The SU(2) bridge midpoint in closed form: the curvature term is exact

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Result, 2026-09-27. Proposition 6 of the series/parallel note described the midpoint of the Brownian bridge on \(SU(2)\), the building block of the non-abelian parallel insertion, only to leading order in the heat time, with the curvature factor \(1/h\), \(h=(d/4)\cot(d/4)\), derived by Laplace’s method and labelled as asymptotics. For the fundamental character the midpoint expectation is computable exactly. For a bridge of duration \(t\) from \(e\) to a group element at geodesic distance \(\theta\in(0,2\pi)\),

\[E\,\chi_{1/2}(m)=e^{-t/32}\Bigl[2\cos\frac\theta4-\frac t{2\theta}\, \sin\frac\theta4\Bigr]\ \ \text{up to image terms of relative size}\ \Bigl(2+\frac{32\pi^2}t\Bigr)e^{-4\pi(2\pi-\theta)/t}\]

(Theorem 1). Expanded to first order it is exactly the prediction of Proposition 6(b), \(2\cos\frac\theta4\,[1-\frac t{32}(1+\frac2{h(\theta)})]\), because \(\frac t{32}\cdot\frac2h=\frac t{4\theta}\tan\frac\theta4\). So the curvature term N2, the side-flux-dependent softening of the mid-face weight, is confirmed with no remainder of order \(t^2\) in the \(w=0\) sector: the only correction to the displayed formula is the exponentially small winding of the bridge around the group. This is the first exact non-abelian statement about the parallel insertion in three dimensions, and it settles the fundamental-representation part of the uniform Laplace remainder that Proposition 7 of the series/parallel note assumes, for the trace in the fundamental representation (the image terms are of the stated order, and not uniformly so as \(\theta\to2\pi\)).

The ingredients are character orthogonality and Poisson summation; the computation is elementary and no novelty is claimed for it.

1. Setting

Group metric as in the series/parallel note §1: \(C_2(j)=j(j+1)\), \(SU(2)\) the three-sphere of radius 2, heat kernel \(k_t=\sum_j(2j+1)e^{-tj(j+1)/2}\chi_j\) with \(\chi_j(\theta)=\sin((2j+1)\theta/2)/\sin(\theta/2)\) at rotation angle \(\theta\) (the geodesic distance from \(e\)). The midpoint \(m\) of the bridge of duration \(t\) from \(e\) to \(g\) has density \(k_{t/2}(m)\,k_{t/2}(m^{-1}g)/k_t(g)\) with respect to Haar measure. Its geodesic midpoint \(m_*=\exp(\frac12\log g)\) has rotation angle \(\theta/2\), so \(\chi_{1/2}(m_*)=\sin(\theta/2)/\sin(\theta/4)=2\cos(\theta/4)\).

2. The exact formula

Theorem 1. Put \(\Theta_\pm(\theta)=\sum_{w\in\mathbb Z}(\pm1)^w e^{-(\theta-4\pi w)^2/(2t)}\) and \(\Xi_\pm(\theta)=\sum_{w\in\mathbb Z}(\pm1)^w(\theta-4\pi w) e^{-(\theta-4\pi w)^2/(2t)}\). For \(\theta\in(0,2\pi)\),

\[E\,\chi_{1/2}(m)=e^{-t/32}\, \frac{2\cos(\theta/4)\,\Xi_-(\theta)-\frac t2\sin(\theta/4)\,\Theta_-(\theta)} {\Xi_+(\theta)}. \tag{1}\]

Keeping only \(w=0\) gives the displayed formula of the summary.

Proof. Expand both heat kernels in characters. Character orthogonality in the form \(\int\chi_a(m)\chi_b(m^{-1}g)\,dm=\delta_{ab}\chi_a(g)/d_a\), together with \(\chi_{j_1}\chi_{1/2}=\chi_{j_1+1/2}+\chi_{j_1-1/2}\), gives

\[k_t(g)\,E\,\chi_{1/2}(m)=\sum_{j_1}\ \sum_{j_2=j_1\pm1/2}(2j_1+1)\, e^{-t[C_2(j_1)+C_2(j_2)]/4}\,\chi_{j_2}(g).\]

