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Jun 23, 2017 at 15:45 history edited Manuel Bärenz CC BY-SA 3.0
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Jun 22, 2017 at 20:19 vote accept Manuel Bärenz
Jun 22, 2017 at 16:31 comment added Marcel Bischoff I guess there are $\omega\neq \omega'$ which are not related by $\mathrm{Aut}(G)$ which cannot be distinguished by any manifolds because of gauge invariance. In 3D the Dijkgraaf-Witten invariant associated with $(G,\omega)$ equals the Reshetikhin-Turaev invariant of $\mathrm{Rep}(D^\omega(G))$, but there are known $\omega\neq \omega'$ with $\mathrm{Rep}(D^\omega(G))$ braided equivalent to $\mathrm{Rep}(D^{\omega'}(G))$ by Goff-Mason-Ng.
Jun 22, 2017 at 13:51 answer added Arun Debray timeline score: 9
Jun 22, 2017 at 12:25 comment added Will Sawin Isn't it true that for pretty much every manifold, the invariant associated to the trivial cocycle is distinct from, because it is larger than, the invariant associated to a nontrivial cocycle? It should happen as long as there is any $\phi$ such that $\phi^*(\omega)$ is nontrivial.
Jun 22, 2017 at 12:23 comment added Will Sawin This is a multiplicative integral, right?
Jun 22, 2017 at 10:45 history asked Manuel Bärenz CC BY-SA 3.0