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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 19th 2011
• (edited Apr 19th 2011)

I have been working on the entry twisted bundle.

Apart from more literature, etc. I have started typing something like a first-principles discussion: first a general abstract definition from twisted cohomology in any cohesive $\infty$-topos, then unwinding this in special cases to obtain the traditional cocycle formulas found in the literature.

Needs more polishing here and there, but I have to pause now.

1. I was thinking of the following: if we think of $\mathbb{C}$ as the $0$-category of $0$-Vector spaces, then a function $X\to \mathbb{C}$ can be thought of as a $0$-vector bundle. Then, if a cocycle $c:X\to \mathbf{B}U(1)$ is given, we can use it to define $c$-twisted functions from $X$ to $\mathbb{C}$. these are a very calssical object: they are the sections of the line bundle on $X$ associated with the cocycle $c$ and the standard representation of $U(1)$ on $\mahtbb{C}$, which we can think of as the standard morphism $\mathbf{B}U(1)\to Vect$.

Then we can go one step higher: we look at a 1-vector bundle on $X$ as to a morphism $X\to Vect$, we take a cocycle $c:X\to \mathbf{B}^2U(1)$, and use it to define $c$-twisted morphisms from $X$ to $Vect$. these are $c$-twisted bundles which, by analogy with the previous case, should be interpreted as sections of the 2-vector bundle on $X$ associated with the cocycle $c$ and something which should be the the standard morphism $\mathbf{B}^2U(1)\to 2-Vect$.

In particular we should have that the sections of the 2-line bundle $X\to \mathbf{B}^2U(1)\to 2Vect$ are a 2-vector space, just as the sections of a 1-line bundle are a vector space. this is relevant for Dijkgraaf-Witten theory, where one has a natural morphism $\mathcal{L}\mathbf{B}G\to \mathbf{B}^2U(1)$ induced by the 3-cocycle $\mathbf{B}G\to \mathbf{B}^3U(1)$ defining the theory.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011

Yes, absolutely. For

$\alpha : X \to \mathbf{B}^n U(1) \to n Vect$

a line $n$-bundle and

$1 : X \to \mathbf{B}^n U(1) \to n Vect$

the trivial such, an $\alpha$-twisted $(n-1)$-vector bundle is a transformation

$V : 1 \Rightarrow \alpha \,.$

For $n = 1$ $V$ is precisely a section of an ordinary line bundle. For $n = 2$ $V$ is precisely a twisted bundle with twist the given gerbe/circle 2-bundle.

By the general recursive definition of n-vector space, we have that the objects of $n Vect$ are algebra objects in $(n-1)Vect$ and the morphisms are bimodule objects in $(n-1)Vect$. Therefore the components of the transformation $V$ over $x \in X$ are $1$-$\alpha(x)$-bimodules, in $(n-1)Vect$. So from the left they look just like 1-modules hence just like ordinary objects in $(n-1)Vect$. In this way the section of an $n$-vector bundle is itself a twisted $(n-1)$-vector bundle.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011
• (edited Jun 2nd 2011)

By the way, let me just remind us that in this kind of game it is good to keep the full generalization by Ando-Gepner-Blumberg in mind. It’s really very beautiful and , in fact, very simple.

I was just reminded of that in the context of the discussion of Thom spectra elsewhere: in their definition 4.1 they identify the “spectrum of twisted A-bundles” with the Thom spectrum of the twisting $A$-line bundle.

I need to better understand this. This should be deep and important.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011

A great place to look at is the companion article

• CommentRowNumber6.
• CommentAuthordomenico_fiorenza
• CommentTimeJun 2nd 2011
• (edited Jun 2nd 2011)

Hi Urs,

thanks for feedback. when you have time could you expand on the definition of the natural morphism $\mathbf{B}^n U(1)\to n\,Vect$, for $n=2,3$?

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJun 2nd 2011
• (edited Jun 2nd 2011)

By iteration the top morphisms of $n Vect$ are morphisms between ordinary vector spaces (which are bimodule objects between algebra objects in algebra objects in etc. )

SInce $\mathbf{B}^n U(1)$ is n-connected, any functor $\mathbf{B}^n U(1) \to n Vect$ necessarily hits in degree $n$ endomorphisms of the 1-dimensional vector space $\mathbb{C}$. Take these to be the evident ones, the standard representation of $U(1)$ on $\mathbb{C}$. This will then also constiture $1$-$1$-bimodule homorphisms.

