14 The normalized quantum-group comparison and the compact limit

14.1 Fixing the Teschner–Vartanov normalization

Write $R^{\mathrm{an}}$ for the analytic $b$-$6j$ symbol and $R$ for the symmetric symbol in [96]. Their equations (4.5)–(4.6) give the standard Virasoro fusion kernel as $F_{st}=(N_s/N_t)m(t)R^{\mathrm{an}}_{st}$, where $$N(z,y,x)= \frac{\Gamma_b(2Q-2z)\Gamma_b(2y)\Gamma_b(2x)} {\Gamma_b(2Q-x-y-z)\Gamma_b(Q-x-y+z) \Gamma_b(x+z-y)\Gamma_b(y+z-x)}.$$(14.1) Here $N_s$ and $N_t$ are products of the two trivalent normalizations in each channel. On the principal contour, $m(a)=|S_b(2a)|^2$ is the Plancherel density. The trivalent factor distinguishing the symmetric symbol is $$D(z,y,x)= \bigl[S_b(2Q-x-y-z)S_b(Q-x-y+z) S_b(x+z-y)S_b(y+z-x)\bigr]^{-1/2}.$$ If $D_s,D_t$ are its channel products, then $R=(D_s/D_t)R^{\mathrm{an}}$. Equivalently, equations (4.10)–(4.11) of that paper read $$G_{st}=\sqrt{\frac{C_s}{C_t}}\,F_{st} =\sqrt{m(s)m(t)}\,R_{st},$$(14.2) where $C_s,C_t$ are the corresponding products of DOZZ three-point functions. These identities specify the scalar factors that were left open in Section 13.

For our ordering of fields, use TV’s external labels $$(p,q,r,d_{\mathrm{out}}) =(\alpha_1,\alpha_3,\alpha_2,Q-\alpha_4).$$(14.3) Reflecting the infinity charge preserves its Virasoro weight. At the one-screening resonance, $$d_{\mathrm{out}}=p+q+r+b,\qquad s_j=p+q+jb,\qquad t_k=q+r+kb,\qquad j,k\in\{0,1\}.$$ Consequently each channel contains one zero-screening and one one-screening trivalent vertex. Explicitly, $$N_{s,j}=N(s_j,q,p)N(d_{\mathrm{out}},r,s_j),\qquad N_{t,k}=N(t_k,r,q)N(d_{\mathrm{out}},t_k,p).$$ The reflection convention matters for the analytic symbol, and is retained throughout the following formulas. Initially the weights and auxiliary factors are generic, with $\delta_0\delta_1\ne0$; balanced limits are taken afterward.

14.2 The double-Gamma factors and the finite sine matrix

The shift identity $$\frac{\Gamma_b(x+b)}{\Gamma_b(x)} =\frac{\sqrt{2\pi}\,b^{bx-1/2}}{\Gamma(bx)}$$ immediately gives $$\begin{aligned} N(x+y,y,x)&=\Gamma_b(Q)^{-1},\\ N(x+y+b,y,x)&= \frac{\Gamma(1+b^2)\Gamma(2bx)\Gamma(2by) \Gamma(2-2b(x+y))} {(2\pi)^2b^{1+b^2}\Gamma_b(Q)}. \end{aligned}$$ Thus $$N_{s,j}=\mathcal K_b n_{s,j},\qquad N_{t,k}=\mathcal K_b n_{t,k},\qquad \mathcal K_b= \frac{\Gamma(1+b^2)} {(2\pi)^2b^{1+b^2}\Gamma_b(Q)^2},$$ with $$\begin{aligned} n_{s,0}&=\Gamma(1-\delta_0)\Gamma(\mu_2)\Gamma(\mu_4), &n_{s,1}&=\Gamma(1+\delta_0)\Gamma(\mu_1)\Gamma(\mu_3), \\ n_{t,0}&=\Gamma(1-\delta_1)\Gamma(\mu_1)\Gamma(\mu_4), &n_{t,1}&=\Gamma(1+\delta_1)\Gamma(\mu_2)\Gamma(\mu_3). \end{aligned}$$ All $b$-dependence in these normalization products is in the same scalar $\mathcal K_b$.

