial_ku(x)\}\ \ \text{for all }x,a,b
\quad(\mathrm{II}\equiv0)\ \Longrightarrow\ \hat R^{(0)}\equiv0 .
\tag{19.11}
$$
It holds in each of the following cases:
- (i) $K$ is affine, $K(x)=K_0+x^aA_a$. Then $\partial_a\partial_bu=\tilde K_{ab}\chi=0$, and $\hat g^{(0)}_{ab}=\operatorname{Cov}_\chi(A_a,A_b)$ is constant.
- (ii) $k=2(d_c-1)$, for example $k=2$ with a qubit medium. Then $k$ independent vectors $\partial_pu$ span $E_\perp$, which contains every $\partial_a\partial_bu$.
- (iii) $k=1$.

### Step 8. Dependence on the body and on $\chi$ (item 2c)

**Body.** (19.4) contains only $K$ and $\chi$; $B$ enters $\hat g$ first at order $\lambda$, through (19.6). Hence $\hat g^{(0)}$, $\mathrm{II}$, $\hat R^{(0)}$ and $\kappa^{(0)}$ are the same with and without a body, for every $B$.

**Medium state $\chi$.** Since $\partial_a\partial_bu=\tilde K_{ab}\chi$, every ingredient of (19.9) is a covariance in $\chi$:
$$
\hat g^{(0)}_{ab}=\operatorname{Cov}_\chi(K_a,K_b),\qquad
\Gamma_{ab|m}=\operatorname{Cov}_\chi(K_{ab},K_m),\qquad
g(\partial_a\partial_mu,\partial_b\partial_nu)=\operatorname{Cov}_\chi(K_{am},K_{bn}).
\tag{19.12}
$$
So $\hat g^{(0)}$ and its curvature depend on $\chi$ only through the symmetrized covariance matrix of the operators $\{K_a(x),K_{ab}(x)\}$ in $\chi$, i.e. through their first and second moments. They are invariant under $\chi\to e^{\mathrm i\theta}\chi$ (M6). Otherwise they change with $\chi$, because $\chi$ selects the immersion $u=\tilde K\chi$. The metric degenerates where $\chi$ is an eigenvector of some $v^aK_a$. For affine $K$ the metric is flat for every $\chi$.

### Step 9. Body-induced curvature (item 3a)

Let $k=2$. By Step 3 and the non-degeneracy of $\hat g^{(0)}$, $\hat\kappa(x;\lambda)=\kappa^{(0)}(x)+\lambda\,\dot\kappa[\hat g^{(1)}](x)+O(\lambda^2)$, where $\dot\kappa[h]:=\frac{\mathrm d}{\mathrm dt}\kappa[\hat g^{(0)}+th]\big|_{t=0}$. We use the standard first variation of the scalar curvature, $\dot S[h]=-\Delta\operatorname{tr}h+\nabla^a\nabla^bh_{ab}-R^{ab}h_{ab}$. In two dimensions $S=2\kappa$ and $R_{ab}=\kappa g_{ab}$, so
$$
\dot\kappa[h]=\tfrac12\bigl(\nabla^a\nabla^bh_{ab}-\Delta\operatorname{tr}h-\kappa^{(0)}\operatorname{tr}h\bigr),
\tag{19.13}
$$
where $\nabla$, $\Delta=\nabla^a\nabla_a$, the trace and the raising of indices all refer to $\hat g^{(0)}$. $\dot\kappa$ is linear in $h$, and $\hat g^{(0)}$ does not depend on the body (Step 8). Therefore
$$
\hat\kappa\big|_{B}-\hat\kappa\big|_{B=0}=\lambda\,\dot\kappa\bigl[\hat g^{(1,B)}\bigr]+O(\lambda^2),\qquad
\hat g^{(1,B)}_{ab}=-\tfrac12\operatorname{Im}\langle\chi\vert B\{\tilde K_a,\tilde K_b\}\vert\chi\rangle .
\tag{19.14}
$$
For the curvature of $g$, the change is $\lambda^{-1}\dot\kappa[\hat g^{(1,B)}]+O(1)$, one order below the leading term $\lambda^{-2}\kappa^{(0)}$. If $\hat g^{(0)}=\delta$ in linear coordinates, (19.13) reduces to $\dot\kappa[h]=\partial_1\partial_2h_{12}-\tfrac12(\partial_2^2h_{11}+\partial_1^2h_{22})$.

