Target-Relative Necessity of Completion: When Readout Loss Obstructs, and What a Sufficient Extension Must Retain

The charged covariant Laplacian and the curvature obstruction

Jeremy RodgersSection 11 of 16Original PDF ↗Publication DOI ↗

The charged covariant Laplacian and the curvature obstruction

Operator, domain, and spectrum

Fix $\theta\in\mathbb{R}$ representing a point of $S_{\mathrm{flat}}$ , and consider a complex scalar field of unit charge on $S^1$ minimally coupled to $A_\theta$ . The gauge-covariant derivative is $D_{x}:=\partial_x-\frac{i\theta}{L},\tag{36}$ and the covariant Laplacian, or gauged kinetic operator, is $H_\theta:=-D_{x}^{\,2} =-\left(\partial_x-\frac{i\theta}{L}\right)^{\!2}.\tag{37}$

We specify the operator precisely. Let $\mathcal{H}:=L^2(S^1)=L^2\bigl([0,L],dx\bigr) \quad\text{with periodic identification } x\sim x+L,\tag{38}$ with inner product $\langle\psi,\phi\rangle=\int_0^L\overline{\psi(x)}\phi(x)\,dx$ , and take the operator domain to be the periodic Sobolev space $\mathcal{D}(H_\theta):=H^2_{\mathrm{per}}(S^1) =\Bigl\{\psi\in L^2(S^1):\ \sum_{n\in\mathbb{Z}}(1+n^2)^2\,|\hat\psi(n)|^2<\infty\Bigr\},\tag{39}$ where $\hat\psi(n)$ are the Fourier coefficients of $\psi$ with respect to the orthonormal basis introduced below. Equivalently, $\mathcal D(H_\theta)$ consists of $\psi\in L^2$ whose distributional derivatives up to order two lie in $L^2$ and which, together with their first derivatives, satisfy the periodic boundary conditions $\psi(0)=\psi(L)$ , $\psi'(0)=\psi'(L)$ ; we write $H^2_{\mathrm{per}}(S^1)$ for this space. With this domain, $H_\theta$ is self-adjoint, and the cleanest justification is Fourier diagonalization rather than a perturbative argument. Let $\mathcal F\colon L^2(S^1)\to \ell^2(\mathbb{Z})$ , $(\mathcal F\psi)(n)=\hat\psi(n)=\langle\psi_n,\psi\rangle$ , be the unitary Fourier transform associated with the orthonormal basis $\{\psi_n\}_{n\in\mathbb{Z}}$ of (41) below. A direct computation, carried out in Proposition 10.1, gives $H_\theta\psi_n= \lambda_n(\theta)\psi_n$ with $\lambda_n(\theta)=\bigl((2\pi n-\theta)/L\bigr)^2$ . Hence $\mathcal F$ conjugates $H_\theta$ into the multiplication operator by the real sequence $\bigl(\lambda_n(\theta)\bigr)_{n\in\mathbb{Z}}$ on $\ell^2(\mathbb{Z})$ : $\bigl(\mathcal F H_\theta\psi\bigr)(n)=\lambda_n(\theta)\,\hat\psi(n), \qquad \mathcal F\bigl(\mathcal D(H_\theta)\bigr) =\Bigl\{\,\hat\psi\in\ell^2(\mathbb{Z}):\ \sum_{n\in\mathbb{Z}}\lambda_n(\theta)^2\,|\hat\psi(n)|^2<\infty\,\Bigr\}.\tag{40}$ Multiplication by a real sequence, taken on its maximal domain, is self-adjoint; and since $\lambda_n(\theta)\asymp (2\pi n/L)^2$ as $|n|\to\infty$ , the maximal domain in (40) coincides with $\mathcal F\bigl(H^2_{\mathrm{per}}(S^1)\bigr)$ , so $\mathcal D(H_\theta)=H^2_{\mathrm{per}}(S^1)$ as claimed. Because the multiplier $\lambda_n(\theta)\to+\infty$ , the operator has compact resolvent and purely discrete spectrum [6, 7]. No relative-boundedness estimate is needed, and none is claimed: the first-order term $\tfrac{2i\theta}{L}\partial_x$ is not bounded on $L^2(S^1)$ , and the diagonalization above avoids any appeal to such a statement.

