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The Pythagorean theorem in hyperbolic 2-space is fundamental, and I discovered the hedronometric 3-space analogue years ago (see here), but what about dimension 4 and beyond?

As discussed in the Heron-like Strategies for Non-Euclidean Tetrahedral Volume post, my investigations into the 4-space case stalls-out because hyperbolic volume is defined in terms of the thorny Derevnin-Mednykh integral.

In the simplest (but hardly simple) case of an “isosceles” right-corner simplex with right-corner leg-faces and a regular hypotenuse-face, one arrives at these formulas for leg volume, \(L\), and hypotenuse volume, \(H\):

$$\begin{align}L &:= \frac{3}{2}\int_{1/3}^{x} \frac{1}{\sqrt{u\left(1-u\right)}} \; \mathrm{atanh}\sqrt{\frac{3u-1}{1-u}} du \\[0.5em] H &:= 6 \int_{1/3}^{x} \frac{1}{\sqrt{1-u^2}} \; \mathrm{atanh}\sqrt{\frac{3u-1}{1-u}} du \end{align}$$

for \( 1/3 \le x \le 1/2 \).  Surely, a Pythagorean relation of some kind exists (right?), but I’m at a loss for what it might be. Given the specific extreme values at \(x=1/2\)

$$L^\star := \frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^2} \sin \frac{\pi k}{2} = 0.45798\dots \qquad H^\star := \sum_{k=1}^{\infty} \frac{1}{k^2} \sin \frac{\pi k}{3} = 1.01494 \dots$$

the relationship is all-but-guaranteed to be non-trivial. (Note: \(L^{\star}\) is half of Catalan’s constant; and \(H^\star\) is, among other things, the maximum value of the Clausen function, \( \mathrm{Cl}_2 \). Maybe mentioning that here will make this blog post show up in web searches, and people who know a little about polylogarithms can give me some insights.)

Here’s a TeX’d-up version of this blind alley: “What is the Pythagorean Theorem for Right-Corner Simplices in Hyperbolic 4-Space?”

From the abstract of my note “Tetrahedra Sharing Volume, Face Areas, and Circumradius: A Hedronometric Approach”:

Volume, face areas, and circumradius sometimes determine multiple —even infinitely-many— non-isomorphic tetrahedra. Hedronometry provides a context for unifying and streamlining previous discussions of this fact.

This was my first attempt to solve someone else’s problem with hedronometry. I’m rather pleased at how well hedronometry fit the circumstances, although I believe some of the subsequent polynomial analysis could use some finessing.

The Open Question:

In the case of “doubly-bisohedral” tetrahedra (Section 2.3), is it ever possible to determine four or more non-isomorphic figures? (The best I can do is three.)

With two Heron-like formulas for tetrahedral volume in Euclidean space, it makes sense to investigate what happens in non-Euclidean —specifically, hyperbolic— space. After all, a number of hedronometric formulas (for instance, the Laws of Cosines) have counterparts in both geometries.

Unfortunately, the search for a hyperbolic hedronometric formula for volume is hindered by the fact that the most-accessible general formula (due to Derevnin and Mednykh) for hyperbolic volume is a thorny integral parameterized by the dihedral angles \(A,B,C,D,E,F\) (with \(A, B, C\) meeting at a vertex and opposing, respectively, \(D, E, F\)):

$$V = -\frac{1}{4}\int_{\theta_0-\phi}^{\theta_0+\phi}\log \frac{\cos\frac{A+B+C+\theta}{2} \cos\frac{A+E+F+\theta}{2} \cos\frac{D+B+F+\theta}{2} \cos\frac{D+E+C+\theta}{2}}{\sin\frac{A+D+B+E+\theta}{2} \sin\frac{A+D+C+F+\theta}{2} \sin\frac{B+E+C+F+\theta}{2} \sin\frac{\theta}{2}} d\theta $$

where \(\theta_0\) and \(\phi\) themselves are given by elaborate expressions in the dihedral angles.

The good news is that, because the integral is parameterized by dihedral angles, and because dihedral angles are related neatly to face and pseudoface areas by a Law of Cosines

$$\cos\frac{Y}{2} \cos\frac{Z}{2} + \sin\frac{Y}{2} \sin\frac{Z}{2} \cos A = \cos \frac{H}{2} = \cos\frac{W}{2} \cos\frac{X}{2} + \sin\frac{W}{2}\sin\frac{X}{2} \cos D$$

with dihedrals \(A, D\) between respective face pairs \(\{Y,Z\}\) and \(\{W,X\}\), we’re pretty close to Pseudo-Heronic formula: we just have to solve for the angles from the areas, and then plug the solutions in as the integral parameters. This turns out to be a symbolic nightmare (for the most part), so that expressing the Derevnin-Mednykh formula in terms of \(W, X, Y, Z, H, J, K\) seems pretty hopeless; but, we can, at least, do the solving and plugging-in numerically for tetrahedra we handle on a case-by-case basis. Of course, that amounts to a strategy, not a formula, for computing volume, but at this point, that’s the best I know how to do.

