My understanding is that the principal graphs are a pair of undirected bipartite graphs, $\Gamma\_+,\Gamma\_-$. They can be calculated from a fusion ring with a given simple object $X$. How to calculate such pair of graphs?


There is related dicussion [An embedding theorem for a fusion ring planar algebra?](https://mathoverflow.net/questions/141955/an-embedding-theorem-for-a-fusion-ring-planar-algebra) , but not enough detail.



How to calculate the principal graphs of a fusion ring with a given simple object?
Probably what you really want is the fusion graph $\Gamma\_X$ with respect to your simple object $X$. It is a directed graph with vertices labelled by the simple objects of your category. Between the vertices labelled by simples $Y$ and $Z$, there are $N^{Y,X}\_Z=\operatorname{dim}(\operatorname{Hom}(Y\otimes X, Z))$ directed edges from $Y$ to $Z$. The fusion graph with respect to $X^\*$ is obtained by reversing the orientation of all the edges, since by semisimplicity and Frobenius reciprocity,
$$
N^{Y,X}\_Z
=
\operatorname{dim}(\operatorname{Hom}(Y\otimes X, Z))
=
\operatorname{dim}(\operatorname{Hom}(Y, Z\otimes X^\*))
=
\operatorname{dim}(\operatorname{Hom}(Z\otimes X^\*, Y))
=
N^{Y,X^\*}\_Z.
$$
This means that if $X$ is self-dual, then you can view this graph as unoriented.


In your question you've referred to a pair of unoriented graphs. These would arise if you were looking at the principal graphs of a subfactor. Given a finite index subfactor $A\subset B$, we have the semisimple tensor category of $A-A$ bimodules generated by $B$. The simples are obtained by decomposing the tensor powers $\bigotimes\_A^n B$ into irreducibles. We can also view $\bigotimes\_A^n B$ as an $A-B$ bimodule and decompose into irreducibles. This gives the semisimple module category of $A-B$ bimodules. Similarly, we have a tensor category of $B-B$ bimodules and a module category of $B-A$ bimodules.


We get the principal graphs $(\Gamma\_+,\Gamma\_-)$ as follows. The graph $\Gamma\_+$ has even vertices the simple $A-A$ bimodules and odd vertices the simple $A-B$ bimodules. There are then $\operatorname{dim}(\operatorname{Hom}\_{A-B}(Y\otimes\_A B, Z))$ edges between the simple $A-A$ bimodule $Y$ and the simple $A-B$ bimodule $Z$. The dual principal graph $\Gamma\_-$ is defined analogously with $B-B$ and $B-A$ bimodules.


Edit after the comments:


Given a unitary fusion category $\mathcal{C}$ and an object $X\in\mathcal{C}$, you can construct a subfactor standard invariant $\mathcal{P}\_\bullet$ from which you can get a pair of principal graphs $(\Gamma\_+,\Gamma\_-)$. First, you let
\begin{align\*}
\mathcal{P}\_{n,+} &= \operatorname{End}(\underbrace{X\otimes X^\*\otimes \cdots \otimes X^\pm}\_{n \text{ copies}})
\\ 
\mathcal{P}\_{n,-} &= \operatorname{End}(\underbrace{X^\*\otimes X\otimes \cdots \otimes X^\mp}\_{n \text{ copies}})
\end{align\*}
where $X^\pm=X$ if $n$ is odd and $X^\*$ if $n$ is even. Now the graph $\Gamma\_+$ has even vertices the simple summands of objects of the form $(X\otimes X^\*)^{\otimes n}$, and odd vertices the simple summands of objects of the form $(X\otimes X^\*)^{\otimes n}\otimes X$. You get the edges similar to before, taking an even $Y$, tensoring with $X$, and decomposing into simples $Z$. The graph $\Gamma\_-$ is defined similarly, but swapping the roles of $X$ and $X^\*$.


So that's the formal explanation of how you get the principal graphs, but there is a simpler way to compute $(\Gamma\_+,\Gamma\_-)$ based on the fusion coefficients $N^{Y,X}\_Z$.


As @Xiao-Gang points out in the comments, we can compute $\Gamma\_+$ as the connected component containing $X$ in the bipartite graph with adjacency matrix
$$
\begin{pmatrix}
0 & N^X
\\
N^{X^\*} & 0
\end{pmatrix}
$$
where $N^X$ is the matrix of fusion coefficients for tensoring on the right by $X$, and similarly for $X^\*$.
The graph $\Gamma\_-$ is then obtained by swapping the roles of $X$ and $X^\*$.


Interestingly, if $X\in \mathcal{C}$ is self-dual and the fusion graph $\Gamma\_X$ in $\mathcal{C}$ is already bipartite (here, $\mathcal{C}$ is $\mathbb{Z}/2\mathbb{Z}$-graded and $X$ is odd), then $(\Gamma\_+,\Gamma\_-) = (\Gamma\_X, \Gamma\_X)$. This occurs for many interesting examples.


