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Recall the notion of Lie algebroid (n Lab, Wikipedia). One motivation for studying Lie algebroids is that they are infinitesimal versions of Lie groupoids, and Lie groupoids present stacks. In particular, Lie groupoids are the objects of a 2-category whose 1-morphisms are (left-principal) biactions of groupoids, and whose 2-morphisms are smooth biequivariant maps thereof. A "smooth stack" is then a Lie groupoid up to equivalence in this category.

I would like to understand the infinitesimal version of this category. I presume that this would require understanding what are:

  • smooth left- and right-actions of Lie algebroids,
  • when a smooth left action by one Lie algebroid on a space and a smooth right action by another Lie algebroid on the same space commute,
  • when a Lie algebroid action is principal.

Hence the question in the title. In particular, I'd like some intuition for whether two Lie algebroids are equivalent. As a consistency check, there should be a functor from the 2-category of groupoids to the 2-category of algebroids that on objects takes a groupoid to its tangent algebroid.

Bonus points: there is a wonderful description of the 1-category whose objects are Lie algebroids as the full subcategory of the category of dg manifolds (morphisms are graded smooth maps that relate the homological vector fields) on the dg manifolds generated in degrees $0$ and $1$. Is there a similar "dg" description of the 2-category?

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Good question! I assume this should be related to the notion of Morita equivalence for Lie algebroids, which you get by taking the infinitesimal version of Morita equivalence for groupoids. The problem, if I recall correctly, is that whereas the definition for groupoids via bibundles is equivalent to the definition via spans of groupoids, the infinitesimal versions are not necessarily the same for Lie algebroids; or more precisely, they are the same for integrable Lie algebroids but not for non-integrable ones. – Santiago Canez Jul 30 '10 at 22:04
@Santiago: Oh, interesting! Can you say more, and/or point me in the right direction? I would certainly accept an answer that spelled out your comment with definitions. – Theo Johnson-Freyd Jul 31 '10 at 0:04
@Santiago: Do you have a reference to this claim about integral/non-integrable Lie algebroids? – David Carchedi Jul 31 '10 at 2:12
@Santiago: Actually, I really want to you say more. I learned at that the 2-category of spans in a 1-category has no more equivalences than the 1-category has isomorphisms. So the 2-category of "spans of groupoids" must be using that groupoids form a 2-category in a naive way: smooth functors and smooth natural transformations. But I don't see how to define "natural transformations of morphisms of Lie algebroids". So the only definition of "spans of Lie algebroids" I can write down is boring. – Theo Johnson-Freyd Jul 31 '10 at 2:48
@Santiago: I've posted below a definition of the "span picture" of the 2-category of Lie algebroids (@Theo: no natural transformations!). You can say more precisely what the infinitesimal version of the "bibundle picture" is? – Konrad Waldorf Jul 31 '10 at 4:37

4 Answers 4

up vote 6 down vote accepted

Here is one answer, but also sort of a no-go observation.

We are interested in creating a bicategory whose objects are Lie algebroids. The 1-morphisms and 2-morphisms are something yet to be determined. We have a couple simple requirements that we are going to demand:

  1. We want a functor from the bicategory of Lie groupoids and bibundles to this bicategory.
  2. This functor should take a Lie groupoid to its underlying Lie algebroid.
  3. The naive notion of morphism of Lie algebroid gives rise to a 1-morphism in this bicateogry, and isomorphism become equivalences.

Now I will describe a bicategory which minimally satisfies these conditions, but I don't think that this bicategory of Lie algebroids is going to be very interesting. The reason is that the above requirements force you to identify too many things.

For example take any space X and any cover U of the space. We can form two groupoids out of this which are Morita equivalent. These are X, viewed as a groupoid with just identities and the Cech groupoid $U \times_X U \rightrightarrows U$. These are equivalent groupoids in the bibundle bicategory and so must be sent to equivalent Lie algebroids. (In a certain sense the bibundle bicategory is what you get when you take Lie groupoids, functors, etc. and force these two types of groupoids to be equivalent. More on this below.)

However their Lie algebroids are very simple and appear very different. They are just the trivial Lie algebroids over X and U, respectively. In particular this second one has no information about the fiber products $$U \times_X U.$$ This means that the tangent bundle of X and the tangent bundle of U must be equivalent objects in this hypothetical bicategory of Lie algebroids, whenever U covers X.

Following through with similar examples allows us to see that any time we pull a Lie algebroid back by a cover of its base we get an "equivalent" Lie algebroid. In particular this means that every Lie algebroid will be equivalent to one on a trivial bundle.