The Open Question:

What’s better than the best I know how to do?

Interestingly, there’s an aspect of tetrahedral volume analysis that has a very nice hedronometric counterpart. The determinant of (angular) Gram matrix of a tetrahedron,

$$G := \left[ \begin{array}{cccc}1 & -\cos A & -\cos B & -\cos F \\ – \cos A & 1 & -\cos C & -\cos E \\ -\cos B & -\cos C & 1 & -\cos D \\ -\cos F & -\cos E & -\cos D & 1 \end{array}\right]$$

is a fundamental quantity in any geometry. (Fun fact: in Euclidean, hyperbolic, and spherical space, the determinant’s value is respectively zero, non-positive, or non-negative.) The quantity actually figures into the Derevnin-Mednykh formula. Hedronometrically, it factors!  Specifically, in hyperbolic space:

$$-4 \overline{W_2}^2\overline{X_2}^2\overline{Y_2}^2\overline{Z_2}^2\cdot \det G = \begin{array}{c}\left( \ddot{H_2} + \ddot{J_2}\ddot{K_2} – \ddot{W_2}\ddot{X_2} – \ddot{Y_2} \ddot{Z_2}\right) \\ \cdot\left( \ddot{J_2} + \ddot{K_2}\ddot{H_2} – \ddot{W_2}\ddot{Y_2} – \ddot{Z_2} \ddot{X_2}\right) \\ \cdot \left( \ddot{K_2} + \ddot{H_2}\ddot{J_2} – \ddot{W_2}\ddot{Z_2} – \ddot{X_2} \ddot{Y_2}\right) \end{array}$$

Here, I’m using my “Morse code” shorthand for sine and cosine:

$$\overline{W_2} := \sin\frac{W}{2} \qquad \ddot{W_2} := \cos\frac{W}{2}$$

Even better, one can show that, for a perfect tetrahedron (where opposite edges are perpendicular), the three factors on the right are nicely proportional in a way that defines a symmetric value, \(M\),
$$\begin{align}
\ddot{H_2} \left( \ddot{H_2} + \ddot{J_2}\ddot{K_2} – \ddot{W_2}\ddot{X_2} – \ddot{Y_2} \ddot{Z_2}\right) &= \ddot{J_2} \left( \ddot{J_2} + \ddot{K_2}\ddot{H_2} – \ddot{W_2}\ddot{Y_2} – \ddot{Z_2} \ddot{X_2}\right) \\
&= \ddot{K_2}\left( \ddot{K_2} + \ddot{H_2}\ddot{J_2} – \ddot{W_2}\ddot{Z_2} – \ddot{X_2} \ddot{Y_2}\right) \\
&=: \ddot{H_2}\ddot{J_2}\ddot{K_2} + M
\end{align}
$$ so that we have

$$-4 \overline{W_2}^2\overline{X_2}^2\overline{Y_2}^2\overline{Z_2}^2\cdot \ddot{H_2}\ddot{J_2}\ddot{K_2} \cdot \det G = \left( \ddot{H_2}\ddot{J_2}\ddot{K_2} + M \right)^3$$

(In theory, a perfect tetrahedron is completely determined by its face areas, so we “could” eliminate references to \(H, J, K\) in this formula. However, the result here isn’t at all pretty.)

This factorization may (or may not) speak to a hedronometric dynamic locked deep within the Derevnin-Mednykh volume integral, but I suspect that the analysis requires more expertise in differential geometry than I possess at this time.

A complete discussion of all of this stuff —including an example deriving a hedronometric volume integral tailored to very specific types of tetrahedra— is in my note from 2009: “Heron-like Strategies for Non-Euclidean Tetrahedral Volume”.

Heron’s formula provides the area, \(A\), of a triangle from the lengths, \(a, b, c\), of its edges:

$$A = s (s-a)(s-b)(s-c) \qquad \text{where}\qquad s := \frac{1}{2}\left(a+b+c\right)$$

The Cayley-Menger determinant generalizes this formula and can provide the “content” of an any-dimensional simplex from the lengths of its edges, but I want something hedronometric.

