20 An exact torus twist calculation and a constant orbit series

20.1 The length function and its integral

Let $X$ be an oriented one-holed hyperbolic torus with geodesic boundary length $b>0$, or a complete finite-area once-punctured torus when $b=0$. Cut along an essential simple curve $\alpha$ of length $a>0$, and choose an essential simple curve $\beta$ crossing $\alpha$ once. Normalize the twist origin so that this crossing curve has minimum length at $\tau=0$. Set $$A^2=\frac{\cosh a+\cosh(b/2)}{\cosh a-1}.$$(20.1) The exact formula is $$\boxed{\cosh\bigl(\ell_\beta(\tau)/2\bigr)=A\cosh(\tau/2).}$$(20.2) It follows from the right-angled-hexagon description of the cut pants; see also [43, Proposition 3.1, pp. 5–6]. Take the positive square root $A>1$. For $b=0$, $A=\coth(a/2)$. The minimum length is $r_0=2\mathop{\mathrm{arcosh}}A$.

There is also a matrix check fixing the signed boundary convention. At zero twist the two axes meet perpendicularly. In an adapted basis put $$M=\begin{pmatrix}e^{a/2}&0\\0&e^{-a/2}\end{pmatrix},\qquad B_u=\begin{pmatrix}e^{u/2}&0\\0&e^{-u/2}\end{pmatrix} \begin{pmatrix}A&\sqrt{A^2-1}\\\sqrt{A^2-1}&A\end{pmatrix}.$$ Then $\det B_u=1$, $\mathop{\mathrm{tr}}B_u=2A\cosh(u/2)$, and direct multiplication gives $$\mathop{\mathrm{tr}}[M,B_u]=2-4\sinh^2(a/2)(A^2-1)=-2\cosh(b/2), \qquad [M,B_u]=MB_uM^{-1}B_u^{-1}.$$ Solving this boundary equation recovers (20.1). Differentiating (20.2), with $Q=A^2\cosh^2(u/2)-1>0$, also gives $$\ell'_\beta(u)=\frac{A\sinh(u/2)}{\sqrt Q},\qquad \ell''_\beta(u)=\frac{A(A^2-1)\cosh(u/2)}{2Q^{3/2}}>0.$$(20.3) In particular $|\ell'_\beta|<1$, consistent with the single crossing in (19.5).

For any nonnegative measurable function $f$, change variables separately on the positive and negative twist axes to obtain $$\int_{\mathbb R}f\bigl(\ell_\beta(u)\bigr)\,\mathop{}\!\mathrm du =2\int_{r_0}^\infty f(r)\frac{\sinh(r/2)}{\sqrt{\cosh^2(r/2)-A^2}}\,\mathop{}\!\mathrm dr.$$(20.4) Indeed, differentiating $\cosh(r/2)=A\cosh(u/2)$ gives $\mathop{}\!\mathrm du/\mathop{}\!\mathrm dr=\sinh(r/2)/\sqrt{\cosh^2(r/2)-A^2}$ for $u>0$. The two signs of $u$ explain the factor $2$. The nonnegative equality allows infinity on both sides; for signed $f$ require absolute integrability. The coarea density’s endpoint singularity at $r_0$ is integrable. For example, $y=\sqrt{\cosh^2(r/2)-A^2}$ converts the right-hand side to $4\int_0^\infty f(2\mathop{\mathrm{arcosh}}\sqrt{A^2+y^2})\,\mathop{}\!\mathrm dy/\sqrt{A^2+y^2}$.

