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A tetrahedron in hyperbolic 3-space can be defined (up to isometry) by the measures of its dihedral angles, $(a, b, c, a^\prime, b^\prime, c^\prime)$, with $a$, $b$, $c$ along edges that meet at a vertex, and $a^\prime$, $b^\prime$, $c^\prime$ along respective opposite edges.

A Regge symmetry generates a new tetrahedron by transforming the dihedral angles of the original. One such symmetry has this effect:

$$\left(a,\frac{-b+c+b^\prime+c^\prime}{2},\frac{b-c+b^\prime+c^\prime}{2}, a^\prime, \frac{b+c-b^\prime+c^\prime}{2}, \frac{b+c+b^\prime-c^\prime}{2}\right)$$

(Regge-symmetric tetrahedra are interesting (to me) because they have the same volume. In fact, any two Regge-symmetric tetrahedra are scissors congruent: you can cut one into polyhedra that re-assemble to form the other.)

I believe it is (or may be) true that we achieve the same result by interpreting the above as a transformation of edge lengths rather than of dihedral angles.

A few bouts of arduous symbol manipulation show that this belief is consistent with a variety of properties of the dihedrally-defined Regge symmetry, but before delving into an epic verification ---fraught with many a sign/sine/sinh error--- I thought it prudent to seek-out a pointer to relevant literature.

Where may I find confirmation (or refutation) of my belief?

Note: Web searches for "Regge symmetry" ---No, Google, not "reggae symmetry"!--- reveal a number of abstract discussions of these transformations as they relate to "$6j$ symbols" and such, but (so far as I have been able to determine) none of these discussions directly addresses the geometry of the angle-edge duality. Maybe I missed one.

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Responding to my own question to get it out of the "Unanswered" queue (after eighteen months there!) ...

Email correspondence with a mathematician in the know confirms that the result of the prescribed transformation of dihedral angles is equivalent to the result of applying the transformation to corresponding edge lengths. Also confirmed: verification invariably seems to involve a long, hard symbolic slog.

Some notes ...

Let $(a_\star, b_\star, c_\star, a^\prime_\star, b^\prime_\star, c^\prime_\star)$ be the edge lengths of a tetrahedron after a transformation of the dihedral angles. It's not too terribly difficult to verify these relations: $$\begin{align} \overline{b_\star}\overline{b^\prime_\star} + \overline{c_\star}\overline{c^\prime_\star} &= \overline{b}\overline{b^\prime} + \overline{c}\overline{c^\prime} \qquad\qquad \overline{b_\star}\overline{b^\prime_\star} - \overline{c_\star}\overline{c^\prime_\star} = -\ddot{b}\ddot{b^\prime} + \ddot{c}\ddot{c^\prime} \\ \ddot{b_\star}\ddot{b^\prime_\star} + \ddot{c_\star}\ddot{c^\prime_\star} &= \ddot{b}\ddot{b^\prime} + \ddot{c}\ddot{c^\prime} \qquad \qquad \ddot{b_\star}\ddot{b^\prime_\star} - \ddot{c_\star}\ddot{c^\prime_\star} = -\overline{b}\overline{b^\prime} + \overline{c}\overline{c^\prime} \\ \ddot{b_\star}\ddot{c^\prime_\star} + \ddot{c_\star}\ddot{b^\prime_\star} &= \ddot{b}\ddot{c^\prime} + \ddot{c}\ddot{b^\prime} \qquad\qquad \ddot{b_\star}\ddot{c_\star} + \ddot{b^\prime_\star}\ddot{c^\prime_\star} = \phantom{-}\ddot{b}\ddot{c} + \ddot{b^\prime}\ddot{c^\prime} \end{align}$$

where $\overline{x} := \sinh{x}$ and $\ddot{x} := \cosh{x}$.

The first two rows are symmetric in the edge pairs $\{b_\star,b^\prime_\star\}$ and $\{c_\star, c^\prime_\star\}$ and cannot by themselves distinguish between those values. At best, they imply $$ \{2 b_\star, 2 b^\prime_\star\} = \pm \left( - b + b^\prime \right) + c + c^\prime \qquad \{2 c_\star, 2 c^\prime_\star\} = b + b^\prime \pm \left(- c + c^\prime \right) $$ The third row breaks some symbolic symmetry, and we can deduce that $b_\star$'s "$\pm$" matches that of $c_\star$ (and that $b^\prime_\star$'s matches $c^\prime_\star$'s).

To resolve the final sign ambiguity, one can verify that $b+b_\star = c+c_\star \;(= b^\prime + b^\prime_\star = c^\prime+ c^\prime_\star)$. One computationally-arduous strategy is to show that $\tanh(b+b_\star) = \tanh(c+c_\star)$. (For me, targeting $\tanh$ turned out to be less difficult than considering $\sinh$ or $\cosh$.)

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It's nice answer +1 –  china math Jun 12 '14 at 5:50

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