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.

  • 2
    $\begingroup$ Have you seen this: mathoverflow.net/questions/239851/… ? $\endgroup$ – Chris Schommer-Pries Jan 17 '17 at 18:17
  • $\begingroup$ @ChrisSchommer-Pries, yes, but that sounds like it's not stronger than homotopy, right? $L(5,1)$ and $L(5,2)$ aren't homotopy equivalent, or am I mixing up stuff? $\endgroup$ – Manuel Bärenz Jan 17 '17 at 18:28
  • $\begingroup$ As far as I know it is an open question whether Turaev-Viro tqfts can distinguish L(7,1) and L(7,2). I was just pointing out that the category that was expected to do this seems not to work. $\endgroup$ – Chris Schommer-Pries Jan 17 '17 at 18:48
  • $\begingroup$ @ChrisSchommer-Pries, is there any other homotopy-equivalent manifolds that it distinguishes then? $\endgroup$ – Manuel Bärenz Jan 17 '17 at 18:51
  • 3
    $\begingroup$ Homotopy equivalent but not homeomorphic 3-manifolds are always given as the connected sum of homeomorphic manifolds with (some collection) of homotopy equivalent but not homeomorphic lens spaces. So you may as well entirely reduce your question to the lens space case. $\endgroup$ – Mike Miller Jan 18 '17 at 10:13

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, $\mathcal{Z}(L(7,1))\neq \mathcal{Z}(L(7,2))$.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.