### Step 10. Sufficient conditions for no change (item 3b)

- **(V1)** $\hat g^{(1,B)}\equiv0$ on $U$. By (19.6) this holds if any of the following is true:
  - (a) $B\chi=\beta\chi$. Then $\langle\chi\vert[B,Z]\vert\chi\rangle=0$ for every $Z$.
  - (b) $[B,K_a(x)]=0$ for all $a$ and $x$.
  - (c) $\{\tilde K_a,\tilde K_b\}\chi=z_{ab}\chi$ for all $a,b,x$. Then $z_{ab}$ is real and $\langle[B,Z]\rangle=z_{ab}\langle B\rangle-z_{ab}\langle B\rangle=0$ for every $B$. In this case $\hat g^{(1,0)}=0$ as well.
- **(V2)** $K$ is affine. Then $\tilde K_a$ is constant, so $\hat g^{(0)}$ and $\hat g^{(1,B)}$ are constant in $x$, $\kappa^{(0)}=0$, and every term of (19.13) vanishes. Also $\hat g^{(1,0)}$ is affine in $x$, so $\hat\kappa=O(\lambda^2)$ altogether.
- **(V3)** $\hat g^{(1,B)}=\mathcal L_\xi\hat g^{(0)}$ for a vector field $\xi$ with $\xi^a\partial_a\kappa^{(0)}=0$. Differentiating $\kappa[\Phi_t^*\hat g^{(0)}]=\kappa^{(0)}\circ\Phi_t$, where $\Phi_t$ is the flow of $\xi$, gives $\dot\kappa[\mathcal L_\xi\hat g^{(0)}]=\xi^a\partial_a\kappa^{(0)}=0$.

$$
\text{(V1) or (V2) or (V3)}\ \Longrightarrow\ \hat\kappa\big|_B-\hat\kappa\big|_{B=0}=O(\lambda^2).
\tag{19.15}
$$

### Step 11. Example 4(a): qubit plane

Without a body, use polar label coordinates $(x,y)=r(\cos\phi,\sin\phi)$. Then $K=r\,n\cdot\sigma$ with $(n\cdot\sigma)^2=\mathbb 1$ and $(\cos\phi\,\sigma_x+\sin\phi\,\sigma_y)\lvert0\rangle=e^{\mathrm i\phi}\lvert1\rangle$, so $\psi_x=\cos t\,\lvert0\rangle+e^{\mathrm i\mu}\sin t\,\lvert1\rangle$ with $t=\lambda r$ and $\mu=\phi-\pi/2$. For this form, $\lVert\mathrm d\psi\rVert^2=\mathrm dt^2+\sin^2t\,\mathrm d\mu^2$ and $\langle\psi\vert\mathrm d\psi\rangle=\mathrm i\sin^2t\,\mathrm d\mu$. The metric is therefore $\mathrm dt^2+\tfrac14\sin^2(2t)\,\mathrm d\mu^2$:
$$
g=\lambda^2\mathrm dr^2+\tfrac14\sin^2(2\lambda r)\,\mathrm d\phi^2,\qquad
g_{ab}=\lambda^2\frac{x^ax^b}{r^2}+\frac{\sin^2(2\lambda r)}{4r^2}\Bigl(\delta_{ab}-\frac{x^ax^b}{r^2}\Bigr),
\tag{19.16}
$$
with $(x^1,x^2)=(x,y)$ and $g_{ab}=\lambda^2\delta_{ab}$ at $r=0$. The metric is non-degenerate except on the circles $r=m\pi/(2\lambda)$, $m=1,2,\dots$. For a metric $\mathrm dr^2+f^2\mathrm d\phi^2$ the Gaussian curvature is $-f''/f$; here $\hat g$ has $f=\sin(2\lambda r)/(2\lambda)$, so
$$
\hat\kappa=4\lambda^2,\qquad \kappa_g=4\qquad\text{wherever $g$ is non-degenerate, for every }\lambda>0 .
\tag{19.17}
$$
**With a body.** Up to phase, every unit vector of $\mathbb C^2$ has the form $\cos t\lvert0\rangle+e^{\mathrm i\mu}\sin t\lvert1\rangle$. The computation above therefore shows that the Fubini–Study metric on $\mathbb{CP}^1$ is a round sphere of radius $1/2$, with curvature $4$. For any $B$, $g=F^*g_{\rm FS}$ with $F(x)=[\psi_x(\lambda)]$. Where $g$ is non-degenerate, $\mathrm dF$ is injective, so $F$ is a local diffeomorphism between 2-manifolds and hence a local isometry. By the Theorema egregium, $\kappa_g=4$.

