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I’d like to formalize in cohesive homotopy theory the Penrose-Ward transform for higher bundles as it appears for 3-bundles in section 4 of
I think I know in principle how it goes. But at one step I need a fiberwise $\flat$-modality. This is something that Mike has kept asking about recently and I didn’t give an answer. I still don’t have the answer, but at least now I have a good motivating example.
So we consider a correspondence
$\array{ && Z \\ & {}^{\mathllap{p_1}}\swarrow && \searrow^{\mathrlap{p_2}} \\ X_1 && && X_2 }$Let $G$ be an $\infty$-group and consider a $G$-principal $\infty$-bundle $P \to X_1$ equipped with a trivialization of its pullback along $p_1$.
The transform in question is supposed to do the following: from the chosen trivialization on $Z$ we are to produce a $p_1$-fiberwise $G$-valued function on $Z$. Associated with this is a $p_1$-fiberwise Maurer-Cartan form on $Z$ which glues to a global form on $Z$ and this we are to push down along $p_2$. (This is an abstract rephrasing of what appears in components in section 4 of the above article.)
Here I’ll ignore the push-forward for the moment and consider only the first step of producing that fiberwise Maurer-Cartan form.
On a single fiber $x \colon \ast \to X$ it works like this: let $g$ be the map that modulates $P \to X_1$. Then the assumed trivialization of the pullback of this $G$-bundle to $Z$ gives the diagram
$\array{ p_1^{-1}(x) &\longrightarrow& Z &\longrightarrow& \ast \\ \downarrow && \downarrow && \downarrow \\ \ast &\stackrel{x}{\longrightarrow}& X_1 &\stackrel{g}{\longrightarrow}& \mathbf{B}G }$and, by looping, this induces a map $p_1^{-1}(x) \longrightarrow G$. This is the $G$-valued function on the fiber. Its fiber MC form is given by the composite
$p_1^{-1}(x) \longrightarrow G \longrightarrow \flat_{dR} \mathbf{B}G \,.$Now we need to generalize this to a construction that does this for all fibers at once.
Need to think about that.
Yes!
What does it mean for a $G$-valued function on $Z$ to be “$p_1$-fiberwise”? And can you do the step of producing that function for all fibers at once, before the $\flat$ gets introduced to make the MC form?
Here are some further thoughts on these questions. I’d think the following should give the right construction for the application described in #1. Whether it contains the seed of an answer to what the fiberwise $\flat$ should be I don’t know, but see below.
First, I think the group object in $\mathbf{H}_{/\mathbf{B}G}$ in question is $G/_{ad}G$ sitting (in $\mathbf{H}$) in the homotopy fiber sequence
$\array{ G &\longrightarrow& G/_{ad}G \\ && \downarrow \\ && \mathbf{B}G }$exhibiting the adjoint action of $G$ on itself. This is defined to be the free loop space object
$G/_{ad}G \coloneqq [\Pi(S^1), \mathbf{B}G]$and this makes manifest the $\infty$-group structure in the slice over $\mathbf{B}G$.
A trivialization of a $G$-principal bundle is in particular a trivialization of its adjoint bundle, and so we may consider generalizing the diagram in #1 to
$\array{ Z &\longrightarrow& G/_{ad} G \\ \downarrow &\swArrow& \downarrow \\ X_1 &\stackrel{g}{\longrightarrow}& \mathbf{B}G } \;\;\; \simeq \;\;\; \array{ Z &\stackrel{\gamma}{\longrightarrow}& P \times^{ad}_G G &\longrightarrow& G/_{ad} G \\ \downarrow^{\mathrlap{p_1}} &\swArrow& \downarrow^{\mathrlap{p_1}} &\swArrow& \downarrow \\ X_1 &\stackrel{=}{\longrightarrow}& X_1 &\stackrel{g}{\longrightarrow}& \mathbf{B}G }$The adjoint bundle $P \times^{ad}_G G$ is the relevant group object in the slice, and $\gamma$ is what I had announced as the “$p_1$-fiberwise function” with values in $G$ (not claiming that this is necessarily the best terminology).
