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Suppose you have a tetrahedron $T$ in Euclidean space with edge lengths $\ell_{01}$, $\ell_{02}$, $\ell_{03}$, $\ell_{12}$, $\ell_{13}$, and $\ell_{23}$. Now consider the tetrahedron $T'$ with edge lengths $$\begin{aligned} \ell'_{02} &= \ell_{02} & \ell'_{13} &= \ell_{13}\\ \ell'_{01} &= s-\ell_{01} & \ell'_{12} &= s-\ell_{12}\\ \ell'_{23} &= s-\ell_{23}& \ell'_{03} &= s-\ell_{03} \end{aligned} $$ where $s = (\ell_{01} + \ell_{12} + \ell_{23} + \ell_{03})/2$. If the edge lengths of $T'$ are positive and satisfy the triangle inequality, then the volume of $T'$ equals the volume of $T$. In particular, if $T$ is a flat tetrahedron in $\mathbb{R}^2$, then $T'$ is as well. This is easily verified by plugging the values $\ell'_{ij}$ above into the Cayley-Menger determinant.

In fact, it's possible to show that the linear symmetries of $\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the Weyl group $D_6$, of order $2^5 * 6! = 23040$. This is a factor of $15$ times larger than the natural geometric symmetries obtained by permuting the vertices of the tetrahedron and negating the coordinates.

The transformations don't always take Euclidean tetrahedra to Euclidean tetrahedra, but they do sometimes. For instance, if you start with an equilateral tetrahedron $T$ with all side lengths equal to $1$, then $T'$ is also an equilateral tetrahedron. Thus if $T$ is a generic Euclidean tetrahedron close to equilateral, $T'$ will also be one, and $T$ and $T'$ will not be related by a Euclidean symmetry.

I can't be the first person to observe this. (In fact, I vaguely recall hearing about this in the context of quantum groups and the Jones polynomial.) What's the history? How to best understand these transformations (without expanding out the determinant)? Are $T$ and $T'$ scissors congruent? Etc.

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    $\begingroup$ The paper at arxiv.org/abs/1709.03936 contains a proof that there are only these 23040 symmetries. It also contains a proof that there are no such "Regge" symmetries for higher dimensions. $\endgroup$
    – Steven Gortler
    Commented Sep 28, 2017 at 16:03
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    $\begingroup$ Our preprint with Arseniy Akopyan arxiv.org/abs/1903.04929 gives a short proof of Regge symmetries for Euclidean and non-Euclidean tetrahedra. $\endgroup$
    – Ivan Izmestiev
    Commented Mar 16, 2019 at 8:28

5 Answers 5

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These are the so-called Regge symmetries, described by T. Regge in a 1970-ish paper. For a bit on it, with references, see the paper

Philip P. Boalch, MR 2342290 Regge and Okamoto symmetries, Comm. Math. Phys. 276 (2007), no. 1, 117--130.\

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    $\begingroup$ It's surprising that this purely classical phenomenon was noticed in the context of representation theory. $\endgroup$
    – Dylan Thurston
    Commented Jan 16, 2017 at 1:45
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    $\begingroup$ @DylanThurston Is there a geometric explanation for this symmetry? $\endgroup$
    – Igor Rivin
    Commented Jan 16, 2017 at 1:50
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    $\begingroup$ The planar case (both T and T' are flat) is equivalent to the Ivory lemma (the diagonals in the quadrilateral formed by confocal conics have equal lengths). $\endgroup$
    – Ivan Izmestiev
    Commented Jan 17, 2017 at 10:24
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    $\begingroup$ @IvanIzmestiev, that's really beautiful. Would you like to expand that comment into a full answer? (If not, I'll do it.) $\endgroup$
    – Dylan Thurston
    Commented Jan 18, 2017 at 18:50
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    $\begingroup$ @IvanIzmestiev I second Dylan's request... $\endgroup$
    – Igor Rivin
    Commented Jan 18, 2017 at 20:56
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To address the scissors congruence question at the end of the post: the Regge symmetries produce tetrahedra which are scissors congruent. This is proved in Section 6 (Theorem 9 and Corollary 10) of

The argument is indirect, proving that the Regge symmetries preserve volume and Dehn invariant, then deducing the statement on scissors congruence from Sydler's theorem.

The corresponding statement for hyperbolic tetrahedra was proved (by construction) in

  • Y. Mohanty. The Regge symmetry is a scissors congruence in hyperbolic space. Alg. Geom. Top. 3 (2003), 1-31. (link to paper on arXiv)
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  • $\begingroup$ In the hyperbolic case, this paper is also very relevant, although also a little hard to read: P. Doyle and G. Leibon, 23040 symmetries of hyperbolic tetrahedra, arXiv:math/0309187, front.math.ucdavis.edu/0309.5187 $\endgroup$
    – Dylan Thurston
    Commented Jan 25, 2017 at 21:24
  • $\begingroup$ That's really great, with hyperbolic tetrahedra! Only strange that nobody found a "cut and paste" argument in the euclidean and spherical cases. $\endgroup$
    – Ivan Izmestiev
    Commented Jan 26, 2017 at 8:33
  • $\begingroup$ @DylanThurston the link in your comment is broken, here's a replacement: arxiv.org/abs/math/0309187 $\endgroup$
    – David Roberts ♦
    Commented Mar 29, 2022 at 1:12
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Let me give a geometric interpretation for the case of tetrahedra of volume zero. The statement becomes as follows:

Given four positive numbers $a,b,c,d$ that satisfy the quadrangle inequalities, denote $$ s=\frac{a+b+c+d}2, \quad a'=s-a \text{ etc.} $$ Take a quadrilateral with side lengths $a,b,c,d$ (in this cyclic order). If its diagonals have lengths $x,y$, then there exists a quadrilateral with side lengths $a',b',c',d'$ and the same diagonal lengths.