Write \(n=2j_1+1\). Then \(C_2(j_1)+C_2(j_2)=\frac14[n^2+(n\pm1)^2-2] =\frac12(n\pm\frac12)^2-\frac38\), so the exponential is \(e^{3t/32}e^{-t(n\pm1/2)^2/8}\). Multiply by \(\sin(\theta/2)\), reindex the \(j_2=j_1-\frac12\) sum by \(n\mapsto n+1\), and write \(s=n+\frac12\) over \(s\in\frac12+\mathbb N\): the summand becomes \((s-\frac12)\sin\frac{(s+\frac12)\theta}2+(s+\frac12)\sin\frac{(s-\frac12)\theta}2 =2s\cos\frac\theta4\sin\frac{s\theta}2-\sin\frac\theta4\cos\frac{s\theta}2\), which is even in \(s\). Hence

\[k_t(g)\sin\frac\theta2\;E\,\chi_{1/2}(m)=\frac{e^{3t/32}}2 \Bigl[2\cos\frac\theta4\,A-\sin\frac\theta4\,B\Bigr],\]

with \(B=\sum_{s\in\frac12+\mathbb Z}e^{-ts^2/8}\cos\frac{s\theta}2\) and \(A=\sum_{s\in\frac12+\mathbb Z}s\,e^{-ts^2/8}\sin\frac{s\theta}2=-2\partial_\theta B\). In the same way \(k_t(g)\sin\frac\theta2=\frac{e^{t/8}}2A_0\) with \(A_0=\sum_{n\in\mathbb Z}n\,e^{-tn^2/8}\sin\frac{n\theta}2\). Poisson summation over the half-integers (which introduces the sign \((-1)^w\)) and over the integers gives \(B=\sqrt{8\pi/t}\,\Theta_-\), \(A=(2/t)\sqrt{8\pi/t}\,\Xi_-\) and \(A_0=(2/t)\sqrt{8\pi/t}\,\Xi_+\). Divide. \(\square\)

The image terms. For \(\theta\in(0,2\pi)\) the pair \(w=\pm1\) in \(\Xi_\pm\) has, relative to the \(w=0\) term \(\theta e^{-\theta^2/(2t)}\), the size \(e^{-8\pi^2/t}\,|2\theta\cosh(4\pi\theta/t)-8\pi\sinh(4\pi\theta/t)|/\theta \le(2+32\pi^2/t)\,e^{-4\pi(2\pi-\theta)/t}\), using \(\sinh x\le x\cosh x\). In \(\Theta_-\) the same pair has relative size at most \(2e^{-4\pi(2\pi-\theta)/t}\). The pairs \(|w|\ge2\) are smaller by a further factor \(e^{-16\pi^2/t}\) up to polynomial factors. This is the large-field term N4, the winding of the bridge around the group, and it is negligible uniformly on \(\theta\le\theta_0<2\pi\).

3. Comparison with Proposition 6

Proposition 6(b) predicts that \(m=m_*e^\xi\) with \(\xi\) Gaussian of mean zero and covariance \(\frac t4{\rm diag}(1,\frac1h,\frac1h)\) in the frame (axis, transverse). In the spin-\(\frac12\) representation \((\xi\cdot T)^2=-\frac14|\xi|^2\), so \(E\,D^{1/2}(e^\xi)=1-\frac18E|\xi|^2+O(t^2)=1-\frac t{32}(1+\frac2h)+O(t^2)\), and \(E\,\chi_{1/2}(m)=2\cos\frac\theta4\,[1-\frac t{32}(1+\frac2h)]+O(t^2)\). The \(w=0\) part of (1) equals \(2\cos\frac\theta4\,e^{-t/32}[1-\frac t{4\theta}\tan\frac\theta4]\), and \(\frac t{32}\cdot\frac2{h(\theta)}=\frac t{16}\cdot\frac{\tan(\theta/4)}{\theta/4} =\frac t{4\theta}\tan\frac\theta4\). The two agree at first order, curvature factor included. As \(\theta\to0\) both give \(2(1-\frac{3t}{32})\), the flat value \(E|\xi|^2=\frac{3t}4\).

Formula (1) says more than the asymptotics. In the \(w=0\) sector the fundamental character of the midpoint is the product of an isotropic factor \(e^{-t/32}\) and the exactly linear curvature correction \(1-\frac t{4\theta}\tan\frac\theta4\); there are no higher-order terms in \(t\). The correction grows as \(\theta\to2\pi\), where \(\tan(\theta/4)\) diverges and the midpoint becomes bimodal, and there the image terms take over. This is the non-abelian counterpart of the parity factor \(\rho_t(\phi)\) of the \(U(1)\) cube (Proposition 3 of the series/parallel note).

