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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeDec 24th 2012
   • (edited Dec 24th 2012)
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   Author: Mirco Richter
   Format: MarkdownItexIs someone patient enough to explain me on a more informal level how a Lie $\infty$-algebroid is the infinitesimal version of an $\infty$-groupoid? I suppose the following: 1.) A Lie $\infty$-algebroid is a graded vector bundle $V\to M$ such that the graded vector space of sections has a Lie $\infty$-algebra structure. 2.) A Lie $\infty$-groupoid (or just smooth $\infty$-groupoid if you like) is a Kan simplicial set K, with some kind of smooth cohesion. (For now, it would be enough to see the Kan simplicial set as internal to the category of smooth maifolds, because I have a better understanding here, than in the general case of a smooth topos.) ... What I don't see is what the groupoid counterpart to the brackets of te algebroid is...

   Is someone patient enough to explain me on a more informal level how a Lie $\infty$-algebroid is the infinitesimal version of an $\infty$-groupoid?

   I suppose the following:

   1.) A Lie $\infty$-algebroid is a graded vector bundle $V\to M$ such that the graded vector space of sections has a Lie $\infty$-algebra structure.

   2.) A Lie $\infty$-groupoid (or just smooth $\infty$-groupoid if you like) is a Kan simplicial set K, with some kind of smooth cohesion. (For now, it would be enough to see the Kan simplicial set as internal to the category of smooth maifolds, because I have a better understanding here, than in the general case of a smooth topos.)

   …

   What I don’t see is what the groupoid counterpart to the brackets of te algebroid is…

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 24th 2012
   • (edited Dec 24th 2012)
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   Author: Urs
   Format: MarkdownItexJust briefly: first for [[smooth ∞-groups]]: regard the [[delooping]] $\mathbf{B}G$ of your smooth $\infty$-group as a [[pointed object|pointed]] [[synthetic differential ∞-groupoid]] $\mathbf{B}G \in SynthDiff \infty Grpd^{*/}$. Then write $\mathbf{B}\mathfrak{g}$ for the $\infty$-presheaf on [[infinitesimally thickened points]] by restriction along the Yoneda embedding $InfThickPoints^{*/} \to SynthDiff\infty Grpd^{*/}$ This $\mathfrak{g}$ is the corresponding $L_\infty$-algebra, under the equivalence of $\infty$-sheaves on infinitesimally thickened points with $L_\infty$-algebras, see at _[synthetic differential oo-groupoid -- Lie differentiation](http://ncatlab.org/nlab/show/synthetic+differential+infinity-groupoid#LieDifferentiation)_. For instance if $G$ is a Lie group you find from this procedure manifestly the synthetic computation of the Lie algebra of $G$ as described in Anders Kock's books on [[synthetic differential geometry]]. Now if $A \in $ [[Smooth∞Grpd]] is a [[smooth ∞-groupoid]]: in order to assign an $\infty$-Lie algebroid to it you first need to choose an atlas in the form of an effective epimorphism $A_0 \to A$. (In the previous case that atlas was the point $* \to \mathbf{B}G$). Then consider the smooth $\infty$-group of [[bisections]] $$ BiSect_A(A_0) := \mathbf{Aut}_A(A_0) $$ The pair consisting of $C^\infty(A_0)$ and the Lie differentiation of $BiSect_A(A_0)$ as above is the $\infty$-[[Lie-Rinehart algebra]] that incarnates the $\infty$-Lie algebroid of $A_0 \to A$.

   Just briefly:

   first for smooth ∞-groups:

   regard the delooping $\mathbf{B}G$ of your smooth $\infty$-group as a pointed synthetic differential ∞-groupoid $\mathbf{B}G \in SynthDiff \infty Grpd^{*/}$. Then write $\mathbf{B}\mathfrak{g}$ for the $\infty$-presheaf on infinitesimally thickened points by restriction along the Yoneda embedding $InfThickPoints^{*/} \to SynthDiff\infty Grpd^{*/}$ This $\mathfrak{g}$ is the corresponding $L_\infty$-algebra, under the equivalence of $\infty$-sheaves on infinitesimally thickened points with $L_\infty$-algebras, see at synthetic differential oo-groupoid – Lie differentiation.

   For instance if $G$ is a Lie group you find from this procedure manifestly the synthetic computation of the Lie algebra of $G$ as described in Anders Kock’s books on synthetic differential geometry.

   Now if $A \in$ Smooth∞Grpd is a smooth ∞-groupoid:

   in order to assign an $\infty$-Lie algebroid to it you first need to choose an atlas in the form of an effective epimorphism $A_0 \to A$.

   (In the previous case that atlas was the point $* \to \mathbf{B}G$). Then consider the smooth $\infty$-group of bisections

   $$BiSect_A(A_0) := \mathbf{Aut}_A(A_0)$$

   The pair consisting of $C^\infty(A_0)$ and the Lie differentiation of $BiSect_A(A_0)$ as above is the $\infty$-Lie-Rinehart algebra that incarnates the $\infty$-Lie algebroid of $A_0 \to A$.

3. • CommentRowNumber3.
   • CommentAuthorMirco Richter
   • CommentTimeDec 25th 2012
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   Author: Mirco Richter
   Format: MarkdownItexThanks. I will try to apply it on a few simple examples for practice and come back to this thread in case there a parsing problems.

