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Is there much known about the dimensions $D$ of $E_{7\frac{1}{2}}$ (that is: $D_6.H_{32}$) beyond $$ 44\otimes44(def)=1\oplus945\oplus99(adj)\oplus891\, ? $$ Generally, does a weight indexing scheme like for the semisimple $E_7$ cousins exist? Particularly, is $D=134805$ an irrep dimension or the sum of two $D_1+D_2$?

Background: In the $E_7$ series, in $def^{\bigotimes4}$ there always pops up a pair of dimensions with equal quadratic Casimir ($C=6(m+1)$, your gauge may vary, I conform to the Hayashi paper on quantum dimensions) but in most cases one partner will be identically zero (one example for each case: $E_7$: $m=8,C=54, D_1=365750,D_2=0$, and $C_3$: $m=1,C=12,D_1=525,D_2=385$). And my "magic" formulae only give the sum.

P.S. In case nothing is known and anyone needs a few dimensions $D$ of $E_{7\frac{1}{2}}$ together with the quadratic Casimirs and the Clebsch–Gordan series, just ask; what I do is not math but it works :-)

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Almost! If $V$ denotes the associated Severi representation, we have:

$$2*V^{(1)} + 2*V^{(2)} + V^{(4)} + V^{(5)} + V^{(6)} = 2*21 + 2*210 + 6930 + 28314 + 99099 = 134805$$

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  • $\begingroup$ Like I said: mathoverflow.net/questions/312657/… :-) In particular, unless all your V have same quadratic Casimir, I would term this "accidental". Do you, perchance, have a list of all irrep dims <134805? I think I can sum them pairwise myself then :-) $\endgroup$ Jan 9, 2019 at 17:58
  • $\begingroup$ In your question you don't ask that the irrep must have the same quadratic Casimir... $\endgroup$
    – Libli
    Jan 10, 2019 at 12:41
  • $\begingroup$ In any case, since $V^{(k)}$ denotes the $k$-th Cartan's power of $V$, it is obvious that the Casimir elements of all these representation are "the same" (in the appropriate sense). $\endgroup$
    – Libli
    Jan 10, 2019 at 12:51
  • $\begingroup$ Ah, should have made that explicit. It's quite possible that my "magic" dimensions further split up due to $E_7\frac{1}{2}$ not being semisimple. They should be the "analog" to 525 and 385 in $C_3$, but again, the analogy probably only works for semisimple groups. $\endgroup$ Jan 13, 2019 at 10:21

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