4. Every spin

Theorem 2. For every spin \(J\) and \(\theta\in(0,2\pi)\), with \(\sigma=0\) for integer \(J\) and \(\sigma=\frac12\) for half-integer \(J\),

\[E\,\chi_J(m)=\frac{\Xi_{(\sigma)}(\theta)}{\Xi_+(\theta)} \sum_{k=-J}^{J}e^{-tk^2/8}\cos\frac{k\theta}2 -\frac t2\,\frac{\Theta_{(\sigma)}(\theta)}{\Xi_+(\theta)} \sum_{k=-J}^{J}k\,e^{-tk^2/8}\sin\frac{k\theta}2,\]

where \(\Xi_{(0)}=\Xi_+\), \(\Theta_{(0)}=\Theta_+\), \(\Xi_{(1/2)}=\Xi_-\), \(\Theta_{(1/2)}=\Theta_-\), and \(k\) runs in unit steps. In the \(w=0\) sector,

\[E\,\chi_J(m)=\sum_{k=-J}^{J}e^{-tk^2/8}\Bigl[\cos\frac{k\theta}2 -\frac t{2\theta}\,k\sin\frac{k\theta}2\Bigr]. \tag{2}\]

For integer \(J\) the first ratio is exactly one, image terms included.

Proof. As for Theorem 1, now with \(\chi_{j_1}\chi_J=\sum_{k=-J}^{J}\chi_{j_1+k}\). This identity holds for every \(j_1\) if \(\chi_j\) is defined by \(\sin((2j+1)\theta/2)/\sin(\theta/2)\) also for \(2j+1\le0\): it is the product-to-sum formula for \(\sin(n\theta/2)\sum_ke^{ik\theta}\). The formal terms with \(2j_2+1\le0\) cancel in pairs carrying the same weight, so the sum may be taken over all \(n=2j_1+1\in\mathbb Z\), halved; the summand is even under \((n,k)\mapsto(-n,-k)\). With \(n_2=n+2k\), \(C_2(j_1)+C_2(j_2)=\frac12(n+k)^2+\frac12(k^2-1)\), so the weight is \(e^{t/8}e^{-tk^2/8}e^{-ts^2/8}\) with \(s=n+k\in\mathbb Z+\sigma\). Expanding \((s-k)\sin\frac{(s+k)\theta}2\) and discarding the terms odd in \(s\) leaves \(\cos\frac{k\theta}2\cdot s\sin\frac{s\theta}2-k\sin\frac{k\theta}2\cos\frac{s\theta}2\). Summing over \(s\) gives \(A_\sigma\) and \(B_\sigma\), Poisson summation gives \(\Xi_{(\sigma)}\) and \(\Theta_{(\sigma)}\), and the factor \(e^{t/8}\) cancels against the denominator. For \(J=\frac12\) this is Theorem 1. \(\square\)

Agreement with Proposition 6 for every spin. The leading term of (2) is \(\sum_k\cos(k\theta/2)=\chi_J(m_*)\), the character at the geodesic midpoint (rotation angle \(\theta/2\)). The Gaussian prediction of Proposition 6(b) at first order is \(\sum_k\cos\frac{k\theta}2\,[1-\frac t8k^2-\frac t8(J(J+1)-k^2)/h]\) with \(1/h=4\tan(\theta/4)/\theta\). The two agree iff, with \(\varphi=\theta/2\),

\[\sum_kk\sin(k\varphi)=\tan\frac\varphi2\sum_k\bigl(J(J+1)-k^2\bigr)\cos(k\varphi).\]

The left side is \(-\chi_J'(\varphi)\) and the sum on the right is \(J(J+1)\chi_J+\chi_J''\). The class-function Laplacian of \(SU(2)\) in this metric gives \(\chi_J''+\cot(\varphi/2)\,\chi_J'=-J(J+1)\chi_J\), which is the identity. (For \(J=\frac12\) and \(J=1\) it can be checked directly: \(\sin(\varphi/2)=\tan(\varphi/2)\cos(\varphi/2)\) and \(2\sin\varphi=\tan(\varphi/2)(2+2\cos\varphi)\).) So the curvature term of Proposition 6 is exact at first order in every representation, and (2) contains no terms of order \(t^2\) beyond the Gaussian factors \(e^{-tk^2/8}\).