Several useful checks are immediate: $$\begin{aligned} \int_{\mathbb R}\mathop{\mathrm{sech}}^2\bigl(\ell_\beta(u)/2\bigr)\,\mathop{}\!\mathrm du &=\frac4{A^2},\\ \operatorname{meas}\{u:\ell_\beta(u)\le R\} &=4\mathop{\mathrm{arcosh}}\left(\frac{\cosh(R/2)}{A}\right),\quad R\ge r_0, \\ \int_{\mathbb R}\cosh^{-s}\bigl(\ell_\beta(u)/2\bigr)\,\mathop{}\!\mathrm du &=\frac{2\sqrt\pi\,\Gamma(s/2)} {A^s\Gamma((s+1)/2)},\quad s>0. \end{aligned}$$ The sublevel measure is zero for $R<r_0$. In the last integral the change of variable $t=\tanh(u/2)$ gives $2A^{-s}B(1/2,s/2)$, which proves the displayed gamma quotient. These are exact twist integrals. For a finite interval $[-U,U]$, $U\geq0$, the omitted tail satisfies $$\int_{|u|>U}\cosh^{-s}(\ell_\beta(u)/2)\,\mathop{}\!\mathrm du \leq\frac{2^{s+2}}{sA^s}e^{-sU/2},\qquad s>0,$$(20.8) because $\cosh(u/2)\geq e^{|u|/2}/2$. For $s=2$ the tail is exactly $4A^{-2}(1-\tanh(U/2))$. These analytic tails do not enclose numerical quadrature or roundoff error.

20.2 A positive partition of unity

Define $$H_a(u)=\frac{\sinh(a/2)}{\cosh u+\cosh(a/2)} =\frac12\left[ \tanh\left(\frac{u+a/2}{2}\right) -\tanh\left(\frac{u-a/2}{2}\right)\right].$$(20.9) The second expression telescopes under translation by $a$, so $$\sum_{n\in\mathbb Z}H_a(\tau+na)=1.$$(20.10) All terms are positive and the convergence is exponential on compact twist intervals. More precisely, for $N\geq0$ put $L_N=(N+1/2)a$ and $P_N(\tau)=\sum_{n=-N}^NH_a(\tau+na)$. Finite telescoping gives $$\begin{aligned} P_N(\tau)&=\tfrac12\left[ \tanh\frac{\tau+L_N}{2}-\tanh\frac{\tau-L_N}{2}\right],\\ 1-P_N(\tau)&=\frac1{1+e^{L_N+\tau}}+ \frac1{1+e^{L_N-\tau}} \leq2\cosh\tau\,e^{-L_N}. \end{aligned}$$ This is an explicit tail estimate, uniform for bounded $\tau$ with $a$ fixed. One may reduce $\tau$ modulo $a$ to $|\tau|\leq a/2$ by relabeling the orbit; no estimate uniform as $a\to0$ is asserted for fixed $N$.

Since (20.2) implies $\cosh r+1=A^2(\cosh u+1)$, eliminate $u$ from (20.9) and set $$K_{a,b}(r)= \frac{A^2\sinh(a/2)} {\cosh r+1+A^2(\cosh(a/2)-1)}.$$(20.12) For $r\geq r_0$, this satisfies $K_{a,b}(r)=H_a(u)$ at the two twists with length $r$. For $r_n=\ell_{D_\alpha^n\beta}(X)$ we obtain the pointwise identity $$\boxed{\sum_{n\in\mathbb Z}K_{a,b}(r_n)=1.}$$(20.13) This is a constant series proved by telescoping, rather than inferred from its average. Its kernel depends on the distinguished cuff length $a$ and the boundary length $b$, and its index set is one twist orbit. It is not an unrestricted universal sum over every curve on the surface. The classes $D_\alpha^n\beta$ are all distinct as unoriented simple classes, since they cross $\alpha$ once and have distinct integral twist coordinates. Equal lengths, which occur at symmetric twist values, do not merge their labels. Equation (20.11) is also the tail estimate for this $K$-series.

There is a simple source of many such examples. If $\rho\in L^1(\mathbb R)$ is an even nonnegative probability density, then $$H_{a,\rho}(u)=\int_{u-a/2}^{u+a/2}\rho(v)\,\mathop{}\!\mathrm dv \quad\text{satisfies}\quad \sum_{n\in\mathbb Z}H_{a,\rho}(u+na)=1.$$ The translated intervals partition the line except for shared endpoints of measure zero, so Tonelli proves the identity. These terms are nonnegative; strict positivity requires further assumptions on $\rho$. Evenness makes $H_{a,\rho}$ a function of $|u|$, hence of the length $r$ through (20.2). The choice $\rho(v)=\frac14\mathop{\mathrm{sech}}^2(v/2)$ gives (20.9). This realizes the proposed “zoo” of identities in a concrete way, with the source of constancy and the parameter dependence both visible.