# How to calculate the principal graphs of a fusion ring with a given simple object?

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? , but not enough detail.

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.

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.

• Thank you for answering the question. I have read the bimodule descriprtion of the principal graphs $(\Gamma_+,\Gamma_−)$. But I do not understand such a description. My question is that given the fusion coefficients $N^{XY}_Z$ for simple objects, how to calculate the principal graphs $(\Gamma_+,\Gamma_−)$? For example, how many even/odd vertices in $\Gamma_\pm$? I know $N^{XY}_Z$, but I am not familar with bimodule, Hom, subfactors, etc. – Xiao-Gang Wen Oct 30 '15 at 1:12
• I asked this question, because I wonder if the conditions on the principal graphs $(\Gamma_+,\Gamma_−)$ can be translated to become the conditions on the fusion coefficients $N^{YX}_Z$ and $N^{YX^*}_Z$ (for fixed $X$, and $Y,Z=1,2,\cdots,N$, where $N$ is the rank of the fusion ring). – Xiao-Gang Wen Oct 30 '15 at 1:24
• The fusion coefficients $N^{Y,X}_Z$ are exactly the dimensions of the hom spaces. I've edited my answer accordingly so it's easier to understand. In the fusion category case, there's just one graph, and the number of vertices is the rank of the category. In the (finite depth) subfactor setting, you have a fusion category $\mathcal{C}$ and an indecomposable module category $\mathcal{M}$. The number of even vertices of $\Gamma_+$ is the rank of $\mathcal{C}$, and the number of odd vertices is the rank of $\mathcal{M}$. – Dave Penneys Oct 30 '15 at 1:47
• It appears that there is no direct way to construct the graph $\Gamma_+$ directly from the data $N^{Y,X}_Z$.(?). Let $N^X$ be the matrix whose elements are $(N^X)_{ZY}=N^{X,Y}_Z$ (here I assume $N^{XY}_Z=N^{YX}_Z$) I wonder if $\Gamma_+$ can be viewed as the graph described by matrix $\begin{pmatrix} 0 &N^X\\ N^{X^*} & 0 \end{pmatrix}$. In this case, the number of even vertices is the rank $N$, and the number of odd vertices is also the rank $N$. Maybe such a graph is not principle (?) – Xiao-Gang Wen Oct 30 '15 at 2:23
• Thank you for answering. It appears that the principal graph $\Gamma_+$ can be constructed in the following way: Let $\Gamma_X$ be the bipartite graph defined by the matrix $\begin{pmatrix} 0 &N^X\\ N^{X^*} & 0 \end{pmatrix}$, then $\Gamma_+$ is the connected subgraph of $\Gamma_X$ that contains $X$ among its odd vertices. – Xiao-Gang Wen Oct 30 '15 at 14:23