15
$\begingroup$

My answer to the "Favorite equations" question was the Pythagorean theorem for right-corner tetrahedra:

Euclidean: $A^2+B^2+C^2=D^2$

Hyperbolic: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}−\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$

Spherical: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$

where $A$, $B$, $C$ are the areas of the "leg-faces" and $D$ is the area of the "hypotenuse-face".

For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex.

I mentioned not knowing the non-Euclidean counterparts of the generalization. After a couple of days of toying, I find these counterparts elusive. (It would probably help if I were better-versed in differential geometry.) So, I turn to MO to ask:

What are the non-Euclidean analogues in higher dimensions?
(I'm particularly interested in the case relating volumes in a hyperbolic 4-simplex.)

While the Pythagorean Theorems for Euclidean Simplices progress in a straightforward manner (just add another leg-content-square), the Pythagorean Theorem for Hyperbolic Tetrahedra already diverges somewhat dramatically from its 2-dimensional counterpart, $\cosh a \cosh b = \cosh c$. So, it's not even clear to me what form the relations would take in general.

Edit to add a couple of references
"The laws of cosines for non-Euclidean tetrahedra" (pdf file) (by me) derives a Law of Cosines for which each non-Euclidean Pythagorean Theorem is a special case. I do not know if these results exist elsewhere in the literature. (BTW: There's a lot of unnecessary equation manipulation shown; I was using this as an opportunity to learn LaTeX. :)

A passage in above jumps into a discussion of "pseudofaces" by making passing reference to the Euclidean case. You can read about Euclidean pseudofaces and how I find them useful here:
"Heron-like results for [Euclidean] tetrahedral volume" (by me).

Edit2: A Special Case (Edit 3: A strikingly-similar pair of volume formulas.)

Consider the special case of a right-corner hyperbolic $4$-simplex whose "legs" are congruent right-corner tetrahedra with isosceles right-triangle faces; the "hypotenuse" is a regular tetrahedron with equilateral faces. Each face of the simplex's hypotenuse is an hypotenuse-face of one of the simplex's legs.

Let the volume of each leg-tetrahedron be $L$ and the volume of the hypotenuse-tetrahedron be $H$. Then we have these formulas:

$$L=3 \int_{\rm{acos}\sqrt{x}}^{\rm{acos}\sqrt{\frac{1}{3}}} \rm{atanh}\sqrt{\frac{3\cos^2t-1}{1-\cos^2t}} \, \mathrm{d}t$$

$$H=6\int_{\rm{acos}x}^{\rm{acos}\frac{1}{3}} \rm{atanh}\sqrt{\frac{3\cos t-1}{1-\cos t}} \, \mathrm{d}t$$

where $\frac{1}{3} \le x = \cos^2{\theta_L} = \cos{\theta_H} \le \frac{1}{2}$, for $\theta_L$ the acute dihedral angle in the leg-tetrahedron and $\theta_H$ the dihedral angle in the hypotenuse-tetrahedron.

The similarity in form is rather intriguing, though not necessarily encouraging: swapping "$\cos^2 t$" for "$\cos t$" in an integrand, or "$\sqrt{x}$" for "$x$" in a limit, can completely change the nature of an integral, so there's no reason to expect a straightforward relationship between these formulas. And yet, Pythagoras beckons: there must be some connection here!

Now, it's possible to write series for $H$ and $L$, then invert the second series, then substitute back in to the first series to arrive at a series for $H$ in terms of $L$.

$$H = \frac{4}{2!\;3!} M^3 + \frac{18}{3!\;5!}M^5 - \frac{ 918 }{4!\;7!}M^7 + \frac{24786}{5!\;9!}M^9 - \frac{ 6018759 }{8 \cdot 6!\;11!} M^{11} - \frac{ 8233607961 }{80\cdot 7!\;13!} M^{13} - \cdots $$ where $M := (6L)^{1/3}$. This isn't the relation I'm seeking, but observe that, for infinitesimal $L$, we have $H \approx 2 L$; that is, $H^2 \approx 4L^2 = L^2 + L^2 + L^2 + L^2$, which is the corresponding Pythagorean relation for Euclidean $4$-simplices. Even so, despite my best efforts of playing with these series, I have yet to get any insights into the nature of a non-infinitesimal connection.

