$\def\ket#1{\vert{#1}\rangle}\def\bra#1{\langle{#1}\vert}\def\Complex{\mathbb{C}}\def\sgn{\operatorname{sgn}}$There is a way to compute $F_{\Phi\Phi\Phi}^{\Phi}$ using the representation theory of the algebra $D(S_3)$. I am computing $F_{\Phi\Phi\Phi}^{\Phi}$ by using just the basic literal definition, so the whole exposition is long-winded. I will first review the definition of $F_{\Phi\Phi\Phi}^{\Phi}$ in the context of $D(S_3)$, and give more concrete details to actually compute $F_{\Phi\Phi\Phi}^{\Phi}$ in the next section.
First, $D(S_3)$ is a finite dimensional algebra and we may consider the representations of $D(S_3)$. In particular, $D(S_3)$ has irreducible representations $1$, $\Lambda$, and $\Phi$, which are 3 of the total 8 irreducible representations. The tensor product of representations are also representations, and the fusion rules give the decomposition of the tensor products of $1$, $\Lambda$, and $\Phi$ into a direct sum of copies of $1$, $\Lambda$, and $\Phi$. I will use $\otimes$ (not $\times$) since it is the actual tensor product. Also, I might use $=$ and $\cong$ interchangeably when it might be technically wrong to do so.
Now, we recall the definition of $F_{\Phi\Phi\Phi}^{\Phi}$, which arises from the different ways of realizing $\Phi$ as a subrepresentation of $\Phi \otimes \Phi \otimes \Phi$. By associativity of the tensor product, there are two ways to decompose $\Phi \otimes \Phi \otimes \Phi$. First, by identifying $\Phi \otimes \Phi \otimes \Phi = (\Phi \otimes \Phi) \otimes \Phi$ and using the fusion rule for $\Phi \otimes \Phi$, we have $$\begin{aligned}
(\Phi \otimes \Phi) \otimes \Phi &= (1 + \Lambda + \Phi) \otimes \Phi \\
&= 1 \otimes \Phi + \Lambda \otimes \Phi + \Phi \otimes \Phi \\
&= 1 \otimes \Phi + \Lambda \otimes \Phi + 1 + \Lambda + \Phi.
\end{aligned}$$ The copies of $\Phi$ we may identify are $1 \otimes \Phi$, $\Lambda \otimes \Phi$, and $\Phi \subseteq \Phi \otimes \Phi$. From this, we can choose 3 linearly independent $D(S_3)$-invariant linear maps that each map $\Phi$ into one of these copies of $\Phi$ in $\Phi \otimes \Phi \otimes \Phi$. These maps form a basis of this space of maps. Note that each of the maps are more precisely maps $\Phi \to 1 \otimes \Phi$, $\Phi \to \Lambda \otimes \Phi$, $\Phi \to \Phi \subseteq \Phi \otimes \Phi$ and so this basis can be indexed by the symbols $1$, $\Lambda$, $\Phi$.
On the other hand, by identifying $\Phi \otimes \Phi \otimes \Phi = \Phi \otimes (\Phi \otimes \Phi)$, we have $$\begin{aligned}
\Phi \otimes (\Phi \otimes \Phi) &= \Phi \otimes (1 + \Lambda + \Phi) \\
&= \Phi \otimes 1 + \Phi \otimes \Lambda + \Phi \otimes \Phi \\
&= \Phi \otimes 1 + \Phi \otimes \Lambda + 1 + \Lambda + \Phi.
\end{aligned}$$ We may again identify 3 copies of $\Phi$: $\Phi \otimes 1$, $\Phi \otimes \Lambda$, and $\Phi \subseteq \Phi \otimes \Phi$. Mapping $\Phi$ to each of these copies gives another basis of $D(S_3)$-invariant linear maps $\Phi \to \Phi \otimes \Phi \otimes \Phi$ indexed by $1$, $\Lambda$, and $\Phi$. There is a change-of-basis matrix between these two bases, and that matrix is by definition the $F$-matrix $F_{\Phi\Phi\Phi}^{\Phi}$. We note that $F_{\Phi\Phi\Phi}^{\Phi}$ is not unique since it depends on the choice of linear maps. For example, for some valid choice of linear map $\Phi \to 1 \otimes \Phi$, any constant multiple of this map gives another valid choice. The choice of these constants is called a gauge.
