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$J$ is a symmetric matrix (built from 6j symbols...it's always knot theory in disguise when I ask :-), $D$ a diagonal matrix, and $B=DJ$. $S$ is a diagonal sign matrix (entries all $+1$ or $-1$). $I$ is the identity matrix. From a bit of diagram juggling, I find the determining equations for $B$ are simply:

$B^2=I$

$(BS)^3=I$

Which I then happily solve by brute force (actually some elements of $J$ are known beforehand) but already for a 6*6 matrix this leads to a big mess.
Is there a more "intelligent" method? (I still need to know all elements of $B$...Sidenote: If $S$ contains only one or two $-1$, I already know the closed form for any size of $B$.) Even better, if yes, would it still work if, in a more general version, $S$ is still diagonal but $S^2=I$ no longer holds?

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    $\begingroup$ Solve ... for what? $J$? $D$? $B$? $S$? $\endgroup$
    – Gerry Myerson
    Sep 11, 2013 at 12:57
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    $\begingroup$ @Gerry: If I were the OP, I would be tempted to answer "yes" ;-) $\endgroup$
    – Johannes Hahn
    Sep 12, 2013 at 0:06
  • $\begingroup$ @Gerry - your question is very justified in the general case, where no variables are known at all (and where I express B in terms of S elements - that's the best I can do). J and D are mostly irrelevant to this question (except they restrict the form of B somewhat). In this special case, I usually solve for B for any sign combination of S. $\endgroup$
    – Hauke Reddmann
    Sep 12, 2013 at 14:32
  • $\begingroup$ @anyone who stumbles over this: It seems (I experimented with MATHEMATICA's NSOLVE command) that if S and D are given, B is uniquely (apart from signs and complex conjugates) defined. Of course I can't prove it. $\endgroup$
    – Hauke Reddmann
    Sep 18, 2013 at 16:27

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Assuming $S^2=I$, your other two relations give you a presentation for the symmetric group of degree 3. So you want to find linear representations of this group (satisfying the extra condition $B=DJ$). In particular, note that $B$ and $S$ have the same eigenvalues. Also, note that the number of -1's in $S$ gives you a constraint on the irreducible subconstituents of your representation.

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