Maybe I am wrong and this is still an interesting bicategory, but it feels like we're loosing too much information. In any event this suggests what your hypothetical bicategory of Lie algebroids actually looks like.

Let's set up some terminology first. Suppose I have a Lie algebroid with base space X. Suppose further that I have a surjective submersion of the base $U \to X$. Then I can pull-back the Lie algebroid on X to one on U. We will call the morphism of Lie algebroids from U to X a *weak equivalence".

So the bicategory you are after should have object Lie algebroids and the morphisms should be spans of Lie algebroids $$X \stackrel{\sim}{\leftarrow} U \rightarrow Y$$ where $U \to X$ is a weak-equivalence.

The 2-morphisms are not just naive morphisms of spans. They are more complicated. This is why your previous MO question doesn't apply. What we are doing is inverting the weak equivalences to make them, well, equivalences. We want to do it in such a way that we still have a bicategory and that it has the obvious universal property (functors out of it are the same as weak equivalence inverting functors out of Lie algebroids). This is sometimes called the derived localization.

Fortunately there is some systematic machinery to accomplish this, devoloped by Dorette Pronk. Some of it is described at the nlab, but the full story is in her paper "Etendues and stacks as bicategories of fractions". Among other things, this paper contains a description of the bicategory of Lie groupoids and bibundles as an example of this sort of derived localization, and so that is a good example to compare with.

From this description it is also clear that the derived localization of Lie algebroids along the weak equivalences is going to be the universal (initial) thing which satisfies the three properties outlined above.

The question remains though: is this an interesting bicategory? I don't have an answer for this. Perhaps you have a use for it?

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Chris, I am admittedly very slow in understanding your answer! Can you explain why the Lie algebroid of the Cech of the cover U is TU? I just see that TU receives an injective map from that Lie algebroid. – Konrad Waldorf Aug 2 '10 at 6:48
Konrad, you're right! That was a mistake. I corrected it to "trivial" Lie algebroid since in that example you get the zero Lie algebroid. Here is a link if people need a reminder about the construction:… . I don't think this effects the rest of the answer significantly. The "pull-back" Lie algebroid is $A \times_{TX} TU$, where we use the map $df:TU \to TX$. This is still a vector bundle since $U \to X$ is a surjective submersion. – Chris Schommer-Pries Aug 2 '10 at 12:19
Thanks for the verification! And the Cech groupoid of a general surjective submersion? For example the one of a principal $G$-bundle $pi:P\to X$. Isn't the Lie algebroid here the vertical tangent space? What I want to say is that you probably get trivial effects when you have a covering with discrete fibres, but can have non-trivial ones when the fibres have dimensions greater than zero. – Konrad Waldorf Aug 2 '10 at 15:45
Right. Let's see, the pair groupoid of M gets associated to the Lie algebroid over M given by TM with anchor map the identity. Since the pair groupoid is equivalent to the point, this Lie algebroid is equivalent to the zero Lie algebroid. This implies, for example, that if our cover is a trivial bundle $M \times X \to X$, and we pull-back the Lie algebroid A over X to $X \times M$, then we get $A \oplus TM$ with the obvious anchor map to $TX \oplus TM$. These will also be equivalent. In fact I think the formula $A \times_{TX} TU$ holds generally for any surjective submersion $U \to X$. – Chris Schommer-Pries Aug 2 '10 at 19:43
This is great, and ultimately I think the correct answer. I realized that I had in my mind two or three functors, and not all of them could possibly be functors. Namely, I had imagined that there was (1) a functor from the bicategory of Lie groupoids to some bicategory of Lie algebroids that on objects took a groupoid to its tangent algebroid; (2) a functor from the bicategory of Lie algebroids to differential chain complexes that on objects took a Lie algebroid to its BRST complex; (3) had a (n?) functor from chain complexes to graded vector spaces taking homology. But (continued) – Theo Johnson-Freyd Aug 3 '10 at 20:49

I'm ashamed to give the following abstract nonsense answer to this excellent question. Also, this is just an expansion of Santiago's first comment above.

We'll have to talk about morphisms between Lie algebroids over different base manifolds (these are, for instance, the objects of the Lie groupoids they come from). A morphism between Lie algebroids is a linear, bracket-preserving and anchor-preserving smooth map between the total spaces covering a smooth map between the base manifolds. An isomorphism would be composed of two diffeomorphisms.

