Root-of-unity skein TQFT for 3-manifolds — Examples and Counterexamples

Here $r=3$, $q=-A^2$ has $q^2$ of order three, and $q+q^{-1}=-1$. Thus $\delta =-1$, $[2]=\delta =-1$, and $[3]=\delta [2]-[1]=(-1)(-1)-1=0$.

The recursion for $p_3$ requires division by $[3]$. Since $[3]=0$ in $\mathbb{C}$, no multiplicative inverse exists, so this recursive expression is undefined. In particular, cancellation cannot assign it a value. The calculation refutes the asserted universal continuation; it makes no claim that a projector at rank three cannot be constructed by a different method.

At $p=4$, $r=2$ and $r-2=0$, so the displayed label set is exactly $\{0\}$. At $p=6$, $r=3$ and $r-2=1$, so it is exactly $\{0,1\}$.

At $p=6$, $\delta =-1$, $p_2=1_2+U_1$, and $e_0=-U_1$ is an idempotent isomorphic to the tensor unit. Since $p_2$ is negligible by [F2], the decomposition $1_2=p_2+e_0$ gives $1\otimes 1\cong 0$ in the quotient. This verifies the stated fusion calculation.

By [F6] the torus dimension is the number of these basis labels, namely $r-1$. Step1.1 gives one label for $p=4$ and two labels for $p=6$, so the dimensions are $1$ and $2$. Together with the retained local fusion calculation in step2.1 this proves every Example clause. No polynomial skein basis before radical quotient is confused with the finite TQFT basis.