So for $n = 2$ it looks like this:

$\rho : \mathbf{B}^2 U(1) \to 2 Vect$

is given by

$\left( \array{ && \stackrel{Id}{\to} \\ & \nearrow && \searrow \\ * &&\Downarrow^{ c }&& * \\ & \searrow && \nearrow \\ && \underset{Id}{\to} } \right) \;\; \mapsto \;\; \left( \array{ && \stackrel{ {}_{ \mathbb{C}}\mathbb{C}_{ \mathbb{C}}}{\to} \\ & \nearrow && \searrow \\ \mathbb{C} &&\Downarrow^{ (-)\cdot c }&& \mathbb{C} \\ & \searrow && \nearrow \\ && \underset{{}_{ \mathbb{C}} \mathbb{C}_{ \mathbb{C}}}{\to} } \right)$
2. clear, thanks. I think I prefer the equivalent simplicial version (just a matter of taste):

$\left( \array{ && * \\ & \nearrow &\Downarrow^{ c }& \searrow \\ * &&\to&& * } \right) \;\; \mapsto \;\; \left( \array{ && \mathbb{C} \\ & {}^{{}_{ \mathbb{C}}\mathbb{C}_{ \mathbb{C}}}\nearrow &\Downarrow^{ \cdot c }&& \searrow{}^{{}_{ \mathbb{C}}\mathbb{C}_{ \mathbb{C}}} \\ \mathbb{C} &&\underset{{}_{ \mathbb{C}} \mathbb{C}_{ \mathbb{C}}}{\to} && \mathbb{C}} \right)$

(where the canonical isomorphism ${}_{ \mathbb{C}} \mathbb{C}_{ \mathbb{C}}\otimes {}_{ \mathbb{C}} \mathbb{C}_{ \mathbb{C}}\cong {}_{ \mathbb{C}} \mathbb{C}_{ \mathbb{C}}$ is implicitly used). so now I just need to work out the morphism $\mathcal{L}\mathbf{B}^3 U(1)\to \mathbf{B}^2 U(1)$ to make my mind completely clear on the 2-vector space the Dijkgraaf-Witten model associates with $S^1$.

• CommentRowNumber9.
• CommentAuthordomenico_fiorenza
• CommentTimeJun 3rd 2011
• (edited Jun 3rd 2011)

mmm… actually, rather than looking at the loop space, the first object to be considered should be the 3-vector space of sections of the line 3-bundle $\mathbf{B}G\to \mathbf{B}^3 U(1)\to 3Vect$: this should coincide with the category $Vect^\tau[G]$ considered in section 4.2 of Topological Quantum Field Theories from Compact Lie Groups

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeJun 3rd 2011
• (edited Jun 3rd 2011)

Yes, where the 3-vector space of sections is the 3-colimit over that 3-functor.

By the way, it is maybe remarkable what happens to this statement as we replace 3-vector spaces by what I guess may be called $(3,1)$-vector spaces or more generally $(\infty,1)$-vector spaces, so if instead of using

$3 Vect \simeq ((k Mod) Mod ) Mod$

for $k$ a ring, we’d use $K Mod$ for $K$ an $\infty$-ring. (That’s of course a drastic change, but let’s just consider it for a minute, just to see what happens.)

Because then we are in the situation of def. 2.22 here, and so that colimit is then the Thom spectrum of the background gauge field.

So there is some interesting connection here. I need to better understand this.

I keep thinking that an interesting sort-of-toy case that would be useful to look at is $\infty$-dimensional extended QFT. Since $Bord_\infty$ is in fact a symmetric monoidal $\infty$-groupoid, in fact the universal Thom spectrum $M O$, or rather the space $\Omega^\infty M O$ underlying it, it should be true that in this case it is okay to work entirely in spectra, hence to use $(\infty,1)$-vector spaces as above instead of $(\infty,n)$-vector spaces. I guess.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeJun 4th 2011
• (edited Jun 4th 2011)

For emphasis, let me expand on this a little.

