# Knot Numerology

EDIT: Ok, I condense it to only that what is needed.

Assume it's possible to use the method described here Matrix decomposition the other way to decompose a $S$ matrix from knot theory. Then each $Ti$ (except the one that corresponds to the trivial tangle) corresponds to a diagram >=< like the one from Kuperbergs $B_2$ spider paper. (Actually, I have no proof for that assertion!)
Example: take the one from his paper. The representation $R_1$ corresponding to the > part is $Sp(4)$, $\lambda_1$. (You can read it off from the unknot value.) The one $R_2$ belonging to = is $Sp(4)$, $\lambda_2$ and the $R_3$ belonging to the other $Ti$ corresponds to $Sp4(2\lambda_1)$. (Haven't proved it, size 10 is too large for my programs.)
In the paper "Algebraic equations determining quantum dimensions" by Masahito Hayashi you'll find the tensor decomposition $R_1\cdot R_1=1+R_2+R_3$ (p. 2417) Moreover $R_2\cdot R_2$ decomposes into 3 summands (i.e. another Kauffman polynomial) and $R_3\cdot R_3$ into 6. This is also reflected by the eigenvalues of the corresponding $S$: $S_1$ and $S_2$ have 3 and $S_3$ 6 different eigenvalues.

Question 1: From where comes the correspondence between the expansion of $S$ into $Ti$ resp. the eigenvalue multiplicity, and representation theory? (The integers showing up are simply the values of the unknot, resp. quantum dimension, at $q=1$.) Especially, why can you almost read off the skein relation for some $S$ (at least it's "height") from the tensor product expansion of the corresponding $R\times R$?

Question 2: The paper also gives the quadratic Casimir for the representations. Why is it (save some dimension factor) equal to the power of $q$ in the writhe normalization factor? The latter can be seen as some closure (i.e. trace) over $S$. Is the Casimir also a kind of trace?

Question 3: While the representation theory experts are here anyway, in which Lie algebra $4\cdot 4=1+3+5+7$ happens? $3\cdot 3=1+3+5$, $5\cdot 5=1+3+5+7+9$. And so on. Quantum dimension associated with 4 $q^6+q^2+q^{-2}+q^{-6}$, for 3 $q^4+1+q^{-4}$ and so on. This is most interesting to me as the corresponding $S$ matrix seems NOT fall into the $ABCDEFG$ classification. Maybe some permutation or rotation group?

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Dear Hauke, please edit your post to make it clear what your question is... – André Henriques Jul 4 '11 at 18:15
Everything with a "?" behind it? :-) Sorry, my thoughts are still confused since I have no knowledge of the area. I hope by deleting superfluous stuff it's clearer now. This all is still "work in progress". – Hauke Reddmann Jul 5 '11 at 11:24
Dear Hauke. Thank you for improving your presentation. I've edited the latex to make it more readable. I still think that there might be some typos left... but as I don't really understand what you wrote, I let you do those last changes yourself (if needed). – André Henriques Jul 5 '11 at 19:56
THX! Now that I (accidentally) saw the "raw" text, I hope I can learn basic LaTeX coding (cursive, powers etc.) from that. It's a bit embarassing to be a dummy, always leaving it to a mod :-) – Hauke Reddmann Jul 6 '11 at 10:13