87
$\begingroup$

Consider a Euclidean tetrahedron with lengths of edges
$$ l_{12}, l_{13}, l_{14}, l_{23}, l_{24}, l_{34} $$ and dihedral angles $$ \alpha_{12}, \alpha_{13}, \alpha_{14}, \alpha_{23}, \alpha_{24}, \alpha_{34}. $$ Consider solid angles \begin{split} &\Omega_1=\alpha_{12}+\alpha_{13}+\alpha_{14}-\pi \\ &\Omega_2=\alpha_{12}+\alpha_{23}+\alpha_{24}-\pi \\ &\Omega_3=\alpha_{13}+\alpha_{23}+\alpha_{34}-\pi \\ &\Omega_4=\alpha_{14}+\alpha_{24}+\alpha_{34}-\pi \\ \end{split} and perimeters of faces \begin{split} &P_1=l_{23}+l_{34}+l_{24} \\ &P_2=l_{13}+l_{14}+l_{34} \\ &P_3=l_{12}+l_{14}+l_{24} \\ &P_4=l_{12}+l_{23}+l_{13}. \\ \end{split} Then the following cross-ratios are equal to each other: $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[P_1, P_2, P_3, P_4]. $$ Question: Is it known? I have found a proof of this statement (see here), but it involves quite tricky algebraic geometry. It will be very interesting to me to see a more elementary approach.

Addition: Similar statements hold in spherical and hyperbolic geometry. For a spherical tetrahedron $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}]. $$ For a hyperbolic tetrahedron $$ [e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{P_1}, e^{P_2}, e^{P_3}, e^{P_4}]. $$

Addition 2: One can prove a more general statement, which I formulate in the spherical case (but it is true in other geometries after appropriate modifications). There exists a $PSL_2(\mathbb{C})-$ transformation, sending eight numbers

$$ 1, e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}, e^{i(\alpha_{12}+\alpha_{23}+\alpha_{34}+\alpha_{14})}, e^{i(\alpha_{12}+\alpha_{24}+\alpha_{34}+\alpha_{13})}, e^{i(\alpha_{13}+\alpha_{23}+\alpha_{24}+\alpha_{14})} $$ to $$ 1, e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}, e^{i(l_{12}+l_{23}+l_{34}+l_{14})}, e^{i(l_{12}+l_{24}+l_{34}+l_{13})}, e^{i(l_{13}+l_{23}+l_{24}+l_{14})}. $$ I know neither an elementary proof of this statement nor any interpretation of the coefficients of the $PSL_2(\mathbb{C})-$ transformation.

$\endgroup$
  • 3
    $\begingroup$ Could you please define "cross ratio" in this context? What specifically does $[a,b,c,d]$ mean? $\endgroup$ – Joseph O'Rourke Jul 18 '19 at 23:24
  • 4
    $\begingroup$ Sure! $[a,b,c,d]=\frac{(a-b)(c-d)}{(a-c)(b-d)}.$ $\endgroup$ – Daniil Rudenko Jul 18 '19 at 23:29
  • 11
    $\begingroup$ That's very cool! $\endgroup$ – Igor Rivin Jul 18 '19 at 23:52
  • 4
    $\begingroup$ I've tried, but no success. The only trivial observation I've made, is that the duality transform (in the spherical geometry) preserves a ratio (tan(a/2)/tan(b/2)), where a and b angles opposite to an edge. If we show, that your cross-ratio depends only from these six ratios, we are done. I played with nice wikipedia-formulas (see the last section), but did not managed to organize it in symmetric form. en.wikipedia.org/wiki/Spherical_trigonometry $\endgroup$ – Arseniy Akopyan Jul 21 '19 at 18:16
  • 7
    $\begingroup$ Empirical observation: the four points $(\cot\frac{\Omega_i}2, \frac1{P_i})$ are collinear. $\endgroup$ – Robin Houston Jul 22 '19 at 9:46
38
$\begingroup$