Now to get the fiberwise MC form, I’d think the obvious step is the following:
define $(\flat_{dR} \mathbf{B}G)/_{ad} G$ to be the homotopy pushout in
$\array{ G &\longrightarrow& G/_{ad}G \\ \downarrow && \downarrow \\ (\flat_{dR} \mathbf{B}G) &\longrightarrow& (\flat_{dR} \mathbf{B}G)/_{ad} G } \,.$One checks, using homotopy pullback stability of homotopy colimits in $\mathbf{H}$, that the notation is not just intuitively but also formally justified in that there is a homotopy fiber sequence
$\array{ (\flat_{dR} \mathbf{B}G) &\longrightarrow& (\flat_{dR}\mathbf{B}G)/_{ad} G \\ && \downarrow \\ && \mathbf{B}G }$and a compatible morphism in the slice
$\array{ G/_{ad}G && \longrightarrow && (\flat_{dR} \mathbf{B}G)/_{ad} G \\ & \searrow && \swarrow \\ && \mathbf{B}G }$which says that the gloabl MC form $\theta \colon G \to \flat_{dR}\mathbf{B}G$ is equivariant under the adjoint $G$-action, just as it should be.
That is then, I’d think, the answer to the application in #1: the fiberwise MC form in question is $\gamma^\ast \theta$ given by the diagram
$\array{ Z &\longrightarrow& G/_{ad} G &\stackrel{\theta}{\longrightarrow}& (\flat_{dR}\mathbf{B}G)/_{ad} G \\ \downarrow &\swArrow& \downarrow &\swArrow& \downarrow \\ X_1 &\stackrel{g}{\longrightarrow}& \mathbf{B}G &\stackrel{id}{\longrightarrow}& \mathbf{B}G } \;\;\; \simeq \;\;\; \array{ Z &\stackrel{\gamma^\ast \theta}{\longrightarrow}& P \times^{ad}_G (\flat_{dR}\mathbf{B}G) &\longrightarrow& (\flat_{dR} \mathbf{B}G)/_{ad} G \\ \downarrow^{\mathrlap{p_1}} &\swArrow& \downarrow^{\mathrlap{p_1}} &\swArrow& \downarrow \\ X_1 &\stackrel{=}{\longrightarrow}& X_1 &\stackrel{g}{\longrightarrow}& \mathbf{B}G } \,.$In conclusion, it seems the above tells us what the fiberwise $\flat_{dR}$ should be in the special cases that the dependent types are $G$-principal bundles. I have no thoughts yet on how to proceed more generally.
In closing, one remark on how $\flat$ itself appears in this context: notice that $\flat \mathbf{B}G$ is also of the form
$\flat \mathbf{B}G \simeq (\flat_{dR}\mathbf{B}G)/G$in that it sits (by definition of $\flat_{dR}$ !) in a homotopy fiber sequence
$\array{ \flat_{dR} \mathbf{B}G &\longrightarrow& \flat \mathbf{B}G \\ && \downarrow \\ && \mathbf{B}G } \,.$The action her is the gauge action on flat differentialforms $A \mapsto g^{-1}\mathbf{d}g + g^{-1} A g$ (i.e. that’s what it is in the standard models, and that’s what we ought to take as the “meaning interpretation” in generality).
I realize that I am unsure about how $\flat \mathbf{B}G$ and $(\flat_{dR} \mathbf{B}G)/_{ad} G$ relate. Are they equivalent?!
In trying to understand this, I noticed that the link “This statement and its proof is spelled out here” at Maurer-Cartan form needs updating; to what I’m not quite sure.
Is the morphism on the LHS of your pushout the ordinary MC form? And is the morphism on the RHS of the pushout the same as your “compatible morphism in the slice”?