The spectrum

Proposition 10.1 (Eigenvalues and multiplicities).

The functions $\psi_n(x)=\frac{1}{\sqrt{L}}\,e^{2\pi i n x/L}, \qquad n\in\mathbb{Z},\tag{41}$ form an orthonormal basis of $\mathcal{H}$ consisting of eigenvectors of $H_\theta$ , and $D_{x}\psi_n=\frac{i\,(2\pi n-\theta)}{L}\,\psi_n, \qquad\qquad H_\theta\psi_n=\lambda_n(\theta)\,\psi_n \quad\text{with}\quad \lambda_n(\theta)=\left(\frac{2\pi n-\theta}{L}\right)^{\!2}.\tag{42}$ Consequently the spectrum of $H_\theta$ is purely discrete and, counted with multiplicity, is the multiset $\operatorname{Spec}(H_\theta)=\Bigl\{\!\!\Bigl\{\,\lambda_n(\theta)=\Bigl(\tfrac{2\pi n-\theta}{L}\Bigr)^{2} \ :\ n\in\mathbb{Z}\,\Bigr\}\!\!\Bigr\},\tag{43}$ where $\{\!\{\cdot\}\!\}$ denotes a multiset: the eigenvalue $\lambda$ occurs in $\operatorname{Spec}(H_\theta)$ with multiplicity $m_\theta(\lambda)=\#\bigl\{\, n\in\mathbb{Z}: \lambda_n(\theta)=\lambda \,\bigr\}.\tag{44}$

Proof. Orthonormality and completeness of $\{\psi_n\}$ in $L^2(S^1)$ is the classical Fourier basis theorem. Each $\psi_n$ is smooth and periodic, hence lies in $H^2(S^1)=\mathcal D(H_\theta)$ . Compute $D_{x}\psi_n=\partial_x\psi_n-\frac{i\theta}{L}\psi_n =\frac{2\pi i n}{L}\psi_n-\frac{i\theta}{L}\psi_n =\frac{i(2\pi n-\theta)}{L}\,\psi_n,$ which is the first identity in (42). Applying $D_{x}$ once more, $D_{x}^{\,2}\psi_n=\frac{i(2\pi n-\theta)}{L}\,D_{x}\psi_n =\left(\frac{i(2\pi n-\theta)}{L}\right)^{\!2}\psi_n =-\left(\frac{2\pi n-\theta}{L}\right)^{\!2}\psi_n,$ so $H_\theta\psi_n=-D_{x}^2\psi_n=\bigl(\tfrac{2\pi n-\theta}{L}\bigr)^2\psi_n$ , which is the second identity. Since $\{\psi_n\}_{n\in\mathbb{Z}}$ is a complete orthonormal system of eigenvectors with real eigenvalues $\lambda_n(\theta)\to+\infty$ as $|n|\to\infty$ , the Fourier transform $\mathcal F$ carries $H_\theta$ with domain (39) onto multiplication by $(\lambda_n(\theta))_{n\in\mathbb{Z}}$ on its maximal domain, as in (40). That operator is self-adjoint with compact resolvent, hence so is $H_\theta$ , and its spectrum, counted with multiplicity, is the multiset (43) with multiplicities (44).

