MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an analogue for the following 2 dimensional fact:

Given 3 angles $\alpha,\beta,\gamma\in (0;\pi)$ there is always a triangle with these prescribed angles. It is spherical/euclidean/hyperbolic, iff the angle sum is smaller than /equal to/bigger than $\pi$. And the length of the sides (resp. their ratio in the Euclidean case) can be computed with the sine and cosine law.

The analogous problem in 3 dimensions would be:

Assign to each edge of a tetrahedron a number in $(0;\pi)$. Does there exists a tetrahedron with these numbers as face angles at those edges. And when is it spherical/euclidean/hyperbolic. Is there a similar Invariant to the angle sum? And are there formulas to compute the length of the edges?

share|cite|improve this question
up vote 16 down vote accepted

The short answer is no - there is no single inequality criterion. Already in $\mathbb{R}^3$ everything is much more complicated. Let me give a sample of inequalities the angles should satisfy. Denote by $\gamma_{ij}, 1\leq i < j \leq 4$ the six dihedral angles of a Euclidean tetrahedron. Then: $$ \gamma_{12}+\gamma_{23} + \gamma_{34}+\gamma_{14} \le 2 \pi $$ $$ 2\pi \le \gamma_{12} + \gamma_{13} + \gamma_{14}+\gamma_{23} + \gamma_{24}+\gamma_{34} \le 3\pi $$ $$ 0 \le \cos \gamma_{12} + \cos\gamma_{13} + \cos\gamma_{14}+ \cos\gamma_{23} + \cos\gamma_{24}+ \cos\gamma_{34} \le 2 $$ (See my book ex. 42.27 for the proofs of these inequalities - they are not terribly difficult, so you might enjoy proving them yourself).

This shows that the set of allowed sixtuples of angles is rather complicated (for spherical/hyperbolic tetrahedra with angles close to $\gamma_{ij}$, these angles will have to satisfy these inequalities as well). The "invariant" you mention corresponds to the unique equation the angles satisfy in the Euclidean space. The latter is also rather delicate: it is the Gauss-Bonnet equation $\omega_1+...+\omega_4=4\pi$, where $\omega_i$ is the curvature of $i$-th vertex - you need to use spherical cosine theorem to compute it from dihedral angles (see e.g. Prop. 41.3 in my book).

Finally, you might like to take a look at this interesting paper by Rivin, to see that a similar generalization of the triangle inequality is just as difficult. To answer your last question (edge lengths from dihedral angles), yes, this is known. I am not an expert on this, but I would start with this recent paper.

share|cite|improve this answer
Your third inequality contains a misprint I guess. – Petya Feb 25 '10 at 17:09
Right. Fixed now. – Igor Pak Feb 25 '10 at 17:54

I am not sure if the 3-dimensional problem formulated in the question is the proper analogue of the 2-dimensional one for triangles - essentially because of the appearance of Dehn invariants and such. At least the following modification of the question can be answered using the results of Dupont and Sah:

Given a combination of side lengths and dihedral angles $\sum l_i\otimes \frac{\theta_i}{2\pi}\in\mathbb{R}\otimes(\mathbb{R}/\mathbb{Z})$, is there a euclidean polytope having this element as Dehn invariant?

The answer is given by an exact sequence which you can find in Section 4 of J.L. Dupont and C.-H. Sah: Homology of euclidean groups of motions made discrete and euclidean scissors congruences. Acta Math. 164 (1990), 1--27:

$$ 0\to \mathcal{P}(\mathbb{R}^3)/\mathcal{Z}_2(\mathbb{R}^3)\stackrel{D}{\longrightarrow} \mathbb{R}\otimes(\mathbb{R}/\mathbb{Z}) \stackrel{J}{\longrightarrow} H_1(SO(3),\mathbb{R}^3)\to 0 $$ In this sequence, $\mathcal{P}(\mathbb{R}^3)$ are scissors congruence classes of polytopes in $\mathbb{R}^3$, $\mathcal{Z}_2(\mathbb{R}^3)$ are the scissors congruence classes of prisms, $D$ is the Dehn invariant and $J(l\otimes \frac{\theta}{2\pi})= \frac{1}{2}l\frac{d\cos\theta}{\sin\theta}$ using the identification $H_1(SO(3),\mathbb{R}^3)\cong\Omega^1_{\mathbb{R}}$ with absolute Kähler differentials.

So, if you are given the six dihedral angles for the tetrahedron, it is at least in principle possible to figure out if there are six side lengths which give a realizable Dehn invariant. Unfortunately the theorem does not tell you if the Dehn invariant will be realizable by a tetrahedron - the theorem generally does not tell you how to construct the polytope realizing the Dehn invariant...

Anyway, there are analoguous exact sequences for hyperbolic and spherical scissors congruence classes. For hyperbolic scissors congruences you get in particular $$ \mathcal{P}(\mathbb{H}^3)\stackrel{D}{\longrightarrow}\mathbb{R}\otimes(\mathbb{R}/\mathbb{Z})\to H_2(SL_2\mathbb{C},\mathbb{Z})^-\to 0 $$ where $\mathcal{P}(\mathbb{H}^3)$ is the group of scissors congruence classes in hyperbolic 3-space, and $H_2(SL_2\mathbb{C},\mathbb{Z})^-$ is the $-1$-eigenspace of complex conjugation on $H_2(SL_2\mathbb{C},\mathbb{Z})$. For spherical scissors congruence the $+1$-eigenspace appears. This can be found in papers of Dupont, Sah, or the book "Scissors congruences, group homology and characteristic classes" by J.L. Dupont. The map $\mathbb{R}\otimes(\mathbb{R}/\mathbb{Z})\to H_2(SL_2\mathbb{C},\mathbb{Z})^-$ can be identified with the reduction $S^2\mathbb{C}^\times\to K_2(\mathbb{C})$ from a symmetric square of the units of $\mathbb{C}$ to $K_2$, though that may not necessarily be considered explicit. At least, this tells you that there is a precise obstruction to realizing a linear combination of side lengths and dihedral angles as the Dehn invariant of some hyperbolic or spherical polytope. It is probably further significant work to produce precise conditions for realizability by tetrahedra.

share|cite|improve this answer

The question seems a little confused, in particular since the OP is asking about dihedral angles but is calling them face angles. In any case, despite the gloom in the accepted answer, a lot is known. In particular, the dihedral angle gram matrix comes from a Euclidean, hyperbolic, spherical tetrahedra if and only if the signature is $(0, 3)$ $(1, 3)$, $(4, 0)$ respectively. Further, in the Euclidean case, the Gauss map maps the tetrahedron onto the surface of the unit sphere. The triangles of the induced cell decomposition have sides equal to the exterior dihedral angles, and their areas can be computed using the spherical theorem of cosines. The exact analogue of the "sum of angles is $\pi$" relation is that the sum of the areas of these triangles is $4\pi.$ This is not a sufficient condition in this case. There are also side conditions to the effect that the sum of the face angles (which can be computed from the dihedral angles) of each face is $\pi.$
In the hyperbolic and spherical cases, the Gaussian image (see my thesis, or its write-up as Rivin-Hodgson, Inventiones) should be a spherical cone manifold, whose angles are either smaller (spherical) or bigger (hyperbolic) than $2\pi,$ plus, in the hyperbolci case, another condition on the length of the shortest closed geodesic (should be longer than $2\pi.$

For the generalization of the Gram matrix condition to arbitrary convex polyhedra, seem Raquel Diaz-Sanchez' thesis:

 A characterization of Gram matrices of polytopes
 R Díaz - Discrete & Computational Geometry, 1999 - Springer
share|cite|improve this answer

There is an article by K. Wirth and A. Dreiding which you might find helpful:

Edge lengths determining tetrahedrons

Elmente der Mathematik, volume 64, (2009) 160-170.

The the title talks about edge lengths, but the approach taken involves taking a triangle drawn in the plane and placing three triangles along its edges to form a "net" with which to try to fold the result into a tetrahedron. The paper discuss circumstances under which this can be done.

share|cite|improve this answer

There has been work on Gram matrices that appears relevant, see e.g. Theorems 14-5 on p. 24-5.

Also of peripheral interest: there is a hyperbolic generalization of the Dehn invariant. But as far as I can tell this sort of thing can't really be a generalization of any 2D construction.

share|cite|improve this answer

To supplement @JosephMalkevitch's net description:

      Fig.25.27 in Geometric Folding Algorithms, p.406.

share|cite|improve this answer

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.