   Thanks. I will try to apply it on a few simple examples for practice and come back to this thread in case there a parsing problems.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeDec 25th 2012
   • (edited Dec 25th 2012)
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   Author: Urs
   Format: MarkdownItexI'd suggest that to warm-up it would be useful to go through the following steps: * read section 6.6 "The Lie algebra of G" of Anders Kock's _[SDG of manifolds](http://home.imf.au.dk/kock/SGM-final.pdf)_; * reformulate what happens in that chapter as the construction of the [[simplicial object|simplicial]] [[smooth locus]] $\mathbf{B}\mathfrak{g} := (\cdots (G\times G)_{(1)} \stackrel{\to}{\stackrel{\to}{\to}} G_{(1)} \stackrel{\to}{\to} *)$ where the subscript means "first order infinitesimal neighbourhood of the neutral element"; * then repeat that story with the Lie group $G$ replaced by a strict Lie 2-group. (We should eventually have all this spelled out on the $n$Lab. If your learning of this eventually leads to such expositions, it would make me happy.)

   I’d suggest that to warm-up it would be useful to go through the following steps:

   • read section 6.6 “The Lie algebra of G” of Anders Kock’s SDG of manifolds;

   • reformulate what happens in that chapter as the construction of the simplicial smooth locus

     $\mathbf{B}\mathfrak{g} := (\cdots (G\times G)_{(1)} \stackrel{\to}{\stackrel{\to}{\to}} G_{(1)} \stackrel{\to}{\to} *)$

     where the subscript means “first order infinitesimal neighbourhood of the neutral element”;

   • then repeat that story with the Lie group $G$ replaced by a strict Lie 2-group.

   (We should eventually have all this spelled out on the $n$Lab. If your learning of this eventually leads to such expositions, it would make me happy.)

5. • CommentRowNumber5.
   • CommentAuthorMirco Richter
   • CommentTimeDec 27th 2012
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   Author: Mirco Richter
   Format: MarkdownItexok. But it definitively will take some time.

   ok. But it definitively will take some time.

6. • CommentRowNumber6.
   • CommentAuthorMirco Richter
   • CommentTimeDec 27th 2012
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   Author: Mirco Richter
   Format: MarkdownItexIs there a definition of an $\infty$-Lie–Rinehart-pair? If we replace the Lie algebra by a Lie $\infty$-algebra we need an infinity version of the Leibniz law and a definition of a Lie $\infty$-algebra module. Is that in the book of Loday and Vallette? (Then most likely they already wrote the stuff down in terms of many brackets, which would safe me time)

   Is there a definition of an $\infty$-Lie–Rinehart-pair? If we replace the Lie algebra by a Lie $\infty$-algebra we need an infinity version of the Leibniz law and a definition of a Lie $\infty$-algebra module.

   Is that in the book of Loday and Vallette? (Then most likely they already wrote the stuff down in terms of many brackets, which would safe me time)

7. • CommentRowNumber7.
   • CommentAuthorcrogers
   • CommentTimeDec 27th 2012
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   Author: crogers
   Format: TextRegarding homotopy Lie-Rinehart etc, you might be interested in Sec. 2 of this preprint: http://arxiv.org/abs/1204.2467 and the references within (e.g. the papers by Huebschmann and Kjeseth).

   Regarding homotopy Lie-Rinehart etc, you might be interested in Sec. 2 of this preprint:
   
   http://arxiv.org/abs/1204.2467
   
   and the references within (e.g. the papers by Huebschmann and Kjeseth).
8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeDec 27th 2012
   • (edited Dec 27th 2012)
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   Author: Urs
   Format: MarkdownItexIf the collection of objects $A_0$ of your smooth $\infty$-groupoid is just a smooth manifold (instead of a more stacky/more derived space) then you get away with much less than a general homotopy Lie-Rinehart algebra. (Recall that already the ordinary concept of [[Lie-Rinehart algebra]] is considerably more general than the standard notion of [[Lie algebroid]]: because it allows $A_0$ to be the formal dual to any associative algebra, hence allows $A_0$ to be a noncommutative space.) But if $A_0$ is just a plain manifold, then all there is in the Lie-Rinehart structure is an action of the Lie algebra on $A_0$, hence simply a Lie algebra map to the vector field Lie algebra. This remains true for higher Lie algebroids. Let me illustrate this further: for $A$ your smooth $\infty$-groupoid and $A_0 \to A$ a choice of its manifold of objects, a [[bisection of a Lie groupoid|bisection]] is a diagram $$ \array{ A_0 &&\stackrel{\phi}{\to}&& A_0 \\ & \searrow &\swArrow_{\rho}& \swarrow \\ && A } $$ where $\phi$ is a diffeomorphism. Under horizontal pasting composition, these diagrams form a (higher) group. The (higher) Lie algebra of that higher group is the Lie algebra part in the Lie-Rinehart pair. There is the evident forgetful (higher) group homomorphism $$ BiSect_A(A_0) \to Diff(A_0) $$ from this (higher) group of bisections to the diffeomorphism group of $A_0$. It is given simply by forgetting the triangle in the above diagram and just remembering the diffeomorphism $\phi$. After differentiation, this is a (higher) Lie algebra homomorphism $$ Lie(BiSect_A(A_0)) \to Vect(A_0) \,. $$ And this is ssentially all you need to make this a (homotopy) Lie-Rinehart pair. (The other information you need is how to restrict to patches of $A_0$, but this is straightforward.)

