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12

My view is that one needs both, left and right invariant vector fields. Some reasons: For $X\in\mathfrak g=T_eG$ the left invariant vector field $L_X$ has a flow consisting of right translations: $$ \text{Fl}^{L_X}_t(x) = x.\exp(tX) $$ and conversely. For a left action $\ell:G\times M\to M$ we need the right invariant vector field $R_X$ so that $R_X\times ...


7

It is possible to define the Lie groupoid of a singular foliation and associates to it its Lie algebroid when it is smooth. This Lie algebroid satisfies the property 2. https://projecteuclid.org/download/pdf_1/euclid.jdg/1090348356 http://users.uoa.gr/~iandroul/AS-holgpd-final.pdf https://en.wikipedia.org/wiki/Lie_algebroid#...


7

These three bicategories ARE equivalent. To go from (2) to (1) it's helpful to introduce a (2)': Let (2)' have the same objects, and morphisms as (2) except each principal bundle $P$ for $G$ over $H$, is equipped with a choice of local sections of the map $P \to H_0$ (so this is equivalent to (2)). If $U$ is the cover of $H_0$ over which these local ...


5

I would like to propose an answer to this question, since 15 year ago I was asking it to myself and was thinking that orbiolds are useless. I read your question (maybe wrongly) as a question in mathematical psychology (or just in psychology). At the present moment I use orbifolds very often. And there was a turning point for me, when I understood one ...


5

I can not say why one studies orbifolds (or e.g. why one studies math at all). However, I can try the approach which might convince your funding agency: There are tons of interesting examples of how orbifolds arise in "applications" (=mathematics): The quotient spaces appearing in symplectic reduction are not always manifolds. If they fail to be manifolds, ...


5

Here is an expansion of my comment, by request. The 2-category of Lie groupoids $\mathrm{LieGpd}$ admits the category of manifolds $\mathrm{Mfld}$ as a full sub-2-category (i.e. $\mathrm{Mfld} \to \mathrm{LieGpd}$ induces an iso on hom-groupoids), but the category of Lie groups $\mathrm{LieGrp}$ only has a functor $\mathrm{LieGrp} \to \mathrm{LieGpd}$, not ...


4

The usual condition that one uses to ensure that all representations of $A$ integrate to representations of $G$ is that $G$ be "source 1-connected", meaning that all of the fibres of the source map $s : G \to M$ are connected and simply connected. That the source 1-connected condition is sufficient is a consequence of a more general theorem which asserts ...


4

Answer #1:There is no real reason for imposing that the base manifold of a groupoid be second countable. Answer #2: You lose some desirable properties if you don't impose second countability: For example, without it, the homotopy type of the geometric realisation of the nerve will no longer be an invariant of the Morita equivalence class of the groupoid.


4

Try p. 17 of notes by Breen (http://math.uchicago.edu/~may/IMA/Breen.pdf) Notes on 1- and 2-gerbes


4

For Stefan-Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample: http://users.uoa.gr/~iandroul/AZsmooth_08DEc2011.pdf Second, regarding the holonomy groupoid of a singular foliation: In fact, it turns out that it is a diffeological space, so one can do differential ...


3

Let me attempt a very simple-minded answer. Say your objects of interest are orbifolds. You have an orbifold V and you want to describe it through a groupoid, usually an action groupoid $G\ltimes M \rightrightarrows M$. This means that you have an action of $G$ on $M$ and that $V\simeq G\backslash M$, i.e. $V$ is the orbit space of the action groupoid. ...


3

There is also a rundown in the more modern book of Mackenzie: General Theory of Lie groupoids and Lie algebroids (p.26 Corollary 1.4.11) Let me summarise the idea: Fix a Lie groupoid $G \rightrightarrows M$ with source map $s$ and target map $t$. Idea: Every fibre $s^{-1} (x)$ is a closed embedded submanifold of $G$ (as $s$ is a submersion). Thus it ...


3

In the same way as groupoids $(\mathrm{src},\mathrm{trg}):X_1\rightrightarrows X_0$ are a simultaneous generalization of group actions $$(\mathrm{pr}_1,\mathrm{act}):G\times X\rightrightarrows X$$ and equivalence relations $$(\mathrm{pr}_1|_R,\mathrm{pr}_2|_R):R\rightrightarrows X\;,\qquad R\subseteq X\times X$$ on sets $X$, Lie groupoids are a ...


3

Perhaps you should look through some of the papers on Ronnie Brown's website. In particular http://www.groupoids.org.uk/pdffiles/bedlewopaper4bcclass.pdf Lie groupoids came from Ehresmann's work on fibre bundles in diff. geom. and were taken forward by Pradines and his ideas on Monodromy and Holonomy groupoids (The exact references to those are in Ronnie ...


3

Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids. Take $G=H=\mathbb{R}$. Define $F(x)=0$ if $x\leq 0$ and $F(x)=exp(−1/x^2)$ if $x>0$. The pullback is not equivalent to a Lie groupoid in this situation: the set-theoretical pullback is $(−\infty,0]\times(−\infty,0]\cup \{(x,x)|x\in \mathbb{R}\}$, which is clearly ...


3

This is more of a long comment, but since the question has been answered in the last edits I will post it as an answer. There are multiple perspectives on the stack presented by a Lie groupoid $\mathcal G$: One is the fibered category of principal bundles which you use, another one is the sheafification of the presheaf of groupoids which sends $X$ to the ...


3

I should update with a mention of some of my own results in http://arxiv.org/abs/1504.02394: There is a proof of Segal's theorem that the classifying space $B\Gamma^q$ of Haefliger's foliation groupoid is homotopy equivalent to classifying space of the discrete monoid of embeddings of $\mathbb{R}^n$ into itself $B\mathbf{Emb}\left(\mathbb{R}^n\right)$ ...


3

Write $F$ your foliation, $M$ its ambiant manifold, $q=dim(M)-dim(F)$ its codimension. The holonomy groupoid $H(F)$, if I'm correct, is the set of classes of triples $(x,\gamma,y)$ where $\gamma$ is a tangential path; and $(x,\gamma,y)~(x,\gamma',y)$ iff $\gamma$ and $\gamma'$ have the same holonomy. I'm not expert enough in groupoids to use the proper ...


3

Let me state a simple observation regarding the fibre of the 2-functor that arose in a conversation with Rui Loja Fernandes. A symplectic groupoid $(\Sigma, \omega) \Rightarrow P$ induces a Poisson bivector $\pi$ on $P$, which is completely determined by the fact that the source and target maps are Poisson and anti-Poisson respectively (beware of the fact ...


3

I'm being thick, but surely you can topologize the space of arrows as a subspace of $M\times M$ where $M$ is the space of objects? It may be a horrible space, but it exists. As far as naming the thing goes, I would call it a presentation of the stack-replacement of the space of leaves or similar. If $L = M/\sim$ is the space of leaves (which admittedly is '...


2

I do not exactly understand your problem but it is possible that some developments and expositions of other ideas of Pradines might help, namely R. Brown and M. E.-S. A.-F. Aof, ``The holonomy groupoid of a locally topological groupoid'', Top. and its Appl., 47 (1992) 97-113. R. Brown and O. Mucuk, ``The monodromy groupoid of a Lie groupoid'', Cah. ...


2

I answer this in http://arxiv.org/abs/1212.2282. An etale stack $\mathscr{X}$ is effective if and only if the substack assigning each manifold $M$ the groupoid of local diffeomorphisms $$M \to \mathscr{X}$$ is actually a sheaf, i.e. if and only if this groupoid is (equivalent to) a set.


2

When one studies equivariant geometry, then it is sometimes the case that one can pass to a smaller space with the action of a smaller group, and everything equivariant you can compute will be the same. In the case that the quotient by the original group action exists, then this just reduces the space to the non-equivariant setting, namely equivariant for ...


2

Think of $BG$ and $BH$ as topological stacks, whereby one can calculate a topological groupoid presenting the stack $BG\times_{BH} BG$, namely the following: the object space is the space underlying $H$ and the morphism space is $G\times H \times G$. The source map $s\colon G\times H \times G \to H$ is the projection on the middle factor; the target map $t\...


2

Edit I just realised your misconception: the source and target maps are automatically surjective since they both have a section, namely the unit map. So asking that they are surjective submersions or just submersions are equivalent. This is the only place where I saw this kind of requirement. It's been literally the definition since the 1980s. If the ...


1

I will give an example as to why composition of bibundles cannot simply be done as pullback, as well as the relation to perhaps more familiar geometric constructions. Firstly, note that if $M$ is a manifold and $\mathbf{B}G$ is a one-object Lie groupoid (where the automorphisms of the one object are the Lie group $G$), a bibundle from $M$ to $\mathbf{B}G$ ...


1

As I mentioned in my other answer, Lie groupoids give rise to stacks, and the stack that a Lie groupoid $X$ gives rise to is the stack of principal $X$-bundles. Then in analogy with the Eilenberg-Watts theorem in the theory of rings, a map of such stacks is given by a bibundle. Eilenberg-Watts says that a suitable functor of module categories $Mod_R \to ...


1

In the following I will use the notation you cited in the problem statement. The point of the proof should be that you want to construct an open neighborhood $N_x$ which is stable under the $\delta$-action of the stabiliser group $G_x$ and (here comes the crucical property!) that $$ s^{-1} (N_x) = \coprod_{\gamma \in G_x} \underbrace{W_\gamma \cap s^{-1} (...


1

If the base manifold M is connected, and A is such a Lie algebroid then the only orbit of A is M itself. It turns out that one can always find a Lie groupoid (called the Weinstein groupoid by Crainic & Fernandes) integrating A and which has the same orbits as A, i.e. which is transitive too. Ref.: Lectures of Integrability of Lie Brackets, Crainic & ...


1

You may find it useful to look at Lectures on the Integrability of Lie Brackets by the same authors. I believe it takes a slower approach to the subject you seem to be reading about. As for the question, from the definition of a Lie algebroid morphism, check that a vector bundle map $F:TI \to A$ covering $\gamma: I \to M$ is equivalent to a path in the ...


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