Euclidean case

Using the formula for the tan of the half solid angle that Robin Houston quotes, and expressing everything in terms of edge lengths by using the cosine law to convert the dot products, I end up with the following linear relationship:

$$\frac{1}{P_i} = \alpha_E \cot\left(\frac{\Omega_i}{2}\right)+\beta_E$$

where:

$$\alpha_E = \frac{12 V}{P_1 P_2 P_3 P_4}$$

$V$ is the volume of the tetrahedron:

$$V = \frac{1}{12\sqrt{2}}\sqrt{\left| \begin{array}{ccccc} 0 & l_{12}^2 & l_{14}^2 & l_{13}^2 & 1 \\ l_{12}^2 & 0 & l_{24}^2 & l_{23}^2 & 1 \\ l_{14}^2 & l_{24}^2 & 0 & l_{34}^2 & 1 \\ l_{13}^2 & l_{23}^2 & l_{34}^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array} \right|}$$

and:

$$\beta_E = \frac{\left(\Sigma_{i \lt j} l_{ij}P_i P_j\right) - 2 \left(\Sigma l_{ij} l_{pq} l_{st} \right)}{P_1 P_2 P_3 P_4}$$

where the second summation is over the 16 choices of distinct triples of edges that do not all share a common vertex. There are $\binom{6}{3}=20$ triples of distinct edges; if we omit the four triples that are incident on each of the four vertices, we obtain the 16 that appear in that sum.

Another expression for $\beta_E$ is:

$$\beta_E = \frac{\text{Permanent}\left( \begin{array}{ccccc} 0 & l_{12} & l_{14} & l_{13} & 1 \\ l_{12} & 0 & l_{24} & l_{23} & 1 \\ l_{14} & l_{24} & 0 & l_{34} & 1 \\ l_{13} & l_{23} & l_{34} & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array} \right) }{2 P_1 P_2 P_3 P_4}$$

Spherical case

In the spherical case, starting from a tetrahedron with specified edge lengths, a similar relationship is given by:

$$\cot\left(\frac{P_i}{2}\right) = \alpha_S \cot\left(\frac{\Omega_i}{2}\right) + \beta_S$$

where:

$$\alpha_S = \frac{\sqrt{\gamma_S}}{4 \sin\left(\frac{P_1}{2}\right)\sin\left(\frac{P_2}{2}\right)\sin\left(\frac{P_3}{2}\right)\sin\left(\frac{P_4}{2}\right)}$$

and:

$$\gamma_S = \left| \begin{array}{cccc} 1 & \cos \left(l_{12}\right) & \cos \left(l_{13}\right) & \cos \left(l_{14}\right) \\ \cos \left(l_{12}\right) & 1 & \cos \left(l_{23}\right) & \cos \left(l_{24}\right) \\ \cos \left(l_{13}\right) & \cos \left(l_{23}\right) & 1 & \cos \left(l_{34}\right) \\ \cos \left(l_{14}\right) & \cos \left(l_{24}\right) & \cos \left(l_{34}\right) & 1 \\ \end{array} \right|$$

A series expansion of $\gamma_S$ to sixth order in the edge lengths yields the expected relationship:

$$\sqrt{\gamma_s} \approx 6 V_E$$

where $V_E$ is the volume computed from the same edge lengths using Cayley's Euclidean formula. (Note that there is a factor of 2 difference between the slopes of the two lines in the limit of small edge lengths, because we are using the cotan of the half perimeter on the $y$ axis in the spherical case.)

$$\beta_S = \frac{\Sigma_i \sin\left(Q_i\right) - \sin\left(\Sigma_{j\lt k} l_{jk}\right) - \sin\left(l_{12}+l_{34}\right) - \sin\left(l_{13}+l_{24}\right) - \sin\left(l_{14}+l_{23}\right)}{8 \sin\left(\frac{P_1}{2}\right)\sin\left(\frac{P_2}{2}\right)\sin\left(\frac{P_3}{2}\right)\sin\left(\frac{P_4}{2}\right)}$$

where:

$Q_i$ is the sum of the edge lengths for the edges incident on vertex $i$.

