# Span of tangle vector space for different Lie groups

In his $G_2$ paper, Kuperberg gives the following numbers of acyclic freeways for n=0...6: 1, 0, 1, 1, 4, 10, 35. (Which is identical to $dim Inv(V^{⊗n}_{1,0})$, the spanning size of the tangle vector space. (??). But that is NOT identical with the number of crossingless trivalent tangle graphs: for n=6, the "hexagon" can't be resolved into the 34 acycling tangle graphs and must be added to the linear independent set.) The $G_2$ numbers (or so I think) are identical for the whole $E_7$ family (except $B_3(\Lambda_3)$, $A_1(3\Lambda_1)$ and $A_1*A_1(\Lambda_1*\Lambda_1)$ where the last number should be 30,34 and 25 if I computed correctly). (But since you now must use 2 irrep colors for any other than $G_2$, 5 of the 34 above graphs are forbidden.)
With some jump of faith, I suppose also for all members of the $E_8$ family the spanning space numbers are the same: 1, 0, 1, 1, 5, 15, 70 (???), except for some "special" groups of that family.

Can you verify my $dim Inv(V^{⊗n}_{1,0})$ values for $E_8$ ? (Maybe with a list of exceptional members ? Directly I can only compute for $A_1(4\Lambda_1)$ and this surely gives less than 70.)

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Here are the dimensions of the space of invariant tensors in the tensor powers of the adjoint representation:

F4: 1 0 1 1 5 16 80 436 2891 ...

E6: 1 0 1 1 5 17 90 542 3962 ...

E7: 1 0 1 1 5 16 80 436 2877 ...

E8: 1 0 1 1 5 16 79 421 2674 ...

The reason the numbers for E6 are too high is that there is a diagram automorphism. The correct category of representations is the representations of the semidirect product of the Lie group and a group of order two. Then these numbers suggest that the sequence starts

1 0 1 1 5 16 80 436 ...

and that for E8 there is an extra relation for n=6. However it is not clear what should happen for n=8 or n>8.

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Thanks! Of course 16, forgot the pentagon ring. Have I botched up notation somewhere? Even more interesting to me are the defining (not the adjoint) irrep tensor powers. Could you list them also? –  Hauke Reddmann Mar 18 '12 at 16:54