**Answer: no.** The Gaussian curvature does not change, for every self-adjoint $B$; the same holds for every $K$ and $\chi$ whenever $d_c=2$. Perturbatively, $\{\sigma_a,\sigma_b\}=2\delta_{ab}$, so (V1c) applies.

### Step 12. Example 4(b): commuting plane

$K_1=\sigma_x^{(1)}$ and $K_2=\sigma_x^{(2)}$ commute with $K$, so (19.3) holds for every $\lambda$. In $\lvert00\rangle$ we have $\langle\sigma_x^{(a)}\rangle=0$, $(\sigma_x^{(a)})^2=\mathbb 1$ and $\langle\sigma_x^{(1)}\sigma_x^{(2)}\rangle=0$. Hence
$$
g=\lambda^2(\mathrm dx^2+\mathrm dy^2),\qquad \hat g=\mathrm dx^2+\mathrm dy^2,\qquad \kappa_g=\hat\kappa=0\qquad\text{for every }\lambda,
\tag{19.18}
$$
and the metric is non-degenerate everywhere. Direct check: $W=\cos^2(\lambda(x-x'))\cos^2(\lambda(y-y'))$, and its $\varepsilon^2$ coefficient gives (19.18) through (19.1).

### Step 13. Example 4(c): curved record map

Since $u\in\operatorname{span}_{\mathbb C}\{e_2,e_3\}\perp e_1=\chi$, we have $K\chi=u$ and $\langle K\rangle=0$: the record map is $u$ itself. The real vectors $f_1=e_2$, $f_2=e_3$, $f_3=\mathrm ie_2$ are $g$-orthonormal ($g(e_2,\mathrm ie_2)=\operatorname{Re}\mathrm i=0$). In them, $u=R(\sin\theta\cos\varphi\,f_1+\sin\theta\sin\varphi\,f_2+\cos\theta\,f_3)$, the round sphere of radius $R$ in $E_3=\operatorname{span}_{\mathbb R}\{f_1,f_2,f_3\}$. By (19.4),
$$
\hat g^{(0)}=R^2\bigl(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2\bigr),\qquad \kappa^{(0)}=\frac1{R^2}.
\tag{19.19}
$$
The same value follows from (19.10). We have $\partial_a\partial_bu\in E_3$, and the normal of the tangent plane inside $E_3$ is $u/R$. Differentiating $g(u,u)=R^2$ twice gives $g(\partial_a\partial_bu,u)=-\hat g^{(0)}_{ab}$, so $\mathrm{II}_{ab}=-\hat g^{(0)}_{ab}u/R^2$ and $\hat R^{(0)}_{1212}=R^{-2}\det\hat g^{(0)}$.

**Body part.** $K_a=\lvert\partial_au\rangle\langle e_1\rvert+\lvert e_1\rangle\langle\partial_au\rvert$ has $\langle K_a\rangle=0$, so $\tilde K_a=K_a$. Moreover $K_b\chi=\partial_bu$ and $K_a\partial_bu=\langle\partial_au\vert\partial_bu\rangle e_1$. Hence
$$
\{\tilde K_a,\tilde K_b\}\chi=2\hat g^{(0)}_{ab}\,\chi\ \Longrightarrow\ \hat g^{(1,B)}=0\ \text{ for every self-adjoint $B$ on }\mathbb C^3,\qquad \hat g^{(1,0)}=0,
\tag{19.20}
$$
by (V1c). So $\hat g=\hat g^{(0)}+O(\lambda^2)$ and $\hat\kappa=R^{-2}+O(\lambda^2)$ for every $B$.

