## Brianchon-Gram-Sommerville and ideal hyperbolic Dehn invariants

A beautiful identity in Euclidean geometry is the Brianchon-Gram relation (also called the Gram-Sommerville formula, or Gram’s equation), which says the following: let $P$ be a convex polytope, and for each face $F$ of $P$, let $\omega(F)$ denote the solid angle along the face, as a fraction of the volume of a linking sphere. The relation then says:

Theorem (Brianchon-Gram relation): $\sum_{F \subset P} (-1)^{\text{dim}F} \omega(F)=0$. In other words, the alternating sum of the (solid) angles of all dimensions of a convex polytope is zero.

Sketch of Proof: we prove the theorem in the case that $P$ is a simplex $\Delta$; the more general case follows by generalizing to pyramids, and then decomposing any polytope into pyramids by coning to an interior point. This argument is due to Shephard.

Associated to each face $F$ is a spherical polyhedron $A(F)$ in $S^{n-1}$; if the span of $F$ is the intersection of a family of half-spaces bounded by hyperplanes $H_i$ with inward normals $n_i$, then $A(F)$ is the set of unit vectors $v \in S^{n-1}$ whose inner product with each $n_i$ is non-negative. Note further that for each $v \in S^{n-1}$ there is some $n_i$ that pairs non-negatively with $v$; consequently to each $v \in S^{n-1}$ one can assign a subset $I(v)$ of indices, so that $n_i$ pairs non-negatively with $v$ if and only if $i \in I(v)$. On the other hand, each subset $J \subset I(v)$ determines a unique face $F(J)$ of dimension $n - |J|$. By the inclusion-exclusion formula, we conclude that $\sum_{F} (-1)^{\text{dim}F}A(F)$ “equals” zero, thought of as a signed union of spherical polyhedra. Since $\omega(F) = \text{vol}(A(F))/\text{vol}(S^{n-1})$, the formula follows. qed.

Another well-known proof starts by approximating the polytope by a rational polytope (i.e. one with rational vertices). The proof then goes via Macdonald reciprocity, using generating functions.

Example: Let $T$ be a triangle, with angles $\alpha,\beta,\gamma$. The solid angle at an interior point is $1$, and the solid angle at each edge is $1/2$. Hence we get $(\alpha + \beta + \gamma)/2\pi - 3/2 + 1 = 0$ and therefore in this case Brianchon-Gram is equivalent to the familiar angle sum identity for a triangle: $\alpha + \beta + \gamma = \pi$.

Example: Next consider the example of a Euclidean simplex $S$. The contribution from the interior is $-1$, and the contribution from the four facets is $2$. There are six edges, with angles $\alpha_i$, that  contribute $\sum \alpha_i/2\pi$. Each vertex contributes one spherical triangle, with (spherical) angles $\alpha_i,\alpha_j,\alpha_k$ for certain $i,j,k$, where each $\alpha_i$ appears as a spherical angle in exactly two spherical triangles. The Gauss-Bonnet theorem implies that the area of a spherical triangle is equal to the angle sum defect: $\text{area}_{ijk} = \alpha_i + \alpha_j + \alpha_k - \pi$ so the vertices contribute $(2\sum \alpha_i - 4 \pi)/4\pi$ and the identity is seen to follow in this case too.

Note in fact that the usual proof of Gauss-Bonnet for a spherical triangle is done by an inclusion-exclusion argument involving overlapping lunes, that is very similar to the proof of Brianchon-Gram given above.

The sketch of proof above just as easily proves an identity in the spherical scissors congruence group. For $X^n$ equal to spherical, Euclidean or hyperbolic space of dimension $n$, the scissors congruence group $\mathcal{P}(X^n)$ is the abelian group generated by formal symbols $(x_0,x_1,\cdots,x_n,\alpha)$ where $x_i \in X^n$ and $\alpha$ is a choice of orientation, modulo certain relations, namely:

1. $(x_0,x_1,\cdots,x_n,\alpha)=0$ if the $x_i$ are contained in a hyperplane
2. an odd permutation of the points induces multiplication by $-1$; changing the orientation induces multiplication by $-1$
3. if $g$ is an isometry of $X^n$, then $(x_0,\cdots,x_n,\alpha) = (gx_0,\cdots,gx_n,g_*\alpha)$
4. $\sum_i (-1)^i (x_0,\cdots,\widehat{x_i},\cdots,x_{n+1},\alpha)$ for any set of $n+2$ points, and any orientation $\alpha$

(Note that this definition of scissors congruence is consistent with that of Goncharov, and differs slightly from another definition consistent with Sah; this difference has to do with orientations, and has as a consequence the vanishing of spherical scissors congruence in even dimensions; whereas with Sah’s definition, $\mathcal{P}(S^{2n}) = \mathcal{P}(S^{2n-1})$ for each $n$)

The argument we gave above shows that for any Euclidean simplex $\Delta$, we have $\sum_F(-1)^{\text{dim}F} A(F) = 0$ in $\mathcal{P}(S^{n-1})$.