   If the collection of objects $A_0$ of your smooth $\infty$-groupoid is just a smooth manifold (instead of a more stacky/more derived space) then you get away with much less than a general homotopy Lie-Rinehart algebra.

   (Recall that already the ordinary concept of Lie-Rinehart algebra is considerably more general than the standard notion of Lie algebroid: because it allows $A_0$ to be the formal dual to any associative algebra, hence allows $A_0$ to be a noncommutative space.)

   But if $A_0$ is just a plain manifold, then all there is in the Lie-Rinehart structure is an action of the Lie algebra on $A_0$, hence simply a Lie algebra map to the vector field Lie algebra.

   This remains true for higher Lie algebroids.

   Let me illustrate this further: for $A$ your smooth $\infty$-groupoid and $A_0 \to A$ a choice of its manifold of objects, a bisection is a diagram

   $$\array{ A_0 &&\stackrel{\phi}{\to}&& A_0 \\ & \searrow &\swArrow_{\rho}& \swarrow \\ && A }$$

   where $\phi$ is a diffeomorphism. Under horizontal pasting composition, these diagrams form a (higher) group. The (higher) Lie algebra of that higher group is the Lie algebra part in the Lie-Rinehart pair.

   There is the evident forgetful (higher) group homomorphism

   $$BiSect_A(A_0) \to Diff(A_0)$$

   from this (higher) group of bisections to the diffeomorphism group of $A_0$. It is given simply by forgetting the triangle in the above diagram and just remembering the diffeomorphism $\phi$.

   After differentiation, this is a (higher) Lie algebra homomorphism

   $$Lie(BiSect_A(A_0)) \to Vect(A_0) \,.$$

   And this is ssentially all you need to make this a (homotopy) Lie-Rinehart pair.

   (The other information you need is how to restrict to patches of $A_0$, but this is straightforward.)

9. • CommentRowNumber9.
   • CommentAuthorjim_stasheff
   • CommentTimeDec 28th 2012
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   Author: jim_stasheff
   Format: TextUrs seems to b e missing the point about homtopy Lie Rinehart as JUST dg algebra - no manifolds in sight. Chris has it right - thanks especially for the link (Vitagliano?)and the references. I'm just in the process of developing a lecture on exactly this topic. Will share the notes/slides when ready.

   Urs seems to b e missing the point about homtopy Lie Rinehart as JUST
   dg algebra - no manifolds in sight. Chris has it right - thanks especially
   for the link (Vitagliano?)and the references. I'm just in the process of developing a lecture on exactly this topic.
   Will share the notes/slides when ready.
10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeDec 28th 2012
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    Author: Urs
    Format: MarkdownItexI don't think I am missing a point. I am trying to make it easier for Mirco. He doesn't need to understand full-blown homotopy Lie-Rinehart pairs in order to understand how to associate a higher Lie algebroid to a higher Lie groupoid by the method that I indicated.

    I don’t think I am missing a point. I am trying to make it easier for Mirco. He doesn’t need to understand full-blown homotopy Lie-Rinehart pairs in order to understand how to associate a higher Lie algebroid to a higher Lie groupoid by the method that I indicated.

11. • CommentRowNumber11.
    • CommentAuthorjim_stasheff
    • CommentTimeDec 30th 2012
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    Author: jim_stasheff
    Format: TextMirco - is the smooth/manifold setting what you were after as opposed to the homological lagebra?

    Mirco - is the smooth/manifold setting what you were after as opposed to the homological lagebra?
12. • CommentRowNumber12.
    • CommentAuthorMirco Richter
    • CommentTimeJan 3rd 2013
    • (edited Jan 3rd 2013)
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    Author: Mirco Richter
    Format: MarkdownItexIf by 'smooth/manifold setting' you mean that the sets $Hom(\Delta[n],G)$ of the oo-groupoid $G$ are just smooth manifolds, then yes ... at least for the moment. In http://ncatlab.org/nlab/show/Lie+infinity-algebroid#ModelsForTheAbstractAxioms is said that a Lie oo-algebroid can be seen in terms of vector bundles and Chevalley-Eilenberg algebras on the section spaces. This is what I want to understand in technical details on a simple situation. My problem is, that I can't see from the general description, how the differential of the C.E. algebra emerges from the oo-groupoid structure. Moreover I never saw an actual (students) exercise, like "here is your oo-groupoid xyz, calculate its differentiation" ... But according to the way my brain works, I have to work out at least one example in all details, to feel comfortable here. So no big math, just someone who tries to follow your way.

    If by ’smooth/manifold setting’ you mean that the sets $Hom(\Delta[n],G)$ of the oo-groupoid $G$ are just smooth manifolds, then yes … at least for the moment.

    In http://ncatlab.org/nlab/show/Lie+infinity-algebroid#ModelsForTheAbstractAxioms is said that a Lie oo-algebroid can be seen in terms of vector bundles and Chevalley-Eilenberg algebras on the section spaces. This is what I want to understand in technical details on a simple situation.