Rescale the screened channel vectors by $e_{s,j}=\mathcal F_{s,j}/N_{s,j}$ and $e_{t,k}=\mathcal F_{t,k}/N_{t,k}$. Their matrix is $\mathsf B_{jk}=(n_{t,k}/n_{s,j})\mathsf M_{jk}$, where $\mathsf M$ is (13.16). Gamma reflection gives the explicit answer $$\boxed{\quad \mathsf B= \frac1{\sin(\pi\delta_1)} \begin{pmatrix} \sin(\pi\mu_2)&-\sin(\pi\mu_4)\\ \sin(\pi\mu_3)&-\sin(\pi\mu_1) \end{pmatrix}. \quad}$$(14.6) For example its upper-left entry is $$\frac{n_{t,0}}{n_{s,0}}\mathsf M_{00} =\frac{\Gamma(1-\delta_1)\Gamma(\delta_1)} {\Gamma(\mu_2)\Gamma(1-\mu_2)} =\frac{\sin(\pi\mu_2)}{\sin(\pi\delta_1)}.$$ This is the induced finite action of $m(t)R^{\mathrm{an}}$ in the specified convention, understood through the exact normalization identity on the screened period space. Identifying it with the residue of the full infinite spectral operator retains the continuation input of Section 13; no pointwise substitution into a singular Barnes integral is being made. Its determinant is $-\sin(\pi\delta_0)/\sin(\pi\delta_1)$.

For reference, put $\sigma_i=\sin(\pi\mu_i)$ and $d_\nu=\sin(\pi\delta_\nu)$. After removing the common $E_b/\mathcal K_b$ from these vectors, their Hermitian matrices are $$H_s^{\mathrm{an}}= \frac{\pi^4}{d_0} \mathop{\mathrm{diag}}\left(\frac1{\sigma_2\sigma_4}, -\frac1{\sigma_1\sigma_3}\right),\qquad H_t^{\mathrm{an}}= \frac{\pi^4}{d_1} \mathop{\mathrm{diag}}\left(\frac1{\sigma_1\sigma_4}, -\frac1{\sigma_2\sigma_3}\right).$$(14.7) They follow from $H_s^{\mathrm{an}}=\mathop{\mathrm{diag}}(h_jn_{s,j}^2)$, and satisfy $\mathsf B^*H_s^{\mathrm{an}}\mathsf B =H_t^{\mathrm{an}}$.

14.3 Symmetric normalization and the residue prescription

The square roots in (14.2) also have a concrete interpretation at a pinch. Write $\alpha_s=s_j+\varepsilon\xi$, with fixed $\xi\ne0,-1$. The product of three-point factors has the form $$C_s\sim\frac{\Lambda_*h_{s,j}} {\varepsilon^2\xi(\xi+1)}.$$ The common $\Lambda_*$ includes the infinity reflection factor and cancels in channel ratios. With square-root branches fixed by continuation, the DOZZ-normalized vector $\mathcal G_s=\sqrt{C_s}\mathcal F_s$ obeys $$\lim_{\varepsilon\to0} \varepsilon\sqrt{\frac{\xi(\xi+1)}{\Lambda_*}}\, \mathcal G_s(s_j+\varepsilon\xi;\varepsilon) =\sqrt{h_{s,j}}\mathcal F_{s,j}.$$(14.8) Hence its finite analytic matrix is $$\mathsf O=\mathop{\mathrm{diag}}(\sqrt{h_s})\mathsf M \mathop{\mathrm{diag}}(\sqrt{h_t})^{-1}, \qquad \mathsf O^T\mathsf O=I.$$ This is complex orthogonality: negative Hermitian coefficients give factors of $i$ in analytic square roots.

