Domenico Fiorenza Dijkgraaf-Witten model

Let c:BGB nU(1)\mathbf{c} : \mathbf{B}G \to \mathbf{B}^n U(1) be a cocycle and assume for simplicity that GG is discrere (Dijkgraaf-Witten type of case), so that for Σ\Sigma a manifold we have

H(Σ,BG)Top(Σ,BG) \mathbf{H}(\Sigma, \mathbf{B}G) \simeq Top(\Sigma, B G)

etc. We had shown that τ ndimΣTop(Σ,B nU(1) disc)B ndimΣU(1) disc\tau_{n-dim \Sigma} Top(\Sigma, B^n U(1)_{disc}) \simeq B^{n-dim \Sigma} U(1)_{disc}. Following Freed-Hopkins-Lurie-Teleman, what we need to show is the following:

for Σ inΣΣ out\Sigma_{in} \to \Sigma \leftarrow \Sigma_{out} a cobordism we need to produce a diagram

Top(Σ,BG) Top(Σ in,BG) Top(Σ out,BG) B ndimΣ in/outU(1). \array{ && Top(\Sigma, \mathbf{B}G) \\ & \swarrow && \searrow \\ Top(\Sigma_{in}, \mathbf{B}G) &&\swArrow&& Top(\Sigma_{out}, \mathbf{B}G) \\ & \searrow && \swarrow \\ && B^{n-dim \Sigma_{in/out}} U(1) } \,.

Here the lower two morphisms should be the ones we already have, and the task is to show that we have the natural transformation filling this diagram.

Now, here is how this should work, and my question would be if you can see the remaining missing step needed to show that it does work:

let’s hom the entire cobordism diagram into the entire background field morphism: this yields

Top(Σ in,BG) Top(Σ,BG) Top(Σ out,BG) Top(Σ in,B nU(1)) Top(Σ,B nU(1)) Top(Σ out,B nU(1)) B ndimΣ inU(1) tau ndimSigma in/outTop(Σ,B nU(1)) B ndimΣ inU(1). \array{ Top(\Sigma_{in}, B G) &\leftarrow& Top(\Sigma, B G) &\to& Top(\Sigma_{out}, B G) \\ \downarrow && \downarrow && \downarrow \\ Top(\Sigma_{in}, B^n U(1)) &\leftarrow& Top(\Sigma, B^n U(1)) &\to& Top(\Sigma_{out}, B^n U(1)) \\ \downarrow && \downarrow && \downarrow \\ B^{n-dim \Sigma_{in}} U(1) &\leftarrow& tau_{n-dim Sigma_{in/out}} Top(\Sigma, B^n U(1)) &\to& B^{n-dim \Sigma_{in}} U(1) } \,.

You see, if we could show that the bottom middle term here, the “relative cohomology term” is also equivalent to B ndimΣU(1)B^{n-dim \Sigma} U(1), then we’d be done. More concretely, imagine the above diagram formed concretely in some 1-categorical model, then we should show that it is equivalent to the following 1-categorical diagram

Top(Σ in,BG) Top(Σ,BG) Top(Σ out,BG) B ndimΣ inU(1) (B ndimΣ in/outU(1)) I B ndimΣ inU(1), \array{ Top(\Sigma_{in}, B G) &\leftarrow& Top(\Sigma, B G) &\to& Top(\Sigma_{out}, B G) \\ \downarrow && \downarrow && \downarrow \\ B^{n-dim \Sigma_{in}} U(1) &\leftarrow& (B^{n-dim \Sigma_{in/out}} U(1))^I &\to& B^{n-dim \Sigma_{in}} U(1) } \,,

where now in the bottom middle is the path object. The middle morphism into that would exhibit the right homotopy/natural transformation that we are after.

So, it seems all to point in the right direction. But can we show the crucial statement that

(B ndimΣ in/outU(1)) Itau ndimΣ in/out((B nU(1)) Σ) (B^{n-dim \Sigma_{in/out}} U(1))^I \simeq tau_{n-dim \Sigma_{in/out}} \left( (B^{n} U(1))^{\Sigma} \right)

?

This might just follow from a slight variation of the argument for

B ndimΣ in/outU(1)tau ndimΣ in/out(B ndimΣ in/outU(1)) Σ in/out B^{n-dim \Sigma_{in/out}} U(1) \simeq tau_{n-dim \Sigma_{in/out}} \left( B^{n-dim \Sigma_{in/out}} U(1) \right)^{\Sigma_{in/out}}

But I am not sure yet how to do it.

Created on May 14, 2011 at 10:53:47 by Domenico Fiorenza