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Martin Bridgeman gave a nice talk at Caltech recently on his discovery of a beautiful identity concerning orthospectra of hyperbolic surfaces (and manifolds of higher dimension) with totally geodesic boundary. The $2$-dimensional case is (in my opinion) the most beautiful, and I would like to take a post to explain the identity, and give a derivation which is slightly different from the one Martin gives in his paper. There are many other things one could say about this identity, and its relation to other identities that turn up in the theory of hyperbolic manifolds (and elsewhere); I hope to get to this in a later post.

Let $\Sigma$ be a hyperbolic surface with totally geodesic boundary. An orthogeodesic is a geodesic segment properly immersed in $\Sigma$, which is perpendicular to $\partial \Sigma$ at its endpoints. The set of orthogeodesics is countable, and their lengths are proper. Denote these lengths by $l_i$ (with multiplicity). The identity is:

$\sum_i \mathcal{L}(1/\cosh^2{l_i/2}) = -\pi^2\chi(\Sigma)/2$

where $\mathcal{L}$ is the Rogers’ dilogarithm function (to be defined in a minute). Treating this function as a black box for the moment, the identity has the form $\sum_i L(l_i) =$ a term depending only on the topology of $\Sigma$. The proof is very, very short and elegant. By the Gauss-Bonnet theorem, the term on the right is equal to $1/8$ of the volume of the unit tangent bundle of $\Sigma$. Almost every tangent vector on $\Sigma$ can be exponentiated to a geodesic on $\Sigma$ which intersects the boundary in finite forward and backward time (eg. by ergodicity of the geodesic flow on a closed hyperbolic surface obtained by doubling). If $v$ is such a tangent vector, and $\gamma_v$ is the associated geodesic arc, then $\gamma_v$ is homotopic keeping endpoints on $\partial \Sigma$ to a unique orthogeodesic (which is the unique length minimizer in this relative homotopy class). The volume of the set of $v$ associated to a given orthogeodesic $\alpha$ can be computed as follows. Lift $\alpha$ to the universal cover, where it is the crossbar of a letter “H” whose vertical lines are lifts of the geodesics it ends on. Any $\gamma_v$ lifts to a unique geodesic segment in the universal cover with endpoints on the edges of the H. So the volume of the set of such $v$ depends only on $\text{length}(\alpha)$, giving rise to the explicit formula for $L$. qed.

That’s it — that’s the whole proof! . . . modulo some calculations, which we now discuss.

The “ordinary” polylogarithms $\text{Li}_k$ are defined by Taylor series

$\text{Li}_k(z) = \sum_{n=1}^\infty \frac {z^n} {n^k}$

which converges for $|z|<1$, and extends by analytic continuation. Taking derivatives, one sees that they satisfy $\text{Li}_k'(z) = \text{Li}_{k-1}(z)/z$, thereby giving rising to integral formulae. $\text{Li}_0(z)$ is the familiar geometric series $z/(1-z)$, so $\text{Li}_1(z) = -\log(1-z)$ and

$\text{Li}_2(z) = -\int \frac {\log(1-z)} {z} dz$

The Rogers dilogarithm is then given by the formula $\mathcal{L}(z) = \text{Li}_2(z) + \frac 1 2 \log(|z|)\log(1-z)$ for real $z<1$. One sees that the Rogers dilogarithm is obtained by symmetrizing the integrand for the integral expression for $\text{Li}_2$ under the involution $z \to 1-z$:

$\mathcal{L}'(z) = -\frac {1}{2} \left(\frac {\log(1-z)}{z} + \frac {\log(z)}{1-z} \right)$

Martin derives his identity by direct calculation, but in fact this calculation can be simplified a bit by some hyperbolic geometry. Consider an ideal quadrilateral $Q$ (whose unit tangent bundle has area $4\pi^2$) with one pair of opposite sides that are distance $l$ apart. Join opposite vertices in pairs to decompose the quadrilateral into four triangles, each with one non-ideal point:

In the (schematic) picture, suppose the two edges of the H are the left and right side (call them $L$ and $R$) and the other two edges are $U$ and $D$. Similarly, call the four triangles $T_L, T_R, T_U, T_D$ depending on which edge of the quadrilateral they bound. The triangle $T_R$ is colored gray in the figure. We secretly identify this figure with the upper half-plane, in such a way that the ideal vertices are (in circular order) $0,x,1,\infty$, where $\infty,0$ are the ideal vertices of the gray triangle. Call $\alpha$ the (hyperbolic) angle of the gray triangle at its vertex, so $x = (1+\cos(\alpha))/2$. Moreover, it turns out that $x = 1/\cosh^2(l/2)$ where $l$ is the distance between $L$ and $R$. We will compute $L$ implicitly as a function of $x$, and show that it is a multiple of the Rogers dilogarithm function, thus verifying Bridgeman’s identity.

Every vector $v$ in $Q$ exponentiates to a (bi-infinite) geodesic $\gamma_v$, and we want to compute the volume of the set of vectors $v$ for which the corresponding geodesic intersects both $L$ and $R$. The point of the decomposition is that for $v$ in $T_L$ (say), the geodesic $\gamma_v$ intersects $L$ whenever it intersects $R$, so we only need to compute the volume of the $v$ in $T_L$ for which $\gamma_v$ intersects $R$. Similarly, we only need to compute the volume of the $v$ in $T_R$ for which $\gamma_v$ intersects $L$. For $v$ in $T_U$, we compute the volume of the $v$ which do not intersect $U$ (since these are exactly the ones that intersect both $L$ and $R$), and similarly for $T_D$.