Again, this can be proved by a direct computation, but Arseniy Akopyan noticed that this is equivalent to the Ivory lemma: diagonals in a curvilinear quadrilateral formed by four confocal conics are equal, see the left half of the picture.

enter image description here

Indeed, with two opposite vertices of an $(a,b,c,d)$-quadrilateral as the foci, draw two ellipses and two hyperbolas through the other two vertices. This gives a curvilinear quadrilateral whose diagonal is also a diagonal of our $(a,b,c,d)$-quadrilateral, see the right half of the picture. The relations between $a,b,c,d$ and $a',b',c',d'$: $$ a+b=c'+d', c-d=d'-c' \text{ etc.} $$ imply that the other two vertices of this curvilinear quadrilateral form together with the foci an $(a',b',c',d')$-quadrilateral.

As a side remark: I studied configuration spaces of planar quadrilaterals in this preprint, but didn't know about Regge symmetries. Thanks to Dylan and Igor for bringing this up!

There is an analog of the Ivory lemma in dimension three, but there is no word about volumes in it. Still, it would be interesting to look at the corresponding configuration of confocal surfaces (one of the families degenerates).

Also note that the Ivory lemma holds in the hyperbolic and spherical geometry as well.

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    $\begingroup$ Thanks for the writeup! Note that the planar argument in the zero-volume case implies (or nearly implies) the result in the non-zero volume case as well: the volume is (by general principles) an algebraic function of the edge lengths, and so if the zero-set goes to the zero-set then the whole volume has to be preserved, possibly with a scale factor; since it's an involution, the factor has to be $\pm 1$. $\endgroup$
    – Dylan Thurston
    Commented Jan 25, 2017 at 21:21
  • $\begingroup$ The Regge symmetries are also symmetries for the volumes of hyperbolic tetrahedra! Matthias Wendt posted a reference (see my comment there as well). The spherical case should also be true (perhaps with some interpretation), but I couldn't find a reference working it out. $\endgroup$
    – Dylan Thurston
    Commented Jan 25, 2017 at 21:29
  • $\begingroup$ Thanks, Dylan! I removed the conjecture in the last lines of my answer. The spherical case must be true. It should be possible to prove it by deformation to a flat spherical tetrahedron (using the Schlaefli formula), but a "cut and paste" argument would be nicer. $\endgroup$
    – Ivan Izmestiev
    Commented Jan 26, 2017 at 10:06
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It may be worth noting that for some choices of six lengths that there exist as many as 30 inequivalent tetrahedra with these six lengths.

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  • $\begingroup$ That's interesting. Most of those tetrahedra won't have the same volume, I presume. However, the computations I alluded to above on symmetries of the Cayley-Menger determinant also show that there are 30 different linear combinations of the edge lengths that yield tetrahedra of the same volume. (There's a group of order $23040$ acting on the signed edge lengths; for any such choice, one of $2^6$ choices will have positive lengths; there are a further $12$ orientation-preserving symmetries of a single tetrahedron. $23040/(2^6*12) = 30$.) Are these related? $\endgroup$
    – Dylan Thurston
    Commented Jan 18, 2017 at 19:05
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The following comment in the question intrigued me:

In fact, it's possible to show that the linear symmetries of $\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the Weyl group $D_6$, of order $2^5 * 6! = 23040$.

In particular, this suggests that it should be possible to perform a change of basis such that the Cayley–Menger determinant is a symmetric polynomial in the six variables, which is also invariant under even numbers of sign changes.

Some experimentation confirmed that the following variables work in the sense that the geometric and Regge symmetries all preserve the 12-element set $\{ \pm u, \pm v, \pm w, \pm x, \pm y, \pm z \}$:

  • $u = \ell_{01} + \ell_{23}$;
  • $v = \ell_{02} + \ell_{13}$;
  • $w = \ell_{03} + \ell_{12}$;
  • $x = \ell_{01} - \ell_{23}$;
  • $y = \ell_{02} - \ell_{13}$;
  • $z = \ell_{03} - \ell_{12}$;

More succinctly, the variables are the three sums of opposite edges and the three differences of opposite edges. The Cayley–Menger determinant can then be shown to be equal to the following 57-term polynomial:

$$ uvwxyz - \dfrac{1}{16} \sum_\text{sym}^{6} u^6 + \dfrac{1}{16} \sum_\text{sym}^{30} u^4 v^2 - \dfrac{1}{8} \sum_\text{sym}^{20} u^2 v^2 w^2 $$

where the ‘symmetric sums’ are over all distinct monomials obtained by permuting the six variables (and the number above the sum is the number of terms). This is visibly invariant under the Weyl group $D_6$ generated by arbitrary permutations and even numbers of sign changes.

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