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1. • CommentRowNumber1.
   • CommentAuthorPraphulla Koushik
   • CommentTimeJan 9th 2019
   • (edited Jan 9th 2019)
   • PermaLink
   Author: Praphulla Koushik
   Format: MarkdownItex> 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.

   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.

2. • CommentRowNumber2.
   • CommentAuthorPraphulla Koushik
   • CommentTimeJan 9th 2019
   • (edited Jan 9th 2019)
   • PermaLink
   Author: Praphulla Koushik
   Format: MarkdownItex---

   ---

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJan 9th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexThe 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](https://arxiv.org/abs/0705.0452).

   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.

4. • CommentRowNumber4.
   • CommentAuthorPraphulla Koushik
   • CommentTimeJan 9th 2019
   • PermaLink
   Author: Praphulla Koushik
   Format: MarkdownItex@Urs Hi, Does it have nothing to do with integrating the 1-form $\omega$ on the path $\gamma$?

   @Urs

   Hi,

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

5. • CommentRowNumber5.
   • CommentAuthorPraphulla Koushik
   • CommentTimeJan 12th 2019
   • PermaLink
   Author: Praphulla Koushik
   Format: MarkdownItexThanks 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.

   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.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJan 12th 2019
   • (edited Jan 12th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexYou 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](https://arxiv.org/abs/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](https://arxiv.org/abs/0802.0663) and [arXiv:0808.1923](https://arxiv.org/abs/0808.1923).)

   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.)