    My problem is, that I can’t see from the general description, how the differential of the C.E. algebra emerges from the oo-groupoid structure. Moreover I never saw an actual (students) exercise, like “here is your oo-groupoid xyz, calculate its differentiation” …

    But according to the way my brain works, I have to work out at least one example in all details, to feel comfortable here. So no big math, just someone who tries to follow your way.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeJan 3rd 2013
    • (edited Jan 3rd 2013)
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    Author: Urs
    Format: MarkdownItexHi Mirco, what I said gives you a way to reduce the problem of differentiating Lie $n$-groupoids to differentiating Lie $n$-groups (namely their groups of bisections). Take a Lie 2-groupoid of your choice, write down the 2-group of bisections. If your original Lie 2-groupoid was given as a strict 2-groupoid in smooth manifolds, then this will be a strict 2-group, and so you can differtiate in each degree as in ordinary Lie theory. This is your student's exercise. Now you just have to do it! :-) If you still feel uneasy about where to start, let me know what your favorite Lie 2-groupoid is which you want to differentiate, and we'll walk through it together. If all this does not help, you could also have a look at * Pavol Severa, _$L_\infty$ algebras as 1-jets of simplicial manifolds (and a bit beyond)_ ([arXiv:0612349](http://arxiv.org/abs/math/0612349)) which does something similar.

    Hi Mirco,

    what I said gives you a way to reduce the problem of differentiating Lie $n$-groupoids to differentiating Lie $n$-groups (namely their groups of bisections).

    Take a Lie 2-groupoid of your choice, write down the 2-group of bisections. If your original Lie 2-groupoid was given as a strict 2-groupoid in smooth manifolds, then this will be a strict 2-group, and so you can differtiate in each degree as in ordinary Lie theory. This is your student’s exercise. Now you just have to do it! :-)

    If you still feel uneasy about where to start, let me know what your favorite Lie 2-groupoid is which you want to differentiate, and we’ll walk through it together.

    If all this does not help, you could also have a look at

    • Pavol Severa, $L_\infty$ algebras as 1-jets of simplicial manifolds (and a bit beyond) (arXiv:0612349)

    which does something similar.

14. • CommentRowNumber14.
    • CommentAuthorMirco Richter
    • CommentTimeJan 3rd 2013
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    Author: Mirco Richter
    Format: MarkdownItexIf I have a given smooth oo-groupoid xyz and a given Lie oo-algebroid abc and I want to proof that abc is the differentiation of xyz, than I have to be able to apply the general abstract method.

    If I have a given smooth oo-groupoid xyz and a given Lie oo-algebroid abc and I want to proof that abc is the differentiation of xyz, than I have to be able to apply the general abstract method.

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeJan 3rd 2013
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    Author: Urs
    Format: MarkdownItexYes, I gave you one. But I heard that > according to the way [your] brain works, [you] have to work out at least one example in all details So let me know which example you'd like to work out.

    Yes, I gave you one. But I heard that

    according to the way [your] brain works, [you] have to work out at least one example in all details

    So let me know which example you’d like to work out.

16. • CommentRowNumber16.
    • CommentAuthorMirco Richter
    • CommentTimeJan 5th 2013
    • (edited Jan 5th 2013)
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    Author: Mirco Richter
    Format: MarkdownItex> If you still feel uneasy about where to start, let me know what your favorite Lie 2-groupoid is which you want to differentiate, and we’ll walk through it together. That is very kind and I would like to take that offer. I think we should consider a cosk^n Kan simplicial set internal to smooth manifolds for some low n at first. This should be general enough without being to tedious. I have some examples in mind and will write one of them down in detail during the next days (including the oo-group of bisections). Loosely speaking, they are given by $Hom_{\mathbf{V}}(\mathbb{R}\otimes \mathbb{Z}(\Delta[n]),\mathbb{R})$ for $\mathbf{V}$ the category of $\mathbb{R}$-vector spaces.

    If you still feel uneasy about where to start, let me know what your favorite Lie 2-groupoid is which you want to differentiate, and we’ll walk through it together.

    That is very kind and I would like to take that offer.

    I think we should consider a cosk^n Kan simplicial set internal to smooth manifolds for some low n at first. This should be general enough without being to tedious.

    I have some examples in mind and will write one of them down in detail during the next days (including the oo-group of bisections). Loosely speaking, they are given by $Hom_{\mathbf{V}}(\mathbb{R}\otimes \mathbb{Z}(\Delta[n]),\mathbb{R})$ for $\mathbf{V}$ the category of $\mathbb{R}$-vector spaces.