I'll close here by mentioning a special-special case: at the extreme, a quadrupally-asymptotic right-corner $4$-simplex has legs that are triply-asymptotic right-corner tetrahedra with doubly-asymptotic right-triangle leg-faces; the simplex's hypotenuse is a quadrupally-asymptotic regular tetrahedron with triply-asymptotic equilateral faces. The volumes of the components (with parameter $x = 1/2$) attain significant values that happen to have interesting series representations of their own:

$$L^\star := \frac{1}{2} \; \Im\left(Li_2\left(\exp\frac{i\pi}{2}\right)\right) = \frac{1}{2} \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{2}} = 0.45798\dots$$

$$H^\star := \Im\left(Li_2\left(\exp\frac{i\pi}{3}\right)\right) = \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{3}} = 1.01494\dots$$

... where $Li_2$ is the dilogarithm.

Here, $L^\star$ is half of Catalan's constant, and $H^\star$ is also known in the literature. (It is, for instance, the maximum of the Clausen function $\mathrm{Cl}_2$.)

If there's going to be a Pythagorean theorem for hyperbolic simplices, then it must apply to this case, ideally relating these values in the non-Euclidean Pythagorean tradition:

$$\text{function}(H^\star) = \text{combination of related functions}(L_1=L^\star;L_2=L^\star;L_3=L^\star;L_4=L^\star)$$

where the right-hand side is symmetric in the four parameters $L_i$ (representing the volumes of the four legs of the simplex), which are all set equal to $L^\star$. (When the formula is populated with infinitesimal quantities, it should collapse to the Euclidean sum-of-squares relation.) However, while the dilogarithm has many interesting properties, the connection between $H^\star$ and $L^\star$ is not obvious (to me). In a separate MO post, I note a hypergeometric series "similar" to a series for $H^\star$ (what I call "$T(1/2)$" there) that has a direct relation to Catalan's constant (and therefore $L^\star$), but this hasn't yet provided appropriate insights into relating $H^\star$ to $L^\star$ directly.

Have I perhaps lost the forest amid a bunch of trees?

$\endgroup$
11
  • 1
    $\begingroup$ Do you have a reference for the spherical and hyperbolic formulae you gave? Also, am I correct in assuming that you are fixing the representative spherical/hyperbolic manifolds to have curvature $\pm 1$, since area deviation general depends on the curvature. And for the simplices, I assume you take them to be the geodesically convex hull of the vertices? $\endgroup$ Jul 18, 2010 at 17:59
  • $\begingroup$ @Willie: (1) I've added a link to my notes deriving the formulas; it's all basic trig. (2) Curvature ±1: Yes. (That always gives the cleanest formulas, right? :) (3) Geodesically convex hull: Yes. $\endgroup$
    – Blue
    Jul 18, 2010 at 22:14
  • 1
    $\begingroup$ @Agol: I think the OP was already talking about orthoschemes? @Don: For the orthoscheme's in arbitrary dimension Euclidean space I found the formula around 10 years ago (5 years too late to be novel) dpmms.cam.ac.uk/~ww278/papers/gp.pdf I'll think about the homogeneous space case for a bit. $\endgroup$ Jul 18, 2010 at 22:32
  • 1
    $\begingroup$ @Willie: "Right-corner" tetrahedra (your "N-rectangular"s) aren't orthoschemes. @Agol: I'm not after the most natural class of simplices study. I'm after a Pythagorean theorem; my class of simplices is pretty much chosen for me. :) @Willie again: I plowed through a similar matrix-based derivation of the generalized Euclidean Pythagorean Theorem in the 80s; I was told then that the result was already well-known. :( $\endgroup$
    – Blue
    Jul 18, 2010 at 23:05
  • 1
    $\begingroup$ Re "dramatic divergence": I am not at all sure that this is relevant for your question, but there seems to be a dichotomy between even-dimensional spaces and odd-dimensional spaces (both in Euclidean and non-Euclidean setting). A classical example is the distinction between the arc length of circle and the area of a spherical belt, which was generalized by V.I.Arnold and V.Vasiliev, see e.g. ams.org/mathscinet-getitem?mr=1024727 $\endgroup$ Jul 19, 2010 at 1:17

1 Answer 1

4
$\begingroup$

I think that the referenced paper has all you might want (it is obviously related to the Derevnyn-Mednykh paper, but is considerably more extensive). They do not specifically talk about orthoschemes, but their results specialize nicely.

A. D. Mednykh and M. G. Pashkevich, "Elementary formulas for a hyperbolic tetrahedron", Sib. Math. J. 47 (2006), 687-695.

$\endgroup$
1
  • $\begingroup$ There's lots of good information in the article, but nothing I see that directly aids my search for a Pythagorean relation. $\endgroup$
    – Blue
    Dec 23, 2011 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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