Connection on a principal bundle $P(M,G)$ giving a functor on $\mathcal{P}_1(M)$


Question : Let $ P(M,G)$ be a principal $ G$ bundle. How does connection on $ P(M,G)$ defines a functor $ \text{Hol}: \mathcal{P}_1(M)\rightarrow BG$ (here $ BG$ is the Lie groupoid whose morphism set is $ G$ whose object set is $ \{*\}$ ).

I have seen in some places that, giving a connection on $ P(M,G)$ is giving a map $ P_1(M)\rightarrow G$ .

Here $ P_1(M)$ are special collection of special types of paths. This is the morphism set of what is called the path groupoid of $ M$ , usually denoted by $ \mathcal{P}_1(M)$ whose objects are elements of $ M$ .

Once this is done, seeing the Lie group $ G$ as a Lie groupoid $ BG$ (I know this is a bad notation but let me use this for this time) whose set of objects is singleton and set of morphisms is $ G$ . This would then give a functor $ \mathcal{P}_1(M)\rightarrow BG$ . They say giving a connection means giving a functor $ \mathcal{P}_1(M)\rightarrow BG$ with some good conditions.

Then, to make sense of $ 2$ -connections, they just have to consider $ \mathcal{P}_2(M)\rightarrow \text{some category}$ .

This is the set up.


I do not understand (I could not search it better) how giving a connection on $ P(M,G)$ gives a map $ \text{Hol}:P_1(M)\rightarrow G$ . For each path $ \gamma$ in $ M$ they are associating an element of $ G$ and calling it to be the holonomy of that path $ \gamma$ . They say it is given by integrating forms on paths.

All I know is, a connection on $ P(M,G)$ is a $ \mathfrak{g}$ valued $ 1$ -form on $ P$ with some extra conditions.

Suppose I have a path $ \gamma$ on $ M$ , how do I associate an element of $ G $ ? Is it $ \int_{\gamma}\omega$ ? How to make sense of this? It is not clear how I should see this as $ \omega$ is a form on $ P$ and $ \gamma$ is a path on $ M$ .

To make sense of this, there are two possible ways I can think of.

  • I have to pull back the path $ \gamma$ which is on $ M$ to a path on $ P$ . So that both the differential form and path are in same space.
  • I have to push forward $ \omega$ to a (collection of) form(s) on $ M$ . So that both the differential form and path are in same space.

Given a path $ \gamma:[0,1]\rightarrow M$ with $ \gamma(0)=x$ , fix a point $ u\in \pi^{-1}(x)$ . Then, connection gives a unique path $ \widetilde{\gamma}$ in $ P$ whose starting point is $ u$ such that projection of $ \widetilde{\gamma}$ along $ \pi$ is $ \gamma$ . The problem here is that we have to fix a point $ u$ . Only then we can get a curve. It can happen that for any two points on $ \pi^{-1}(x)$ may give same result but I am not sure if that is true. I mean, let $ \widetilde{\gamma}_u,\widetilde{\gamma}_v$ be lifts of $ \gamma$ fixing $ u\in \pi^{-1}(x)$ and $ v\in \pi^{-1}(x)$ respectively. Does it then happen that $ \int_{\widetilde{\gamma}_u}\omega=\int_{\widetilde{\gamma}_v}\omega$ ?

Even if this is the case, what does it mean to say integrating a $ \mathfrak{g}$ valued $ 1$ -form on a path? How is it defined? I guess it should give an element $ A$ of $ \mathfrak{g}$ (just like integrating a $ \mathbb{R}$ valued $ 1$ -form along a path gives an element of $ \mathbb{R}$ ). Do we then see image of $ A$ under $ \text{exp}:\mathfrak{g}\rightarrow G$ to get an element of $ G$ ? We can declare this to be $ \int_{\gamma}\omega$ .

Is this how we associate an element of $ G$ to a path $ \gamma$ in $ M$ ??

Otherwise, given $ \omega$ on $ P$ , using trivialization, we can get an open cover $ \{U_i\}$ of $ M$ and get forms $ \mathfrak{g}$ valued $ 1$ -forms $ \omega_i$ on $ U_i$ with some compatibility on intersections.

We can consider $ \gamma_i:[0,1]\bigcap \gamma^{-1}(U_i)\rightarrow U_i$ . These $ \gamma_i$ are paths on $ U_i$ and $ \omega_i$ are $ 1$ -forms on $ U_i$ . So, $ \int_{U_i}\gamma_i$ makes sense. This gives a collection of elements $ \{A_i\}$ of $ \mathfrak{g}$ and may be all these comes from a single element $ A\in \mathfrak{g}$ and seeing its image under $ \text{exp}:\mathfrak{g}\rightarrow G$ gives an element in $ G$ . We can then declare it to be $ \int_{\gamma}\omega$ .

Is this how we associate an element of $ G$ to a path $ \gamma$ in $ M$ ??

Any comments are welcome.

I could not find a place where this is discussed in detail. So, asking here.