17. • CommentRowNumber17.
    • CommentAuthorMirco Richter
    • CommentTimeJan 15th 2013
    • (edited Jan 15th 2013)
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    Author: Mirco Richter
    Format: MarkdownItexOk. I tried to come up with an example of a non trivial weak $2$-groupoid and here is what I have got so fare. (I have a lot of other stuff to do this month, so hopefully you don't mind if it will take a little longer) Ideally this example could be used to give a reader, unfamiliar with higher differential geometry, a first impression. Consequently I choose the smooth cohesion on the objects and (higher) morphisms to be the most easiest: They are just cartesian spaces. If we managed this in detail, we can step on to more elaborated smooth spaces afterwards. (Which I'm really looking forward to) The $2$-groupoid is given by the following data: $G_0:=\mathbb{R}^1$, $G_1:= \mathbb{R}^3$, $G_2:=\mathbb{R}^6$, $G_{3}:=\mathbb{R}^{10}$ and the simplicial structure maps are: $$s_0: \mathbb{R}^{1}\to \mathbb{R}^{3}\; ; \; x_{0} \mapsto (x_{0},x_{0},x_{0})$$ $$ d_0: \mathbb{R}^{3} \to \mathbb{R}^{1}\; ; \; (x_0,x_1,x_{2}) \mapsto (x_{2})$$ $$d_1: \mathbb{R}^{3} \to \mathbb{R}^{1}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0})$$ $$s_0: \mathbb{R}^{3} \to \mathbb{R}^{6}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0},x_{0},x_{1},x_{0},x_{1},x_{2})$$ $$s_1: \mathbb{R}^{3} \to \mathbb{R}^{6}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0},x_{1},x_{1},x_{2},x_{2},x_{2})$$ $$d_0: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\; (x_0,...x_{5})\mapsto (x_{3},x_{4},x_{5})$$ $$d_1: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\;(x_0,...x_{5})\mapsto (x_{0},x_{2},x_{5})$$ $$d_2: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\;(x_0,...x_{5})\mapsto (x_{0},x_{1},x_{3})$$ $$s_0: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{0},x_{1},x_{2},x_{0},x_{1},x_{2},x_{3},x_{4},x_{5})$$ $$s_1: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{1},x_{1},x_{2},x_{3},x_{3},x_{4},x_{3},x_{4},x_{5})$$ $$s_2: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{1},x_{2},x_{2},x_{3},x_{4},x_{4},x_{5},x_{5},x_{5})$$ $$d_0: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\; (x_0,...x_{9})\mapsto (x_{4},x_{5},x_{6},x_{7},x_{8},x_{9})$$ $$d_1: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{2},x_{3},x_{7},x_{8},x_{9})$$ $$d_2: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{1},x_{3},x_{4},x_{6},x_{9})$$ $$d_3: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{1},x_{2},x_{4},x_{5},x_{7})$$ Tedious, but at least it is a workable and technically easy example of a non trivial, weak 2-groupoid. We'll see, whether or not this will work as a good teaching example. To get a simplicial cartesian space, we can apply the cosk functor from $3$-truncated simplicial objects to simplicial objects, but there shouldn't be anything new in higher dimensions. At first lets see if this is (the truncation of) a $2$-groupoid and check the Kan-conditions: Write $\bigwedge(m,j)$ for the horns in dimension $m$, i.e for the cartesian space $$\lbrace \left(x_0,\ldots, \widehat{x_k},\ldots, x_n \right)\;|\; x_j\in G_m \; and \; d_i(x_j)=d_{j-1}(x_i)\; for \; i \lt j \; and \; i,j \neq k \rbrace \;.$$ In particular $$\bigwedge(2,0)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_0=x_0\}$$ $$\bigwedge(2,1)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_2=x_0\}$$ $$\bigwedge(2,2)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_2=x_2\}$$ Given any horn $((x_0,x_1,x_2),(x_0,y_1,y_2))\in \bigwedge(2,0)$ we can define a lifting by $(x_0,y_1,x_1,y_2,z,x_2)\in G_2$, since $d_1(x_0,y_1,x_1,y_2,z,x_2)=(x_0,x_1,x_2)$ as well as $d_2(x_0,y_1,x_1,y_2,z,x_2)=(x_0,y_1,y_2)$. As suspected the lift is not unique but is parametrized by the single $z$. Similar given a horn $((x_0,x_1,x_2),(y_0,y_1,x_0))\in \bigwedge(2,1)$ we can define a lifting by $(y_0,y_1,z,x_0,x_1,x_2)\in G_2$, since $d_0=(x_0,x_1,x_2)$ and $d_2=(y_0,y_1,x_0)$. If we see $((x_0,x_1,x_2),(y_0,y_1,x_0))$ as a composable pair of $1$-morphisms, a composition is given for any $z\in \mathbb{R}$ by the face $d_1(y_0,y_1,z,x_0,x_1,x_2)=(y_0,z,x_2)$. This shows, that the 'weakness' of the composition appears here in the form of the parameter $z$. A similar calculation holds for $\bigwedge(2,2)$ and together this shows, that $G$ is at least not a $1$-groupoid. Now lets look at the Kan conditions in dimension $3$. For $G$ to be a $2$-groupoid there must be unique lifts here. Anyway the calculations are pretty easy and I just do it for $\bigwedge(3,1)$. The horn cartesian space are like $$\bigwedge(3,1)=\{((x_0,\ldots,x_5),(y_0,\ldots,y_5),(z_0,\ldots,z_5))\;;\;$$ $$(y_{3},y_{4},y_{5})=(x_{0},x_{2},x_{5}), (z_{3},z_{4},z_{5})=(x_{0},x_{1},x_{3}), (z_{0},z_{1},z_{3})=(y_{0},y_{1},y_{3})\}$$ For any horn $((x_0,x_1,x_2,x_3,x_4,x_5), (y_0,y_1,y_2,x_0,x_2,x_5), (y_0,y_1,z_2,x_0,x_1,x_3)) \in \bigwedge(3,1)$ (i.e any three composable $2$-morphism) let $m:=(y_0,y_1,z_2,y_2,x_0,x_1,x_2,x_3,x_4,x_5)$. Then $d_0(m)=(x_0,x_1,x_2,x_3,x_4,x_5)$, $d_2(m)=(y_0,y_1,y_2,x_0,x_2,x_5)$ and $d_3(m)=(y_0,y_1,z_2,x_0,x_1,x_3)$, so $m$ is the unique lift of the horn and the appropriate composition is given by $d_1(m)=(y_0,z_2,y_2,x_3,x_4,x_5)$. This is done for the other horn space, in a similar manner. Consequently, the Kan conditions are satisfied for any relevant dimensions and if we apply the coskeletal functor we get a Kan simplicial cartesian space. Moreover the lifts are unique for $k$-morphisms if $k\gt 1$ and consequently, we have a $2$-groupoid. Ok. That's it for today. Before going to differentiate, I would like to be sure, that we agree here. P.S.: Looks like my 'greater than' and 'less then' signs are not recognized here. Don't know how to fix it...