Using absolute square roots instead gives a real matrix with an indefinite pairing: $$\boxed{\quad \mathsf U= \frac{\operatorname{sgn}(d_1)}{\sqrt{|d_0d_1|}} \begin{pmatrix} \sqrt{\sigma_1\sigma_2}&-\sqrt{\sigma_3\sigma_4}\\ \sqrt{\sigma_3\sigma_4}&-\sqrt{\sigma_1\sigma_2} \end{pmatrix}. \quad}$$(14.9) Indeed, $$\sigma_1\sigma_2-\sigma_3\sigma_4=d_0d_1,\qquad \mathsf U^T J_s\mathsf U=J_t.$$ $$J_s=\operatorname{sgn}(d_0)\mathop{\mathrm{diag}}(1,-1),\quad J_t=\operatorname{sgn}(d_1)\mathop{\mathrm{diag}}(1,-1).$$ Equations (14.6) and (14.9) are fully normalized finite operator comparisons. The logarithmic basis of Section 13 supplies the balanced limits; one cannot set $d_0=d_1=0$ in the displayed matrices.

There is a separate reason to retain the operator prescription. The analytic continuation of the Plancherel factor is $$\begin{aligned} m(a)&=S_b(2a)S_b(2Q-2a)\\ &=-4\sin\bigl(\pi b(2a-Q)\bigr) \sin\bigl(\pi(2a-Q)/b\bigr). \end{aligned}$$ The two $S_b$ shift equations prove the second equality. At the screened labels this becomes $$m(s_j)= 4\sin\bigl(\pi[\delta_0-(2j-1)b^2]\bigr) \sin(\pi\delta_0/b^2),$$(14.11) with the analogous formula for $t_k,\delta_1$. These factors oscillate for small real $b$ and vanish at balanced weights. The trivalent symmetric normalizations also encounter $S_b(Q+jb)=0$. Thus dividing a finite matrix by pointwise Plancherel values does not define a bare-symbol limit.

14.4 The actual screened Gram matrix has rank two at the limit

The tetrahedral dictionary of [96, Section 3] is $$A_e=\exp(\pi i-2\pi ib\alpha_e)=e^{i\eta_e}.$$ The vertex triples are $(1,2,s),(3,4,s),(1,4,t),(2,3,t)$; compare the original convention in [83]. Here $A_e$ is an exponentiated angle, distinct from surface area. Put $$\vartheta_i=\pi\mu_i,\qquad h=2\pi b^2,\qquad \sum_i\vartheta_i=2\pi.$$ With the reflected labels (14.3), a face ordering gives the formal angle Gram matrix $$\mathbf G_{jk}(h)= \begin{pmatrix} 1&\cos(\vartheta_2+\vartheta_3+kh)&\cos\vartheta_2& \cos(\vartheta_4-h)\\ \cos(\vartheta_2+\vartheta_3+kh)&1&\cos\vartheta_3&\cos\vartheta_1\\ \cos\vartheta_2&\cos\vartheta_3&1& \cos(\vartheta_1+\vartheta_3+jh)\\ \cos(\vartheta_4-h)&\cos\vartheta_1& \cos(\vartheta_1+\vartheta_3+jh)&1 \end{pmatrix}.$$(14.12) Calling this a formal Gram matrix does not assert that the angle data are realized by a convex tetrahedron.

At $h=0$ it is the Euclidean planar Gram matrix $$(\mathbf G_0)_{ab}=\cos(\phi_a-\phi_b),\qquad (\phi_1,\phi_2,\phi_3,\phi_4)= (0,\vartheta_2+\vartheta_3,\vartheta_2, \vartheta_1+\vartheta_2+\vartheta_3).$$(14.13) Since $0<\vartheta_2<\pi$, these unit vectors span a plane. Therefore $$\boxed{\qquad \operatorname{rank}\mathbf G_0=2.\qquad}$$(14.14) For equal half-weights it is $$\mathbf G_0= \begin{pmatrix}1&-1&0&0\\-1&1&0&0\\ 0&0&1&-1\\0&0&-1&1\end{pmatrix}.$$ By contrast, the outward normals of a nondegenerate Euclidean tetrahedron span three dimensions. This difference holds for every four-weight choice on the screening locus, and remains under varying such weights because rank at most two is closed.

There is already a finite-$b$ obstruction. If $\mathcal C_{ab}$ denotes the cofactor matrix, then $$\mathcal C_{1+j,1+j}=0,\qquad \mathcal C_{4-k,4-k}=0.$$(14.15) Each identity is the determinant of three planar unit vectors. At a finite vertex of a compact hyperbolic tetrahedron, the three incident face normals lie in the positive-definite orthogonal complement of a timelike vertex vector and are independent. Their Gram determinant is positive. Thus (14.15) excludes compact tetrahedra already before taking the limit.

For completeness, the determinant and orientation checks are also explicit. Each $\mathbf G_{jk}$ is a planar rank-two Gram matrix perturbed in a single symmetric off-diagonal entry. The entry, its increment $\epsilon$, and a positive factor $q$ are $$\begin{array}{c|c|c|c} (j,k)&\text{entry}&\epsilon&q\\ \hline (0,0)&14&\cos(\vartheta_4-h)-\cos\vartheta_4& \sin^2\vartheta_3\\ (0,1)&13&\cos\vartheta_2-\cos(\vartheta_2+h)& \sin^2\vartheta_1\\ (1,0)&24&\cos\vartheta_1-\cos(\vartheta_1+h)& \sin^2\vartheta_2\\ (1,1)&23&\cos\vartheta_3-\cos(\vartheta_3+h)& \sin^2(\vartheta_4-h) \end{array}$$ In every row, $\det\mathbf G=-q\epsilon^2$ and the cofactor at that off-diagonal entry is $-q\epsilon$. For fixed interior weights and small positive $h$, $\epsilon>0$. Hence the matrices have Lorentz signature but fail the positive off-diagonal cofactor condition for an ordinary convex hyperbolic tetrahedron; see [47, Theorem 2.1]. The zero principal cofactors therefore do not justify calling these configurations tetrahedra with two ideal vertices.

14.5 The screened classical saddle also degenerates

The dilogarithmic potential gives an independent check. At $b=0$, write $X=A_1,Y=A_2,Z=A_3$. The reflected screening labels satisfy $$A_4=XYZ,\qquad A_s=-XY,\qquad A_t=-YZ.$$ Set $P=X^2Y^2=e^{2\pi i\delta_0}$ and $R=Y^2Z^2=e^{2\pi i\delta_1}$. Substitution into [96, eq. (3.1)] cancels two pairs of dilogarithms and leaves $$U(u)=\operatorname{Li}_2(u)+\operatorname{Li}_2(PRu) -\operatorname{Li}_2(Pu)-\operatorname{Li}_2(Ru).$$(14.16) The exponentiated saddle equation $uU_u=-2\pi i$ becomes $$(1-Pu)(1-Ru)=(1-u)(1-PRu), \qquad (1-P)(1-R)u=0.$$ For nonbalanced weights there is no nonzero regular saddle. Before cancellation, the equation contains the factor $(1-(XYZ)^2u)^2$; its double root is at the very branch factors that cancel in (14.16). This is not a regular double critical point of the reduced phase. If $P=1$ or $R=1$, the reduced potential itself vanishes on a consistent local branch. Thus the finite screening operator survives in a singular pole/branch collision regime, rather than in the ordinary compact Euclidean saddle regime.

14.6 A genuine compact family: the regular tetrahedron

For comparison, prescribe the actual six interior dihedral angles $\eta_e$ of a small compact tetrahedron and set $$\alpha_e=\frac{\pi-\eta_e}{2\pi b}.$$(14.17) These quantum labels encode edge dihedrals. The $\mu_i$ of the flat boundary metric encode intrinsic vertex deficits, so the two kinds of data must be related geometrically before they can be identified.

