I finally managed to compute the dimensions of some higher irreps of my super-duper-generalized $E_7$ series. Given $Dim(V)=i,Dim(J)=j$ (where $V$ and $J$ are defining and adjoint irrep, and $i,j$ are of course integer; $i$ may be negative, $j$...maybe), then the Clebsch-Gordan series of $V\otimes{J}$ contains an irrep with dimensions $D=\frac{-8*i*(2+i)*(-1+i-j)*j)}{3*(10*i-11*i^2+i^3+16*j+2*i*j)}$ and another expansion contains one with $d=\frac{i*(2+i)*(2-3*i+i^2-2*j)*(-i+i^2-2*j)}{
3*(i-2*i^2+i^3-8*j-i*j)}$.
(Copypaste check: Lie group $D_6,i=-32,j=66,D=-352,d=-8800$).
One would think that now the problem is dead and buried, but no, there are still loads of spurious nontrivial solutions with $D$ and $d$ integer. Are there at least only finitely many (plus trivial families like $j=i-1,j=0$ etc.)?
I think the spurious solutions must "mean" something (even if the quantum dimensions at $q\ne{1}$ are no longer Laurent polynomials in $q$)...
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$\begingroup$ Is the question about these Diophantine equations, or on these representations ? Perhaps I should know, but what do you ask about knots ? $\endgroup$– Dietrich BurdeJul 16, 2013 at 21:15
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$\begingroup$ What have you tried, and what counts as "spurious" or "trivial"? There are infinitely many $j$ for each of $i = -2, 0, 2, 4, 10$, plus a few (finitely many but always some) for each other $i$. $\endgroup$– Noam D. ElkiesJul 16, 2013 at 21:16
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$\begingroup$ @Dietrich: It's just this diophantine at the moment. The solutions will increase my understanding about knot polynomials (or not...and so I didn't tag "knot"). $\endgroup$– Hauke ReddmannJul 17, 2013 at 11:26
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$\begingroup$ @Noam: Here are all dimensions I computed (Mathematica copypaste ready): {i,j,(-i+i^2-2*j)/2,((-1+i)*(2+i))/2,((-4+i)^3*i*(2+i)*(-i+i^2-2*j)*j)/((i-2*i^2+i^3-8*j-ij)*(10*i-11*i^2+i^3+16*j+2*ij)),(i*(2+i)*(2-3*i+i^2-2*j)*(-i+i^2-2*j))/(3*(i-2*i^2+i^3-8*j-ij)),(-8*i*(2+i)*(-1+i-j)*j)/(3*(10*i-11*i^2+i^3+16*j+2*ij)),0,-(i*(-1+i-j)*(-3*i+3*i^2-16*j+ij))/(3*(i-2*i^2+i^3-8*j-ij)),(i*(2-3*i+i^2-2*j)*(10*i-11*i^2+i^3+32*j-2*ij))/(6*(10*i-11*i^2+i^3+16*j+2*ij)),((-1+i)*i*(4+i))/6}. Any (additional) "0" I would count as "trivial", everything else as "spurious". $\endgroup$– Hauke ReddmannJul 17, 2013 at 11:27
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$\begingroup$ still @Noam: I restrained 0<j<i(i-1)/4 because otherwise the second or third dimension on this list will get negative, and I'm fairly sure that's bad. But I'm not sure. Your list: i=-2,0,1,4 (2 was a typo?) fall under the "additional 0" verdict; i=10 is an interesting case. With the additional j condition only j=9 remains. (Does {10, 9, 36, 54, 80, 240, 0, 0, 0, 30, 210} ring a bell, irrep-dim-wise?) $\endgroup$– Hauke ReddmannJul 17, 2013 at 11:40
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