those boundary faces of the edges of $c$ fixed. The notation with the sum and products in eq. (\[eq:transitionamplitudequantumgroup-5\]) is to indicate that one needs to use all possible configurations allowed.

Fibonacci Fusion Hilbert space
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We now consider the topological data of $SU(2)^5_q$. The admissible spin representations are $j=0,\frac{1}{2},1,\frac{3}{2}$, and their quantum dimension are given by $$d_{j}^{q}=sin\left(\frac{\pi(2j+1)}{5}\right)/sin\left(\frac{\pi}{5}\right).
\label{eq:quantumdimension}$$ The composition of representations follows the following fusion rules: $$\begin{aligned}
0\otimes j & =j\nonumber \\
\frac{3}{2}\otimes j & =\frac{3}{2}-j\nonumber \\
\frac{1}{2}\otimes\frac{1}{2} & =0\oplus1\nonumber \\
\frac{1}{2}\otimes1 & =\frac{1}{2}\oplus\frac{3}{2}\nonumber \\
1\otimes1 & =0\oplus1.\label{eq:fusionrulessu23}\end{aligned}$$ Remember that a segment of the triangulation is dual to a face on the dual 2-complex where there is associated a representation of $SU(2)^5_q$. So more fundamentally there are 2 states on the segments, one with quantum dimension $1$ and one with quantum dimension $\phi$, and we have a notion of a 2-dimensional fusion Hilbert space [@ZhenghanWang; @Pachos] as the superposition of these 2 states. The space $H_{j_{q1}}\times H_{j_{q2}}\times H_{j_{q3}}$ is in this way 3-dimensional and is related to the triangular faces. The tetrahedron $H_{j_{q1}}\times H_{j_{q2}}\times H_{j_{q3}}\times H_{j_{q4}}$ is 5 dimensional. The dimension grows with increasing number of representations, following the well known Fibonacci sequence, which gives the name of 