Background for Computing $F_{\Phi\Phi\Phi}^{\Phi}$
To compute $F_{\Phi\Phi\Phi}^{\Phi}$, we will use concrete descriptions of the representations of $D(S_3)$. For a concise source on $D(S_3)$ and its representations, see [1]. For a more in-depth mathematical source on the quantum double $D(G)$ of a finite group $G$, see [2]. We write $S_3$ using cycle notation, so $S_3 = \{1,(12),(13),(23),(123),(132)\}$.
Quantum Doubles and $D(S_3)$ Representations
Given a finite group $G$, the (complex) quantum double of $G$ is a complex algebra $D(G)$ with a basis of elements of the form $gh^\ast$ for $g, h \in G$ with the multiplication rule $$(g_1h_1^\ast)(g_2h_2^\ast) = \delta_{h_1,g_2^{-1}h_2g_2} (g_1g_2)h_2^\ast.$$ $D(G)$ also contains the group algebra of $\Complex[G]$ as a sub-algebra since $D(G)$ contains "group elements", i.e. for each $g \in G$, there exists an element $g \in D(G)$ defined as $g := \sum_{h \in G} gh^\ast$. $D(G)$ also contains "dual group elements", i.e. for each $h \in G$, there exists $h^\ast \in D(G)$ defined as $h^\ast := 1h^\ast$ where $1$ is the identity element of $G$.
The representations $1$, $\Lambda$, and $\Phi$ can be described as follows:
- $1$ has a single basis vector $\ket{1,1}$. The action of group/dual group elements of $D(G)$ is given by $g\ket{1,1} = \ket{1,1}$ and $h^\ast \ket{1,1} = \delta_{1,h} \ket{1,1}$.
- $\Lambda$ has a single basis vector $\ket{\Lambda,1}$. The action of group/dual group elements is given by $g\ket{\Lambda,1} = \sgn(g)\ket{\Lambda,1}$ and $h^\ast \ket{\Lambda,1} = \delta_{1,h} \ket{\Lambda,1}$.
- $\Phi$ has two basis vectors $\ket{\Phi,1}$ and $\ket{\Phi,2}$. The action of group elements where $g \in \{(12),(23),(13)\}$ is given by $g \ket{\Phi,1} = \ket{\Phi,2}$ and $g\ket{\Phi,2} = \ket{\Phi,1}$, meaning $g$ switches the two basis vectors. $g \in \{1,(123),(132)\}$ acts as the identity $g\ket{\Phi,m} = \ket{\Phi,m}$. The dual group elements on basis vectors is given by $h^\ast \ket{\Phi,1} = \delta_{(123),h} \ket{\Phi,1}$ and $h^\ast \ket{\Phi,2} = \delta_{(132),h} \ket{\Phi,2}$, meaning only $(123)$ and $(132)$ have nonzero actions, each on a different basis vector.
In [1], these representations are denoted $A$, $B$, and $F$ respectively. Also, the notation here for the basis vectors and actions is simplified in comparison to the notation in [1]. Lastly, note that in physics terminology, $1$ is the vacuum, $\Lambda$ is pure electric charge, and $\Phi$ is the pure magnetic charge.
Next we define the tensor product representations. $D(S_3)$ is a Hopf algebra with comultiplication given by $$\Delta(gh^\ast) = (g \otimes g) \sum_{k \in G} (k^{-1}h)^{\ast} \otimes k^\ast$$ and this formula allows us to define the tensor product of representations. Basically, given representations $J_1$ and $J_2$ of $D(S_3)$, the vector space tensor product $J_1 \otimes J_2$ is a representation of $D(S_3)$ via the formula $$(gh^\ast)\ket{J_1,m_1}\ket{J_2,m_2} := \Delta(gh^\ast)\ket{J_1,m_1}\ket{J_2,m_2}$$ where the multiplication on the right is done component-wise.
