TITLE: Unusual symmetries of the Cayley-Menger determinant for the volume of tetrahedra
QUESTION [54 upvotes]: 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.

REPLY [3 votes]: 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.