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Notation abuse warning: I will use the E7 series irrep names. You'll soon see why.

In the general Lie algebra, the irrep you "start" with is $J$, the adjoint. From the series for $J\bigotimes{J}=1+J+A...$ ($1$ is the 1-dimensional irrep) you get $A$, the antisymmetric one. (The quantum dimensions up to here are all given e.g. in Westbury's "R-matrices and the magic square" paper.) The next step would be $J\bigotimes{A}=J+A+S...$ where you can "fish out" the symmetric irrep $S$ from the Clebsch Gordan series. (Does anybody have the quantum dimension $Dim(S)=S(\alpha,\beta,\gamma)$ handy, just as for $J$ and $A$? In principle I could compute it myself, but this would speed up my work considerably.)
Now comes the silly part :-) In the E7 series, $V\bigotimes{V}=1+J+S+A$ and I can now backwards compute the dimension of the defining irrep $V$ and see what comes out for, say, E8. Well, nonsense is coming out, which is hardly surprising. Still, I would have expected integer nonsense or even better a zero ($324-273-52+1=0$ for F4 -the minus signs are only very mild cheating. :-)! Irrational nonsense is baaaad.
So: Is it possible to bring all irreps, even from different rows of the magic square, in an 1:1 correspondence (if neccessary by inventing fictive irreps which still have integer dimension) or am I completely barking up the wrong Lie?

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Dear Hauke. I'm sorry, but your question is incomprehensible. Maybe you should be more specific with your E7 example: can you then tell us what $J$, $A$, and $S$ are (e.g. by specifying the highest weight, or by telling us the dimenison of the irrep). But most importantly, it's not clear what you're asking. What do you mean by "bring all irreps in a 1:1 correspondence"??? –  André Henriques Aug 23 '12 at 13:33
    
I took the names from "Birdtracks". OK, let's just take E7 and E8 (and hope weights are standardized): E7: V=0000001, J=1000000, A=0000010, S=0000002, 6=0010000, 7=0 (accidentally), 8=2000000. E8: V nope, J=00000001, A=10000000, S=0 (a.), 6=00000010, 7=0 (a.), 8=000000002. J*J=E+J+A+6+7+8. This works for E7 as well as E8, and you have the closed quantum dimension formula of Vogel for J,A,6,7,8, and counting terms in the Clebsch-Gordan you know that 6 corresponds to 6, 7 to 7 and 8 to 8 (duh :-). J*A=J+A+S+... Dito. You can say which irrep of E8 corresponds to S of E7. (It's a 0.) (cont.) –  Hauke Reddmann Aug 24 '12 at 9:38
    
Thus you can bring at least all irreps that pop up in some $J\bigotimes{n}$ of E8 in an 1:1 correspondence with those of E7. (No, I can't guarantee that you can REALLY say A is A and 8 is 8. Just from symmetry arguments A(E7) could be 8(E8) and vice versa. But from the lowest irrep CG expansions I checked, the correspondence seemed convincing to me.) ... And I wanted to do something for those irreps who feel lonesome now :-) –  Hauke Reddmann Aug 24 '12 at 9:49
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up vote 1 down vote accepted

The (quantum) dimensions are all given in Vogel's papers (all but one unpublished). There may be some errors in my paper you refer to as I confused Vogel's parameters with the values of the quadratic Casimir.

I don't follow your "silly part". The representation $V$ is not defined for $E_8$ and I don't see how you get a proposed quantum dimension.

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Nor do I :-) Yes, that's exactly the point: V is not defined for E8, but still you can compute the quantum dimension of this undefined dingbat (assuming V^2=1+J+S+A, of course). So I thought that maybe a whole series of fictive irreps could be defined such that all the dimension equations are fulfilled. (Hey, the square root of -1 doesn't exist either :-) But this is "fun" only with integer dimensions. –  Hauke Reddmann Aug 24 '12 at 9:27
    
This comment does explain what is behind your question. In my opinion this does not seem like a profitable line of enquiry but I would be happy to be proved wrong. –  Bruce Westbury Aug 26 '12 at 15:05
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