# Which manifolds are sensitive to the cocycle in the Dijkgraaf-Witten model?

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.

• This is a multiplicative integral, right? – Will Sawin Jun 22 '17 at 12:23
• 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. – Will Sawin Jun 22 '17 at 12:25
• 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. – 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

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