Why is there a unique hyperbolic simplex of largest area?

Question

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)?

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.