$\endgroup$
  • 2
    $\begingroup$ It would be interesting to see if a similar formula holds in the spherical case (as Akopyan hints at). If so, it might have a more symmetric form (between a spherical tetrahedron and its dual). $\endgroup$ – Ian Agol Jul 23 '19 at 16:45
  • 2
    $\begingroup$ Look on the formula from Theorem 3. It is hyperbolic, but can be calculated. homepages.warwick.ac.uk/~masgar/Seminar/WGTS/Abrosimov.pdf $\endgroup$ – Arseniy Akopyan Jul 25 '19 at 9:16
  • 2
    $\begingroup$ @ArseniyAkopyan In the Euclidean case, it was just a direct computation based on the formula for the tan of half the solid angle that Robin Houston quoted. I turned the dot products between edge vectors in that formula into expressions depending only on the edge lengths, using the cosine law. In the spherical case, I constructed the vertices of a tet with arbitrary edge lengths, and computed all the geometrical features needed to find the four points $(\cot(\Omega_i/2),\cot(P_i/2))$, and the line containing them all. $\endgroup$ – Greg Egan Jul 25 '19 at 12:13
  • 2
    $\begingroup$ Thanks. According to the presentation above we can not catch spherical volume from a Gram matrix. and therefore from $\alpha$ and $\beta$. Miracle did not happen :-( $\endgroup$ – Arseniy Akopyan Jul 25 '19 at 12:30
  • 2
    $\begingroup$ Cool (re sphere slope). Considering the dual equation, this gives another relation between the $l_{ij}$ and $\alpha_{ij}$. $\endgroup$ – Ian Agol Jul 27 '19 at 2:07
30
$\begingroup$

This is not an answer to the question, but an experimental observation that suggests a sharper conjecture: it’s only written as an answer because I’d like to flesh it out a bit more than there’s room for in a comment.

The image of $e^{i\Omega}$ under the Möbius transformation $z\mapsto\frac{i-zi}{1+z}$ is $\tan(\frac\Omega2)$, so an equivalent statement of Rudenko’s theorem is that $$ [\tan(\frac{\Omega_1}2), \tan(\frac{\Omega_2}2), \tan(\frac{\Omega_3}2), \tan(\frac{\Omega_4}2)]=[P_1, P_2, P_3, P_4]. $$

Experimentally, something stronger seems to be true: the four points

  • $(1/\tan(\frac{\Omega_1}2), 1/P_1)$
  • $(1/\tan(\frac{\Omega_2}2), 1/P_2)$
  • $(1/\tan(\frac{\Omega_3}2), 1/P_3)$
  • $(1/\tan(\frac{\Omega_4}2), 1/P_4)$

are collinear.

It’s fairly easy to check this numerically for particular examples, because $\tan(\frac{\Omega}2)$ has a nice expression that can be computed easily from coordinates:

$$ \tan \left( \frac{\Omega}{2} \right) = \frac{\left|\vec a\cdot(\vec b\times\vec c)\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a} $$

where $\vec a$, $\vec b$ and $\vec c$ are the vectors from the vertex in question to the other three vertices, and $a$, $b$ and $c$ are the lengths of these vectors.

Since the expression in the numerator is six times the volume of the tetrahedron, it is the same at all four vertices, so to check collinearity it suffices to compute the denominator.

I haven’t played with the spherical or hyperbolic cases at all.

