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

    Question : Let P(M,G)P(M,G) be a principal GG bundle. How does connection on P(M,G)P(M,G) defines a functor Hol:𝒫 1(M)BG\text{Hol}: \mathcal{P}_1(M)\rightarrow BG (here BGBG is the Lie groupoid whose morphism set is GG whose object set is {*}\{*\}).

    I have seen in some places that, giving a connection on P(M,G)P(M,G) is giving a map P 1(M)GP_1(M)\rightarrow G.

    Here P 1(M)P_1(M) are special collection of special types of paths. This is the morphism set of what is called the path groupoid of MM, usually denoted by 𝒫 1(M)\mathcal{P}_1(M) whose objects are elements of MM.

    Once this is done, seeing the Lie group GG as a Lie groupoid BGBG (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 GG. This would then give a functor 𝒫 1(M)BG\mathcal{P}_1(M)\rightarrow BG. They say giving a connection means giving a functor 𝒫 1(M)BG\mathcal{P}_1(M)\rightarrow BG with some good conditions.

    Then, to make sense of 22-connections, they just have to consider 𝒫 2(M)some category\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)P(M,G) gives a map Hol:P 1(M)G\text{Hol}:P_1(M)\rightarrow G. For each path γ\gamma in MM they are associating an element of GG and calling it to be the holonomy of that path γ\gamma. They say it is given by integrating forms on paths.

    All I know is, a connection on P(M,G)P(M,G) is a 𝔤\mathfrak{g} valued 11-form on PP with some extra conditions.

    Suppose I have a path γ\gamma on MM, how do I associate an element of GG ? Is it γω\int_{\gamma}\omega? How to make sense of this? It is not clear how I should see this as ω\omega is a form on PP and γ\gamma is a path on MM.

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

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

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

    Even if this is the case, what does it mean to say integrating a 𝔤\mathfrak{g} valued 11-form on a path? How is it defined? I guess it should give an element AA of 𝔤\mathfrak{g} (just like integrating a \mathbb{R} valued 11-form along a path gives an element of \mathbb{R}). Do we then see image of AA under exp:𝔤G\text{exp}:\mathfrak{g}\rightarrow G to get an element of GG? We can declare this to be γω\int_{\gamma}\omega.

    Is this how we associate an element of GG to a path γ\gamma in MM??

    Otherwise, given ω\omega on PP, using trivialization, we can get an open cover {U i}\{U_i\} of MM and get forms 𝔤\mathfrak{g} valued 11-forms ω i\omega_i on U iU_i with some compatibility on intersections.

    We can consider γ i:[0,1]γ 1(U i)U i\gamma_i:[0,1]\bigcap \gamma^{-1}(U_i)\rightarrow U_i. These γ i\gamma_i are paths on U iU_i and ω i\omega_i are 11-forms on U iU_i. So, U iγ i\int_{U_i}\gamma_i makes sense. This gives a collection of elements {A i}\{A_i\} of 𝔤\mathfrak{g} and may be all these comes from a single element A𝔤A\in \mathfrak{g} and seeing its image under exp:𝔤G\text{exp}:\mathfrak{g}\rightarrow G gives an element in GG. We can then declare it to be γω\int_{\gamma}\omega.

    Is this how we associate an element of GG to a path γ\gamma in MM??

    Any comments are welcome.

    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 γ\gamma the parallel transport operation of the given connection along that path.

    For details you could try arXiv:0705.0452.

  1. @Urs


    Does it have nothing to do with integrating the 1-form ω\omega on the path γ\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.)

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