Scissors congruence satisfies several fundamental properties:

1. $S^n = 0$ in $\mathcal{P}(S^n)$. To see this, “triangulate” the sphere as a pair of degenerate simplices, whose vertices lie entirely on a hyperplane.
2. There is a natural multiplication $\mathcal{P}(S^{a-1}) \otimes \mathcal{P}(S^{b-1}) \to \mathcal{P}(S^{a+b-1})$; to define it on simplices, think of $S^{a+b-1}$ as the unit sphere in $\mathbb{R}^{a+b}$. A complementary pair of subspaces $\mathbb{R}^a$ and $\mathbb{R}^b$ intersect $S^{a+b-1}$ in a linked pair of spheres of dimensions $a-1,b-1$; if $\Delta_a,\Delta_b$ are spherical simplices in these subspaces, the image of $\Delta_a \otimes \Delta_b$ is the join of these two simplices in $S^{a+b-1}$.

It follows that the polyhedra $A(F)=0$ in $\mathcal{P}(S^{n-1})$ whenever $F$ is a face of dimension at least $1$; for in this case, $A(F)$ is the join of a spherical simplex with a sphere of some dimension, and is therefore trivial in spherical scissors congruence. Hence the identity above simplifies to $\sum_v A(v)=0$ in $\mathcal{P}(S^{n-1})$.

One nice application is to extend the definition of Dehn invariants to ideal hyperbolic simplices. We recall the definition of the usual Dehn invariant. Given a simplex $P \in X^n$, for each face $F$ we let $\angle(F)$ denote the spherical polyhedron equal to the intersection of $P$ with the link of $F$. Then $D(P) = \sum_F F\otimes \angle(F) \in \oplus_i \mathcal{P}(X^{n-i})\otimes \mathcal{P}(S^{i-1})$. Ideal scissors congruence makes sense for ideal hyperbolic simplices, except in dimension one (where it is degenerate). For ideal hyperbolic simplices (i.e. those with some vertices at infinity), the formula above for Dehn invariant is adequate, except for the $1$-dimensional faces (i.e. the edges) $e$. This problem is solved by the following “regularization” procedure due to Thurston: put a disjoint horoball at each ideal vertex of $P$, and replace each infinite edge $e$ by the finite edge $e'$ which is the intersection of $e$ with the complement of the union of horoballs; hence one obtains terms of the form $e' \otimes \angle(e)$ in $D(P)$. This definition apparently depends on the choice of horoballs. However, if $H,H'$ are two different horoballs, the difference is a sum of terms of the form $c \otimes \angle(e)$ where $c$ is constant, and $e$ ranges over the edges sharing the common ideal vertex. The intersection of $P$ with a horosphere is a Euclidean simplex $\Delta$, and the $\angle(e)$ are exactly the spherical polyhedra $A(v)$ as $v$ ranges over the vertices of $\Delta$. By what we have shown above, the sum $\sum_v A(v)$ is trivial in scissors congruence; it follows that $D(P)$ is well-defined.

For more general ideal polyhedra (and finite volume complete hyperbolic manifolds) one first decomposes into ideal simplices, then computes the Dehn invariant on each piece and adds. A minor variation of the usual argument on closed manifolds shows that the Dehn invariant of any complete finite-volume hyperbolic manifold vanishes.

Update(7/29/2009): It is perhaps worth remarking that the Brianchon-Gram relation can be thought of, not merely as an identity in spherical scissors congruence, but in the “bigger” spherical polytope group, in which one does not identify simplices that differ by an isometry. Incidentally, there is an interesting paper on this subject by Peter McMullen, in which he proves generalizations of Brianchon-Gram(-Sommerville), working explicitly in the spherical polytope group. He introduces what amounts to a generalization of the Dehn invariant, with domain the Euclidean translational scissors congruence group, and range a sum of tensor products of Euclidean translational scissors congruence (in lower dimensions) with spherical polytope groups. It appears, from the paper, that McMullen was aware of the classical Dehn invariant (in any case, he was aware of Sah’s book) but he does not refer to it explicitly.