Finally we check that (43) is well defined on the reduced source space, i.e. invariant under $\theta\mapsto\theta+2\pi k$ : the substitution $n\mapsto n+k$ gives $\lambda_{n+k}(\theta+2\pi k)=\bigl(\tfrac{2\pi(n+k)-\theta-2\pi k}{L}\bigr)^2=\lambda_n(\theta)$ , and $n\mapsto n+k$ is a bijection of $\mathbb{Z}$ , so the multiset (43) is unchanged, with all multiplicities preserved. Thus $\operatorname{Spec}(H_\theta)$ depends only on $\theta$ mod $2\pi$ , i.e. only on the point of $S_{\mathrm{flat}}$ . ◻

Remark 10.2 (Multiplicities are part of the answer).

We record (43) as a multiset, not as a subset of $\mathbb{R}$ , because the multiplicity data is physically real and is not recoverable from the underlying set. For generic $\theta$ , distinct $n$ give distinct $\lambda_n(\theta)$ and every eigenvalue is simple; but at $\theta=0$ the eigenvalues $\lambda_n$ and $\lambda_{-n}$ coincide for all $n$ , so every nonzero eigenvalue has multiplicity two while $\lambda_0=0$ is simple; and at $\theta=\pi$ one finds $\lambda_n(\pi)=\bigl(\tfrac{2\pi n-\pi}{L}\bigr)^2=\bigl(\tfrac{\pi(2n-1)}{L}\bigr)^2$ , so $\lambda_n(\pi)=\lambda_{1-n}(\pi)$ and every eigenvalue has multiplicity two. Discarding multiplicities would erase this structure and, in particular, would change the target: “the set of eigenvalues” and “the spectrum with multiplicities” are different correct-answer maps and may have different completions. We take the latter throughout.

The spectral target and the curvature obstruction

Definition 10.3 (Spectral target).

Let $A_{\mathrm{spec}}$ be the set of multisets of nonnegative reals arising as spectra of the operators $H_\theta$ , and define the correct-answer map $a_{\mathrm{spec}}\colon S_{\mathrm{flat}}\longrightarrow A_{\mathrm{spec}}, \qquad a_{\mathrm{spec}}(\theta):=\operatorname{Spec}(H_\theta)\tag{45}$ with $\operatorname{Spec}(H_\theta)$ as in (43). By the last paragraph of the proof of Proposition 10.1, $a_{\mathrm{spec}}$ is well defined on $S_{\mathrm{flat}}$ , i.e. on $\theta$ mod $2\pi$ . We denote this target by $Q_{\mathrm{spec}}$ .

Theorem 10.4 (The spectral target is obstructed by the curvature readout).

$Q_{\mathrm{spec}}$ is not exactly solvable from the curvature readout $p_F$ of (34). Explicitly, the reduced source states $\theta=0$ and $\theta=\pi$ satisfy $p_F(0)=p_F(\pi)=0, \qquad\text{but}\qquad \operatorname{Spec}(H_0)\neq\operatorname{Spec}(H_\pi).\tag{46}$ Indeed $0\in\operatorname{Spec}(H_0)$ while $0\notin\operatorname{Spec}(H_\pi)$ , and moreover $\min\operatorname{Spec}(H_0)=0<\pi^2/L^2=\min\operatorname{Spec}(H_\pi)$ .

Proof. First, $\theta=0$ and $\theta=\pi$ are distinct points of $S_{\mathrm{flat}}=\mathbb{R}/2\pi\mathbb{Z}$ , since $\pi\not\equiv 0 \pmod{2\pi}$ ; equivalently their holonomies $h(0)=1$ and $h(\pi)=-1$ are distinct elements of $\mathrm{U}(1)$ . They are therefore genuinely gauge-inequivalent physical configurations, not two labels for one state.

Both are flat: by (27), $F_0=F_\pi=0$ , so $p_F(0)=p_F(\pi)=0$ and the two states lie in the same (indeed the only) fiber of $p_F$ .

Now compute. By (42) with $\theta=0$ , $\lambda_n(0)=\left(\frac{2\pi n}{L}\right)^{\!2},$ and taking $n=0$ gives $\lambda_0(0)=0$ , so $0\in\operatorname{Spec}(H_0)$ ; the corresponding eigenvector is the constant function $\psi_0=L^{-1/2}$ , and indeed $D_{x}\psi_0=\partial_x\psi_0=0$ when $\theta=0$ .