By Twists of K-theory and TMF we know that we have canonical representation on “$(\infty,1)$-lines”

$B^2 U(1) \to B GL_1 KU$

and

$B^3 U(1) \to B GL_1 tmf$

This is in terms of discrete $\infty$-groupoids. But at least for the first one we know how to refine this to smooth $\infty$-groupoids, I think. Not for the second, though.

In any case, it makes sense to regard

$\Pi X \to B^2 U(1) \to B GL_1 KU \to KU Mod$

as an associated/linearized version of a background field for a 2-dimensional $\sigma$-model, and

$\Pi X \to B^3 U(1) \to B GL_1 tmf \to tmf Mod$

as the associated/linearized version of a background field for a 3-dimensional $\sigma$-model. In some sense.

3. the 3-vector space of sections is the 3-colimit over that 3-functor.

shouldn’t global sections be a limit rather than a colimit? I know in Topological Quantum Field Theories from Compact Lie Groups there is an ubiquitous presence of colimits, but they also write (above remark 3.10): “To guarantee that this formula describes a well-defined functor from $Fam_n(\mathcal{C})\to \mathcal{C}$, we need to make certain assumptions on $\mathcal{C}$: namely, that it is additive in a strong sense which guarantees that the colimit ${\underset{\to}{\lim}}_{x\in X}\chi(x)$ exists and coincides with the limit ${\underset{\leftarrow}{\lim}}_{x\in X}\chi(x)$

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJun 4th 2011

Yes, exactly. So the colimit computes “dual sections” (for a vector bundle: not morphisms from the trivial line bundle to the vector bundle, but the other way round) and under suitable finiteness assumptions that’s isomorphic to genuine sections.

4. for eventual future reference and pasting into nLab let me sketch here the description of a global section of a line 2-bundle $\alpha:\mathcal{L}\mathbf{B}G\to \mathbf{B}^2 U(1)\to 2Vect$. Since $\mathcal{L}\mathbf{B}G\cong G//_{Ad}G$, a natural transformation from the trivial line 2-bundle to the given line 2-bundle is the datum, for every $x\in G$ of a $\mathbb{C}$-$\mathbb{C}$-bimodule (aka complex vector space) $W_x$, together with morphisms $\varphi_{g,x}:W_x\to W_{g x g^{-1}}$ for any $g$ in $G$, such that

$\varphi_{h,g x}\varphi_{g,x}=\alpha_x(g,h)\cdot \varphi_{h g,x}$

which (with slightly different notations) is the 2-vector space described in proposition 4.9 of Topological Quantum Field Theories from Compact Lie Groups

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeJun 5th 2011

Yes, and we can say more: the 2-cocycle $\alpha(-,-)$ is the transgression of the 3-cocycle $\mathbf{B}G \to \mathbf{B}^3 U(1)$ to loops, which is literally given by the internal hom out of the circle. That’s this prop. 9.1 in my article with Zoran here.

So this means that the 2-vector bundle on $\mathcal{L}\mathbf{B}G$ that you are looking at above is exactly the one obtained by the 3-bundle on $\mathbf{B}G$ by our abstract transgression formula at infinity-Chern-Simons theory (schreiber).

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeJun 5th 2011
• (edited Jun 5th 2011)

Hey Domenico,

I am still absorbed with looking into string topology, and its realization as an HQFT – that’s why I have to be postponing other tasks, such as working on our article and joining in your activity here more forcefully.

But, similarly to my indication above, I think there is an interesting relation to the discussion here:

As you know, what is called “string topology” is some kind of sigma-model for target space a smooth oriented compact manifold $X$. In order to make this precise, one needs to say what the background field is. Now, this must be such that its space of sections is the homology of $X$.

But let’s say this in a more complicated/general way to work out the structure implied here: as I was wondering about the other day but as luckily Cohen and Godin already noticed in 2004 in their A polarized view of string topology (right at the beginning) we can think of string topology as having “spaces of states” given by any generalized (co)homology theory, if only it is oriented over the given manifold.

So ket $K$ be some ring spectrum with this orientability property over $X$. Using the work on twisted generalized cohomology mentioned above, we then have this statement:

string topology on $X$ has as background field the trivial $K$-line bundle

$const_K : X \to K Line \hookrightarrow K Mod$

hence its space of states over the point, the space of sections of $const_K$, is the space of $K$-chains on $X$: $K$ (co)-homology. (Give or take some shift in degree, possibly).

