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• CommentRowNumber1.
• CommentAuthorPraphulla Koushik
• CommentTimeJan 9th 2019
• (edited Jan 9th 2019)

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 $M$??

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

• CommentRowNumber2.
• CommentAuthorPraphulla Koushik
• CommentTimeJan 9th 2019
• (edited Jan 9th 2019)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJan 9th 2019

The short answer is: Assign to a path $\gamma$ the parallel transport operation of the given connection along that path.

For details you could try arXiv:0705.0452.

1. @Urs

Hi,

Does it have nothing to do with integrating the 1-form $\omega$ on the path $\gamma$?

2. Thanks for that arxiv notes. I can see that you have discussed in first three pages. I will try to understand and write down here if I have any question.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeJan 12th 2019
• (edited Jan 12th 2019)

You will have already seen the answer to #4 now, but just briefly, for the record:

Yes, the parallel transport of a connection along a path may be expressed as a path-ordered integral of differential 1-form representatives of the connection along the path, possibly with insertion of transition functions where the path crosses local patches of a local trivialization.

I should say that these are classical constructions. But arXiv:0705.0452 should give a fully detailed account.

(This article was originally just meant to be a review preface to the higher-dimensional generalization, which itself was then broken up further into arXiv:0802.0663 and arXiv:0808.1923.)