## 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 $$M$$??