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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeFeb 15th 2013
   • (edited May 3rd 2014)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSome comments on the implementation of [Atiyah- and Courant Lie n-algebroids](http://ncatlab.org/nlab/show/Courant+algebroid#RelationToAtiyahGroupoids) in [[cohesive homotopy type theory]], and some general thoughts on "0-strict" $\infty$-groupoids ($\infty$-groupoids equipped with a 1-epimorphism out of a 0-truncated object). Let $\mathbf{H}$ be some [[cohesive (infinity,1)-topos]]. The [[homotopy Lie-Rinehart pair]]-perspective on an $\infty$-groupoid $\mathcal{X} \in \mathbf{H}$ which is equipped with a [[1-epimorphism]] $\phi \colon X \to \mathcal{X}$ is to consider the pair consisting of $X$ and of the [[infinity-group]] [[group of bisections|of bisections]] $$ BiSect(\mathcal{X},X) \coloneqq \mathbf{Aut}_{\mathcal{X}}(X) \coloneqq \underset{\mathcal{X}}{\prod} Equiv_{/\mathcal{X}}(X)_ \,. $$ (Traditionally in a (homtopy-)[[Lie-Rinehart algebra]] of course one only remembers the [[L-infinity algebra]] $Lie( \mathbf{Aut}_{\mathcal{X}}(X) )$ obtained from this under [[Lie differentiation]], but here I will stick to the complete Lie integrated picture.) Now, it turns out that famous homotopy Lie-Rinehart algebras out there are constructed (secretly, but one can see that this is what happens) by starting with a map $$ \chi \colon X \to \mathbf{F} $$ to some [[moduli infinity-stack]] $\mathbf{F}$ and then taking the group of bisections to be the automorphism group of this $\chi$ over $\mathbf{F}$. For instance * the [[Atiyah Lie algebroid]] assigned to a circle principal bundle modulated by $\nabla^0 \colon X \to \mathbf{B}U(1)$ is the Lie differentiaton of $(X, \mathbf{Aut}_{\mathbf{B}U(1)}(\nabla^0))$; * the [[Courant Lie 2-algebroid]] assigned to a map $\nabla^1 \colon X \to \mathbf{B}^2 U(1)_{conn^1}$ modulating a "bundle gerbe with connective data but no curving" is the Lie differentiation of $(X, conc \mathbf{Aut}_{\mathbf{B}^2 U(1)_{conn^1}}(\nabla^1))$. (here $conc$ stands for "differential concretification", a technical subtlety related to the right cohesive structure on these objects, which for the purpose of the present discussion one should ignore) Now given such an "integrated homotopy Lie-Rinehart pair" consisting of an object $X$ and an automorphism $\infty$-group of a map $\chi : X \to \mathbf{F}$ over $\mathbf{F}$, can we canonically find for it the corresponding $\infty$-groupoid, hence the $\mathcal{X} \in \mathbf{H}$ such that 1. $X \to \mathcal{X}$ is a 1-epimorphism; 1. $\mathbf{Aut}_{\mathbf{F}}(\chi) \simeq \mathbf{Aut}_{\mathcal{X}}(X)$ ? Well, that's just the [[1-image]] of $\chi$: $$ \mathcal{X} \simeq im_1(\chi) $$ isn't it? Let's look at the first example in the above series: Let $\mathbf{H} = $ [[SmoothGrpd]], let $X \in SmthMfd \hookrightarrow \mathbf{H}$ be a smooth manifold, and let $\nabla^0 \colon X \to \mathbf{B}U(1)$ be the map modulating a circle principal bundle $P \to X$. Then what is $$ X \to \mathcal{X}\coloneqq im_1(\nabla^0) \,. $$ To check this, remember that the 1-image of a map may be computed as the homotopy colimit over its homotopy Cech nerve. Doing so here, we find that $$ im_1(\nabla^0) \simeq ( P \times_{U(1)} P \stackrel{\to}{\to} X) \,. $$ This is the [[Lie integration]] of the [[Atiyah Lie algebroid]] of the circle bundle $P \to X$ modulated by $\nabla^0$.

   Some comments on the implementation of Atiyah- and Courant Lie n-algebroids in cohesive homotopy type theory, and some general thoughts on “0-strict” -groupoids (-groupoids equipped with a 1-epimorphism out of a 0-truncated object).

   Let be some cohesive (infinity,1)-topos.

   The homotopy Lie-Rinehart pair-perspective on an -groupoid which is equipped with a 1-epimorphism is to consider the pair consisting of and of the infinity-group of bisections

   (Traditionally in a (homtopy-)Lie-Rinehart algebra of course one only remembers the L-infinity algebra obtained from this under Lie differentiation, but here I will stick to the complete Lie integrated picture.)

   Now, it turns out that famous homotopy Lie-Rinehart algebras out there are constructed (secretly, but one can see that this is what happens) by starting with a map

   to some moduli infinity-stack and then taking the group of bisections to be the automorphism group of this over .

   For instance

   • the Atiyah Lie algebroid assigned to a circle principal bundle modulated by is the Lie differentiaton of ;

   • the Courant Lie 2-algebroid assigned to a map modulating a “bundle gerbe with connective data but no curving” is the Lie differentiation of .

     (here stands for “differential concretification”, a technical subtlety related to the right cohesive structure on these objects, which for the purpose of the present discussion one should ignore)

   Now given such an “integrated homotopy Lie-Rinehart pair” consisting of an object and an automorphism -group of a map over , can we canonically find for it the corresponding -groupoid, hence the such that

   1. is a 1-epimorphism;

   ?

   Well, that’s just the 1-image of :

   isn’t it?

   Let’s look at the first example in the above series:

   Let SmoothGrpd, let be a smooth manifold, and let be the map modulating a circle principal bundle .

   Then what is

   To check this, remember that the 1-image of a map may be computed as the homotopy colimit over its homotopy Cech nerve. Doing so here, we find that

   This is the Lie integration of the Atiyah Lie algebroid of the circle bundle modulated by .

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeFeb 19th 2013
   • (edited Feb 20th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have now started to write out a bunch of details along the above lines in a new entry * _[[higher Atiyah groupoid]]_ (Not proof-read yet, as it is getting too late for me now. Will fix typos and furher refine tomorrow...)

   I have now started to write out a bunch of details along the above lines in a new entry

   • higher Atiyah groupoid

   (Not proof-read yet, as it is getting too late for me now. Will fix typos and furher refine tomorrow…)