## Result

1. **Item 1.** The background metric is the infinitesimal witness angle, $\alpha(x,x+\varepsilon v)^2=\varepsilon^2g(v,v)+O(\varepsilon^3)$ (19.1).
   - Exact form: $\hat g_{ab}=\operatorname{Cov}_\chi(\bar K_a,\bar K_b)$ (19.2); $\hat g=\hat g^{(0)}$ for all $\lambda$ in the commuting case (19.3).
   - Leading order: $\hat g^{(0)}=u^*g$ (19.4).
   - First order: $\hat g^{(1)}=\frac{\mathrm i}{4}\langle[K+B,\{\tilde K_a,\tilde K_b\}]\rangle$, split into the body-free part (19.5) and the body part (19.6).
   - Relation to earlier packages: (19.7). $\hat g^{(0)}$ is the infinitesimal $d_0$ of (5.10); $-\hat g^{(1)}$ is the infinitesimal $c_3^{(\beta_0)}$ of (6.10)–(6.11), part by part.
2. **Item 2.** The curvature of $\hat g^{(0)}$ is given by the Gauss equation of the record surface $u(U)\subset(\mathcal H_c,g)$, (19.8)–(19.10).
   - Flatness holds if $\mathrm{II}\equiv0$ (19.11), in particular for affine $K$, for $k=2(d_c-1)$, or for $k=1$.
   - $\hat g^{(0)}$ and its curvature do not depend on the body. They depend on $\chi$ only through the covariances (19.12).
3. **Item 3.** The body changes $\hat\kappa$ at order $\lambda$ by $\lambda\,\dot\kappa[\hat g^{(1,B)}]$, given by (19.13)–(19.14). This change vanishes under (V1), (V2) or (V3), see (19.15).
4. **Item 4.**
   - (a) Without a body, $g=\lambda^2\mathrm dr^2+\frac14\sin^2(2\lambda r)\mathrm d\phi^2$, with $\kappa_g=4$ and $\hat\kappa=4\lambda^2$ (19.16)–(19.17). With a body, the curvature does not change, for any $B$.
   - (b) $g=\lambda^2(\mathrm dx^2+\mathrm dy^2)$, flat for every $\lambda$ (19.18).
   - (c) $\hat g^{(0)}=R^2(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\varphi^2)$ with $\kappa^{(0)}=R^{-2}$ (19.19), and $\hat g^{(1,B)}=0$ for every $B$ (19.20).

## Consistency checks

1. **Scaling and dimensions.** Under $(K,B)\to(sK,sB)$ we have $\psi_x(\lambda)\to\psi_x(s\lambda)$, so $\hat g(\lambda)\to s^2\hat g(s\lambda)$. This requires $\hat g^{(0)}\to s^2\hat g^{(0)}$ and $\hat g^{(1)}\to s^3\hat g^{(1)}$; indeed (19.4) is quadratic and (19.5)–(19.6) are cubic in $(K,B)$. In (19.10), $\mathrm{II}\sim s$ and $\det\hat g^{(0)}\sim s^4$, so $\kappa^{(0)}\sim s^{-2}$. This matches $R\to sR$ in (19.19), and in 4(a) $\kappa_g=4$ is scale-invariant. Dimensions: $\lambda K$ is dimensionless, so $\hat g^{(0)}$ and $\lambda\hat g^{(1)}$ both carry $[K]^2$.
2. **Special cases.** For $k=1$, (19.9) vanishes by antisymmetry, as every 1-dimensional metric must. For $B=\beta\mathbb 1$, the body only adds a global phase to $\psi_x$; consistently, (19.6) gives $\hat g^{(1,B)}=0$.
3. **Exact versus perturbative (4a).** Expanding (19.16) gives $\hat g=\delta+O(\lambda^2)$. This matches (19.4), since $\operatorname{Cov}_{\lvert0\rangle}(\sigma_a,\sigma_b)=\delta_{ab}$, and (19.5), which vanishes because $\{\sigma_a,\sigma_b\}=2\delta_{ab}$. Correspondingly $\hat\kappa=4\lambda^2$ has no terms of order $1$ or $\lambda$, in line with (19.11) and (19.15).

## Open issues

- The order-$\lambda^2$ terms of $\hat g$ and $\hat\kappa$ are not computed. In 4(a) and 4(c) the body affects $\hat\kappa$ at most at order $\lambda^2$; in 4(a) it has no effect at all. No example with a nonzero first-order body-induced curvature is exhibited. Such an example needs non-affine $K$, $d_c\ge3$, and violation of (V1)–(V3).
- The conditions (V1)–(V3) and (19.11) are sufficient, not necessary.
- Global questions are not addressed: the degeneracy sets (for example the circles in 4(a)), and the relation between the geodesic distance of $g$ and $\alpha$.

## Methods used

- Fubini–Study metric, Taylor expansion of overlaps, polarization
- Duhamel formula, adjoint series, symmetrized covariances
- Induced metric of an immersion, Gauss equation, second fundamental form
- First variation of scalar curvature, Lie-derivative (diffeomorphism) invariance
- Theorema egregium, curvature of $\mathrm dr^2+f^2\mathrm d\phi^2$, Bloch parametrization of $\mathbb{CP}^1$