The minimal completion for the spectral target

Proof. $h$ is a bijection $S_{\mathrm{flat}}\to\mathrm{U}(1)$ by (32) and (33), hence surjective, and $p_F=u\circ h$ with $u\colon\mathrm{U}(1)\to\{0\}$ the constant map, so $h$ is a readout extension in the sense of Definition 6.1. Define $b\colon\mathrm{U}(1)\to A_{\mathrm{spec}}$ by $b(e^{i\theta}):=\operatorname{Spec}(H_\theta)$; this is well defined because, as shown at the end of the proof of Proposition 10.1, $\operatorname{Spec}(H_\theta)$ depends only on $\theta$ mod $2\pi$, i.e. only on $e^{i\theta}$. Then $a_{\mathrm{spec}}=b\circ h$, so $h$ is $Q_{\mathrm{spec}}$-sufficient by Definition 6.3. ◻

Proof. ($\Leftarrow$) Reverse direction. Suppose first $\theta'\equiv\theta\pmod{2\pi}$. Then $\operatorname{Spec}(H_{\theta'})=\operatorname{Spec}(H_\theta)$ by the invariance established at the end of the proof of Proposition 10.1.

Suppose next $\theta'\equiv-\theta \pmod{2\pi}$. By the same invariance it suffices to treat $\theta'=-\theta$. Then for every $n\in\mathbb{Z}$, $$\lambda_{-n}(-\theta) =\left(\frac{2\pi(-n)-(-\theta)}{L}\right)^{\!2} =\left(\frac{-(2\pi n-\theta)}{L}\right)^{\!2} =\left(\frac{2\pi n-\theta}{L}\right)^{\!2} =\lambda_n(\theta).\tag{49}$$ The map $n\mapsto -n$ is a bijection of $\mathbb{Z}$ onto itself. Hence (49) exhibits a bijection between the indexing set of $\operatorname{Spec}(H_{-\theta})$ and that of $\operatorname{Spec}(H_\theta)$ carrying each eigenvalue to an equal eigenvalue. A multiset is determined by its indexed family up to bijective reindexing, so the two multisets are equal, with all multiplicities: $$m_{-\theta}(\lambda)=\#\{n:\lambda_n(-\theta)=\lambda\} =\#\{n:\lambda_{-n}(-\theta)=\lambda\} =\#\{n:\lambda_{n}(\theta)=\lambda\}=m_\theta(\lambda).$$ So $\operatorname{Spec}(H_{-\theta})=\operatorname{Spec}(H_\theta)$. Concretely, the reindexing $n\mapsto -n$ is implemented on the Hilbert space by complex conjugation, $\psi_n\mapsto\overline{\psi_n}=\psi_{-n}$, which is an antiunitary intertwiner between $H_\theta$ and $H_{-\theta}$; but the elementary reindexing argument above is already complete and requires nothing further.

($\Rightarrow$) Forward direction. Suppose $\operatorname{Spec}(H_\theta)=\operatorname{Spec}(H_{\theta'})$. We recover $\theta$ up to sign and $2\pi$ from the spectrum, using the lowest eigenvalue.

Define, for any $\theta\in\mathbb{R}$, the principal representative $$\alpha(\theta)\in[0,\pi] \qquad\text{the unique number in } [0,\pi] \text{ with } \theta\equiv\alpha(\theta) \ \text{ or }\ \theta\equiv-\alpha(\theta) \pmod{2\pi}.\tag{50}$$ Existence and uniqueness of $\alpha(\theta)$: reduce $\theta$ mod $2\pi$ to a representative $\tilde\theta\in[0,2\pi)$. If $\tilde\theta\in[0,\pi]$ set $\alpha:=\tilde\theta$; if $\tilde\theta\in(\pi,2\pi)$ set $\alpha:=2\pi-\tilde\theta\in(0,\pi)$, and then $\theta\equiv\tilde\theta=-\alpha \pmod{2\pi}$. In both cases $\alpha\in[0,\pi]$ and $\theta\equiv\pm\alpha$. For uniqueness, suppose $\alpha,\beta\in[0,\pi]$ both work; then $\alpha\equiv\pm\beta\pmod{2\pi}$. If $\alpha\equiv\beta$ then $\alpha-\beta\in 2\pi\mathbb{Z}$ and $|\alpha-\beta|\leq\pi<2\pi$, forcing $\alpha=\beta$. If $\alpha\equiv-\beta$ then $\alpha+\beta\in2\pi\mathbb{Z}$ with $0\le\alpha+\beta\le 2\pi$, so $\alpha+\beta\in\{0,2\pi\}$, giving either $\alpha=\beta=0$ or $\alpha=\beta=\pi$. In all cases $\alpha=\beta$.