It seems that this construction can be done in much more generality. Given any map $f:A\to B$ of pointed types (such as $\flat \mathbf{B} G \to \mathbf{B} G$), we can consider the pushout of $F = fib(f)$ and the free loop space $L B$ over $\Omega B$. The same stability argument should show that we have a fiber sequence $F \to F +_{\Omega B} L B \to B$ alongside the obvious one $F\to A\to B$, and we might wonder about the relationship of $F+_{\Omega B} L B$ to $A$.
If $B$ is connected (like $\mathbf{B} G$) and if we had a map $F +_{\Omega B} L B \to A$ over $B$ and under $F$, then the five-lemma ought to imply that this map would be an equivalence. There is an obvious map $F +_{\Omega B} L B \to A$ under $F$ that is null on $L B$, but it doesn’t commute with the maps to $B$.
Are you sure about that stability-of-colimits argument? It seems right to me too offhand, but the end result is suspicious; it feels like we have a group $G$ acting on a space $F$, and we have some way of constructing some other $G$-action that “feels free” relative to just remembering the images of the basepoint of $F$ under the original action, so why should the new action also be on $F$ rather than on some bigger space?
(Where by “this construction” I meant the construction of $\flat_{dR} \mathbf{B} G/_{ad} G$.)
Wait, now I don’t believe the stability argument any more. At least, the one I had in mind will I think only show that the fiber of $F+_{\Omega B} L B\to B$ is the pushout of $\Omega B$ and $F\times \Omega B$ along $\Omega B \times \Omega B$, so the fiber of $\flat_{dR}\mathbf{B} G/_{ad} G$ would be the pushout of $G$ and $\flat_{dR}\mathbf{B} G \times G$ along $G\times G$. But maybe you had a different argument in mind, could you spell it out?
I have fixed some of the links, thanks for the alert. The general cohesive defininition is at cohesive infinity-topos – structures – Maurer-Cartan forms and curvature characteristic forms, the smooth discussion is at smooth infinity-groupoid – structures – Maurer-Cartan forms and curvature characteristic forms.
(These links broke when I split off the “…structures” entries. Apparently more such links are broken, will need to fix them eventually.)
[I see I am overlapping with your comments. I’ll retain what I typed, but the first bit you just reproduced in #9.]
Let’s see. So we want to compute the homotopy fiber of the bottom vertical map in
$\array{ G &\longrightarrow& G/_{ad}G \\ \downarrow && \downarrow \\ \flat_{dR} \mathbf{B}G &\longrightarrow& Q \\ && \downarrow^{\mathrlap{p}} \\ && \mathbf{B}G }$where $Q$ is the colimit under the top left span diagram. The separate pullbacks are clearly these
$\array{ G \times G &\longrightarrow& G \\ \downarrow && \downarrow \\ G \times \flat_{dR} \mathbf{B}G &\longrightarrow& fib(p) \\ && \downarrow \\ && \ast }$The point to be careful about is what the top map is. Chasing through the diagrams (which I won’t draw here), it is projection on the second factor. So we may write the span as the pointwise product of two spans
$\array{ G \times G &\stackrel{(\ast,id)}{\longrightarrow}& \ast \times G \\ \downarrow^{\mathrlap{(id,\theta)}} && \downarrow \\ G \times \flat_{dR} \mathbf{B}G &\longrightarrow& fib(p) \\ && \downarrow \\ && \ast }$and then all maps in the span are product maps.
[ okay, it seems we agree to this point. Maybe there is disagreement about the next step, where I am claiming that: ]
From this I conclude that $fib(p) \simeq \flat_{dR} \mathbf{B}G$.
Let me check my reasoning…
From #6 (probably clear by now):
Is the morphism on the LHS of your pushout the ordinary MC form?
