Often, TQFTs are defined in families, parametrised by some algebraic data. For example, the Turaev-Viro-Barrett-Westbury TQFTs are parametrised by spherical fusion categories, the Crane-Yetter TQFTs are parametrised by ribbon fusion categories, and the $n$-dimensional Dijkgraaf-Witten theory is parametrised by a finite group $G$ and an $n$-cocycle $\omega$.

Instead of regarding TQFTs with a fixed datum as an invariant of manifolds (and cobordisms), one can also fix a manifold and regard a family of TQFTs as invariant of the parametrising data. For example, the Crane-Yetter invariant of $\mathbb{CP}^2$ is the "Gauss sum" $\sum_X d(X)^2 \theta_X$ of the ribbon fusion category, where $X$ ranges over simple objects and $\theta$ is the twist eigenvalue. (I'm thanking Ehud Meir for making me appreciate this viewpoint.)

From this viewpoint, my question is: In the Dijkgraaf-Witten TQFT, which manifolds give invariants that are sensitive to the cocycle?

In details, let us define the following invariant: $$ DW_{G,\omega}(M) = \sum_{\phi\colon \pi_1(M) \to G} \int_M \phi^* (\omega)$$ Here, $M$ is an $n$-manifold, $\omega \in H^n(G,U(1))$ is an element of the $n$-th group cohomology, and $\phi^*\colon H^n(G,U(1)) \to H^n(M,U(1))$ is induced by the flat $G$-connection $\phi$.

I'm looking for a manifold $M$ such that $DW_{G,\omega}(M) \neq DW_{G,\omega'}(M)$ for some $\omega \neq \omega'$. Ideally, the example would be in 3 or 4 dimensions.

  • $\begingroup$ This is a multiplicative integral, right? $\endgroup$ – Will Sawin Jun 22 '17 at 12:23
  • 2
    $\begingroup$ 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. $\endgroup$ – Will Sawin Jun 22 '17 at 12:25
  • $\begingroup$ 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. $\endgroup$ – Marcel Bischoff Jun 22 '17 at 16:31

The example $G = \mathbb Z/2$ and $M = \mathbb{RP}^3$ works.

The inclusion $\mathbb Z/2\to\{\pm 1\}\subset\mathrm U(1)$ induces an isomorphism $H^3(B\mathbb Z/2, \mathbb Z/2)\to H^3(B\mathbb Z/2, \mathrm U(1))$, so we can pull the cocycles back to $\mathbb Z/2$ cohomology and evaluate on the $\mathbb Z/2$ fundamental class.

$\mathbb{RP}^3$ has two isomorphism classes of principal $\mathbb Z/2$-bundles, the trivial bundle $\varepsilon$ and the connected double cover $\xi$. Each determines a classifying map to $B\mathbb Z/2$, and hence a map in cohomology $\phi^*\colon H^*(B\mathbb Z/2; \mathbb Z/2)\cong \mathbb Z/2[\alpha]\to H^*(\mathbb{RP}^3; \mathbb Z/2)\cong \mathbb Z/2[x]/(x^4)$. For $\varepsilon$, this is the zero map; for $\xi$, this is the ring homomorphism induced by $\alpha\mapsto x$.

Since $H^3(B\mathbb Z/2;\mathbb Z/2)\cong\mathbb Z/2$, there are two cohomology classes of cocycles. Let $\omega$ be a coboundary, so that $\phi^*\omega = 0$ for all principal $\mathbb Z/2$-bundles, and hence the Dijkgraaf-Witten partition function is

$$\mathrm{DW}_{\mathbb Z/2, 0}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \xi} = 1.$$

Let $\omega$ be a cocycle in the other cohomology class (in the notation above, $\alpha^3$). Then, $\varepsilon$ pulls it back to $0$, but $\xi$ pulls it back to $x^3$, and $\langle x^3, [\mathbb{RP}^3]\rangle = 1$, so

$$\mathrm{DW}_{\mathbb Z/2, \alpha^3}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(1)}}{2}}_{\text{from } \xi} = \frac 12 - \frac 12= 0.$$

| cite | improve this answer | |

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.