At q squared of order three, TL2 is semisimple with a degenerate Jones trace

Take $q=e^{i\pi \mathrm{/}3}$, so $q^2$ has order $\mathrm{\ell }=3$, $\delta =q+q^{-1}=1$, and $n=2=\mathrm{\ell }-1$. The algebra ${\mathrm{T}\mathrm{L}}_2(\mathbb{C},1)$ is semisimple, but its normalized closure trace form is degenerate.

In the left panel the solid top cap and bottom cup are the basis diagram $u=d_1$. The right panel adds the exterior closure arcs in dashed blue; each joins top point $i$ to bottom point $i$. Thus the four arcs form one closed component, so $c(u)=1$ in the closure-trace definition. The drawing records the matching graph only; the algebra and trace calculations below prove the claims independently.

\begin{tikzpicture}[x=1cm,y=1cm,line width=0.8pt,baseline=(current bounding box.center)]
  \fill (0,1) circle (1.5pt); \fill (1,1) circle (1.5pt);
  \fill (0,-1) circle (1.5pt); \fill (1,-1) circle (1.5pt);
  \draw (0,1) to[out=90,in=90] (1,1);
  \draw (0,-1) to[out=-90,in=-90] (1,-1);
  \node at (0.5,0) {$u$};
  \node at (0.5,-1.65) {basis diagram};

  \fill (3,1) circle (1.5pt); \fill (4,1) circle (1.5pt);
  \fill (3,-1) circle (1.5pt); \fill (4,-1) circle (1.5pt);
  \draw (3,1) to[out=90,in=90] (4,1);
  \draw (3,-1) to[out=-90,in=-90] (4,-1);
  \draw[dashed,blue] (3,1) to[out=180,in=180] (3,-1);
  \draw[dashed,blue] (4,1) to[out=0,in=0] (4,-1);
  \node at (3.5,-1.65) {closed graph: one component};
\end{tikzpicture}