By (42) with $\theta=\pi$ , $\lambda_n(\pi)=\left(\frac{2\pi n-\pi}{L}\right)^{\!2} =\frac{\pi^2}{L^2}\,(2n-1)^2 .$ Since $2n-1$ is odd for every $n\in\mathbb{Z}$ , we have $(2n-1)^2\geq 1$ , hence $\lambda_n(\pi)\geq\frac{\pi^2}{L^2}>0 \qquad\text{for every } n\in\mathbb{Z},\tag{47}$ with equality precisely when $2n-1=\pm1$ , i.e. $n\in\{0,1\}$ . Therefore $0\notin\operatorname{Spec}(H_\pi)$ and $\min\operatorname{Spec}(H_\pi)=\pi^2/L^2$ , attained with multiplicity two.

Since $0\in\operatorname{Spec}(H_0)$ and $0\notin\operatorname{Spec}(H_\pi)$ , the two multisets differ, so $a_{\mathrm{spec}}(0)\neq a_{\mathrm{spec}}(\pi)$ while $p_F(0)=p_F(\pi)$ . This is exactly the hypothesis (6) of Corollary 3.2, whence $Q_{\mathrm{spec}}$ is not exactly solvable from $p_F$ : no map $b\colon\{0\}\to A_{\mathrm{spec}}$ , that is, no single spectrum assigned to the curvature value $0$ , is correct for both $\theta=0$ and $\theta=\pi$ . ◻

Corollary 10.5 (Holonomy is target-active for the spectrum).

Let $r=h\colon S_{\mathrm{flat}}\to\mathrm{U}(1)$ , $h(\theta)=e^{i\theta}$ , be the holonomy relation, and let $\Phi_{\mathrm{spec}}\bigl(0,e^{i\theta}\bigr):=\operatorname{Spec}(H_\theta)$ , which is well defined by Proposition 10.1, so that $a_{\mathrm{spec}}=\Phi_{\mathrm{spec}}\circ(p_F,h)$ in the sense of (7). Then $h$ is active for $Q_{\mathrm{spec}}$ on the single fiber $F_0$ of $p_F$ .

Proof. Take $s_0=\theta=0$ and $s_1=\theta=\pi$ in Definition 4.2. Both lie in $F_0$ by (35), and by Theorem 10.4, $\Phi_{\mathrm{spec}}(0,h(0))=\operatorname{Spec}(H_0)\neq\operatorname{Spec}(H_\pi)=\Phi_{\mathrm{spec}}(0,h(\pi))$ , which is (8). ◻

The contrast with Example 9.2 is the whole point. The readout $p_F$ is the same in both cases, and the structural loss , the collapse of all of $\mathrm{U}(1)$ to a point , is the same. For the flatness target the loss is inactive and the readout is fully adequate. For the spectral target the same loss is active, and by Theorem 4.3 it obstructs. Whether completion is necessary is a fact about the pair, never about the readout alone.

Remark 10.6 (Physical reading).

The physical content of Theorem 10.4 is the Aharonov–Bohm phenomenon in spectral form [1, 11]. Along the circle the field strength vanishes , identically, and for dimensional reasons , so there is no local electromagnetic field anywhere on the particle’s accessible configuration space for it to feel. Nevertheless the charged particle’s energy levels depend on $\theta$ : the covariant derivative (36) twists the periodicity condition felt by the wavefunction, shifting the Fourier momenta from $2\pi n/L$ to $(2\pi n-\theta)/L$ . At $\theta=\pi$ the shift is maximal, the momentum lattice is displaced by half a spacing, no mode has zero covariant momentum, and the ground-state energy is lifted from $0$ to $\pi^2/L^2$ . The obstruction is not an artifact of the formalism; it is a measurable spectral shift produced by information that curvature does not record.