For a regular tetrahedron of edge length $\ell$, let $\beta$ be a face angle and $\eta$ an interior dihedral. Hyperbolic trigonometry gives $$\cos\beta=\frac{\cosh\ell}{1+\cosh\ell},\qquad \cos\eta=\frac{\cosh\ell}{1+2\cosh\ell},\qquad A=4\pi-12\beta.$$(14.18) At the Euclidean limit $\eta_{\mathrm E}=\arccos(1/3)$. Its angle Gram has $$\det G=(1+\cos\eta)^3(1-3\cos\eta),$$ and rank three at $\eta_{\mathrm E}$. All six limiting heavy edge weights are $$2b\alpha_e=1-\eta_{\mathrm E}/\pi\simeq0.608173.$$ In the reflected screening convention, $2bd_{\mathrm{out}}\to2-\mu_4>1$, whereas this geometric prescription gives the same number below one for all six labels. Moreover the screened internal charges are sums of pairs of external charges. The rank obstruction is independent of reflected representatives; these explicit prescriptions already differ at leading order $1/b$, beyond an order-$b$ normalization shift.

Expansion of (14.18) gives $$\begin{aligned} A&=\sqrt3\,\ell^2-\frac{\sqrt3}{8}\ell^4+O(\ell^6), \\ t:=\eta_{\mathrm E}-\eta &=\frac{\ell^2}{12\sqrt2} -\frac{23\ell^4}{1152\sqrt2}+O(\ell^6) =\frac{A}{12\sqrt6}+O(A^2). \end{aligned}$$ Using $\mathop{}\!\mathrm dW=-3\ell\,\mathop{}\!\mathrm d\eta$ and $W(0)=0$, $$\boxed{\quad W(A)=\frac{A^{3/2}}{6\sqrt2\,3^{3/4}} \left(1-\frac{A}{10\sqrt3}+O(A^2)\right). \quad}$$(14.20) In edge length this is $W=\ell^3/(6\sqrt2)-23\ell^5/(480\sqrt2)+O(\ell^7)$.

14.7 The regular classical fold and its scale

Put all six exponentiated angles equal to $a=e^{i\eta}$. The TV potential becomes $$U(u,\eta)=\operatorname{Li}_2(u) +3\operatorname{Li}_2(a^4u) -4\operatorname{Li}_2(-a^3u).$$ Exponentiating the saddle equation gives $(1+a^3u)^4=(1-u)(1-a^4u)^3$. After removing $u(a+1)^2$, the quadratic is $$P_\eta(u)=a^8(a^2-2a+3)u^2 -3a^4(a-1)^2u+(3a^2-2a+1).$$(14.21) Its discriminant is $$-a^8(a+1)^2(3a^2-2a+3),\qquad 3a^2-2a+3=2a(3\cos\eta-1).$$ Thus the saddles merge simply at the Euclidean angle. The double root and its real logarithmic lift are $$a_{\mathrm E}=\frac{1+2\sqrt2\,i}{3},\qquad u_{\mathrm E}=a_{\mathrm E}^{-5} =\frac{241+22\sqrt2\,i}{243},\qquad v_{\mathrm E}=2\pi-5\eta_{\mathrm E}.$$ None of the dilogarithm arguments is $1$ at this point.

For $u=e^{iv}$ the action, including a term independent of $v$, is $S(v,\eta)=U(e^{iv},\eta)-2\pi v+\text{const.}$ Choose the local branches satisfying the saddle equation. Direct differentiation at the fold yields $$S_v=S_{vv}=0,\qquad S_{vvv}=54i,\qquad S_{v\eta}=18\sqrt2\,i.$$ For example, $$S_{v\eta}=-12\left(\frac{a^3u}{1+a^3u} +\frac{a^4u}{1-a^4u}\right).$$ Writing $x=v-v_{\mathrm E}$ gives the concrete normal form $$\boxed{\quad S(v_{\mathrm E}+x,\eta_{\mathrm E}-t) -S(v_{\mathrm E},\eta_{\mathrm E}-t) =i(9x^3-18\sqrt2\,t x) +O(x^4+t x^2+t^2|x|). \quad}$$(14.22)