Or maybe (co)homology instead of (co)chains. Present technology only constructs string topology as a QFT in (co)homology. Lurie at the end of his TFT article roughly indicates that there should be a chain version. He mentions that this should be twisted by a 2-gerbe = circle 3-bundle, but gives no indication which that should be. But so maybe for the chain-level story the background field is rather some nontrivial

$X \to \mathbf{B}^3 U(1) \to K Line \to K Mod$

and then states are the sections of that, hence the corresponding “twisted $K$-bundles” (hence the connection to the discussion here).

Sorry, a bit telegraphic. Need to go offline now.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeJun 6th 2011
• (edited Jun 6th 2011)

I have another minute now. I can say the above again much shorter and more systematically:

By theorem 4.5 of Ando-Blumberg-Gepner (here) we have the following:

for $X \to S Line$ the trivial $S$-line bundle ($S$ the sphere spectrum), $A$ any ring spectrum and

$\alpha : X \to S Line \to A Line \hookrightarrow A Mod$

the corresponding trivial “$A$-$\infty$-vector bundle” we have that the $A$-$(\infty,1)$-vector space of sections of $\alpha$ is

$\lim_\to \alpha \simeq (\Sigma^{\infty + dim X} X) \wedge_S A \,.$

This is indeed the $A$-homology spectrum of $X$, whose homology groups are the $A$-homology groups of $X$

$H_\bullet(X,A) = \pi_\bullet( \Sigma^\infty X \wedge_S A)$

So we have:

• for a trivial background field $(\infty,1)$-vector bundle over an $\infty$-ring $A$;

• the $(\infty,1)$-vector space of states of the corresponding $\sigma$-model is the space of the corresponding (trivially-twisted) twisted $\infty$-bundles, which is the $A$-homology spectrum of $X$, hence the space of states of the $A$-string topology HQFT.

(There may be shifts in degree by $dim X$ in this story that I am not properly accounting for here.)

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeJun 6th 2011

I have now written this out in a little bit more detail in a new entry (infinity,1)-vector bundle.

5. Yes, and we can say more: the 2-cocycle $\alpha(-,-)$ is the transgression of the 3-cocycle $\mathbf{B}G \to \mathbf{B}^3 U(1)$ to loops, which is literally given by the internal hom out of the circle. That’s this prop. 9.1 in my article with Zoran here.

Yes, that’s exactly what I had in mind, thanks! Now there’s only a little step I’m missing: bot in your paper with Zoran and in my mind, the computations leading to the 2-cocycle $\alpha$consist in an eveluation of the 3-cocycle $\mathbf{B}G\to \mathbf{B}^3 U(1)$ on a prism. This is a 3-manifold with boundary and the computation is carried out by a triangulation of the prism. Then the fact we are integrating a cocylce ensures the final result will be independent of the triangulation. This easily generalizes to arbitrary dimensions: if $\Sigma$ is a $k$-dimensional oriented manifold with corners, then a $j$-simplex in $Maps(\Sigma,\mathbf{B}^n U(1))$ is a morphism $\Sigma\times \Delta^{j} \to \mathbf{B}^n U(1)$, and so if $j=n-k$, a triangulation of this morphism gives an element of $U(1)$, which is independent of the triangulation and is to be thought of as an integral over $\Sigma\times \Delta^j$. This defines the morphism $Maps(\Sigma,\mathbf{B}^n U(1)) \to \mathbf{B}^{n-k}U(1)$. But clearly this is a rough description: in order to get a correct suitable constains should be added. For instance I suspect one should consider maps from manifolds with boundary relative to the boundary, in order to go in a well defined and nontrivial way onto homotopy classes.

So my main problem here is: in a concrete situation I know which computation to do, but I still miss the correct abstract framework in which doing them (I need to understand transgression better)

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeJun 6th 2011

Wait, the prism does not have a boundary, top and bottom are identified. (“closed prism”). Because we are transgressing to the loop space.

• CommentRowNumber21.
• CommentAuthordomenico_fiorenza
• CommentTimeJun 6th 2011
• (edited Jun 6th 2011)

doesn’t it have three “vertical” faces?

