Classification of flat U(1) connections on an oriented circle

Proof. Step 1: holonomy is gauge invariant. Let $u\in C^\infty(S^1_L,U(1))$. Pull $u$ back to the universal cover: the composite $\tilde u:=u\circ\mathrm{pr}\colon[0,L]\to U(1)$, where $\mathrm{pr}(s)=s\bmod L$, lifts along the covering $\mathbb{R}\xrightarrow{\,t\mapsto e^{it}\,}U(1)$ to a smooth $\varphi\colon[0,L]\to\mathbb{R}$ with $u(x)=e^{i\varphi(x)}$ for $x\in[0,L]$, because $[0,L]$ is simply connected. Single-valuedness of $u$ on $S^1_L$ forces $$\varphi(L)-\varphi(0)\;=\;2\pi n, \qquad n=\deg(u)\in\mathbb{Z}.\tag{46}$$ Now $-i\,u^{-1}\mathrm{d}u = -i\,e^{-i\varphi}\cdot i\varphi'\,e^{i\varphi}\,\mathrm{d}x =\varphi'\,\mathrm{d}x=\mathrm{d}\varphi$, so the declared action (40) gives $$A^{u}\;=\;A-i\,u^{-1}\mathrm{d}u\;=\;A+\mathrm{d}\varphi, \qquad\text{i.e.}\qquad a^{u}=a+\varphi' .\tag{47}$$ Hence the period changes by $+2\pi n$: $$\int_0^L a^{u} =\int_0^L a+\bigl(\varphi(L)-\varphi(0)\bigr) =\int_0^L a+2\pi n ,$$ and therefore $$\operatorname{Hol}(A^{u}) =\exp\!\Bigl(i\int_0^L a + 2\pi i n\Bigr) =\exp\!\Bigl(i\int_0^L a\Bigr) =\operatorname{Hol}(A) .$$ The gauge group can shift the period integral, but only by integer multiples of $2\pi$, which $\exp(i\,\cdot\,)$ does not see. (The sign of the shift is $+2\pi n$ with the declared convention; either sign would of course leave the holonomy unchanged, but we record the correct one.)

Step 2: equal holonomy implies gauge equivalence. Suppose $\operatorname{Hol}(A)=\operatorname{Hol}(A')$. Then $\exp\bigl(i\int_0^L(a'-a)\bigr)=1$, so there is $n\in\mathbb{Z}$ with $$\int_0^L\bigl(a'(x)-a(x)\bigr)\,\mathrm{d}x \;=\; 2\pi n .\tag{48}$$ Define $$\chi(x)\;:=\;\int_0^x\bigl(a'(y)-a(y)\bigr)\,\mathrm{d}y, \qquad x\in[0,L],\tag{49}$$ so that $\chi$ is smooth with $$\chi'=a'-a \qquad\text{and}\qquad \chi(L)-\chi(0)=2\pi n\tag{50}$$ by (48). The function $\chi$ itself need not descend to $S^1_L$, but $$u(x)\;:=\;e^{\,i\chi(x)}\tag{51}$$ does: $u(L)=e^{i\chi(0)}e^{2\pi i n}=e^{i\chi(0)}=u(0)$, and all derivatives match at the seam because $a,a'$ are smooth and $L$-periodic, so $\chi'$ and all its derivatives are $L$-periodic. Hence $u\in C^\infty(S^1_L,U(1))$ is a bona fide gauge transformation, of degree $n$. By exactly the computation of Step 1 applied to this $u$ (with $\varphi=\chi$), $$-i\,u^{-1}\mathrm{d}u \;=\; \mathrm{d}\chi \;=\; (a'-a)\,\mathrm{d}x \;=\; A'-A ,\tag{52}$$ using (50). Therefore, directly from the declared action (40) in the form (47), $$A^{u}\;=\;A-i\,u^{-1}\mathrm{d}u\;=\;A+(A'-A)\;=\;A' .$$ So $u$ itself carries $A$ to $A'$, and the two connections are gauge equivalent.

Step 3: the bijection. By Steps 1 and 2, $\operatorname{Hol}$ descends to an injection $\mathcal{C}_{\mathrm{geo}}^{\,\circ}/\operatorname{Gau}(P)\hookrightarrow U(1)$. It is surjective: for $\vartheta\in\mathbb{R}$ the constant connection $$A_\vartheta:=\frac{\vartheta}{L}\,\mathrm{d}x \qquad\text{satisfies}\qquad \operatorname{Hol}(A_\vartheta)=e^{i\vartheta},\tag{53}$$ and every element of $U(1)$ is of the form $e^{i\vartheta}$. Hence (45) is a bijection.

Step 4: the homeomorphism. Give $\mathcal{C}_{\mathrm{geo}}^{\,\circ}\cong C^\infty(S^1_L,\mathbb{R})$ its usual Fréchet ($C^\infty$) topology and the quotient $\mathcal{M}:=\mathcal{C}_{\mathrm{geo}}^{\,\circ}/\operatorname{Gau}(P)$ the quotient topology, with projection $\mathrm{pr}_{\mathcal{M}}\colon\mathcal{C}_{\mathrm{geo}}^{\,\circ}\to\mathcal{M}$. Write $\Theta\colon\mathcal{M}\to U(1)$ for the bijection (45).

$\Theta$ is continuous: the map $\mathcal{C}_{\mathrm{geo}}^{\,\circ}\to U(1)$, $A\mapsto\exp(i\int_0^L a)$, is continuous (indeed it is continuous already in the $L^1$ topology, hence in $C^\infty$), and it is constant on $\operatorname{Gau}(P)$-orbits by Step 1; the universal property of the quotient topology then makes $\Theta$ continuous.

For the inverse, we do not attempt to choose a global continuous argument $\vartheta$ on $U(1)$, no such choice exists. Instead define $$\sigma\colon\mathbb{R}\longrightarrow\mathcal{M}, \qquad \sigma(\vartheta):=\mathrm{pr}_{\mathcal{M}}\bigl(A_\vartheta\bigr) =\bigl[\tfrac{\vartheta}{L}\,\mathrm{d}x\bigr] .\tag{54}$$ The map $\mathbb{R}\to\mathcal{C}_{\mathrm{geo}}^{\,\circ}$, $\vartheta\mapsto A_\vartheta$, is continuous (it is linear into the constants, and the $C^\infty$ seminorms of $A_\vartheta$ depend continuously on $\vartheta$), so $\sigma$ is continuous as a composite with $\mathrm{pr}_{\mathcal{M}}$.

$\sigma$ is $2\pi$-periodic. Indeed $A_{\vartheta+2\pi}-A_\vartheta=\tfrac{2\pi}{L}\mathrm{d}x$, and by Step 2 (with $a=\vartheta/L$, $a'=(\vartheta+2\pi)/L$, so that $\int_0^L(a'-a)=2\pi$, i.e. $n=1$) the connections $A_\vartheta$ and $A_{\vartheta+2\pi}$ are gauge equivalent; explicitly the gauge transformation is $u(x)=e^{2\pi i x/L}$, of degree $1$. Hence $\sigma(\vartheta+2\pi)=\sigma(\vartheta)$ for all $\vartheta$.

Therefore $\sigma$ factors through the quotient $\mathbb{R}\to\mathbb{R}/2\pi\mathbb{Z}$, and since $\mathbb{R}/2\pi\mathbb{Z}\cong U(1)$ via $\vartheta\bmod 2\pi\mapsto e^{i\vartheta}$ (a homeomorphism), we obtain a continuous map $$\bar\sigma\colon U(1)\longrightarrow\mathcal{M}, \qquad \bar\sigma\bigl(e^{i\vartheta}\bigr)=\sigma(\vartheta) ,\tag{55}$$ well defined and continuous by the universal property of the quotient $\mathbb{R}\to\mathbb{R}/2\pi\mathbb{Z}$ (which is an open, hence quotient, map). Finally $\Theta\circ\bar\sigma=\mathrm{id}_{U(1)}$ by (53), and $\bar\sigma\circ\Theta=\mathrm{id}_{\mathcal{M}}$ because both sides are the identity on the classes $[A_\vartheta]$, which exhaust $\mathcal{M}$ by Step 3 together with the injectivity of $\Theta$. Hence $\bar\sigma=\Theta^{-1}$ is continuous, and $\Theta$ is a homeomorphism. ◻

Proof. $\operatorname{Im}(p_F,J_{\mathrm{hol}})=\{0\}\times\operatorname{Im}J_{\mathrm{hol}}=\{0\}\times U(1)\cong U(1)$, by the surjectivity established in Step 3 of Theorem 6.5. The family $\{p_F\}\cup\{J_{\mathrm{hol}}\}$ separates orbits by the injectivity established there, so Proposition 5.5 gives injectivity of $e_{\mathrm{hol}}$; surjectivity is Theorem 5.3(1). ◻

Proof. Let $f$ be an orientation-preserving diffeomorphism of $S^1_L$ and let $A$ be a connection. Since $f$ preserves the orientation, it fixes the fundamental generator $c=[\gamma_0]$ of $H_1(S^1_L;\mathbb{Z})$. Because $G=U(1)$ is abelian and every connection on $S^1_L$ is flat by (39), the holonomy of a connection around a loop depends on that loop only through its class in $H_1(S^1_L;\mathbb{Z})$; the pushforward $f_*A$ therefore has the same holonomy around $c$ as $A$ does: $$\operatorname{Hol}\bigl(f_*A\bigr)=\operatorname{Hol}(A) .\tag{59}$$ (Concretely, $\int_{S^1_L} f_*A=\int_{S^1_L}A$ by the change-of-variables formula for the orientation-preserving $f$, the integrand being a $1$-form.)

(1). By (59) and the sufficiency direction of Theorem 6.5, equality of holonomy already implies gauge equivalence; hence $f_*A$ is gauge equivalent to $A$ for every orientation-preserving $f$ and every $A$. The equivalence relation generated by the gauge group together with the orientation-preserving diffeomorphisms therefore coincides with gauge equivalence alone, so the quotient (37) is unchanged by adjoining them, and (58) holds.

(2). Let $A$ and $A'$ be connections related either by a gauge transformation or by the pushforward along an orientation-preserving diffeomorphism. In the second case part (1) supplies a gauge transformation achieving the same result, so in either case there is $u\in\operatorname{Gau}(P)$ with $A'=A^{u}$. The operator $\mathcal{D}$ is defined with respect to the fixed background metric and measure of Convention 6.1, which are not transformed; the diffeomorphism therefore plays no further role, and only the gauge covariance established in Convention 6.3 is needed. There we verified $D_{A^{u}}\phi^{u}=u\,D_A\phi$ for $\phi^{u}=u\phi$; iterating, $$\mathcal{D}_{A^{u}}\bigl(u\,\phi\bigr) =u\,\mathcal{D}_{A}\phi \qquad\text{for all }\phi\in C^\infty(S^1_L,\mathbb{C}).$$ Thus the multiplication operator $\mathcal{U}_u\phi:=u\phi$ intertwines $\mathcal{D}_A$ with $\mathcal{D}_{A^{u}}$. Since $|u(x)|=1$ pointwise, $\mathcal{U}_u$ is an isometry of $L^2(S^1_L,\mathrm{d}x)$ with inverse $\mathcal{U}_{u^{-1}}$, hence unitary; and it preserves $C^\infty(S^1_L,\mathbb{C})$, hence carries the domain of $\mathcal{D}_A$ onto that of $\mathcal{D}_{A^{u}}$. Therefore $\mathcal{D}_{A}$ and $\mathcal{D}_{A'}=\mathcal{D}_{A^{u}}$ are unitarily equivalent, and their spectral multisets coincide. Since $A$ and $A'$ were an arbitrary pair related by a gauge transformation or by an allowed diffeomorphism, the spectral multiset is a well-defined function on $S_{\mathrm{geo}}$, which is the assertion. ◻

Proof. The spectrum. By Theorem 6.5 every class has the representative $A_\vartheta=(\vartheta/L)\,\mathrm{d}x$ of (53), and by Example 4.3(ii) the spectrum is unchanged along the gauge orbit, so we may compute with $A_\vartheta$. With $a\equiv\vartheta/L$ constant, $\mathcal{D}_{A_\vartheta}$ has the complete orthonormal system of eigenfunctions $\psi_k(x)=L^{-1/2}e^{2\pi i k x/L}$, $k\in\mathbb{Z}$ (complete in $L^2(S^1_L,\mathrm{d}x)$ by Fourier theory). By the convention (42), $$\Bigl(\partial_x-i\tfrac{\vartheta}{L}\Bigr)\psi_k =\Bigl(\tfrac{2\pi i k}{L}-\tfrac{i\vartheta}{L}\Bigr)\psi_k =i\,\frac{2\pi k-\vartheta}{L}\,\psi_k ,$$ and hence, squaring and inserting the overall minus sign of (60), $$\mathcal{D}_{A_\vartheta}\psi_k =-\left(i\,\frac{2\pi k-\vartheta}{L}\right)^{\!2}\psi_k =\left(\frac{2\pi k-\vartheta}{L}\right)^{\!2}\psi_k .$$ Since the $\psi_k$ are complete, the spectrum is purely discrete and is exactly the multiset (61). Note that $\vartheta$ is defined only modulo $2\pi$, and (61) is invariant under $\vartheta\mapsto\vartheta+2\pi$ via the relabelling $k\mapsto k+1$; the multiset is therefore a well-defined function of $[A]$, as it must be.

“If”: inversion-related holonomies are spectrally equal. Replacing $\vartheta$ by $-\vartheta$ in (61) and relabelling $k\mapsto -k$ carries the multiset to itself, since $(2\pi(-k)-(-\vartheta))^2=(-(2\pi k-\vartheta))^2=(2\pi k-\vartheta)^2$. Hence $[A_\vartheta]$ and $[A_{-\vartheta}]$, which by Theorem 6.5 are distinct classes whenever $e^{i\vartheta}\neq e^{-i\vartheta}$, i.e. whenever $\vartheta\not\equiv0,\pi\pmod{2\pi}$, have the same spectrum.

“Only if”: equal spectra force inversion-related holonomies. We prove this directly rather than by citation. Suppose $\operatorname{spec}\mathcal{D}_{A_\vartheta} =\operatorname{spec}\mathcal{D}_{A_{\vartheta'}}$ as multisets. Normalize $\vartheta,\vartheta'\in[0,2\pi)$, which is permissible by the periodicity just noted. By (61) every eigenvalue of $\mathcal{D}_{A_\vartheta}$ is a nonnegative real number, so we may form the set of both real square roots of the eigenvalues, rescaled by $L/2\pi$: $$\begin{aligned} \Sigma(\vartheta) &:=\Bigl\{\pm\tfrac{L}{2\pi}\sqrt{\lambda}\;:\; \lambda\in\operatorname{spec}\mathcal{D}_{A_\vartheta}\Bigr\} \\ &\phantom{:}= \Bigl\{k-\tfrac{\vartheta}{2\pi}\;:\;k\in\mathbb{Z}\Bigr\} \;\cup\; \Bigl\{-k+\tfrac{\vartheta}{2\pi}\;:\;k\in\mathbb{Z}\Bigr\} \\ &\phantom{:}= \Bigl(-\tfrac{\vartheta}{2\pi}+\mathbb{Z}\Bigr) \;\cup\; \Bigl(\tfrac{\vartheta}{2\pi}+\mathbb{Z}\Bigr), \end{aligned}$$ which is determined by the spectral multiset alone. Thus equality of spectra gives $$\Bigl(\tfrac{\vartheta}{2\pi}+\mathbb{Z}\Bigr)\cup\Bigl(-\tfrac{\vartheta}{2\pi}+\mathbb{Z}\Bigr) \;=\; \Bigl(\tfrac{\vartheta'}{2\pi}+\mathbb{Z}\Bigr)\cup\Bigl(-\tfrac{\vartheta'}{2\pi}+\mathbb{Z}\Bigr)$$ as subsets of $\mathbb{R}$. Intersecting both sides with the interval $[0,1)$ and comparing, the unordered pair of residues $\{\tfrac{\vartheta}{2\pi}\bmod 1,\;-\tfrac{\vartheta}{2\pi}\bmod 1\}$ equals the pair $\{\tfrac{\vartheta'}{2\pi}\bmod 1,\;-\tfrac{\vartheta'}{2\pi}\bmod 1\}$. Hence $\vartheta'\equiv\vartheta$ or $\vartheta'\equiv-\vartheta \pmod{2\pi}$, that is, $\operatorname{Hol}(A')=\operatorname{Hol}(A)^{\pm1}$, as claimed.

The completion. By the two directions just proved, the relation $J_{\mathrm{spec}}$ has exactly the level sets of the map $U(1)\to U(1)/(z\sim z^{-1})$. Since the readout $p_F$ is constant and contributes nothing, Definition 5.2 gives $E_{\mathrm{spec}}=\operatorname{Im}(p_F,J_{\mathrm{spec}})\cong U(1)/(z\sim z^{-1})$. The map $z=e^{i\vartheta}\mapsto\tfrac12(z+z^{-1})=\cos\vartheta$ is a continuous bijection from this quotient onto $[-1,1]$ (it is well defined and injective on the quotient, and surjective onto $[-1,1]$), and it is a homeomorphism since the quotient is compact and $[-1,1]$ is Hausdorff. This gives (62). Non-injectivity of the completion map and the description of its fibres now follow from Proposition 5.5 applied to the coarser relation family $\{p_F\}\cup\{J_{\mathrm{spec}}\}$. ◻