Yes.
re #7, thanks for this comment here:
If $B$ is connected (like $\mathbf{B} G$) and if we had a map $F +_{\Omega B} L B \to A$ over $B$ and under $F$, then the five-lemma ought to imply that this map would be an equivalence. There is an obvious map $F +_{\Omega B} L B \to A$ under $F$ that is null on $L B$, but it doesn’t commute with the maps to $B$.
I had been wondering about this. Thanks for mentioning the use of the five-lemma here. My problem was that I was unable (unable, that is, while sitting on a RER B train from Paris Charles-de-Gaulle to downtown) to produce the sensible map from $G/_{ad}G$ to $\flat \mathbf{B}G$.
re #9, #11
right, I suppose my last step there lacks justification. Hm.
But I’ll have to call it quits now. Very late here.
Here is a naive question: since $\flat_{dR}\mathbf{B}G$ is a fiber of a map to $\mathbf{B}G$, it automatically comes with a “tautological” action by $G$. You seem to be saying that this action is not (or not obviously) the adjoint action, since you’re trying to exhibit $\flat_{dR}\mathbf{B}G$ as a fiber of a different map to $\mathbf{B}G$ and calling that the adjoint action. What is this tautological action then? I would have thought that the adjoint action is the only available action of $G$ on $\flat_{dR}\mathbf{B}G$.
Yes, I am after just that canonical action, but I didn’t see how to produce a map
$G/_{ad} G \longrightarrow \flat \mathbf{B}G = (\flat_{dR}\mathbf{B}G)/G \,.$This map exists in the models at least for ordinary groups, but I didn’t see how to producte it abstractly. So instead I tried to force such kind of map. But my attempt via the pushout was no good, as you pointed out.
I am thinking of another way to do it, but not sure how to make that work:
Assume that $G$ is at least braided, i.e. that it is $\Omega^2 \mathbf{B}^2 G$. Then consider the map of cospan diagrams
$\array{ \mathbf{B}G &\stackrel{\Delta}{\longrightarrow}& \mathbf{B}G \times \mathbf{B}G &\stackrel{\Delta}{\longleftarrow}& \mathbf{B}G \\ \downarrow^{} && \downarrow && \downarrow \\ \mathbf{B}G &\stackrel{(\mathbf{B}\theta, \mathbf{B}\theta)}{\longrightarrow}& \mathbf{B} (\flat_{dR}\mathbf{B}G) \times \mathbf{B} (\flat_{dR}\mathbf{B}G) & \stackrel{(\mathbf{B}\theta, \mathbf{B}\theta)}{\longleftarrow} & \mathbf{B}G }$Forming pullbacks gives a map $G/_{ad}G \to somewhere$ over $\mathbf{B}G$, where “somewhere” knows something about a $G$-action on $\flat_{dR}\mathbf{B}G$.
If we could go from there to
$\array{ \mathbf{B}G &\stackrel{\mathbf{B}\theta}{\longrightarrow}& \mathbf{B} (\flat_{dR}\mathbf{B}G) & \stackrel{}{\longleftarrow} & \ast }$we’d be done, as the limit over that last diagram is $\flat \mathbf{B}G$.
Hm.
Quick remark before I need to rush off:
that diagram in #16 gives a map from $G/_{ad}G$ to what deserves to be written $G \backslash (\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G)/G$, hence an adjoint-equivariant map from $G$ to $(\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G)/G$.
In case $G$ is 0-truncated and in the standard models, then that $G$-action on flat forms is transitive and so in that case $(\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G)/G \simeq \flat_{dR}\mathbf{B}G$. In general I suppose this equivalence won’t hold. But maybe that extra quotient construction then is the next closest thing that exists in general.
What are the two actions of $G$ on $\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G$?
In case $G$ is 0-truncated and in the standard models
Do you mean by this “for a 0-truncated $G$ in the standard models”? Do you know whether the map $G/_{ad} G \to \flat \mathbf{B}G$ exists in the standard models when $G$ is not 0-truncated?