We claim $$\lambda_{\min}(\theta):=\min\operatorname{Spec}(H_\theta)=\frac{\alpha(\theta)^2}{L^2}.\tag{51}$$ To see this, note $\lambda_n(\theta)=\bigl(\tfrac{2\pi n-\theta}{L}\bigr)^2$, so $$\min_{n\in\mathbb{Z}}\lambda_n(\theta) =\frac{1}{L^2}\min_{n\in\mathbb{Z}}\,(2\pi n-\theta)^2 =\frac{1}{L^2}\Bigl(\operatorname{dist}\bigl(\theta,\,2\pi\mathbb{Z}\bigr)\Bigr)^{\!2},$$ the minimum being attained since $(2\pi n-\theta)^2\to\infty$ as $|n|\to\infty$. It remains to identify $\operatorname{dist}(\theta,2\pi\mathbb{Z})=\alpha(\theta)$. Write $\theta\equiv\pm\alpha$ with $\alpha=\alpha(\theta)\in[0,\pi]$; since $2\pi\mathbb{Z}$ is invariant under negation and under translation by $2\pi$, we get $\operatorname{dist}(\theta,2\pi\mathbb{Z})=\operatorname{dist}(\pm\alpha,2\pi\mathbb{Z}) =\operatorname{dist}(\alpha,2\pi\mathbb{Z})$. For $\alpha\in[0,\pi]$ the nearest point of $2\pi\mathbb{Z}$ is $0$, since the next candidates are $\pm2\pi$ at distance $2\pi-\alpha\ge\pi\ge\alpha$ and $2\pi+\alpha>\alpha$. Hence $\operatorname{dist}(\alpha,2\pi\mathbb{Z})=\alpha$, proving (51).

Now assume $\operatorname{Spec}(H_\theta)=\operatorname{Spec}(H_{\theta'})$. Equal multisets have equal minima, so by (51) $$\frac{\alpha(\theta)^2}{L^2}=\lambda_{\min}(\theta)=\lambda_{\min}(\theta') =\frac{\alpha(\theta')^2}{L^2},$$ whence $\alpha(\theta)^2=\alpha(\theta')^2$. Since $\alpha(\theta),\alpha(\theta')\in[0,\pi]$ are both nonnegative, $\alpha(\theta)=\alpha(\theta')=:\alpha$. By (50), $\theta\equiv\pm\alpha$ and $\theta'\equiv\pm\alpha\pmod{2\pi}$. Combining the four sign cases: if $\theta\equiv\alpha$ and $\theta'\equiv\alpha$ then $\theta'\equiv\theta$; if $\theta\equiv\alpha$ and $\theta'\equiv-\alpha$ then $\theta'\equiv-\theta$; if $\theta\equiv-\alpha$ and $\theta'\equiv\alpha$ then $\theta'\equiv-\theta$; if $\theta\equiv-\alpha$ and $\theta'\equiv-\alpha$ then $\theta'\equiv\theta$. In every case $\theta'\equiv\pm\theta\pmod{2\pi}$, as required. ◻

Proof. Immediate from Theorem 11.2 and Definition 10.3. ◻

Proof. (1) $\cos$ is $2\pi$-periodic, so $c$ descends to $\mathbb{R}/2\pi\mathbb{Z}$; its image is $[-1,1]$ since $\cos$ restricted to $[0,\pi]$ is already onto $[-1,1]$.

(2) The function $\cos$ is strictly decreasing on $[0,\pi]$, hence injective there. Given $\theta,\theta'$, write $\alpha=\alpha(\theta),\alpha'=\alpha(\theta')\in[0,\pi]$ as in (50); since $\cos$ is even and $2\pi$-periodic, $\cos\theta=\cos\alpha$ and $\cos\theta'=\cos\alpha'$. Therefore $$c(\theta)=c(\theta') \iff \cos\alpha=\cos\alpha' \iff \alpha=\alpha' \iff \theta'\equiv\pm\theta \pmod{2\pi} \iff \theta\sim_{\mathrm{spec}}\theta',$$ where the second equivalence uses injectivity of $\cos$ on $[0,\pi]$, and the third is the argument at the end of the proof of Theorem 11.2 (equality of principal representatives is equivalent to $\theta'\equiv\pm\theta$).

(3) $c$ is surjective by (1), and $p_F=u\circ c$ with $u\colon[-1,1]\to\{0\}$ constant, so $c$ is a readout extension. For sufficiency, define $b_{\mathrm{spec}}\colon[-1,1]\to A_{\mathrm{spec}}$ by $$b_{\mathrm{spec}}(\kappa):=\operatorname{Spec}\bigl(H_{\arccos\kappa}\bigr), \qquad \arccos\kappa\in[0,\pi].\tag{54}$$ If $c(\theta)=\kappa$ then by (2) $\theta\sim_{\mathrm{spec}}\arccos\kappa$ (both have principal representative $\arccos\kappa$), so by Corollary 11.3 $\operatorname{Spec}(H_\theta)=\operatorname{Spec}(H_{\arccos\kappa})=b_{\mathrm{spec}}(c(\theta))$. Hence $a_{\mathrm{spec}}=b_{\mathrm{spec}}\circ c$, and $c$ is $Q_{\mathrm{spec}}$-sufficient.

(4) Write $\varphi:=\varphi_{[-1,1]}$ and $E_{\mathrm{spec}} :=E_{Q_{\mathrm{spec}}}$ for brevity. By Theorem 8.3, applied to the sufficient extension $c$, there is a unique surjection $\varphi\colon[-1,1]\to E_{\mathrm{spec}}$ with $j_{Q_{\mathrm{spec}}}=\varphi\circ c$, given by $$\varphi(\kappa) =\bigl(u(\kappa),\,b_{\mathrm{spec}}(\kappa)\bigr) =\bigl(0,\,b_{\mathrm{spec}}(\kappa)\bigr), \qquad \kappa\in[-1,1].\tag{55}$$ It remains to prove injectivity. Suppose $\varphi(\kappa)=\varphi(\kappa')$. Comparing second coordinates in (55) gives $b_{\mathrm{spec}}(\kappa)=b_{\mathrm{spec}}(\kappa')$, which by (54) says $$\operatorname{Spec}\bigl(H_{\arccos\kappa}\bigr)=\operatorname{Spec}\bigl(H_{\arccos\kappa'}\bigr).\tag{56}$$ By Theorem 11.2, (56) forces $\arccos\kappa'\equiv\pm\arccos\kappa \pmod{2\pi}$. Both $\arccos\kappa$ and $\arccos\kappa'$ lie in $[0,\pi]$, so each is its own principal representative in the sense of (50); by the uniqueness established there, they are equal. Applying $\cos$ yields $\kappa=\kappa'$.

Hence $\varphi$ is a bijection, and $c$ carries precisely the distinctions of $E_{\mathrm{spec}}$ , no more and no fewer.

The final assertion follows from Corollary 8.4: if $e$ is $Q_{\mathrm{spec}}$-sufficient and $c(\theta_0)\neq c(\theta_1)$, then by (2) $\theta_0\not\sim_{\mathrm{spec}} \theta_1$, so by Corollary 11.3 $a_{\mathrm{spec}}(\theta_0)\neq a_{\mathrm{spec}} (\theta_1)$, and Theorem 7.1 forces $e(\theta_0)\neq e(\theta_1)$. ◻