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?


7 Answers 7


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.

  • $\begingroup$ Your third inequality contains a misprint I guess. $\endgroup$
    – Petya
    Feb 25, 2010 at 17:09
  • $\begingroup$ Right. Fixed now. $\endgroup$
    – Igor Pak
    Feb 25, 2010 at 17:54

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

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.


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.

  • $\begingroup$ My understanding is that if the faces each satisfy the triangle inequality and the Cayley-Menger determinant fir the volume squared with all six edges gives a positive number, we will make the tetrahedron. $\endgroup$ Sep 13, 2022 at 15:30

To supplement @JosephMalkevitch's net description:

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


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.


One observation to be made is a tetrahedron does not necessarily exist with an arbitrary set of dihedral angles. As the exercise shows below, we must have the dihedral angles at each vertex sum to at least $180°$. They must also satisfy the triangle inequality-like relation: at each vertex the sum of any two dihedral angles must be equal to or less than the third plus $180°$. Equality with these criteria is acceptable if we allow degenerate cases.

We think of the tetrahedron, which may be non-Euclidean, as being formed from a space enclosed by four planes or four congruent spheres. For such an enclosure to occur there must be a correct combination of dihedral angles.

Consider the case of all dihedral angles being equal. We model this case as four congruent spheres whose radius is taken as one unit and whose centers are at the vertices of a Euclidean regular tetrahedron. If we place the spheres so they just touch each other, a distance of two units between sphere centers, that corresponds to zero dihedral angles ... but the tetrahedron does not form. There are triangular gaps through which a small bug initially trapped between the spheres can crawl or fly out.

To really trap the bug and enclose a tetrahedron, we must move the centers closer allowing the spheres to overlap. We finally close off the gaps when the center-to-center distance is reduced to $\sqrt3$ units. If we then consider the dihedral angles formed locally along the edge of any small circle where two spheres intersect, we discover that this minimal dihedral angle for a tetrahedron to close is $60°$ or $\pi/6$ radians. (We note that each vertex has three of these dihedral angles coming together and these add up to $180°$, a sum we shall encounter later on.) Clearly the faces of this tetrahedron curve away from each other and the angles of the triangular faces are decreased as if in a hyperbolic space; this is a hyperbolic tetrahedron.

If we continue to push the sphere centers closer together increasing the dihedral angles beyond $60°$, the curvature of the faces and decrement of the vertex angles become less; the tetrahedron tends towards Euclidean. The Euclidean shape is obtained as a limit when the enclosed region shrinks towards a point (the overall center of the set of spheres); we may recover the Euclidean tetrahedron with dihedral angles approximately $70°32'$ (actually, $\cos^{-1}(1/3)$) by regarding the spheres as infinitesimally removed from the overall center and looking close-up at the resultant limiting shape.

If we go on so that now the spheres overlap and the enclosed region now curves positively around the collective center, then we have an elliptic tetrahedron, in this case with equal dihedral angles exceeding the $70°32'$ figure cited above. We maintain an enclosed tetrahedron all the way until the centers of the individual spheres coincide, in which case the dihedral angles hit their maximum of $180°$ and the tetrahedron is now the whole sphere (which may be considered a degenerate case). We conclude that with equal dihedral angles, a tetrahedron exists only for angle values between $60°$ and $180°$, including endpoints if we allow degenerate cases.

How do the planar angles at vertices on the faces correlate with the dihedral angles and the hyperbolic/Euclidean/elliptic character of the tetrahedron as cited above? Let's examine a hyperbolic case from above. We set the dihedral angles all to $65°$ and then imagine each vertex as the center of a small sphere. This small sphere will intersect the tetrahedron to form a spherical triangle, whose angles are the given dihedral angles around the vertex and whose arcs measure the planar angles at the vertices.

Before solving, we first note that the dihedral angles, each rendered as an angle of a spherical triangle, must add up to more than $180°$ (exactly $80°$ would he acceptable in the bounding, degenerate case). That checks. We also note that each dihedral angle of $65°$ plus $180°$ is properly greater than the third such angle plus ($2×65°=130°$). So we have a soluble case at every vertex according to spherical geometry.

With all dihedral angles set to $65°$ in this regular example, we apply the Spherical Law of Cosines and infer that the arcs of the small spherical triangle around each vertex measures approximately $42°57'$. Therefore the planar angkes at the vertices are also $\approx42°57'$. Since this applies at all three corners of any face, the angles of the triangles sum to less than $180°$, as in hyperbolic geometry. To get the sum up to $180°$ (three $60°$ arcs in the small sphere surrounding each vertex) and realize a Euclidean tetrahedron by this method, we find that the dihedral angles must be bumped up to $\cos^{-1}(1/3)\approx70°32'$, corresponding exactly to the enclosed region becoming infinitesimal and its faces simultaneously Euclidean-planarized. Once beyond this point the sums of angles of all triangular faces simultaneously go above $180°$ as the entire tetrahedron goes over to elliptic. For instance, in the regular case if all dihedral angles are $90°$ we find the same is true of the planar angles at the vertices, so the sum of these angles at every face is $270°>180°$.

Let us look more briefly at a slightly more complicated example. The dihedral angles at $\overline{AB},\overline{AC},\overline{AD}$ are fixed at $90°$ and the remaining (basal) dihedral angles are equal. The spherical model for this contains three fixed unit spheres with center-to-center distances $\sqrt2$ forming the fixed $90°$ dihedral angles with their pair wise intersections; the fourth sphere moves in with its center along thectgreefold axis. The tetrahedron first closes when the basal dihedral angles reach $45°$, where all vertices initially satisfy the $180°$ sum requirement. It becomes infinitesimal and Euclidean, with all triangular faces simultaneously achiving planar-angle suns of $180°$, at basal dihedral angles of $\cos^{-1}(1/\sqrt3)\approx54°44'$. It then becomes elliptic for larger basal dihedral angles ... up to $135°$ where the basal vertices have two dihedral angles adding up to the third one plus $180°$ ($135°+135°=90°+180°$). If we try to increase the dihedral angke further, the tetrahedron breaks; the enclosed region suddenly needs a fifth face. Thereby a $C_{3v}$ tetrahedron with $90°$ lateral dihedral angles exists for basal angles between $45°$ (where the basal vertices vertices first satisfy the $\ge180°$ sum requirement) and $135°$ (where the basal vertices hit the limit of two dihedral angkes summing to no more than the third plus $180°$).


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