I've heard that Reshetikhin-Turaev (RT) is stronger than homotopy, and it can distinguish certain homotopy-equivalent, but non-homeomorphic Lens spaces (I think $L(7,1)$ and $L(7,2)$). Now the Turaev-Viro-Barrett-Westbury (TVBW) invariant for a spherical fusion category $\mathcal{C}$ is the Reshetikhin-Turaev invariant for $\mathcal{Z}(\mathcal{C})$, which is a restriction, so in principle, RT could be stronger than TVBW.


Are there calculations that show explicitly how TVBW is stronger than homotopy? What category do you have to use to achieve this?


Otherwise, I'm very happy to hear expert opinions stating that this is an open problem.



Is Turaev-Viro-Barrett-Westbury stronger than homotopy?
I think the following example seems to work: consider a special class of TVBW, the Dijkgraaf-Witten invariant whose input are a finite $G$ and a 3-cocycle from $H^3(G, C^\*)$ (together they define a spherical fusion category). More specifically, let $G$ be an Abelian group. The invariant evaluated on $L(p,q)$ is given by the following:


$\displaystyle\mathcal{Z}(L(p,q))=\frac{1}{|G|}\sum\_{g\in G, g^p=1}\prod\_{j=1}^{p-1}\omega(g, g^j, g^n)$,


here $n=q^{-1}$ mod $p$.


Set $G=\mathbb{Z}/14$, and take a representative cocycle:


$\omega(a,b,c)=\exp\Big[\frac{2\pi i}{14^2}[a]([b]+[c]-[b+c]) \Big]$,


where $[a]\equiv a \text{ mod } 14$. 


One can check that with this cocycle, $\mathcal{Z}(L(7,1))\neq \mathcal{Z}(L(7,2))$.