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeJun 6th 2011

Yes, ah, these you mean. Wait, let me think again about what you said.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeJun 6th 2011

Domenico,

sorry, I am maybe not concentrating sufficiently, since I am a bit absorbed in another process. Please bear with me.

I believe what I tried to say is that for the case of the transgression to loop space the $\Sigma$ in your previous comment has no boundary. It’s the circle.

I thought you were getting at the point where we are transgression to the mapping space over a $\Sigma$ which does have a boundary. That case, I’d agree, requires extra care, and while we have talked about it a bit every now and then (“relative cohomology”) we should write that out in more detail.

But how is that related to the discussion of the transgression of $\alpha : B G \to B^3 U(1)$ to $\mathcal{L} B G$?

6. I believe what I tried to say is that for the case of the transgression to loop space the $\Sigma$ in your previous comment has no boundary. It’s the circle.

yes, that’s what I was interested in. my problem was, roughtly: “I have a 2-simplex in $\mathcal{L}\mathbf{B}G$ and I want to associate with it an element of $U(1)$. how do I do this? well, a 2-simplex in $\mathcal{L}\mathbf{B}G$ is nothing but a “closed prism” in $\mathbf{B}G$, and I know how to assign an element in $U(1)$ to a 3-simplex in $\mathbf{B}G$, so let me just split my closed prism in 3-simplices and add on (i.e., multiply) the results”. which is what is actually done, but this involves at an intermediate step considering a 3-manifold with boundary mapped to $\mathbf{B}G$. from this I thought that the natural thing to do was to consider boundaries and corners from the very beginning.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeJun 6th 2011

Ah, so maybe I misunderstood you. I thought you said that some discussion of constraints is missing in the discussion of the transgression to $\mathcal{L} B G$.

But you rather mean that more generally it is natural to consider transgression to mapping spaces for $\Sigma$ having a boundary. This of course I agree with. Just let me know if you think there is some gap in the specific argument for the transgression of $\alpha$ to $\mathcal{L} B G$

7. Just let me know if you think there is some gap in the specific argument for the transgression of $\alpha$ to $\mathcal{L} B G$

This is fine. But I need understand it better. Our current argument is: we start with $\mathbf{B}G\to \mathbf{B}^3 U(1)$; then we take loop spaces $\mathcal{L}\mathbf{B} G\to \mathcal{L}\mathbf{B}^3 U(1)$; finally there is the step $\mathcal{L}\mathbf{B}^3 U(1) \to \mathbf{B}^2 U(1)$. This last step I understand by the truncation argument we have at Dijkgraaf-Witten theory, but concretely it seems to be something very simple: just split a prism and split it into 3-simplices, so I’d like to really convince myself that these two points of view are really the same thing.

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeJun 7th 2011

Hi,

okay, let’s write it out in more detail. The way I understand it, the computation with the prism shows what the morphism of simplicial sets

$\Delta^1/\partial \Delta^1 \times \Delta^n \to B G \to B^3 U(1)$

is like for $n=2$. On the non-degenerate 3-cells of $\Delta^1/\partial \Delta^1 \times \Delta^2$ this assigns values in $U(1)$. One needs to argue that these values don’t change under the truncation

$[\Delta^1 / \partial \Delta^1, B^3 U(1)] \to B^2 U(1).$

Since the truncation is at one level higher, it identifies morphisms between the 2-cells that we are after. Clearly the map $\{\Delta^1/\partial \Delta^1 \times \Delta^2 \to B^3 U(1)\} \to \{\Delta^2 \to B^2 U(1)\}$ is surjective, so it remains to show that it is injective.

Let’s see. In fact I think we have generally that

$[S^1, B^n U(1)] \simeq [U(1) \stackrel{const_1}{\to} U(1) \to \cdots \to 1]$

and so the truncation just chops of the top $U(1)$ contribution without changing anything beneath.

I think this is easy to see in a pedestrian way for low $n$. I should think of a formal way to say it for all $n$.

• CommentRowNumber28.
• CommentAuthorUrs
• CommentTimeNov 28th 2011

following discussion with Chris Rogers, I have added to twisted bundle a References-section As 2-sections of 2-bundles.