Since I come from theoretical physics background, my terminology may look a little bit odd. I will try my best to explain in plain simple English what I mean.
In case of $SU(2)$, the algebra of (tensor products of) finite-dimensional irreducible representations can be understood with help of recoupling theory (originally developed by Penrose under the name of "Binor calculus") in the following two easy steps:
It can be shown that spin-$j$ irrep is equal to the symmetrized tensor product of $2j$ fundamental (spin-half) irreps. Therefore we can represent irreps graphically as $2j$ lines with a "symmetrizer" box attached to them. Though because we choose to assign an additional minus sign to any crossing in the graphical calculus, this is actually an "antisymmetrizer" box, technically.
Now that we have a convenient way of representing arbitrary irreps, we have to impose a fundamental rule of this graphical calculus which basically says that we are dealing with $SU(2)$: the skein relation
Now we can start constructing tensor products of arbitrary irreps, intertwining tensors, $\{6j\}$ symbols, etc.
My question is about how and to which extent one can repeat this construction for $SU(3)$. In particular, I was able to reproduce the following:
- $SU(3)$ has 2 fundamental irreps ($\mathbf{3}$ and $\mathbf{\bar{3}}$), which means that we have to have two "colors" of the wires.
- But since the two fundamentals are related by conjugation, these are better represented with arrows on the wires. Changing the direction of the arrow corresponds to conjugation. E.g. the Kronecker delta tensor has one index in the $\mathbf{3}$ irrep and another in $\mathbf{\bar{3}}$. Graphically this corresponds to joining the arrows and preserving the orientation.
- I was able to reproduce representations with highest weight $(a,0)$ and $(0,b)$ (for integer $a$ and $b$) by taking $a$ and $b$ wires and attaching an "antisymmetrizer" box to them, just like with $SU(2)$. The type of the irrep $(a, 0)$ or $(0, a)$ is determined by the direction of the arrow.
However, the rest still remains a mystery to me. In particular,
- Can this construction be extended to the general case of $(a, b)$? I tried tensoring $\mathbf{3}^a \otimes \mathbf{\bar{3}}^b$ but it doesn't seem to contain $(a, b)$. Apparently I was wrong and it does contain $(a, b)$. I still don't know how to extract it using graphical calculus though. See update.
- What would be the skein relation for $SU(3)$ recoupling?
P.S. in my terminology, $(a,b)$ is the highest weight irrep with non-negative integer $a$ and $b$ – the two highest weights of $SU(3)$. E.g. this is a weight diagram for the $(4,2)$ irrep:
In this notation: $\mathbf{1} = (0, 0)$; $\mathbf{3} = (1, 0)$; $\mathbf{\bar{3}} = (0, 1)$; $\mathbf{8} = (1, 1)$; $ \dots $.
Update:
In the comments it was suggested that $$ (a,b) = \text{ker} T,$$ where $$ T: S(\mathbf{3}^a) \otimes S(\mathbf{\bar{3}}^b) \rightarrow S(\mathbf{3}^{a-1}) \otimes S(\mathbf{\bar{3}}^{b-1}) $$ is a surjective map of special kind.
I was able to reproduce the map $T$ with recoupling wires. Here's an example for $a=4,\,b=2$ :
Please excuse my artwork. Red lines denote (idempotent) "antisymmetrizer" boxes. However, I am not sure how one could reproduce the kernel $\text{ker}T$ of this map with recoupling wires.