    Ok. I tried to come up with an example of a non trivial weak $2$-groupoid and here is what I have got so fare. (I have a lot of other stuff to do this month, so hopefully you don’t mind if it will take a little longer)

    Ideally this example could be used to give a reader, unfamiliar with higher differential geometry, a first impression. Consequently I choose the smooth cohesion on the objects and (higher) morphisms to be the most easiest: They are just cartesian spaces. If we managed this in detail, we can step on to more elaborated smooth spaces afterwards. (Which I’m really looking forward to)

    The $2$-groupoid is given by the following data:

    $G_0:=\mathbb{R}^1$, $G_1:= \mathbb{R}^3$, $G_2:=\mathbb{R}^6$, $G_{3}:=\mathbb{R}^{10}$ and the simplicial structure maps are:

    $$s_0: \mathbb{R}^{1}\to \mathbb{R}^{3}\; ; \; x_{0} \mapsto (x_{0},x_{0},x_{0})$$ $$d_0: \mathbb{R}^{3} \to \mathbb{R}^{1}\; ; \; (x_0,x_1,x_{2}) \mapsto (x_{2})$$ $$d_1: \mathbb{R}^{3} \to \mathbb{R}^{1}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0})$$ $$s_0: \mathbb{R}^{3} \to \mathbb{R}^{6}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0},x_{0},x_{1},x_{0},x_{1},x_{2})$$ $$s_1: \mathbb{R}^{3} \to \mathbb{R}^{6}\;;\; (x_0,x_1,x_{2}) \mapsto (x_{0},x_{1},x_{1},x_{2},x_{2},x_{2})$$ $$d_0: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\; (x_0,...x_{5})\mapsto (x_{3},x_{4},x_{5})$$ $$d_1: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\;(x_0,...x_{5})\mapsto (x_{0},x_{2},x_{5})$$ $$d_2: \mathbb{R}^{6} \to \mathbb{R}^{3}\;;\;(x_0,...x_{5})\mapsto (x_{0},x_{1},x_{3})$$ $$s_0: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{0},x_{1},x_{2},x_{0},x_{1},x_{2},x_{3},x_{4},x_{5})$$ $$s_1: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{1},x_{1},x_{2},x_{3},x_{3},x_{4},x_{3},x_{4},x_{5})$$ $$s_2: \mathbb{R}^{6} \to \mathbb{R}^{10}\;;\; (x_0,...x_{5}) \mapsto (x_{0},x_{1},x_{2},x_{2},x_{3},x_{4},x_{4},x_{5},x_{5},x_{5})$$ $$d_0: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\; (x_0,...x_{9})\mapsto (x_{4},x_{5},x_{6},x_{7},x_{8},x_{9})$$ $$d_1: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{2},x_{3},x_{7},x_{8},x_{9})$$ $$d_2: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{1},x_{3},x_{4},x_{6},x_{9})$$ $$d_3: \mathbb{R}^{10} \to \mathbb{R}^{6}\;;\;(x_0,...x_{9})\mapsto (x_{0},x_{1},x_{2},x_{4},x_{5},x_{7})$$

    Tedious, but at least it is a workable and technically easy example of a non trivial, weak 2-groupoid. We’ll see, whether or not this will work as a good teaching example.

    To get a simplicial cartesian space, we can apply the cosk functor from $3$-truncated simplicial objects to simplicial objects, but there shouldn’t be anything new in higher dimensions.

    At first lets see if this is (the truncation of) a $2$-groupoid and check the Kan-conditions: Write $\bigwedge(m,j)$ for the horns in dimension $m$, i.e for the cartesian space

    $$\lbrace \left(x_0,\ldots, \widehat{x_k},\ldots, x_n \right)\;|\; x_j\in G_m \; and \; d_i(x_j)=d_{j-1}(x_i)\; for \; i \lt j \; and \; i,j \neq k \rbrace \;.$$

    In particular

    $$\bigwedge(2,0)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_0=x_0\}$$ $$\bigwedge(2,1)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_2=x_0\}$$ $$\bigwedge(2,2)=\{((x_0,x_1,x_2),(y_0,y_1,y_2))\;;\;y_2=x_2\}$$

    Given any horn $((x_0,x_1,x_2),(x_0,y_1,y_2))\in \bigwedge(2,0)$ we can define a lifting by $(x_0,y_1,x_1,y_2,z,x_2)\in G_2$, since $d_1(x_0,y_1,x_1,y_2,z,x_2)=(x_0,x_1,x_2)$ as well as $d_2(x_0,y_1,x_1,y_2,z,x_2)=(x_0,y_1,y_2)$. As suspected the lift is not unique but is parametrized by the single $z$.