What the trace does not give. By Proposition 6(a), \(E\,D^J(m)=D^J(m_*)\Lambda^J\) with \(\Lambda^J\) real diagonal in the axis basis, and (2) fixes only \(\sum_\mu e^{i\mu\theta/2}\Lambda^J_{\mu\mu}\), one equation for \(2J+1\) entries. The isolated cube needs the entries, because its four cut faces have different axes.

5. The fundamental sector of the isolated cube is exact

Theorem 3. (a) In spin \(\frac12\) the midpoint’s matrix expectation is a real multiple of a unitary: \(E\,D^{1/2}(m)=\lambda(\theta,t)\,D^{1/2}(m_*)\) with

\[\lambda(\theta,t)=e^{-t/32}\,\frac{\Xi_-(\theta)-\frac t4\tan\frac\theta4\,\Theta_-(\theta)}{\Xi_+(\theta)} \ \overset{w=0}{=}\ e^{-t/32}\Bigl[1-\frac t{4\theta}\tan\frac\theta4\Bigr], \qquad\theta\in(0,2\pi).\]

2. For one refined cube (series/parallel note, §4), whose four cut side faces have heat times \(t_j\) and side angles \(\theta_j\), the spin-\(\frac12\) term of the character sum for \(\Psi\) is exactly

\[d_{1/2}\,e^{-3t_m/8}\,{\rm tr}_{1/2}\prod_jE\,D^{1/2}(m_j^{\pm1}) =2\,e^{-3t_m/8}\Bigl(\prod_{j=1}^4\lambda(\theta_j,t_j)\Bigr)\chi_{1/2}(h_{\rm int}),\]

where \(h_{\rm int}\) is the holonomy of the geodesic interpolation around the mid-face. In this sector the interpolation term N1 is exact (it is \(h_{\rm int}\)), the curvature term N2 is exact (it is \(\prod\lambda\)), and the commutator term N3 vanishes.

Proof. (a) By Proposition 6(a) of the series/parallel note, \(E\,D^{1/2}(m)=D^{1/2}(m_*)\Lambda\) with \(\Lambda={\rm diag}(\Lambda_+,\Lambda_-)\) real in the axis basis, where \(D^{1/2}(m_*)={\rm diag}(e^{i\theta/4},e^{-i\theta/4})\). Its trace \(e^{i\theta/4}\Lambda_++e^{-i\theta/4}\Lambda_-\) equals \(E\,\chi_{1/2}(m)\), which is real because the midpoint density and \(\chi_{1/2}\) are real. The imaginary part gives \(\sin(\theta/4)(\Lambda_+-\Lambda_-)=0\), so \(\Lambda_+=\Lambda_-=\lambda\), and \(\lambda=E\,\chi_{1/2}(m)/(2\cos(\theta/4))\) by Theorem 1. (b) The bridge for a cut face runs from \(Q_j^{-1}\) to \(P_j\); left translation by \(Q_j^{-1}\) maps it to a bridge from \(e\), so \(E\,D^{1/2}(m_j)=\lambda_jD^{1/2}(m_{*j})\) with \(m_{*j}\) the geodesic midpoint, and \(E\,D^{1/2}(m_j^{-1})=\lambda_jD^{1/2}(m_{*j})^{-1}\) because \(\lambda_j\) is real. The four midpoints are independent, since each is fixed by its own cut face and an isolated cube has no shared mid-edges, so the trace of the product of expectations is \(\prod_j\lambda_j\) times the trace of the product of the unitaries, which is \(\chi_{1/2}(h_{\rm int})\). The prefactor is \(d_{1/2}e^{-t_mC_2(1/2)/2}\) with \(C_2(\frac12)=\frac34\). \(\square\)

For \(J\ge1\) the diagonal entries differ: the Gaussian prediction is \(\Lambda^J_{\mu\mu}\simeq1-\frac t8[\mu^2+(J(J+1)-\mu^2)/h]\), which depends on \(\mu\) unless \(h=1\). The Weyl reflection combined with time reversal gives \(\Lambda^J_{\mu\mu}=\Lambda^J_{-\mu,-\mu}\) for every \(J\): with \(w\) in the normaliser of the torus, \(\operatorname{Ad}(w)X=-X\), the map \(g\mapsto e^Xwgw^{-1}\) carries the bridge from \(e\) to \(e^X\) to the bridge from \(e^X\) to \(e\), whose midpoint law is the same by reversibility, and conjugation by \(D(w)\) exchanges \(\mu\) and \(-\mu\) on a diagonal matrix (reviewer’s argument; the reality of the trace alone gives only one relation). The trace (2) then fixes \(\lfloor J\rfloor+1\) unknowns by one equation, so for \(J=1\) one relation remains between \(\Lambda^1_{11}\) and \(\Lambda^1_{00}\). The matrix-element version of the orthogonality computation of Theorem 1, with the Clebsch–Gordan coefficients of \(j_1\otimes J\), is the tool for them.

6. The spin-1 matrix and isolated cube

Theorem 4 (2026-09-27, GPT-6 Astra; refereed by Fable the same day, all fifteen items verified by hand, no errors). Notation of this section: \(E\) in the image bounds, \(A_0\), \(A_1\), \(B(x)\) and \(P_j\) are local to Theorem 4 and differ from the quantities of the same names in Theorems 1 and 3. The spin-1 midpoint matrix is exactly determined by two real entries \(\lambda_0=\Lambda^1_{00}\) and \(\lambda_1=\Lambda^1_{11}=\Lambda^1_{-1,-1}\). The formulas below give both entries, including winding, for every \(t>0\) and \(0<\theta<2\pi\). They involve a Gaussian tail, which retains higher-order curvature corrections beyond the character formula.

Put \(x=\theta/2\), \(\eta=e^{t/8}\), and define

\[G_t(z)=e^{-2z^2/t},\qquad T_t(z)=\int_z^\infty(1-\eta\cos u)e^{-2u^2/t}\,du, \qquad S_t(x)=\sum_{w\in\mathbb Z}T_t(x-2\pi w). \tag{3}\]

The Gaussian integral of \(1-\eta\cos u\) over the real line is zero; thus \(T_t\) is odd and decays at both ends. In particular the image sum in (3) is absolutely convergent. A special-function closed form is

\[T_t(z)=\sqrt{\frac{\pi t}{8}}\left[ \operatorname{erfc}\!\left(\sqrt{\frac2t}\,z\right) -\operatorname{Re}\operatorname{erfc}\!\left( \sqrt{\frac2t}\,z-i\sqrt{\frac t8}\right)\right]. \tag{4}\]

With \(\Theta=\Theta_+(\theta)\) and \(\Xi=\Xi_+(\theta)\) of Theorem 1,

\[\boxed{\lambda_0=1+\frac{t\eta^{-1}}{\sin x\,\Xi} \left[(\eta\cos x-1)\Theta-\cot x\,S_t(x)\right],} \tag{5}\]

\[\boxed{\lambda_1=\eta^{-1}\left\{1-\frac{t}{2\sin x\,\Xi} \left[(\eta-\cos x)\Theta-\frac{S_t(x)}{\sin x}\right]\right\}.} \tag{6}\]

These expressions are regular at \(\theta=\pi\). The endpoint \(\theta=0\) is obtained by continuity and has \(\lambda_0=\lambda_1\) by conjugation invariance. The denominator \(\Xi\) is positive on the stated open interval, since it is proportional to \(k_t(g)\sin x\).

Matrix-element computation. Write \(c_{a,b}(\alpha,\mu)=|\langle a\alpha;1\mu\mid b,\alpha+\mu\rangle|^2\). The spin-1 Clebsch–Gordan squares are

Here \(c_{a,b}(\alpha,-1)=c_{a,b}(-\alpha,1)\), and only actual summands of \(a\otimes1\) are included: \(a=0\) has just \(b=1\), and \(a=1/2\) has \(b=1/2,3/2\). These squares follow by lowering the highest vector of spin \(a+1\) and taking the orthogonal spin-\(a\) and spin-\((a-1)\) vectors at each total weight. The lowering rule used is \(J_-|a,\alpha\rangle=\sqrt{(a+\alpha)(a-\alpha+1)} |a,\alpha-1\rangle\); normalization gives the denominators in the table.

Matrix-element orthogonality gives, with \(C_a=a(a+1)\),

\[k_t(g)E[D^1(m)]_{\mu\mu} =\sum_{a,b}(2a+1)e^{-t(C_a+C_b)/4} \sum_{\alpha=-a}^{a}c_{a,b}(\alpha,\mu)e^{i(\alpha+\mu)\theta}. \tag{7}\]

The factor \(2b+1\) in the second heat kernel cancels its orthogonality denominator. For \(\mu=0\), put \(N_0=k_t(g)\lambda_0\) and

\[B(x)=\sum_{n\in\mathbb Z}e^{-tn^2/8}\cos(nx),\qquad H(x)=\sum_{n\ne0}e^{-tn^2/8}\frac{\sin(nx)}n, \qquad H(x)=\int_0^x(B(y)-1)\,dy.\]

We compute the \(b=a\pm1\) and \(b=a\) parts of (7) separately. In the first two parts multiplication by \(\alpha^2\) becomes \(-\partial_\theta^2\). Write \(n=2a+1\) and shift to \(s=n\pm1\); the common heat weight is \(e^{-ts^2/8}\). Combining the two parts gives

\[\begin{aligned} N_{\rm off} &=\sum_{s\ge1}e^{-ts^2/8} \left(s+\frac1s\partial_x^2\right)[\cot x\sin(sx)]\\ &=\frac{\cot x\,H(x)-B(x)+1}{\sin^2x}. \end{aligned}\tag{8}\]

The formal lower-spin term at \(a=1/2\) vanishes because \((a^2+\partial_\theta^2)\chi_a=0\); the missing upper partner at \(s=1\) has \(\sin((s-1)x)=0\). This accounts for the endpoints of the reindexing. The cancellation in (8) follows directly from

\[\partial_x^2[\cot x\sin(sx)] =2\csc^2x\cot x\sin(sx)-2s\csc^2x\cos(sx) -s^2\cot x\sin(sx).\]

For the diagonal coupling \(b=a\), set \(f_n(x)=\sin(nx)/\sin x\). Its elementary differential identity is \(f_n''+2\cot x\,f_n'=-(n^2-1)f_n\). Hence

\[N_{\rm diag}=k_t(g)-1 +2\eta\cot x\left(\frac{V(x)}{\sin x}\right)',\qquad V(x)=\sum_{n\ge2}\frac{n e^{-tn^2/8}}{n^2-1}\sin(nx).\]

The omitted \(n=1\) term in the heat kernel is exactly 1. Moreover, \(V''+V=B'/2+\eta^{-1}\sin x\) and \(V(0)=0\), so the Wronskian identity gives

\[\left(\frac{V}{\sin x}\right)' =\frac1{\sin^2x}\int_0^x\sin y \left(\frac{B'(y)}2+\eta^{-1}\sin y\right)dy.\]

Adding (8), integrating \(\sin y B'(y)\) once, and using \(2\int_0^x\sin^2y\,dy=x-\sin x\cos x\) yields

\[N_0=k_t(g)+\frac1{\sin^2x}\left[ (\eta\cos x-1)B(x)+\cot x\int_0^x(1-\eta\cos y)B(y)\,dy\right]. \tag{9}\]

Every series above and its displayed derivatives converge absolutely for \(t>0\). Poisson summation, with the normalization of Theorem 1, gives

\[B(x)=\sqrt{\frac{8\pi}t}\sum_wG_t(x-2\pi w),\qquad k_t(g)=\frac{\eta}{t\sin x}\sqrt{\frac{8\pi}t}\,\Xi.\]

Periodicity of the cosine and oddness of \(T_t\) give

\[\int_0^x(1-\eta\cos y)B(y)\,dy =-\sqrt{\frac{8\pi}t}\,S_t(x):\]

the integration constants are \(\sum_wT_t(-2\pi w)=0\), paired in \(w\). Substitution into (9) proves (5).

Weyl reflection with time reversal, as proved in §5, gives \(\Lambda^1_{11}=\Lambda^1_{-1,-1}\). Theorem 2 therefore reads

\[\lambda_0+2\cos x\,\lambda_1 =1+2\eta^{-1}\cos x-t\eta^{-1}\sin x\frac{\Theta}{\Xi}. \tag{10}\]

Inserting (5), the numerator for \(2\cos x\,\lambda_1\) factors by \(\cos x\), giving (6). This first proves (6) when \(\cos x\ne0\); continuity proves it also at \(x=\pi/2\). Here (6) is obtained from (5) through (10), and (10) itself follows from (7) because \(\sum_\mu c_{a,b}(m-\mu,\mu)=1\); so the trace is a consistency of the framework. Reality and the equality of the opposite weights are explicit. \(\square\)

