Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$J$ 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?

share|improve this question
4  
Solve ... for what? $J$? $D$? $B$? $S$? –  Gerry Myerson Sep 11 '13 at 12:57
1  
@Gerry: If I were the OP, I would be tempted to answer "yes" ;-) –  Johannes Hahn Sep 12 '13 at 0:06
    
@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. –  Hauke Reddmann Sep 12 '13 at 14:32
    
@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. –  Hauke Reddmann Sep 18 '13 at 16:27

1 Answer 1

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.

share|improve this answer
    
You mean "of degree 3". The order of $S_3$ is 6, not 3. –  Johannes Hahn Sep 12 '13 at 0:07
    
indeed. I fixed the typo, thanks. –  Dima Pasechnik Sep 12 '13 at 1:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.