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.

onlythese 23040 symmetries. It also contains a proof that there are no such "Regge" symmetries for higher dimensions. $\endgroup$ – Steven Gortler Sep 28 '17 at 16:03