Volume form on a hyperbolic manifold with geodesic boundary

Let $M$ be a compact connected orientable Riemannian $n$-manifold with boundary $\partial M\ne\emptyset$. Since $H^n(M,\mathbb R)=0$, the connecting morphism $\delta: H^{n-1}(\partial M,\mathbb R)\to H^n(M,\partial M,\mathbb R)$ of the exact sequence of the pair $(M,\partial M)$ is an epimorphism, implying that there is an $(n-1)$-form $\alpha$ on $\partial M$ such that volume form of $M$ is $\delta \alpha$. Then $\operatorname{vol}(M)=\delta \alpha([M])=\alpha([\partial M])$.

Let $M$ be hyperbolic with a geodesic boundary now. Did someone figure out some nice general formula for $\alpha$ in that case? $\alpha$ cannot be the volume form on $\partial M$, at least for $n=2$, since the area of $M$ in that case is a multiple of $\pi$, while the length of $\partial M$ can be arbitrary. In fact, for $n=2$ this is the subject of Gauss-Bonnet theorem, so I guess I am asking for a generalization of it to hyperbolic manifolds with geodesic boundary for $n>2.$

• There is something wrong with the sequence you wrote. You presumably mean that there is an epimorphism $$H^{n-1}(\partial M)\to H^n(M,\partial M),$$ which would also fit with the remainder of the question. – ThiKu Dec 13 '15 at 8:55
• More importantly, the above epimorphism $\delta$ is not given by the differential $d$, but it is the connecting morphism of the long exact sequence. You may visualize it by observing that – ThiKu Dec 13 '15 at 9:01
• application to a relative cycle $z$ yields $$\delta \alpha(z)=\alpha(\delta z}$$, which in particular implies your equality to $Vol(\partial M)$ when $\alpha$ is the preimage of the volume form. – ThiKu Dec 13 '15 at 9:03
• Thanks, ThiKu! I reflected your comments in my edit. I dont think though that $\alpha$ can be taken to be the volume form on $\partial M.$ Cut a closed hyperbolic surface by a non-seperating geodesic. The surface area will remain a multiple of $\pi,$ while the length of the geodesic may be arbitrary. – Adam Dec 13 '15 at 15:23
• The point is that you can not apply Stokes' Theorem to $\delta\alpha$ because this is not $d\alpha$. – ThiKu Dec 13 '15 at 16:05

I don't know how to answer your question in the spirit posed, but what you're asking for is a function $f:\partial M\to \mathbb{R}$ such that $\int_{\partial M} fdA = vol(M)$, since any $(n-1)$-form on $\partial M$ is of the form $fdA$.

Such a function must be determined globally. I can imagine many ways to write a formula for such a function. One way is to form a Voronoi region about the boundary, by taking the open neighborhood of points around the boundary which have a unique closest point on the boundary. Then this region may be written as a function $g:\partial M \to \mathbb{R}$ determining the length of the longest segment starting perpendicularly from the point and lying in the Voronoi cell, and then $f$ may be computed from $g$ by elementary calculus and Fermi coordinates.

There should be another formula determined from the orthospectrum. See this paper for a volume formula in terms of orthospectrum. This may be turned into an integral over $\partial M$ by a sum over the orthospectrum of certain functions which are rotationally symmetric and centered at the endpoints of orthogeodesics. Each orthogeodesic determines a pair of planes in the universal cover of M. A point on one plane determines a region in the unit tangent bundle, which is the set of geodesics emanating from the point which hit the other boundary plane. Then the function is the volume of this region, which is determined just by the length of the orthogeodesic and the distance from the point to the foot of this orthogeodesic. Then one sums over such functions.

• Note that Basmajian-stye formulas work for geodesic boundary only (and, by the way, here is a fairly elementary approach to this and related results: arxiv.org/abs/1404.1583) – Igor Rivin Dec 15 '15 at 4:54

This is a partial answer for $n=2k$ even. Let $e(\nabla^{TM})$ be the Gauss-Bonnet-Chern integrand on the hyperbolic manifold $M$. By Hirzebruch proportionality, $e(\nabla^{TM})=(-1)^k\frac 2{\mathrm{vol}(S^n)}\,d\mathrm{vol}_M$.

Next, you need a nowhere vanishing vector field $V$ on $M$. To construct it, start with an arbitrary vector field $M\to TM$ that crosses the zero section $M\subset TM$ transversely. Then it has finitely many zeros. Assuming that $M$ is connected, we can pull each zero $p$ of $V$ out of $M$ along a path linking $p$ to $\partial M$.

There is the so-called Mathai-Quillen current $\psi$ on the total space of $\pi\colon TM\to M$ such that $d\psi=\pi^*e(\nabla^{TM})-\delta_M$. It can be computed locally using only the connection $\nabla^{TM}$. Hence, $V^*\psi$ is a smooth form on $M$ satisfying $d(V^*\psi)=e(\nabla^{TM})$. By Stokes' theorem, the form $\beta=V^*\psi|_{\partial M}$ satisfies $$\delta[\beta]=\bigl[e(\nabla^{TM})\bigr]\in H^n(M;\mathbb R)\;.$$

In other words, you get $$\alpha=(-1)^k\frac{\mathrm{vol}(S^n)}2\,\beta =(-1)^k\frac{\mathrm{vol}(S^n)}2\,V^*\psi|_{\partial M}$$ from a section of $TM|_{\partial M}\cong T(\partial M)\oplus\mathbb R$ without zeros that can be continued to $M$ without zeros, and from the local geometry of $\partial M$. In particular, the form $\alpha$ will depend on the manifold $M$, not on $\partial M$ alone. You can try to draw pictures for $V$ on hyperbolic surfaces with boundary to see this.