Use the saddle exponential preceding [96, eq. (3.11)], namely $e^{S/(ih)}$ with $h=2\pi b^2$. Rescaling $x=(h/27)^{1/3}y$ gives $y^3/3-\zeta y$ as the leading exponent, where $$\boxed{\quad \zeta=\frac{6\sqrt2\,t}{h^{2/3}} =\frac{A}{2\sqrt3\,h^{2/3}} +O\left(\frac{A^2}{h^{2/3}}\right). \quad}$$(14.23) This proves the cubic normal form of the classical phase and derives the scale $A\asymp b^{4/3}$ with its normalization. A smooth-amplitude integral on a specified Airy contour has the corresponding uniform asymptotic; see [54, Section 36.12]. Section 15 studies the corresponding assertion for the exact regular-family integral with an explicit scalar regularization. The local classical calculation here does not establish the required uniform contour and tail estimates for that quantum integral.

With TV’s choice $\Delta-5\pi^2/3$ for the term independent of $v$, the volume formula of [83], in TV’s doubled potential convention, gives $\operatorname{Im}S_\pm=\pm2W$. Thus the two saddle exponentials contain $e^{\pm2W/h}$. If the volume exponent is written as $W/\hbar_{\mathrm{vol}}$, its parameter is $$\hbar_{\mathrm{vol}}=\pi b^2=h/2.$$(14.24) The leading cubic saddle values agree: $(2/3)\zeta^{3/2}\sim2W/h$.

14.8 Correction of the integration-coordinate dictionary

An earlier version of this monograph placed a dense pole lattice at the classical fold. That conclusion was incorrect: the logarithmic coordinate of the dilogarithm potential was matched to the wrong shifted integration variable. The exact integral fixes the dictionary without an asymptotic choice of branch.

With all six charges $\alpha=(\pi-\eta)/(2\pi b)$, write the original integration variable in [96, eq. (2.27)] as $w$. Its integrand is $$I_b(w,\eta)=S_b(w-3\alpha)^4 S_b(4\alpha-w)^3S_b(2Q-w).$$(14.25) Set $a=e^{i\eta}$. Directly, $$e^{2\pi ib(w-3\alpha)}=-a^3e^{2\pi ibw}.$$ Thus the variable $u$ in the classical potential is $$u=e^{2\pi ibw},\qquad v=2\pi bw.$$(14.26) The $Q/2$-shift used previously is incompatible with this identification. A second check is the exact shift quotient $$\begin{aligned} \frac{I_b(w+b,\eta)}{I_b(w,\eta)} &= \frac{[2\sin\pi b(w-3\alpha)]^4} {[2\sin\pi b(4\alpha-w-b)]^3 [2\sin\pi b(2Q-w-b)]}\\ &\longrightarrow \frac{(1+a^3u)^4}{(1-a^4u)^3(1-u)}. \end{aligned}$$ The limit reproduces the saddle polynomial (14.21).

The lift of the double root lying in the separating contour strip is $$v_*=4\pi-5\eta_{\mathrm E},\qquad w_*=\frac{v_*}{2\pi b}.$$ It differs by $2\pi$ from the logarithmic lift used in the classical calculation; all its local derivatives remain the same. At this lift, $$\begin{aligned} w_*-3\alpha_{\mathrm E} &=\frac{\pi-2\eta_{\mathrm E}}{2\pi b},\\ 4\alpha_{\mathrm E}-w_*&=\frac{\eta_{\mathrm E}}{2\pi b},\\ 2Q-w_*&=\frac{5\eta_{\mathrm E}}{2\pi b}+2b. \end{aligned}$$(14.28) All three arguments lie in $0<\operatorname{Re}z<Q$ for sufficiently small $b$, a fixed scaled distance from its boundary. There is therefore no integration-pole accumulation at the fold. The external trivalent factors do have parameter poles; their regularization and the further analytic requirements for a uniform transition are treated in the next section. Pole separation in this local strip alone does not control the complete integration contour or its tails.