Concrete Fusion Rules
Now we want to concretely realize the basic fusion rules by decomposing some tensor product of representations into irreducible parts.
$\Phi \otimes \Phi$. Recall that $\Phi \otimes \Phi \cong 1 + \Lambda + \Phi$. Within $\Phi \otimes \Phi$, we first look for $1$, which has a single basis vector invariant under group element actions. Since group elements of $D(S_3)$ swap the basis vectors of $\Phi$, an experienced quantum information theorist may try to take a superposition so that group elements swap the summands. Indeed, the Bell state $$\ket{\Phi^2;1,1} = \frac{1}{\sqrt{2}}(\ket{\Phi,1}\ket{\Phi,2} + \ket{\Phi,2}\ket{\Phi,1})$$ is one that works and also matches the action of dual group elements. Likewise, looking for $\Lambda$, we want a superposition where swapping the summands introduces a negative. The Bell state $$\ket{\Phi^2;\Lambda,1} = \frac{1}{\sqrt{2}}(\ket{\Phi,1}\ket{\Phi,2} - \ket{\Phi,2}\ket{\Phi,1})$$ works and matches the action of the dual group elements. The two remaining orthogonal vectors of $\Phi \otimes \Phi$ constitute $\Phi$, and a choice of basis vectors that take the role of $\ket{\Phi,1}$ and $\ket{\Phi,2}$ is $$\ket{\Phi^2;\Phi,1} = \ket{\Phi,2}\ket{\Phi,2}$$ $$\ket{\Phi^2;\Phi,2} = \ket{\Phi,1}\ket{\Phi,1}$$
$1 \otimes \Phi$ and $\Phi \otimes 1$: Since $1$ is one-dimensional, there's really only one choice of basis vectors for $\Phi$ in $1 \otimes \Phi$: $$\ket{1 \times \Phi;\Phi,1} = \ket{1,1}\ket{\Phi,1}$$ $$\ket{1 \times \Phi;\Phi,2} = \ket{1,1}\ket{\Phi,2}$$ One must simply verify with the action of dual elements that this ordering matches $\ket{\Phi,1}$ and $\ket{\Phi,2}$. Likewise, for $\Phi \otimes 1$ just swap the tensor components of the basis vectors of $1 \otimes \Phi$.
$\Lambda \otimes \Phi$ and $\Phi \otimes \Lambda$: Since $\Lambda$ is also one-dimensional, one natural choice of basis vectors for $\Phi$ is like the one for $1 \otimes \Phi$ but replacing $\ket{1,1}$ with $\ket{\Lambda,1}$. However, this choice is not compatible with the group action since $g \in \{(12),(13),(23)\}$ introduces a negative from the action on $\ket{\Lambda,1}$. Thus, we must make a small change: $$\ket{\Lambda \times \Phi;\Phi,1} = \ket{\Lambda,1}\ket{\Phi,1}$$ $$\ket{\Lambda \times \Phi;\Phi,2} = -\ket{\Lambda,1}\ket{\Phi,2}$$
Computing $F_{\Phi\Phi\Phi}^{\Phi}$
Now we want to find ways of identifying $\Phi$ within $$(\Phi \otimes \Phi) \otimes \Phi \cong 1 \otimes \Phi + \Lambda \otimes \Phi + 1 + \Lambda + \Phi,$$ which amounts to finding a basis for $\Phi$ within $1 \otimes \Phi$, $\Lambda \otimes \Phi$, $\Phi \otimes \Phi$ as subrepresentations of $(\Phi \otimes \Phi) \otimes \Phi$. We also do the same for $$\Phi \otimes (\Phi \otimes \Phi) \cong \Phi \otimes 1 + \Phi \otimes \Lambda + 1 + \Lambda + \Phi.$$
