Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Why is there a unqiue ideal $n$-simplex in $\mathbb H^n$ with largest volume for $n\geq 3$?

For $n=3$, this is a standard calculation, and for larger dimensions is much harder (see Haagerup and Munkholm); I made the mistake of originally stating that this is easy for all $n$. Anyway, I'm wondering if there is a deeper reason for this fact (one which does not rely on calculation); let me say what type of answer I am looking for.

Recall that this is a crucial step in Gromov's proof of Mostow Rigidity, a (very rough) sketch of which is as follows. If we have two cocompact subgroups $\Gamma_1,\Gamma_2\subseteq\operatorname{Isom}(\mathbb H^n)$ which are isomorphic as abstract groups, then such an isomorphism must extend to a homeomorphism of their boundaries (as hyperbolic groups). In other words, we get a self-homeomorphism of $\partial\mathbb H^n$. But now since simplicial volume is a homotopy invariant, this homeomorphism must preserve the $(n+1)$-tuples of points giving simplices of maximal volume. Then one proves that any self-homeomorphism of $\partial\mathbb H^n$ with this property is in fact induced by an element of $\operatorname{Isom}(\mathbb H^n)$, so $\Gamma_1,\Gamma_2$ are conjugate (via this element) in $\operatorname{Isom}(\mathbb H^n)$.

Is there a high-brow proof of the fact in the title of this question? (that is, one which uses some rigidity results in Lie groups). A first step at answering this question would be to realize that the configuration space of $n+1$ points in $\partial\mathbb H^n$ modulo isometries is nontrivial iff $n\geq 3$. Now volume is some real-analytic function on this moduli space. Is there a nice explanation for why it miraculously has a unique global maximum (in fact, a unique local maximum!) (which thus implies Mostow rigidity)?

share|cite|improve this question
Jungreis' paper may be relevant, although he indirectly is appealing to Haagerup-Munkholm. –  Ian Agol Dec 30 '11 at 1:24
Note that the volume is not a continuous function on the entire moduli space of the $n+1$ points: you need to remove the degenerate configurations. –  Bruno Martelli Dec 30 '11 at 8:22
(that is, those where the $n+1$ points are contained in a hyperplane) –  Bruno Martelli Dec 30 '11 at 8:22
Actually, this is more or less the only problem that can occur. If you restrict yourself to $(n+1)$-uples of points where at least 3 of them are distinct, then the volume function is continuous, see Luo, Proposition 4.1 (or Ratcliffe). However, the configuration space is not compact once you removed these bad configurations, so it is not even a priori obvious that there exists a simplex of maximum volume. –  Bruno Martelli Dec 30 '11 at 11:10
There is a proof of the maximality of regular simplices using symmetrization due to Peyerimhoff: One may give a purely synthetic argument for the maximality of ideal regular hyperbolic 3-simplices this way. I don't know if this counts as a "high-brow" proof though. –  Ian Agol Dec 30 '11 at 19:29

1 Answer 1

Firstly, this is not something you "can easily prove by calculation": The proof (by Haagerup and Munkholm) was published in Acta.

Secondly, in three dimensions, the set of ideal simplices are parametrized by positive triples $\alpha, \beta, \gamma$ such that $\alpha + \beta + \gamma = \pi$ -- these are the dihedral angles. One can then show that the volume is a convex function of the dihedral angles (this is actually true for any convex ideal polyhedron, but the tetrahedron case is at the base of the proof, and is proved differently). This immediately implies that the regular ideal simplex is the one of maximal volume (by symmetry considerations) -- this result (in greater generality) is in a paper of mine in the early nineties:

Euclidean structures on simplicial surfaces and hyperbolic volume I Rivin The Annals of Mathematics 139 (3), 553-580

In higher dimensions the argument does not quite work, but the fact that a simplex is determined by its codimension-two areas, together with the Schlafli differential formula implies that the volume (as the function of dihedral angles) should have the same signature everywhere, and since the volume function is proper on the set of ideal simplices, it should have a maximum, and so should be concave. This, however, does not quite prove Haagerup-Munkholm, since the set of possible dihedral angles is much more complicated in dimensions bigger than three, and is not obviously convex, so the symmetry argument breaks down.

share|cite|improve this answer
Oops, looks like I'm too focused on $n=3$ and have forgotten that things might be different in higher dimensions . . . I've edited the question which now hopefully gives a more accurate picture. –  John Pardon Dec 29 '11 at 20:57
I think Schlafli's formula for 4-D hyperbolic simplices actually gives an affine function of the sum of the dihedral angles, since the volume of the codim 2 faces is always $\pi$. But I'm not sure how one uses this to deduce the regular ideal 4-simplex is maximal volume, since as you say, the moduli space of dihedral angles is complicated. –  Ian Agol Dec 30 '11 at 17:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.