Independent check through the \(\mu=1\) column (Fable referee, 2026-09-27). Reducing (7) directly at \(\mu=1\), with total weight \(m\), \(p=2m\) and \(q=p-1\): the \(b=a\pm1\) parts combine to the summand \((s+p)(s+p-2)+(s-p)(s-p+2)=2s^2+2q^2-2\), where \(q\) runs over the weights of spin \((s-1)/2\), so the off-diagonal part is \(\sum_{s\ge1}(e^{-ts^2/8}/s)[(s^2-1)f_s+\cot x\,f_s']\) with \(f_s\) the character sum of spin \((s-1)/2\); the diagonal part has bracket \(2f_n'(i-\cot x)=-2f_n'e^{-ix}/\sin x\). Assembling with \(H'=B-1\) and the Wronskian of the diagonal step reproduces (6) exactly. Since (6) was derived from (5) and (10), this confirms (5) as well. (Before (9), the \(-1\) of the diagonal part is absorbed by \(1-\cos^2x=\sin^2x\).) At first order \(\lambda_0-\lambda_1=-\frac t8(\frac1h-1)<0\), since \(h<1\) on \(0<\theta<2\pi\).

The zero image and explicit image bounds. Define \(R_t(x)=e^{2x^2/t}T_t(x)\) and

\[A_0=\eta\cos x-1-\cot x\,R_t(x),\qquad A_1=\eta-\cos x-\frac{R_t(x)}{\sin x}.\]

Keeping \(w=0\) in both numerator and denominator of (5)–(6) gives

\[\lambda_0^{(0)}=1+\frac{t\eta^{-1}}{\theta\sin x}A_0, \qquad \lambda_1^{(0)}=\eta^{-1}\left[1-\frac{t}{2\theta\sin x}A_1\right]. \tag{11}\]

For explicit bounds on all remaining images, put

\[q=e^{-4\pi(2\pi-\theta)/t},\qquad r=e^{-16\pi^2/t},\qquad E=\frac{2q}{1-r},\]

\[ \delta=q\left[\frac2{1-r}+\frac{32\pi^2}{t} \frac{1+r}{(1-r)^3}\right],\qquad M=\frac{(1+\eta)t}{4(2\pi-x)}.\]

Whenever \(\delta<1\),

\[|\lambda_0-\lambda_0^{(0)}| \le\frac{t\eta^{-1}}{\theta\sin x(1-\delta)} \left\{E\bigl[|\eta\cos x-1|+|\cot x|M\bigr] +\delta|A_0|\right\}, \tag{12}\]

\[|\lambda_1-\lambda_1^{(0)}| \le\frac{t\eta^{-1}}{2\theta\sin x(1-\delta)} \left\{E\left[\eta-\cos x+\frac M{\sin x}\right] +\delta|A_1|\right\}. \tag{13}\]

To verify these bounds, write \(G=G_t(x)\). The pair \(w=\pm n\) has Gaussian ratio bounded by \(2q_n\), where \(q_n=e^{-8\pi^2n^2/t+8\pi nx/t}\le q r^{n-1}\). Its contribution to \(\Xi/(\theta G)-1\) is at most \((2+32\pi^2n^2/t)q_n\), by the same \(\sinh u\le u\cosh u\) estimate as in Theorem 1. Consequently

\[\left|\frac\Theta G-1\right|\le E,\qquad \left|\frac\Xi{\theta G}-1\right|\le\delta.\]

For \(z\ne0\), oddness and the Gaussian tail bound give

\[|T_t(z)|\le(1+\eta)\int_{|z|}^\infty G_t(u)\,du \le\frac{(1+\eta)t}{4|z|}G_t(z).\]

All nonzero images have \(|x-2\pi w|\ge2\pi-x\), whence \(|S_t(x)/G-R_t(x)|\le ME\). Dividing by \(\Xi/(\theta G)\) proves (12)–(13), including the denominator error. The formulas (5)–(6) hold at all \(t>0\); the explicit small-image estimates require the displayed condition \(\delta<1\). They are exponentially small uniformly on each compact subinterval of \(0<\theta<2\pi\) as \(t\downarrow0\). Their displayed prefactors become loose at \(\theta=0\), where the exact expressions have removable singularities; their exponential separation degenerates at the cut locus \(\theta=2\pi\).