We cannot expect to compute the volume, \(V\), of a tetrahedron from the areas, \(W, X, Y, Z\), of its faces. A tetrahedron has six degrees of freedom —for instance, you can leave any five edges as-is and still adjust the sixth— so that its nature cannot be captured by just four parameters. For a hedronometric “Heron’s formula” to exist, then, we have two avenues to explore:

• Introduce more faces
• Handle fewer tetrahedra

The “introduce more faces” route may sound ridiculous —a tetrahedron only has four faces!— but a hedronometrist knows that a tetrahedron has three pseudofaces, with areas \(H, J, K\), that can join the fun. This ostensibly gives us fully seven degrees of freedom, although the Sum of Squares identity imposes a dependency that reduces this to the proper six:

$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$

Lo and behold! We can derive a face-based volume formula,

$$\begin{align}81 V^4 &= 2 W^2 X^2 Y^2 + 2 W^2 Y^2 Z^2 + 2 W^2 Z^2 X^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\[0.5em] &- H^2 \left( W^2 X^2 + Y^2 Z^2 \right) – J^2 \left( W^2 Y^2 + Z^2 X^2 \right) – K^2 \left( W^2 Z^2 + X^2 Y^2 \right)\end{align}$$

which I call “Pseudo-Heron”. The form of the right-hand side suggests that it’s the determinant of some Cayley-Menger-like matrix (and I have other reasons to suspect Cayley-Menger-like ties), but I have so far been unable to construct an appropriate matrix.

An Open Question here:

What is the Cayley-Menger-like matrix whose determinant gives the Pseudo-Heron volume formula?

(By the way: I won’t go into it here, but this formula bears some resemblance to the Third Law of Cosines for non-Euclidean tetrahedra.)

The other route to a hedronometric Heron’s formula —“handle fewer tetrahedra”— raises lots of possibilities, since there are lots of families of tetrahedra admitting just four degrees of freedom.

For instance, a “right-corner” tetrahedron —with three edges meeting at a right angles at a vertex opposite “hypotenuse-face” \(W\)— has pseudoface areas computable from face areas

$$H^2 = X^2 + Y^2 \qquad J^2 = Y^2 + Z^2 \qquad K^2 = Z^2 + X^2$$

so that the Pseudo-Heron formula collapses to a single term:

$$9 V^2 = 2 X Y Z$$

Nice. Now, we could present all sorts of contrived families with face-only volume formulas, but we want something fairly general. What’s a good choice?

Early on in my hedronometric studies, I fixated on a class of tetrahedron with a special property: opposite edges determined orthogonal vectors. The property simplified a number of equations and generally made things nicer, so I called such tetrahedra “perfect”. (The mathematical literature has the less-honorific term “orthogonal”.) As it turns out, perfect tetrahedra have an interesting conceptual tie to Heron’s original formula.

(Too-)Briefly: Heron’s formula for (non-obtuse) triangle area is equivalent to the Pythagorean Theorem for right corner tetrahedra; when you prop-up a triangle in 3-space, with one vertex on each coordinate axis (you can only do this with non-obtuse triangles), then that triangle is the hypotenuse-face of a tetrahedron with a right corner at the origin. We can express the lengths of that tetrahedron’s three perpendicular edges, and the areas of its leg-faces, in terms of the lengths of the edges about the original triangle, and then use the relation \(W^2 = X^2 + Y^2 + Z^2\) to get Heron’s formula (or vice-versa). Easy.

The point is that Heron’s area formula is an application of the tetrahedral Pythagorean Theorem when we can prop-up a triangle against the coordinate axes in 3-space. It’s perhaps not-unreasonable, then, to seek a Heron’s volume formula for tetrahedra we can prop-up against the coordinate axes in 4-space. In 4-space, any pair of coordinate axes —and thus, the plane containing them— is perpendicular to the remaining pair —and the plane containing them. A proppable tetrahedron has opposing edges in such perpendicular planes: the edges determine orthogonal vectors, so a proppable tetrahedron is perfect!

Perfect tetrahedra admit only four degrees of freedom: the original six are reduced by the conditions that opposite edges are perpendicular. (If two pairs are perpendicular, then the last pair must be as well. That’s why we lose only two degrees of freedom from perfection.) As such, their volume can be determined by the areas of their faces. Unfortunately, the resulting “Heron’s formula” for that volume is nowhere-near as clean at Heron’s triangle formula (or the Pseudo-Heron tetrahedron formula); at best, we can relate volume and face areas in an enormous polynomial … essentially a quartic in \(V^3\).

The “Heron Quartic” is too complicated to give here, so I’ll just defer discussion to my note (from 2005) covering all this stuff: “Heron-like Results for Tetrahedral Volume”.

The Open Question(s) here:

What is the best way to express the Heron Quartic, and what is it trying to tell us?