What are the two actions of $G$ on $\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G$?
So generally for $G\to H$ a group homomorphism then
$\array{ H &\longrightarrow& \ast \\ \downarrow && \downarrow \\ H/G &\longrightarrow& \mathbf{B}G \\ \downarrow && \downarrow \\ \ast &\longrightarrow& \mathbf{B}H }$exhibits the induced $G$-action on $H$.
Here the group homomorphism in question is now $G \longrightarrow \flat_{dR}\mathbf{B}G$ (since I am assuming $G$ to be braided). The action in question is hence via that homomorphism, or rather its diagonal.
$\array{ ((\flat_{dR} \mathbf{B}G) \times (\flat_{dR} \mathbf{B}G))/G &\longrightarrow& G \backslash ((\flat_{dR} \mathbf{B}G) \times (\flat_{dR} \mathbf{B}G))/G &\longrightarrow& \mathbf{B}G \\ \downarrow && \downarrow && \downarrow \\ \ast &\longrightarrow& \mathbf{B}G &\longrightarrow& \mathbf{B}(\flat_{dR} \mathbf{B}G)\times \mathbf{B}(\flat_{dR} \mathbf{B}G) }$Do you mean by this “for a 0-truncated G in the standard models”?
Yes
Do you know whether the map $G/_{ad} G \to \flat \mathbf{B}G$ exists in the standard models when $G$ is not 0-truncated?
I don’t even know that, unfortunately.
…that $G$-action on flat forms is transitive and so in that case $(\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G)/G \simeq \flat_{dR}\mathbf{B}G$
Doesn’t that require the action to be free as well as transitive?
Wait a minute, I am confused. $G/_{ad}G$ is the free loop space $L\mathbf{B}G$. But the free loop space of any object $L X \to X$ always has a section, assigning to each $x:X$ the constant path. So if there is a map $G/_{ad}G \to \flat \mathbf{B}G$ over $\mathbf{B} G$, then $\flat \mathbf{B}G \to \mathbf{B}G$ also has a section; but I don’t think that can happen unless $\mathbf{B}G$ is discrete.
Doesn’t that require the action to be free as well as transitive?
That’s true, right. In fact the part acting non-freely is just $\flat G$ again, and this is what gives $(\flat_{dR}\mathbf{B}G)/G \simeq \ast /\flat G \simeq \flat \mathbf{B}G$. So I was wrong in the last equivalence in #17. Instead of the quotient of the diagonal action being one copy of $\flat_{dR}\mathbf{B}G$ it’s more something like the product of that with $\flat \mathbf{B}G$.
I don’t think that can happen unless BG is discrete.
Right, that’s also true. In components the reason is that the conjugation action on $G$ goes to the inhomogeneous transformation $A \mapsto g^{-1}A g + g^{-1}d g$ on differential forms, which prevents the 0-form from being the invariant section.
So I need to think more about all this…
the conjugation action on $G$ goes to the inhomogeneous transformation $A \mapsto g^{-1}A g + g^{-1}d g$
Ah, right. So that means that we don’t want the “obvious” adjoint action on $\flat_{dR}\mathbf{B}G$, we want this sort of “twisted” one. So we do need some other fibration over $\mathbf{B} G$ whose fiber is $\flat_{dR}\mathbf{B}G$ and which admits a map from $G/_{ad}G$. Is there a way to say “$g^{-1} dg$” synthetically?
Is there a way to say “$g^{-1} dg$” synthetically?
Asked this way, the answer would be: sure, this is again just the MC form $g^\ast \theta$, i.e. the image of g under $\theta \colon G \to \flat_{dR}\mathbf{B}G$.
Something tautological or other needs to be done here to make all this fall into place…
Maybe we don’t actually want the fiber to be $\flat_{dR}\mathbf{B}G$. Maybe the variant $(\flat_{dR}\mathbf{B}G\times \flat_{dR}\mathbf{B}G)/_{diag} G$ is actually “right”.
In the motivating example of #1 there is in fact an extra condition: the MC forms in arXiv:1305.4870 are produced patchwise in the fibers in (4.13); but then above (4.15) and above (4.17) further cohomological conditions are used to produce, in (4.18), (4.23) (4.24), an actual fiberwise MC form.
Ah, I think I have the answer. It’s super-easy. But my insistence above on having a group object in the slice was misled.
Instead of using the adjoint action on $G$, we should use the right action. That’s just
$\array{ G &\longrightarrow& \ast \\ && \downarrow \\ && \mathbf{B}G }$from there we want to map to
$\array{ \flat_{dR}\mathbf{B}G &\longrightarrow& \flat \mathbf{B}G \\ && \downarrow \\ && \mathbf{B}G }$Well, that’s easy, there is actually an essentially unique map
$\array{ \ast && \stackrel{\alpha}{\longrightarrow} && \flat\mathbf{B}G \\ & \searrow && \swarrow \\ && \mathbf{B}G }$Now while that does certainly induce a map $G\to \flat_{dR}\mathbf{B}G$ on the homotopy fibers, one tends to worry that this will be the trivial map. But, no, it is the MC form. This follows from the pasting law (and it will be trivial to you, but let me spell it out anyway):
the morphism on homotopy fibers which is induced from the previous diagram is that from the pullback diagram
$\array{ G &\longrightarrow& \ast \\ \downarrow && \downarrow \\ \ast & \longrightarrow& \mathbf{B}G }$to the pullback diagram
$\array{ \flat_{dR}\mathbf{B}G &\longrightarrow& \flat \mathbf{B}G \\ \downarrow && \downarrow \\ \ast &\longrightarrow & \mathbf{B}G }$via the universal map $\phi$ in
$\array{ G &\longrightarrow& \ast \\ \downarrow^{\mathrlap{\phi}} && \downarrow^{\mathrlap{\alpha}} \\ \flat_{dR}\mathbf{B}G &\longrightarrow& \flat\mathbf{B}G \\ \downarrow && \downarrow \\ \ast &\longrightarrow& \mathbf{B}G }$Here the bottom square and the total rectangle are homotopy pullbacks by assumption. Hence so is the top square. Hence
$\phi \simeq \theta$is indeed the MC form.
And so to come back to #1, the answer is that given a $G$-principal $\infty$-bundle on $X_1$ (classified by some $g \colon X_1\to \mathbf{B}G$) with a trivialization of its pullback to some correspondence space $Z \to X_1$, then this induces the fiberwise flat $\mathfrak{g}$-valued differential form that is denoted $\underset{x \colon X_1}{\sum} \theta_x$ in this pasting diagram of homotopy pullbacks:
$\array{ G && \longrightarrow && Z && \longrightarrow && \ast \\ & \searrow^{\mathrlap{\theta_x}} && && \searrow^{\mathrlap{\underset{x\colon X_1}{\sum} \theta_x}} && && \searrow^{\mathrlap{\theta/G}} \\ \downarrow && \flat_{dR}\mathbf{B}G && \longrightarrow && && \longrightarrow && \flat \mathbf{B}G \\ & \swarrow && && \swarrow && && \swarrow \\ \ast && \stackrel{x}{\longrightarrow} && X_1 && \stackrel{g}{\longrightarrow} && \mathbf{B}G }$Or rather, this is half the answer for #1. Next I need to sum up these fiberwise flat forms to a single non-flat form on the total space…
Re: #27, that’s basically just the definition of $\phi$, right? I actually thought of that, but you really seemed to want the adjoint action on $G$ so I assumed it wasn’t right. (-:
Yeah, sorry, I was misled. For the absolute MC form I needed a group object, so I thought I should look for one in the slice, too. My bad, sorry for the wild goose chase.
No problem, I learned some things. (-:
@Urs, what’s missing in the middle of the diagram in #28?
Whatever name you want to give to the pullback given by the bottom right square.
Ah, I see. I thought it might be something that already had a name.
Right, I should have said more about it, sorry. So let’s write $P$ for the $G$-principal $\infty$-bundle which is modulated by that map $g\colon X_1 \to \mathbf{B}G$. Then one name of that thing without name is the associated $\flat_{dR}\mathbf{B}G$-bundle
$P \underset{G}{\times} (\flat_{dR}\mathbf{B}G)$Moreover, a more evocative notaton for $\flat_{dR}\mathbf{B}G$ is $\Omega_{flat}(-,\mathfrak{g})$, the sheaf of $L_\infty$-algebra valued differential forms, so this is
$P \underset{G}{\times} \Omega_{flat}(-,\mathfrak{g})$and that makes most manifest how this is a fiberwise varying collection of such differential form coefficients. One might consider the étalification of this associated bundle to a concrete bundle, and that would be then something like a $\mathfrak{g}$-valued cotangent bundle of $X_1$.
Or sections of the adjoint bundle P×_G g, such as appears in the left of the algebroid Atiyah sequence? But you said that we shouldn’t be thinking of the adjoint action (though I may be thinking of the wrong bit, I can’t read back over the thread to unravel it now)
Right, the action is that by gauge transformations. If $G$ is a Lie 1-group, then
$g \in G(U) = C^\infty(U,G)$ takes $A \in \Omega_{flat}(U,\mathfrak{g})$ to
$A \mapsto g^{-1}A g + g^{-1} d g \,.$In the dcct pdf this is prop. 5.4.44, see also example 3.9.37.
(And the statement about the $G$-equivariance of $\theta$ is now also on the $n$Lab here).
I may still be confused. We have the construction in #28
$\array{ G && \longrightarrow && Z && \longrightarrow && \ast \\ & \searrow^{\mathrlap{\theta_x}} && && \searrow^{\mathrlap{\underset{x\colon X_1}{\sum} \theta_x}} && && \searrow^{\mathrlap{\theta/G}} \\ \downarrow && \flat_{dR}\mathbf{B}G && \longrightarrow && P \underset{G}{\times}(\flat_{dR}\mathbf{B}G) && \longrightarrow && \flat \mathbf{B}G \\ & \swarrow && && \swarrow && && \swarrow \\ \ast && \stackrel{x}{\longrightarrow} && X_1 && \stackrel{g}{\longrightarrow} && \mathbf{B}G }$and my next task would be to find from these fiberwise flat differential forms a globally defined differential form on $Z$. From looking at the model one expects this not to be necessarily flat anymore, as the horizontal part of the differential will see how the fiberwise flat forms change as one moves along the fibers. My expectation was that I’d have to first pass from fiberwise flat to fiberwise general $\mathfrak{g}$-valued forms $\Omega(-,\mathfrak{g})$ and then observe the vanishing of some cocycle in order to find a section of a trivial $\Omega(-,\mathfrak{g})$-bundle over $Z$ which would be the globally defined form in question.
But what confuses me now is that I seem to get all this for free and for flat forms: namely the map $\sum_{x\colon X_1}\theta_x$ in the above diagram is equivalently a section of the pullback of the bundle $P \underset{G}{\times}(\flat_{dR}\mathbf{B}G)$ to $Z$. But by the commutativity of the diagram, that is equivalently the pullback to $\flat \mathbf{B}G$ along $Z\to \ast \to \mathbf{B}G$, which is the trivial flat form bundle $Z \times (\flat_{dR}\mathbf{B}G)$. Hence the above gives a section of that hence a flat form
$Z \to \flat_{dR}\mathbf{B}G$That worries me a bit. It would be good, but somehow it seems too good to be true. Maybe it is an indication that I am missing something and my intended formalization of #1 still does not work.
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