I am working in the "span picture" and not in the "bibundle picture". I define what a Span of Lie algebroids is, yielding the 1-morphisms of the 2-category Theo is asking for. Let $E_1$ and $E_2$ be Lie algebroids with base manifolds $M_1$ and $M_2$, respectively. A span $E_1 \to E_2$ is a third Lie algebroid $A$ over some base manifold $B$ together with a Lie algebroid isomorphism $A \to E_1$ covering a diffeomorphism $B \to M_1$ and a Lie algebroid morphism $A \to E_2$ covering a smooth map $B \to M_2$. The 2-morphisms are just Lie algebroid morphisms between these "correspondence Lie algebroids". Without having checked all the details, it looks pretty clear to me that this forms a 2-category, with the composition of 1-morphisms the ususal "span composition".

EDIT: One can also make up a version with undirected spans by simply ommiting the condition of being an isomorphism.

Of course, this doesn't answer any of Theo's interesting "$\bullet$"-questions, but principally one could now possibly translate this "span picture" into a "bialgebroid-picture".

Finally, the consistency check: recall that there is a functor from Lie groupoids + smooth functors to Lie algebroids + algebroid morphisms, the latter in the above sense. Applying this functor to spans of Lie groupoids produces immediately a span of Lie algebroids, both in the directed and undirected version.

EDIT II: as Theo points out below, that's not totally correct because this functor does not send weak equivalences between Lie groupoids to isomorphisms between Lie algebroids.

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This certainly defines a 2-category, but as I mentioned in the comments, it will not lead to non-isomorphic equivalent Lie algebroids, and in particular will not accept a 2-functor from the bibundle category of groupoids that on objects takes a groupoid to its tangent-along-the-identity-section algebroid. So you have defined a 2-category whose 0-objects are Lie algebroids, but not the 2-category whose 0-objects are Lie algebroids :) – Theo Johnson-Freyd Jul 31 '10 at 13:21
Ah, interesting! Right, I must not require that one of the Lie algebroid morphisms is an iso. So the actual question is: if F is a smooth essentially surjective full and faithful functor between Lie groupoids, which properties characterize the induced map on Lie algebroids? – Konrad Waldorf Jul 31 '10 at 18:08
I posted an answer that builds on Konrad's here. Basically you want to replace "isomorphism" with a certain class of Lie algebroid homomorphisms. These homomorphisms come from Morita equivalences of Lie groupoids, and the bicategory you get is a variation on the bicategory of spans. It formally inverts these special homomorphisms. – Chris Schommer-Pries Aug 1 '10 at 16:08

Chris above explained a "no go" result: there is not a good 2-category whose objects are Lie algebroids, where by "good" I mean that it receives a functor from the bibundle category of Lie groupoids but doesn't make too many things equivalent. In this answer, I will outline something that is not a 2-category, but some sort of infinitesimal version thereof. It doesn't capture the correct notion of bibundle, so isn't quite what I'm looking for; maybe someone will see this and comment.

First, consider groupoids. There is a (strict) 2-category whose objects are Lie groupoids, 1-morphisms are smooth functors of Lie groupoids, and 2-morphisms are smooth natural transformations. In fact, this category is enriched in (generally infinite-dimensional) Lie groupoids. For example, given two groups $A,B$, thought of as groupoids, $B$ acts on $\operatorname{Hom}(A\to B)$ by conjugation, and the corresponding action groupoid is the groupoid of functors-and-natural-transformations $A\to B$. The point, obvious for groups, is that when $A,B$ are smooth, this action groupoid can be given a smooth structure.

Applying the Lie functor to just the hom groupoids, we get a category enriched in (infinite-dimensional) Lie algebroids with objects Lie groupoids.

Similarly, I expect that there is a category whose objects are Lie algebroids and so that between any two Lie algebroids there is a "hom Lie algebroid". Beyond "category enriched in Lie algebroids", I don't know what to call such an object. "2-infinitesimal smooth 2-category"?

Recall that Lie algebroids (and the morphisms between them) are the same as "dg" or "Q" (Z-graded super)manifolds generated in degrees 0 and 1 (a full subcategory of all dg manifolds). It's probably obvious to the experts, although not to me, that the category of (infinite-dimensional) Q-manifolds can be enriched over itself so that it is Cartesian-closed. Then perhaps it's a fact that the Q-manifold between two things generated in degrees 0 and 1 is also generated in degrees 0 and 1.

Alternately, there is probably a direct description.

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I stumbled on your question because I was asking myself the same. I have seen the following online. It does not formally directly address you question, but if you combine it with this should give an answer.

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Unfortunately, the papers make a rather strong assumption, namely that the Lie algebroids be integrable. I think the questioner asked for a stand-alone category of Lie algebroids not necessarily integrable (it seems the idea to isolate the natural Lie Poisson structure on the duals of algebroids is a step in the right direction). – santker heboln Jul 26 at 21:57

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