A triangle is regular, provided it is equilateral, or, also, equiangular. How these conditions generalize to characterizations of regularity of simplices? In particular, it turns out that

a simplex, all of whose facets meet with the same angle, is regular.

I think I have a proof this fact, but it's not so nice and direct as one would like it to be (I'll sketch it at request, though I hesitate to do it now, as it could be somehow misleading). I suspect there is a more elegant argument, e.g. based on duality. Can you see a proof, or do you have a reference, for this and other regularity results for simplices? (A counterexample of the above statement is also very welcome, of course, provided it is false).


I am not sure what you mean by a direct proof, but here is a reasonably straightforward argument. Consider your equi-angular simplex $S$, and consider the link $L(v)$ of a vertex $v$ (the intersection of a small sphere centered at $v$ with your simplex, scaled so that the sphere has radius $1.$ The dihedral angles of $L(v)$ are equal to the corresponding dihedral angles of $S,$ so the links of all vertices are equiangular. We can show that an equiangular spherical simplex is regular by polar duality (as you had suspected): The dual simplex $T^\ast$ of a spherical simplex $T$ is one whose vertices are the face normals of $T.$ This polar duality interchanges distances with exterior dihedral angles, and transforms the Gram matrix in a very simple way, described in this paper by Kokkendorff (2005). The regular case is particularly simple, since the gram matrix has the form $I + c J,$ (where $J$ is the matrix of all ones). $J$ is idempotent, so $(I + c J)^-1 = I + c_1 J,$ by expanding the inverse in a power series (notice that in the spherical case the Gram matrix is positive definite, so $c < 1,$ so we are allowed to do this).

Anyway, when the smoke clears, we see that all of the links of our original simplex $S$ are regular spherical simplices, congruent amongst themselves. The facets of the links correspond to the links of the facets of $S,$ and the result follows by induction (the base case is the two dimensional case which you already know how to deal with).

  • $\begingroup$ Very good, thanks! This is what I hoped. Btw, the proof I had in mind started with the same construction of L(v), using then an inductive argument on the dimension (for this, however, one has to extend the initial statement to include n-dimensional "spherical simplices" that is, intersections of (n+1) n-dimensional hemispheres). $\endgroup$ – Pietro Majer Aug 23 '12 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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