    Similar given a horn $((x_0,x_1,x_2),(y_0,y_1,x_0))\in \bigwedge(2,1)$ we can define a lifting by $(y_0,y_1,z,x_0,x_1,x_2)\in G_2$, since $d_0=(x_0,x_1,x_2)$ and $d_2=(y_0,y_1,x_0)$. If we see $((x_0,x_1,x_2),(y_0,y_1,x_0))$ as a composable pair of $1$-morphisms, a composition is given for any $z\in \mathbb{R}$ by the face $d_1(y_0,y_1,z,x_0,x_1,x_2)=(y_0,z,x_2)$. This shows, that the ’weakness’ of the composition appears here in the form of the parameter $z$.

    A similar calculation holds for $\bigwedge(2,2)$ and together this shows, that $G$ is at least not a $1$-groupoid.

    Now lets look at the Kan conditions in dimension $3$. For $G$ to be a $2$-groupoid there must be unique lifts here. Anyway the calculations are pretty easy and I just do it for $\bigwedge(3,1)$. The horn cartesian space are like

    $$\bigwedge(3,1)=\{((x_0,\ldots,x_5),(y_0,\ldots,y_5),(z_0,\ldots,z_5))\;;\;$$ $$(y_{3},y_{4},y_{5})=(x_{0},x_{2},x_{5}), (z_{3},z_{4},z_{5})=(x_{0},x_{1},x_{3}), (z_{0},z_{1},z_{3})=(y_{0},y_{1},y_{3})\}$$

    For any horn $((x_0,x_1,x_2,x_3,x_4,x_5), (y_0,y_1,y_2,x_0,x_2,x_5), (y_0,y_1,z_2,x_0,x_1,x_3)) \in \bigwedge(3,1)$ (i.e any three composable $2$-morphism) let $m:=(y_0,y_1,z_2,y_2,x_0,x_1,x_2,x_3,x_4,x_5)$. Then $d_0(m)=(x_0,x_1,x_2,x_3,x_4,x_5)$, $d_2(m)=(y_0,y_1,y_2,x_0,x_2,x_5)$ and $d_3(m)=(y_0,y_1,z_2,x_0,x_1,x_3)$, so $m$ is the unique lift of the horn and the appropriate composition is given by $d_1(m)=(y_0,z_2,y_2,x_3,x_4,x_5)$.

    This is done for the other horn space, in a similar manner. Consequently, the Kan conditions are satisfied for any relevant dimensions and if we apply the coskeletal functor we get a Kan simplicial cartesian space. Moreover the lifts are unique for $k$-morphisms if $k\gt 1$ and consequently, we have a $2$-groupoid.

    Ok. That’s it for today. Before going to differentiate, I would like to be sure, that we agree here.

    P.S.: Looks like my ’greater than’ and ’less then’ signs are not recognized here. Don’t know how to fix it…

18. • CommentRowNumber18.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 15th 2013
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexJust use \lt and \gt. So, $\lt$ and $\gt$.

    Just use \lt and \gt.

    So, $\lt$ and $\gt$.

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeJan 15th 2013
    • (edited Jan 15th 2013)
    • PermaLink
    Author: Urs
    Format: MarkdownMirco, hm, that's an unenlightning example. And I gather you are not really interested in this 2-groupoid in itself, are you? Don't you have some examples of 2-groupoids that you ran into in your research, that mean something? I thought you must have, given that you wonder about how to differentiate them. Maybe you ran into some gerby extension of some Lie groupoid somewhere? Or were looking at quotients of actions of a 2--group? Or something like this? Of course if you insist on the above example, you should do it. Is it clear how to compute the 2-group of [[bisection of a Lie groupoid|bisections]]? Then go ahead!

    Mirco, hm, that's an unenlightning example. And I gather you are not really interested in this 2-groupoid in itself, are you?

    Don't you have some examples of 2-groupoids that you ran into in your research, that mean something? I thought you must have, given that you wonder about how to differentiate them.

    Maybe you ran into some gerby extension of some Lie groupoid somewhere? Or were looking at quotients of actions of a 2--group? Or something like this?

    Of course if you insist on the above example, you should do it. Is it clear how to compute the 2-group of bisections? Then go ahead!

20. • CommentRowNumber20.
    • CommentAuthorMirco Richter
    • CommentTimeJan 15th 2013
    • (edited Jan 15th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexRight, this groupoid itself it not of much interest to me. Its advantage is, that it is so easy and that similiar kinds of $n$-groupoids can be computed for any $n \in \mathbb{N}$. However, it is a $2$-groupoid generalization of the pair groupoid and the latter is a basic example in the ordinary Lie theory of groupoids. Why I choose this one to start with, is because I want to see how the Lie $2$-algebra appears here at the end. So its more like a cartoon-version, reduced to the algebraic structure. I'm not in a hurry and as I said, I would like to do some more elaborate stuff later on. If this works out, then we get a sequence of examples with increasing complexity and I'm pretty sure, that this will be fruitful for the understanding of other people, too.

    Right, this groupoid itself it not of much interest to me. Its advantage is, that it is so easy and that similiar kinds of $n$-groupoids can be computed for any $n \in \mathbb{N}$. However, it is a $2$-groupoid generalization of the pair groupoid and the latter is a basic example in the ordinary Lie theory of groupoids.

    Why I choose this one to start with, is because I want to see how the Lie $2$-algebra appears here at the end. So its more like a cartoon-version, reduced to the algebraic structure. I’m not in a hurry and as I said, I would like to do some more elaborate stuff later on.

    If this works out, then we get a sequence of examples with increasing complexity and I’m pretty sure, that this will be fruitful for the understanding of other people, too.

