# Every principal bundle is flat, in the infinity world

###### 5th February, 2022

Earlier today, Mahmoud Zeinalian explained something to me that Dennis Sullivan once explained to him, and it’s been sitting in my brain ever since then. In an attempt to empty out my thoughts, and also preserve what little understanding I currently believe to have of the story, I thought I’d write a little blog post about it. It’s going to move quite quickly, because I don’t want to spend time developing the prerequisites — the main purpose is for this to jog my brain two weeks down the line when I forget all the details!

The rather “provocative” statement that I’m going to try to justify today is the following:

Every principal bundle is flat, from the (\infty,1)-point-of-view.

Let’s give some setup and explain why, first of all, this should sound rather bizarre, if not completely incorrect.

# The setup

Pick some principal G-bundle \pi\colon E\to B, endowed with a connection \nabla. We’re going to look at what sort of information we can associated to singular cells in the base space B, using the things that we have available to us.

We know that the fibre E_b=\pi^{-1}(b) of E at any point b\in B “looks like” the group G, so, using the axiom of choice if we have to (or whatever really — we’re playing fast and loose here!), pick an element in each fibre to be the identity element of G. That is, very discontinuously think of all the fibres as copies of G.

Now let’s look at what a 1-simplex (i.e. a line) in B gives us. Say the line goes from the point b to the point b'. We get a G-equivariant map \Gamma_b^{b'}\colon E_b\to E_{b'} by using the parallel transport offered to us from the connection \nabla. But, using the fact that we have already identified our fibres with G, and a nice technical lemma about Lie groups (saying (roughly) that, if a map commutes with all left actions, then it must be given by right action, i.e. by an element of the Lie group itself), we see that this is exactly the data of an element of G, say g_{b,b'}.

The next step is to think about what happens with 2-simplices (i.e. triangles) in B. If we call the three vertices 0, 1, and 2, and label the edge between i and j by ij, then, by the above, the three edges will give us three elements g_{01},g_{12},g_{02}\in G. But here is where things “break”: it is not necessarily the case that g_{12}g_{01}=g_{02}; this identity that the notation suggestively implies is only true if \nabla is flat! (Recall: flatness of the connections corresponds to path-independence of the associated parallel transport).

So we’ve come to the conclusion that a bundle is flat if and only if… it is flat. Great!

# Some cubical wizardry

Trying to be ever so slightly more precise, we can formalise what we were doing in the above construction: for p=1,2, to each singular p-chain (i.e. each p-simplex) in B we were associating a (p-1)-cube with values in G. That is, were were constructing an element of the singular cochain complex on B with values in “cubical chains with values in G”, which we could write as \varphi_\nabla \in \mathrm{C}^{\bullet,\mathrm{sing}}(B,\mathrm{C}_{\bullet-1}^\mathrm{cube}(B,G)). The fact that our connection is not flat is reflected in the fact that this element does not satisfy the Maurer–Cartan equation: \mathrm{d}\varphi_\nabla+\varphi_\nabla^2 \neq 0 where defining the product turns out to be really easy, because cubes, unlike simplices, actually satisfy “the product of two cubes is a cube”.

So here’s the question: we’ve constructed \varphi_\nabla to be concentrated only in two (low) degrees; can we “extend” it to give an element that does satisfy Maurer–Cartan? The answer is YES, and this is exactly what we mean by our opening claim that “every bundle is flat in the (\infty,1)-sense”. I won’t spell out the details on how exactly this works (since I’ve already been very informal up until now anyway), but I’ll show you some very lovely pictures, and then afterwards tell you where you can go to read about this in proper detail.

Thinking about what we need to “construct” in order to extend our element, we can guess that we need some sort of “homotopy” between g_{12}g_{01} and g_{02}. But how exactly can we do this? And, further, how can we do it in a cubical way?

Well we know that every path in B gives us an element of G, and we know that the path 02 gives us g_{02}, and the composite path 01;12 gives us the element g_{12}g_{01}, so let’s draw a bunch of intermediate paths! We can do this by starting at 0, heading along the line 01 for some time 0\leq t_0\leq1, and then heading towards the point 2 for the rest of the time t_0-1. When t_0=0 we recover exactly the line 02, and when t_0=1 we recover exactly the composite 01;12; for intermediate values, we get something like in the (very hastily drawn) picture below: