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If you ever wanted to construct the tangent bundle of a differentiable stack, it's relatively simple:

First, if $\mathbf{X}$ is a stack coming from a Lie groupoid $\mathcal{G}$, you could just say $\mathbf{TX}$ is the stack associated to the tangent groupoid $\mathcal{TG}$.

A more formal way is the following:

Consider the composite

$\text{Mfd} \stackrel{T}{\rightarrow} \text{VectorBundles} \stackrel{\text{forget}}{\rightarrow} \text{Mfd} \stackrel{\text{yoneda}}{\rightarrow} \text{St}(\text{Mfd}).$

by restricting its weak left Kan extension to stacks we get a 2-functor (by abuse of notation)

$\text{St}(\text{Mfd}) \stackrel{T}{\rightarrow} \text{St}(\text{Mfd}).$

Now, suppose $\mathbf{X}$ is a stack coming from a Lie groupoid $\mathcal{G}$. Then $\mathbf{X}$ is the weak colimit of the composition

$\Delta^{\rm op} \stackrel{N(\mathcal{G})}{\rightarrow}Mfd\stackrel{\text{yoneda}}{\rightarrow}\text{St}(\text{Mfd})$

(in fact even of its 2-truncation.)

Since $T$ is weak-colimit preserving, it follows that $T \mathbf{X}$ is the weak colimit of

$\Delta^{\rm op} \stackrel{N(\mathcal{TG})}{\rightarrow}\text{Mfd}\stackrel{\text{yoneda}}{\rightarrow}\text{St}(\text{Mfd})$

which is in turn just the stack associated to the tangent groupoid $\mathcal{TG}$. So both definitions agree.

Now suppose you wanted to define the cotangent stack. The first line of attack, done naively, seems to fail. The cotangent functor $T^*$ is contravariant so it doesn't send groupoid objects to groupoid objects. However, less naively, in the literature (for instance here:, one can define the cotangent groupoid of $\mathcal{G}$ to be a certain symplectic groupoid $\mathcal{T^{*}G}_1 \rightrightarrows \text{Lie}(\mathcal{G})^{*}$, where $\text{Lie}(\mathcal{G})^{*}$ is the dual Lie algebroid of $\mathcal{G}$. (Is this invariant under Morita equivalence?)

The second way would seem to be:

Consider the opposite of the composite

$\text{Mfd}^{\rm op} \stackrel{T^*}{\rightarrow} \text{VectorBundles} \stackrel{\text{forget}}{\rightarrow} \text{Mfd} \stackrel{\text{yoneda}}{\rightarrow} \text{St}(\text{Mfd}),$

$\text{Mfd} \to \text{St}(\text{Mfd})^{\rm op}$

and take its weak left Kan extension. (So $\mathbf{T^{*}X}$ is the weak limit of $T^{*} M$ over all $M \to \mathbf{X}$.)

The opposite of this 2-functor goes $\text{St}(\text{Mfd})^{\rm op} \to \text{St}(\text{Mfd})$.

How do these two notions of cotangent stack relate to each other? Is either reasonable? Is there a better notion than either of these?

I would like, for instance, if you have a Riemannian metric on an orbifold, to get an equivalence between its tangent stack and its cotangent stack.

Along these lines, the underlying space of the cotangent bundle of an orbifold is an orbifold (or is it a manifold, like its frame bundle?), so it should be represented by a orbifold groupoid. Is this the same as the cotangent groupoid?

Keep in mind, the correct definition of tangent stack should produce the correct notion of differentiable forms on a stack, and the cotangent stack should somehow have a symplectic structure.

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I've cleaned up some TeX. The main trick: jsMath and Markdown have different ideas about special characters, so you should always protect TeX on MathOverflow by backticks (see "How to write math" on the sidebar for details for how this works). I also moved most of things to "text" mode, because TeX formats spacing differently for $Mfd$ and $\text{Mfd}$. But feel free to roll back whatever I messed up. (In particular, I forget that jsMath gets it wrong when asked to do text-mode in scriptstyle, so the words "yoneda" are too large right now, sorry.) – Theo Johnson-Freyd May 13 '10 at 19:14
Oh, +1 for anything on Lie groupoids :) – Theo Johnson-Freyd May 13 '10 at 19:14
Thanks for fixing my Latex Theo. I didn't know why it wasn't showing up correctly... – David Carchedi May 13 '10 at 21:59
I wanted to ask you today, but then the elevator door separated us :-): could you say what you believe you NEED the cotangent bundle for? Because I have this suspicion that whatever that is, it may have a good answer that does not involve a stacky cotangent bundle explicitly. – Urs Schreiber May 14 '10 at 19:23
@Urs: I don't explicitly NEED it right now. I just noticed while discussing with Ivan that defining the cotangent stack wasn't as easy as its tangent-brother. But, one thing it would be useful for would be to define differentiable forms, and another is that we'd expect cotangent stacks to be "symplectic stacks". – David Carchedi May 15 '10 at 16:51

This is a question near and dear to my heart, since it is actually what my thesis will be about. Let me denote the Lie algebroid of a groupoid $G$ with base $M$ by $A$.

If you're viewing the stack presented by $G$ as a smooth model for the quotient space $M/G$, then the first proposal doesn't really seem to give the right answer. Namely, the Lie algebroid structure on $A$ induces a Poisson structure on $A^*$, and the quotient space of the groupoid $T^*G$ you mention is just the space of symplectic leaves for this Poisson structure --- this space doesn't give the cotangent bundle of $M/G$ and doesn't have a natural symplectic structure of its own. (Of course, these quotients aren't necessarily smooth, but even if they are you don't get the right answer.) Indeed, if you start with a trivial groupoid $M$, the space you get is just $M$.

Probably the underlying question you first have to answer is: in what sense is $T^*$ a functor? As Andre says, if all you care about is etale maps there's no problem. In general, you still get something interesting if you change the target category for the $T^*$ functor to the so-called symplectic category, where objects are symplectic manifolds, and a morphism $M \to N$ is a lagrangian submanifold of $\overline{M} \times N$, where the bar denotes changing the sign of the symplectic structure. (The usual caveat applies: this is not really a category since compositions are not always well-defined, but we'll just ignore this for now.) Composition is just the usual composition of relations.

Now to a smooth map $f: X \to Y$ you can assign the lagrangian submanifold

$T^*f := \{ (x, df_x^*\xi, f(x), \xi) \} \subseteq \overline{T^*X} \times T^*Y$,

where $df$ is the tangent map. This then gives a nice functor from the category of smooth manifolds to the symplectic category. In particular, if $f: X \to Y$ is a diffeomorphism (or even just etale), then $T^*f$ is in fact the graph of an actual map.

The upshot is that if you start with a groupoid, you can apply this functor to everything in sight to get a "groupoid" in the symplectic category. (This idea is due to Alan Weinstein. I'm calling this a "groupoid" since it is not clear in what sense this object is actually a groupoid. In particular, the symplectic category doesn't have fiber products, so the notion of a groupoid object isn't defined.) If your groupoid is etale, this recovers Andre's suggestion. So, you can instead try to formulate the notion of the cotangent stack of a stack in terms of this object. This post has gone on long enough, so I won't go into details here, but numerous examples suggest that this is indeed the right thing (or at least "a" right thing) to consider. In particular, from this point of view the "symplecticness" of cotangent stacks comes from the fact that they live in the symplectic category (this can be made more precise).

I should also say that this construction is somehow related to the groupoid structure on $T^*G$ you mentioned, but it contains more information that just that.

Edit: Here's one example illustrating some of the above. Say $G$ is a group acting smoothly on $M$. Then the action lifts to a Hamiltonian action on $T^*M$ with momentum map $\mu$. When the action is free, the resulting symplectic quotient $\mu^{-1}(0)/G$ is exactly the cotangent bundle of $M/G$. Even when the action is not free, it makes sense to call the stack $[\mu^{-1}(0)/G]$ the cotangent stack of $[M/G]$. The "groupoid" resulting by performing the above construction on the action groupoid $G \times M$ indeed encodes this quotient.

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Thanks for your answer Santiago. What you say about using the symplectic category is indeed very interesting. (You can probably fix the no groupoid object problem by using micro-geometry). However, I'm not sure how to fit this with the stacky picture in a way that does what I want. However, your example in your edit seems to give me some hope. I'll look into it. Thanks! – David Carchedi May 14 '10 at 8:03

This here may or may not be what you need, but maybe it deserves to be mentioned anyway:

there exists a very general abstract nonsense about what "cotangent bundles" are in great generality.

The story begins with the notion of Kähler differentials:

which provide a general recipe for obtaining the cotangent bundles of spaces that are formally dual to algebraic structures. A half-way generalization of this to higher categorical contexts is the notion of cotangent complex

This derives the cotangent complex functor and hence gives a recipe for constructing the "cotangent bundle" of "derived spaces", say whose function algebras are simplicial algebras.

But the thing is that this construction is still a special case of something much more general:

one observes that the assignment of the module of Kähler differentials to a ring is a left adjoint to one of the two forgetful functors $Mod \to CRing$. This fibration may be thought of as the fibered category incarnation of the stack of "algebraic vector bundles" (quasi-coherent sheaves, really). It is an old observation due to Quillen, that this fibration has a more abstract definition: it is the "tangent category" over $CRing$ in the sense described in detail here: .

Now, this definition of "tangent category" has a very elegant generalization to higher category theory: the "tangent (oo,1)-category"-construction

This achieves the following: for ANY (locally presentable) (oo,1)-category $C$ whose objects you want to think of as generalized spaces (for instance stacks) it provides a notion of the fibration of "generalized vector bundles" over these spaces. And the left adjoint to this fibration is the generalized cotangent complex assignment: it sends each object of $C$ to something that plays the role of the cotangent complex over that $C$.

This is developed in Jacob Lurie's "Deformation Theory"

There it is applied to figuring out what "vector bundles" and cotangent bundles etc over spaces are that are formal duals of E-infinity ring spectra.

But the construction is entirely general. In particular you could feed in the locally presentable (oo,1)-category of stacks on some site.

Or, possibly better suited for what you might need, one could extend it from a given site to the stacks on that site.

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Thanks Urs, I'll take a look at this when I get back from London. – David Carchedi May 15 '10 at 16:53

If your stack comes from an etale groupoid, then you can use the same construction as for the tangent bundle. That's because the cotangent bundle behaves covariantly w.r.t. etale morphisms.

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Thanks, that's a goood point Andre. – David Carchedi May 13 '10 at 21:43

Here is a second answer, given that in the secondary comments you now said what you are interested in might be not so much the cotangent stack itself, as the definition of differential forms on a stack. From the classical theory one might expect that knowing one means knowing the other, but maybe for oo-stacks the two notions decouple a bit more.

There are useful ways to express differential forms on a stack without (explicitly?) realizing them as sections of a stacky cotangent bundle.

The two main approaches are this:

  1. Via de Rham stacks

    One can form the "de Rham stack" of an (oo-)stack, the higher analog of the de Rham space

    An insightful discussion of what this means for oo-stacks on algebraic sites is in

    Carlos Simpson and Constantin Teleman, "de Rham theorem for oo-stacks"

    I think their construction has a good analog for more general sites, such as the site of C^oo-rings which one can use to model differentiable oo-stacks. I talk about this in the context of the "infinitesimal path oo-groupoid" here

    with more details on how it relates to differential forms on oo-stacks here:

    Effectivelly what is happening here is the following: given an oo-groupoid/oo-stack, one forms the oo-groupoid/oo-stack of infintiesimal paths in it. Then functions on this infinitesimal path oo-groupoid are differential forms.

    a very explicit discussion of how this works for ordinary manifolds is here

    The generalization of this to oo-stacks is essentially nothing but the (derived) Yoneda extension.

  2. via derived loop spaces

    Another route that one can go is to pass from a site of just algebra or C^oo-algebras to one of simplicial algebras / simplicial C^oo-algebras. This leads to some serious magic: when you compute the abstract free loop space object of an ordinary scheme of C^oo-scheme (the latter including smooth manifolds) inside the oo-topos of oo-stacks on such a "derived" site, the resulting "derived loop space" object behaves effectively as the space of infinitesimal loops of the original scheme. Functions on this loop space object encode thhe Hochschild homology of the function algebra of the (C^oo)-scheme. Under mild conditions, this is effectively the algebra of differential forms.

    The perspective of differential forms on oo-stacks as functions on derived loop space objects is discussed very nicely in

    David Ben-Zvi, David Nadler, "Loop spaces and connections",

    Using the kind of computations on simplicial C^oo-rings that Herman Stel sent around recently in Utrecht, one can generalize these constructions from plain simplicial algebras to simplicial C^oo-rings and hence adapt them to the smooth framework, which also accomodates differentiable stacks.

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