21. • CommentRowNumber21.
    • CommentAuthorMirco Richter
    • CommentTimeJan 15th 2013
    • (edited Jan 15th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexThe bisection stuff is new to me. As the zero truncated object, I guess we should just take $U:=cosk^0(G)$. Then the effective epimorphism is an object in the hom Kan simplicial set $hom_{oo-grpd}(U,G)$ and this in turn is just a simplicial morphism (natr. trans. from a simpl. set into another). Is that right?

    The bisection stuff is new to me. As the zero truncated object, I guess we should just take $U:=cosk^0(G)$. Then the effective epimorphism is an object in the hom Kan simplicial set $hom_{oo-grpd}(U,G)$ and this in turn is just a simplicial morphism (natr. trans. from a simpl. set into another). Is that right?

22. • CommentRowNumber22.
    • CommentAuthorMirco Richter
    • CommentTimeJan 22nd 2013
    • (edited Jan 22nd 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItex@Urs: If you don't think the slice (oo,1)-category $\mathbf{grpd}_{/\mathbf{G}}$ (the slice over our groupoid $\mathbf{G}$) in terms of fibrant simplicial categories, then in which terms do you think of it? I thought of the above example as in the (oo,1)-category $\mathbf{grpd}$ seen as enriched over Kan simpl. sets. To apply your proposed def of bisections, I should better present the example in terms of the quasi-category picture of $\mathbf{grpd}$?

    @Urs: If you don’t think the slice (oo,1)-category $\mathbf{grpd}_{/\mathbf{G}}$ (the slice over our groupoid $\mathbf{G}$) in terms of fibrant simplicial categories, then in which terms do you think of it?

    I thought of the above example as in the (oo,1)-category $\mathbf{grpd}$ seen as enriched over Kan simpl. sets. To apply your proposed def of bisections, I should better present the example in terms of the quasi-category picture of $\mathbf{grpd}$?

23. • CommentRowNumber23.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 24th 2013
    • (edited Feb 24th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexCan someone tell a little more about bisections in the simplicial setting? Unfortunately, the entry [[bisections]] isn't that helpful to me. - In what picture of infinity-groupoids are we? - If $G$ is a Kan complex, you write $G_0 \to G$ for the required epimorphism. But what is that? $G_0$ isn't even a simplicial set. - The diagram of bisections. Below there is a reference to a map $\Phi: X \to X$, but this doesn't appear in the diagram at all. - Moreover what kind of diagram is it and what is the $\eta$ here? (A 2-morpism between simplicial sets? Then what is that?) I don't think I can calculate even a simple example, because there are so many 'silent assumptions' to overcome. Moreover looking for bisections on the web in the higher context gives nothing. But I'm pretty sure, I'm not the only one. Right now I think, that there is littarary no-one who ever did a real world calculation applying that stuff. (I DON'T speak of group(oi)s or strict higher groups,I mean a full siced weak! higher groupoid, like my example above) At least that is my impression now. Under normal circumstances I'm very effective in doing calculations. For those of you who are researching in this field: You must have already calculated some easy examples, or not? At least one of your students should have. Otherwise how can you decide, that the stuff is practically applicable at all. Could it really be the case, that there is such a huge theory without a single concrete example? Sorry,I know this is a bit trolling, but I put quite some time in this now, without any result...

    Can someone tell a little more about bisections in the simplicial setting?

    Unfortunately, the entry bisections isn’t that helpful to me.

    • In what picture of infinity-groupoids are we?

    • If $G$ is a Kan complex, you write $G_0 \to G$ for the required epimorphism. But what is that? $G_0$ isn’t even a simplicial set.

    • The diagram of bisections. Below there is a reference to a map $\Phi: X \to X$, but this doesn’t appear in the diagram at all.

    • Moreover what kind of diagram is it and what is the $\eta$ here? (A 2-morpism between simplicial sets? Then what is that?)

    I don’t think I can calculate even a simple example, because there are so many ’silent assumptions’ to overcome. Moreover looking for bisections on the web in the higher context gives nothing.

    But I’m pretty sure, I’m not the only one. Right now I think, that there is littarary no-one who ever did a real world calculation applying that stuff. (I DON’T speak of group(oi)s or strict higher groups,I mean a full siced weak! higher groupoid, like my example above)

    At least that is my impression now. Under normal circumstances I’m very effective in doing calculations.

    For those of you who are researching in this field: You must have already calculated some easy examples, or not? At least one of your students should have. Otherwise how can you decide, that the stuff is practically applicable at all.

    Could it really be the case, that there is such a huge theory without a single concrete example?

    Sorry,I know this is a bit trolling, but I put quite some time in this now, without any result…

24. • CommentRowNumber24.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 24th 2013
    • (edited Feb 24th 2013)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexIn the above example, the only real obstacle is the bisection 'oo'- group. If someone could calculate that, the rest is easy. I think it's worth to spend the time on it (mainly @URS), because after that we could write the example in te Lie differentialtion entry or so and it will guide others through that stuff, too! And I'm pretty sure that is what you want, too beause being overly abstract is the major argument against te nPOV...

    In the above example, the only real obstacle is the bisection ’oo’- group. If someone could calculate that, the rest is easy.

    I think it’s worth to spend the time on it (mainly @URS), because after that we could write the example in te Lie differentialtion entry or so and it will guide others through that stuff, too! And I’m pretty sure that is what you want, too beause being overly abstract is the major argument against te nPOV…