$\endgroup$
  • 2
    $\begingroup$ The four expressions for $1/\tan(\Omega_i/2)$ all have the same denominator, namely the volume of the tetrahedron, which should make the linearity easier to establish. $\endgroup$ – Matt F. Jul 22 '19 at 20:08
  • 5
    $\begingroup$ I've checked it for spherical tetrahedra. Your observation holds there as well! $\endgroup$ – Arseniy Akopyan Jul 22 '19 at 22:02
  • $\begingroup$ Wonderful! I did not know the formula for tan. $\endgroup$ – Daniil Rudenko Jul 22 '19 at 22:46
  • $\begingroup$ I will add to the post one more comment, there are a few other points on the same line. But I don't know any way to interpret its coefficients. $\endgroup$ – Daniil Rudenko Jul 22 '19 at 22:47
  • 1
    $\begingroup$ @ArseniyAkopyan Sorry, I just realised that the cross-ratios mentioned in the question make it clear that it’s $(\cot(\Omega_i/2),\cot(P_i/2))$ that are colinear — and I do see that numerically for arbitrary sizes. $\endgroup$ – Greg Egan Jul 24 '19 at 6:02
16
$\begingroup$

Here goes the elementary proof of the claim by Robert Houston that the quadraples $(P_1^{-1},P_2^{-1},P_3^{-1},P_4^{-1})$ and $(\cot \frac{\Omega_1}2,\cot \frac{\Omega_2}2,\cot \frac{\Omega_3}2,\cot \frac{\Omega_4}2)$ are affinely equivalent. In the spherical case the first quadraple should be replaced to $\{\cot\frac{P_i}2\}$, in the hyperbolic case to hyperbolic cotangents. Further the tetrahedron $A_1A_2A_3A_4$ is assumed to be generic (that implies the general case automatically.) I write the solution in Euclidean case, but the changes in spherical/hyperbolic case are almost straightforward, since the basic construction works in all three geometries.

At first, we note that if the quadraples $(a_1,a_2,a_3,a_4)$ and $(b_1,b_2,b_3,b_4)$ of reals satisfy $(a_1-a_2):(a_3-a_4)=(b_1-b_2):(b_3-b_4)$ and two other analogous relations, then they are affinely equivalent. Indeed, without loss of generality $a_4=b_4=0,a_3=b_3=1$, $a_1\leqslant a_2$ and we have $a_1-a_2=b_1-b_2$, $1+(a_1-a_2-1)/a_2=(a_1-1)/a_2=(b_1-1)/b_2=1+(b_1-b_2-1)/b_2$, $a_2=b_2$, $a_1=b_1$. So we have to prove (call it equation $(\star)$) that $$ \frac{P_1^{-1}-P_3^{-1}}{\cot \frac{\Omega_1}2-\cot \frac{\Omega_3}2}= \frac{P_3-P_1}{\sin \frac{\Omega_3-\Omega_1}2}\cdot\frac{\sin \frac{\Omega_1}2 \sin \frac{\Omega_3}2}{P_1P_3} $$ equals to $$ \frac{P_4-P_2}{\sin \frac{\Omega_4-\Omega_2}2}\cdot\frac{\sin \frac{\Omega_2}2 \sin \frac{\Omega_4}2}{P_2P_4}. $$ In order to prove it we draw three planes: through $A_2$ orthogonal to the external bisector of $\angle A_1A_2A_3$, through $A_3$ orthogonal to the external bisector of $\angle A_2A_3A_4$, through $A_4$ orthogonal to the external bisector of $\angle A_3A_4A_1$.