Gaussian and central-endpoint checks. Integration by parts in (3), at fixed \(x>0\), gives

\[R_t(x)=\frac{t}{4x}(1-\eta\cos x)+O_x(t^2).\]

Thus, with \(h=(x/2)\cot(x/2)\),

\[\lambda_0=1-\frac{t}{4h}+O_\theta(t^2),\qquad \lambda_1=1-\frac t8\left(1+\frac1h\right)+O_\theta(t^2). \tag{14}\]

Here the exponentially small image errors are absorbed into the remainder. This is exactly \(1-\frac t8[\mu^2+(2-\mu^2)/h]\) for \(\mu=0,\pm1\). At \(\theta=0\), conjugation invariance and (10) give the independent check

\[\lambda_0(0,t)=\lambda_1(0,t) =\frac13\left[1+2\eta^{-1}-\frac{t\eta^{-1}}2 \frac{\Theta_+(0)}{\partial_\theta\Xi_+(0)}\right].\]

The zero-image limit of either expression (11) is \([1+2\eta^{-1}-(t/2)\eta^{-1}]/3=1-t/4+O(t^2)\), in agreement with \(h(0)=1\).

Consequence for the isolated cube. For cut face \(j\), denote the bridge endpoints by \(A_j,B_j\), put \(X_j=\log(A_j^{-1}B_j)\), and set \(m_{*j}=A_je^{X_j/2}\). Let \(P_j\) project onto weight zero along \(X_j\) in spin 1. In the three-dimensional vector realization it is the rank-one projector \(\hat X_j\hat X_j^{\mathsf T}\). With (5)–(6) at \((\theta_j,t_j)\), put

\[L_j=\lambda_{1j}I+(\lambda_{0j}-\lambda_{1j})P_j,\qquad U_j=D^1(m_{*j}),\qquad F_j^+=U_jL_j,\quad F_j^-=L_jU_j^{-1}.\]

The four independent bridges of one isolated cube give its exact spin-1 contribution to \(\Psi\):

\[\boxed{\Psi_1=3e^{-t_m}\operatorname{tr}_1 \left(F_1^{\varepsilon_1}F_2^{\varepsilon_2} F_3^{\varepsilon_3}F_4^{\varepsilon_4}\right),\qquad \varepsilon_j\in\{+,-\},} \tag{15}\]

in the boundary order and orientations of the mid-face. The exponent is \(-t_m C_2(1)/2=-t_m\). Left translation reduces each bridge to (5)–(6); inversion takes the adjoint and explains the order in \(F_j^-\). The independence is the same isolated-cube hypothesis as in Theorem 3.

This sector retains an exact anisotropy \(\lambda_0-\lambda_1=(t/8)(1-1/h)+O_\theta(t^2)\). After transporting axes to a common frame, two such factors obey

\[[L_i,L_j]=(\lambda_{0i}-\lambda_{1i}) (\lambda_{0j}-\lambda_{1j})[P_i,P_j],\qquad \|[P_i,P_j]\|=|c|\sqrt{1-c^2}\le\tfrac12,\]

where \(c\) is the scalar product of the transported unit axes. Consequently the axis-dependent matrix product in (15) supplies the spin-1 noncommuting correction explicitly; generic oblique axes give a commutator of order \(t_it_j\) at fixed nonzero side angles. The scalar factorization of Theorem 3 is special to spin \(1/2\).

For an explicit winding bound on (15), let \(\epsilon_j\) be the maximum of the right sides of (12)–(13) for face \(j\), assuming each \(\delta_j<1\), and replace each \(L_j\) by its zero-image value to obtain \(\Psi_1^{(0)}\). Since an expectation of unitary matrices has norm at most 1, telescoping the four products gives

\[|\Psi_1-\Psi_1^{(0)}| \le9e^{-t_m}\left[\prod_{j=1}^4(1+\epsilon_j)-1\right]. \tag{16}\]

The factor 9 is the character coefficient 3 times the trace-norm bound \(|\operatorname{tr}_1 A|\le3\|A\|\).

7. Consequence for STATE

Atlas cell 1 is filled through spin 1 by Theorem 4, with exact matrix entries, explicit image bounds and the isolated-cube contribution. For \(J>1\), (7) with the spin-\(J\) Clebsch–Gordan squares is still an exact spectral representation. The finite Gaussian-tail reduction and its bounds have been completed here for \(J=1\); the higher-spin entries and estimates permitting the full character sum remain the next obligation before the isolated-cube Laplace claim of Proposition 7. The full mid-plane and iteration remain the subsequent atlas cells.