The reasoning is much like what was done before for finding $1$, $\Lambda$, and $\Phi$ within $\Phi \otimes \Phi$ and finding the basis for $\Phi$ within $1 \otimes \Phi$ and $\Lambda \otimes \Phi$ so I will skip the details. Here are valid choices of identifying $\ket{\Phi,1}$ and $\ket{\Phi,2}$ in each representation:
- $1 \otimes \Phi$: $\{\ket{\Phi^2;1,1}\ket{\Phi,1},\ket{\Phi^2;1,1}\ket{\Phi,2}\}$
- $\Lambda \otimes \Phi$: $\{\ket{\Phi^2;\Lambda,1}\ket{\Phi,1},-\ket{\Phi^2;\Lambda,1}\ket{\Phi,2}\}$
- $\Phi \otimes \Phi$: $\{\ket{\Phi^2;\Phi,2}\ket{\Phi,2},\ket{\Phi^2;\Phi,1}\ket{\Phi,1}\}$
Writing these as linear maps mapping $\ket{\Phi,1}$ and $\ket{\Phi,2}$ to each basis vector, this gives:
- $\ket{\Phi^2;1,1}\ket{\Phi,1}\bra{\Phi,1} + \ket{\Phi^2;1,1}\ket{\Phi,2}\bra{\Phi,2}$
- $\ket{\Phi^2;\Lambda,1}\ket{\Phi,1}\bra{\Phi,1}-\ket{\Phi^2;\Lambda,1}\ket{\Phi,2}\bra{\Phi,2}$
- $\ket{\Phi^2;\Phi,2}\ket{\Phi,2}\bra{\Phi,1} + \ket{\Phi^2;\Phi,1}\ket{\Phi,1}\bra{\Phi,2}$
For $\Phi \otimes (\Phi \otimes \Phi)$:
- $\Phi \otimes 1$: $\{\ket{\Phi,1}\ket{\Phi^2;1,1},\ket{\Phi,2}\ket{\Phi^2;1,1}\}$
- $\Phi \otimes \Lambda$: $\{\ket{\Phi,1}\ket{\Phi^2;\Lambda,1},-\ket{\Phi,2}\ket{\Phi^2;\Lambda,1}\}$
- $\Phi \otimes \Phi$: $\{\ket{\Phi,2}\ket{\Phi^2;\Phi,2},\ket{\Phi,1}\ket{\Phi^2;\Phi,1}\}$
As linear maps:
- $\ket{\Phi,1}\ket{\Phi^2;1,1}\bra{\Phi,1} +\ket{\Phi,2}\ket{\Phi^2;1,1}\bra{\Phi,2}$
- $\ket{\Phi,1}\ket{\Phi^2;\Lambda,1}\bra{\Phi,1}-\ket{\Phi,2}\ket{\Phi^2;\Lambda,1}\bra{\Phi,2}$
- $\ket{\Phi,2}\ket{\Phi^2;\Phi,2}\bra{\Phi,1}+\ket{\Phi,1}\ket{\Phi^2;\Phi,1}\bra{\Phi,2}$
Lastly, to compute $F_{\Phi\Phi\Phi}^{\Phi}$, we must compute the change-of-basis matrix between these two bases of linear maps. These two bases turn out to be orthogonal, thus we may compute the inner products between elements of the first basis and the second basis. To compute the inner products, you can turn the bras into kets and take the bra-ket inner product. For these choices of bases, you will get $$F_{\Phi\Phi\Phi}^{\Phi} = \begin{pmatrix}
1 & 1 & \sqrt{2} \\
-1 & -1 & \sqrt{2} \\
\sqrt{2} & -\sqrt{2} & 0
\end{pmatrix}.$$ To match the $F_{\Phi\Phi\Phi}^{\Phi}$ in the question, you can make the gauge transformation of multiplying the second element of the first basis by $-1$, and multiplying the second and third elements of the second basis by $-1$.