enter image description here

Let them meet at point $I$, and let $Q_1,Q_2,Q_3,Q_4$ be projections of $I$ onto lines $A_1A_2,A_2A_3,A_3A_4,A_4A_1$ respectively (see the picture). We have $IQ_1=IQ_2=IQ_3=IQ_4=:r$ and also $A_2Q_1=A_2Q_2$, $A_3Q_2=A_3Q_3$, $A_4Q_3=A_4Q_4$ by construction (the segments are directed in the natural sense: either $Q_1,Q_2$ both belong to rays $A_2A_1,A_2A_3$ respectively, or both do not, and so on). Also the right triangles $\triangle A_1IQ_1,\triangle A_1IQ_4$ are equal by hypotenuse and cathetus. Thus $A_1Q_1=A_1Q_4$. It implies that $$A_1A_2+A_3A_4=A_1Q_1+A_2Q_1+A_3Q_3+A_4Q_3=\\A_1Q_4+A_2Q_2+A_3Q_2+A_4Q_4=A_1A_4+A_2A_3,$$ which is not quite what you need but also an interesting relation. Well, to be serious, of course $Q_1Q_4$ is parallel to the internal, not external, bisector of $\angle A_2A_1A_4$:

enter image description here

Denoting $x_i=A_iQ_i$ we find (on this picture, in general we should consider directed segments) $x_2-x_1=l_{12},x_2+x_3=l_{23},x_3+x_4=l_{34},x_4+x_1=l_{14}$. Thus $2x_{3}=l_{23}+l_{34}-l_{12}-l_{14}=P_1-P_3$, $2x_1=l_{14}+l_{23}-l_{12}-l_{34}$.

Analogously for dihedral angles we have four relations like $\angle(A_1A_2A_3,A_1A_2I)=\angle(A_1A_2A_3,A_2A_3I)=:\beta_4$, $\angle(A_2A_3A_4,A_2A_3I)=\angle(A_2A_3A_4,A_3A_4I)=:\beta_1$, $\angle(A_3A_4A_1,A_3A_4I)=\angle(A_3A_4A_1,A_4A_1I)=:\beta_2$, $\angle(A_1A_2A_4,A_4A_1I)=\pi-\angle(A_1A_2A_4,A_1A_2I)=:\beta_3$

(suppose that $I$ lies inside the tetrahedron, otherwise carefully use directed angles: I am going to divide by 2 that is dangerous for the angels which are considered remainders modulo $\pi$, or consider several cases or use "generic case" abstract nonsense reasoning or whatever).

We get $\beta_4+\pi-\beta_3=\alpha_{12}$,$\beta_1+\beta_4=\alpha_{23}$,$\beta_2+\beta_1=\alpha_{34}$,$\beta_3+\beta_2=\alpha_{41}$; $2\beta_1=\alpha_{23}+\alpha_{34}+\pi-\alpha_{12}-\alpha_{41}=\pi+\Omega_3-\Omega_1$, $2\beta_3=\alpha_{23}+\alpha_{14}+\pi-\alpha_{12}-\alpha_{34}$.