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### 4 Responses to Brianchon-Gram-Sommerville and ideal hyperbolic Dehn invariants

1. Gil Kalai says:

Dear Danny, This is a very interesting post and I wish I could understand better the hyperbolic things. Peter McMullen wrote quite a few papers on valuations on convex sets (include the Dehn Invariants) and on dussections
(e.g. P. McMullen, Valuations and dissections, in Handbook of Convex Geometry (P.M. Gruber and J.M. Wills, eds.), North-Holland, Amsterdam, 1993, 933–988. MR1243000 (95f:52018).) He had a well known conjecture on characterization of translation invariant valuations (those referred to in the update) that was settled by Alesker.
(Somehow it was not clear to me if Dehn invariants are important for hyperbolic manifolds (as volume is) or they all vanish for all of them. )

(There are some nice questions I know regarding Dehn’s invariants: 1) (due to Bob Connelly is if all Dehn’s invariant are fixed for a flexing of a flexible polyhedron; this is known for the volume by Sabitov; 2) If the Dehn invariants (or some of them) can be read from the eigenvalues of the Laplacian like the volume can.)

2. Danny Calegari says:

Dear Gil – thank you for the very interesting pointer to work of McMullen and Alesker. It appears that some of their work is very close to things that I care a lot about, and yet – 1. I was completely unaware of it, and – 2. it is apparently very well-known. I wonder what else is so close but completely out of my field of vision!

If I understand the situation correctly, the Hadwiger invariants are sufficient to determine translational scissors congruence for polyhedra (maybe this was shown by Hadwiger?) and Sah determined all the relations between Hadwiger invariants. What Alesker apparently does (his paper is very interesting) is to characterize translation-invariant continuous (complex-valued) valuations on the space of *all* convex compact subsets (of some Euclidean space, where continuity is with respect to Hausdorff distance).

Again, if I understand the situation correctly, Sabitov’s proof of the bellows conjecture works by showing that one can compute the volume enclosed by a Euclidean polyhedron as a root of a certain polynomial whose coefficients are determined by the edge lengths; the volume is therefore determined up to finite ambiguity, and is “locally constant” on the space of flexings of the polyhedra (if any). What is (very) clever about this is that it nowhere uses the hypothesis that the polyhedron is flexible. I think a dimension count, and Schlafli’s formula shows that “formally” the Dehn invariant of a polyhedron depends on as many parameters as there are edges (maybe the polyhedron needs to be a sphere?) so I guess a similar proof for Dehn invariants is conceivable, but such a dimension count also suggests that most polyhedra are not flexible, so it is probably not much of a guide to intuition. Are there flexible hyperbolic/spherical polyhedra in dimension 3? The same dimension count holds there . . .

The fact that Dehn invariants (other than volume) vanish for hyperbolic manifolds is, in fact, very important, since the kernel of the Dehn invariant on hyperbolic scissors congruence is closely related to the Bloch group, which is an important object in algebraic K-theory (of course, this is one of the themes of Goncharov’s paper). Some aspects of this theory are worked out for K_3 (of number fields) by Walter Neumann and Jun Yang, and others.

Best,

Danny

3. Gil Kalai says:

“If I understand the situation correctly, the Hadwiger invariants are sufficient to determine translational scissors congruence for polyhedra (maybe this was shown by Hadwiger?) and Sah determined all the relations between Hadwiger invariants.”

The information is wiki article on Hilbery 3rd problem (which is consistent more or less with my memory)
Hilbert’s third problem
is that this is still open in dimensions d>4 . It was proved in dimension 3 by Sydler in in dimension 4 by Jessen.

4. Danny Calegari says:

Dear Gil – I fixed the link to the wikipedia article in your comment.

I have been a bit sloppy with definitions above. There are in fact several variations of what is meant by scissors congruence. Given a space X in which one has polyhedra (usually, Euclidean, hyperbolic or spherical space) there are several different groups of equivalence classes of polyhedra that one calls scissors congruence groups. One always allows cut and paste of polyhedra (the scissors part) as equivalence, but one also identifies pairs of polyhedra that differ by the action of an element of some fixed (pseudo)-group G acting on X (usually by isometries). The case of classical scissors congruence (that Dehn studied) is when one takes G to be the full group of isometries. You’re quite right that the question of whether Dehn’s invariants are a complete set of invariants of classical scissors congruence is only known in Euclidean space in dimension up to 4 (the Dehn-Sydler-Jessen theorem) and in hyperbolic/spherical space in dimension up to 2 (!). Rich Schwartz has some great notes on this, and a Java applet illustrating a key step in the proof, at his webpage.

One also studies, in Euclidean space, the translational scissors congruence, where one is only allowed to move polyhedra around by translation (and, in some versions, by an antipodal “flip” x -> -x). In this context, Hadwiger defined his invariants, and showed that they are a complete invariant (in every dimension). Sah later found (I think) the complete set of relations between Hadwiger’s invariants, thus explicitly calculating the translational scissors congruence groups. I mentioned translational scissors congruence in my comment, because I think this is closest to what Alekser studies.

Incindentally, in my post, I also distinguished between two flavors of spherical scissors congruence — the “classical” spherical scissors congruence group, where one allows all isometries, and the spherical polytope group, where one allows no isometries (except the identity).

Best,

Danny