**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?

questionis... – André Henriques Jul 4 '11 at 18:15