Let $H$ be a projection of $I$ onto the plane $A_2A_3A_4$, then $A_3H$ is the internal bisector of $\angle A_2A_3A_4$. We get $Q_2H=r\cos\beta_1=r\sin \frac{\Omega_1-\Omega_3}2$, $\frac{P_1-P_3}2=x_3=Q_2H\cot \frac{\angle A_2A_3A_4}2$. Therefore $$ \frac{P_1-P_3}{2\sin \frac{\Omega_1-\Omega_3}2}=r\cot\frac{\angle A_2A_3A_4}2, $$ analogously considering the vertex $A_1$ instead of $A_3$ we get $$ \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}=r\tan \frac{\angle A_2A_1A_4}2. $$ So we may exclude $r$ from the formulae and get $$ \frac{P_1-P_3}{2\sin \frac{\Omega_1-\Omega_3}2}= \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}\cdot \cot \frac{\angle A_2A_3A_4}2 \cot \frac{\angle A_2A_1A_4}2. $$ Analogously $$ \frac{P_2-P_4}{2\sin \frac{\Omega_2-\Omega_4}2}= \frac{l_{14}+l_{23}-l_{12}-l_{34}}{2\sin \frac{\alpha_{12}+\alpha_{34}-\alpha_{23}-\alpha_{14}}2}\cdot \cot \frac{\angle A_1A_2A_3}2 \cot \frac{\angle A_1A_4A_3}2. $$ Therefore $(\star)$ reads as $$ \frac{\sin \frac{\Omega_1} 2\sin \frac{\Omega_3} 2}{P_1P_3}\cot \frac{\angle A_2A_3A_4}2 \cot \frac{\angle A_2A_1A_4}2= \frac{\sin \frac{\Omega_2} 2\sin \frac{\Omega_4} 2}{P_2P_4}\cot \frac{\angle A_1A_2A_3}2 \cot \frac{\angle A_1A_4A_3}2. $$ This immediately follows from the following "sine theorem" for the tetrahedron: the expression $$ \frac{\sin \frac{\Omega_1} 2 \sqrt{\cot \frac{\angle A_2A_1A_3}2 \cot \frac{\angle A_2A_1A_4}2\cot \frac{\angle A_3A_1A_4}2} } {P_1\sqrt{\tan \frac{\angle A_2A_3A_4}2 \tan \frac{\angle A_2A_4A_3}2 \tan \frac{\angle A_3A_2A_4}2}} $$ (denote it $\eta_1$) equals to analogous expressions for other indices. For proving it we denote $S_1$ and $r_1=2S_1/P_1$ the area and inradius of $\triangle A_2A_3A_4$, then we have $$ P_1^2\tan \frac{\angle A_2A_3A_4}2\tan \frac{\angle A_3A_2A_4}2 \tan \frac{\angle A_2A_4A_3}2=P_1^2\cdot\frac{r_1^3}{(P_1/2-l_{24}) (P_1/2-l_{34})(P_1/2-l_{23})}=\\ \frac{(2S_1)^3}{4(P_1/2)(P_1/2-l_{24}) (P_1/2-l_{34})(P_1/2-l_{23})}=\frac{8S_1^3}{4S_1^2}=2S_1 $$ by Heron formula. Next, using the Cagnoli formula (see p. 101 here) $$\sin \frac{\Omega_1}2=\sin \alpha_{12}\frac{\sin \frac{\angle A_3A_1A_2}2\sin \frac{\angle A_4A_1A_2}2}{\cos \frac{\angle A_3A_1A_4}2}$$ we get $$\sin^2 \frac{\Omega_1}2 \cot \frac{\angle A_2A_1A_4}2\cot \frac{\angle A_2A_1A_3}2\cot \frac{\angle A_3A_1A_4}2 =\sin^2 \alpha_{12}\frac{\sin \angle A_3A_1A_2\sin \angle A_4A_1A_2}{2\sin \angle A_3A_1A_4}.$$

So, $\eta_1^2=\eta_2^2$ reads as $$ \frac{\sin \angle A_3A_1A_2\sin \angle A_4A_1A_2}{S_1\cdot \sin \angle A_3A_1A_4}= \frac{\sin \angle A_3A_2A_1\sin \angle A_4A_2A_1}{S_2\cdot \sin \angle A_3A_2A_4}. $$ Substituting $2S_2=l_{13}\cdot l_{14}\cdot \sin \angle A_3A_1A_4$, $2S_1=l_{23}\cdot l_{24}\cdot \sin \angle A_3A_2A_4$ this reduces to a product of two sine laws, and everything is proved.

Note that we had the expression $l_{12}-l_{23}+l_{34}-l_{41}$ above, and if you replace the internal bisectors to external bisectors you may also get $l_{12}+l_{23}+l_{34}+l_{41}$ that should answer to your extended question.

$\endgroup$
14
$\begingroup$

Here one can find an algebra-geometric proof of the trigonometric relations. The main point is that given a (say, spherical) tetrahedron $T$ one can construct a rational elliptic surface $X_T$ with the property that surfaces corresponding to a tetrahedron and its dual are isomorphic. The cross-ratios in the question are some invariants of $X_T$ computed in two different ways. These invariants are analogs of Cayley invariants of cubic surfaces: cross-ratios of four points in which four lines intersect the fifths.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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