These volumes can be expressed in terms of integrals of harmonic functions. Let $\chi_L$ denote the harmonic function on the disk which is $1$ on the arc of the circle bounded by $L$, and $0$ on the rest of the circle. This function at each point is equal to $1/2\pi$ times the visual angle (i.e. the length in the unit tangent circle) subtended by the given arc of the circle, as seen from the given point in the hyperbolic plane. Define $\chi_R,\chi_U,\chi_D$ similarly. Then the total volume we need to compute is equal to

$4\pi \left( (\int_{T_L} 2\chi_R) + (\int_{T_U} 1 - 2\chi_U) \right)$

(here we have identified $\int_{T_L} \chi_R = \int_{T_R} \chi_L$ by symmetry, and similarly for the other pair of terms). Let us approach this a bit more systematically. If $\alpha$ denotes the angle at the nonideal vertex of triangle $T_R$, we denote $\int_{T_R} \chi_R = A(\alpha)$, $\int_{T_R} \chi_U = B(\alpha)$ and $\int_{T_R} \chi_L = C(\alpha)$. The integral we want to evaluate can be expressed easily in terms of explicit rational multiples of $\pi$, and the function $A,B,C$. These functions satisfy obvious identities:

$C(\alpha) = \int_{T_R} 1 - A(\alpha) - 2B(\alpha) = \pi-\alpha - A(\alpha) - 2B(\alpha)$

and

$A(\alpha) + B(\pi - \alpha) = \pi/3$

where the last identity comes by observing that we are integrating a certain function over an ideal triangle, and observing that the average of this function under the symmetries of the ideal triangle is equal to the constant function $1/3$. In particular, we see that we can express everything in terms of $A$. After some elementary reorganization, we see that the contribution $V(\alpha)$ to the volume of the unit tangent bundle of the surface associated to this particular orthogeodesic is

$V(\alpha) = \pi^2(8 - 16/3) - 4\pi\alpha - 8\pi(A(\alpha) - A(\pi - \alpha))$

To compute $A(\alpha)$, it makes sense to move to the upper half-space model, and move the endpoints of the interval to $0$ and $\infty$. The harmonic function is equal to $1$ on the negative real axis, and $0$ on the positive real axis. It takes the value $\theta/\pi$ on the line $\text{arg}(z) = \theta$. The area form in the hyperbolic metric is proportional to the Euclidean area form, with constant $1/\text{Im}(z)^2$. In other words, we want to integrate $\text{arg}(z)/\pi\text{Im}(z)^2$ over the region indicated in the figure, where the nonideal angle is $\alpha$, and the base point is $0$:

If we normalize so that the circular arc is part of the semicircle from $0$ to $1$, then the real projection of the vertical lines in the figure are $0$ and $x$. There is no elementary way to evaluate this integral, so instead we evaluate its derivative as a function of $x$ where as before, $x = (1+\cos(\alpha))/2$. This is the definite integral

$A'(x) = \int_{y = \sqrt{x-x^2}}^\infty (\tan^{-1}(y/x)/\pi y^2) dy$

Integrating by parts gives $(\alpha/\pi\sin{\alpha}) + 1/\pi \int_{y = \sqrt{x-x^2}}^\infty xdy/y(y^2+x^2)$. This evaluates to

$A'(x) = (\alpha/\pi\sin{\alpha}) - 1/\pi ( \log(1-x)/2x)$

Thinking of $V(\alpha)$ as a function of $x$, we get

$V'(x) = -4\pi d\alpha/dx - 8\pi(A'(x) + A'(1-x)) = 8\mathcal{L}'(x)$

Comparing values at $x=0$ we see that $V=8\mathcal{L}$ and the identity is proved.

Well, OK, this is not terribly simple, but a posteriori it gives a way to express the Rogers dilogarithm as a sum of integrals of very simple harmonic functions over hyperbolic triangles, which is a nice geometric way to think of it.

(Update 10/30): This paper by Dupont and Sah relates Rogers dilogarithm to volumes of $\text{SL}(2,\mathbb{R})$-simplices, and discusses some interesting connections to conformal field theory and lattice model calculations. I feel like a bit of a dope, since I read this paper while I was in graduate school more than a dozen years ago, but forgot all about it until I was cleaning out my filing cabinet this morning. They cite an older paper of Dupont for the explicit calculations; these are somewhat tedious and unenlightening; however, he does manage to show that the Rogers dilogarithm is characterized by the Abel identity. In other words,

Lemma A.1 (Dupont): Let $f:(0,1) \to \mathbb{R}$ be a three times differentiable function satisfying

$f(s_1) - f(s_2) + f(\frac{s_2}{s_1}) - f(\frac{1-s_1^{-1}}{1-s_2^{-1}}) + f(\frac{1-s_1}{1-s_2})=0$

for all $0 < s_2 < s_1 < 1$. Then there is a real constant $\kappa$ such that $f(x) = \kappa L(x)$ where $L(x)$ is the Rogers dilogarithm (up to an additive constant).

Nevertheless, they don’t seem to have noticed the formula in terms of integrals of harmonic functions over hyperbolic triangles. Perhaps this is also well-known. Do any readers know?