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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeJan 30th 2012
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexThis is a partial continuation of a conversation between me and David Roberts about path spaces of Lie groupoids. The conversation has so far taken place by email and MathOverflow comments, but was getting unwieldy for either so we're moving it here. As always with nForum discussions, anyone else is welcome to join in. There was a MathOverflow question: <http://mathoverflow.net/questions/85960/induced-map-on-path-manifolds-is-it-a-submersion> which really had two parts. One was about defining a manifold of suitable paths that aren't quite the usual paths, the other was about regular maps. Let me try to outline the setup - this will test to see if I've truly understood it. We start with a Lie groupoid $X$. So that should be a groupoid internal to the category of (finite dimensional) smooth manifolds, so my instinct says: 1. Smooth manifold of objects, $X_0$ 2. Smooth manifold of morphisms (total space of *all* morphisms), $X_1$ 3. All maps are regular, more specifically the source and target maps should be submersions and the identity map an embedding. Let's check: [[Lie groupoid]] is a bit confused (the *Idea* section isn't even a sentence!), but the above seems a reasonable starting point. Okay, so now we want a "path space" (or loop space) of the above. The idea is to take a smooth path in the set of objects, but allow for modification by morphisms. However this construction ends up, a step along the way is the following construction. Take a finite open cover of $S^1$, let's say by intervals (not sure if that's *necessary* or just *nice*), $\{I_1,\dots,I_k\}$, and let's assume that these are such that the intersections, $I_{ij}$, are empty for non-successive intervals. Each interval is diffeomorphic to $(0,1)$ so we can consider the smooth spaces (**not** manifold) $C^\infty(I_j, X_0)$ and take their product. On overlaps, $I_{i j}$, we want the paths to match by a path of morphisms. So for a "path" $(\alpha_j)$" there should be a smooth map $\beta_{i j} \colon I_{i j} \to X_1$ with $\beta_{i j} \alpha_j = \alpha_i$ on $I_{i j}$. We consider a smooth path in $X$ to be a collection $\{\alpha_i, \beta_{i j}\}$ satisfying these conditions. I'm pretty confident that this is a Frechet manifold. The difficulty with open paths isn't present here because the openness of the intervals is a Chimera. All you really need is for the intervals to non-trivially overlap so you could use a closed cover of $S^1$ with the property that the intersections were either empty or had non-void interior. This, I think, is the clearest way to see that the issue with open intervals is a non-issue. Then there is a question about submersions. The maps $s, t \colon X_1 \to X_0$ are submersions. The question is whether or not the induces mappings $C^\infty(I, X_1) \to C^\infty(I,X_0)$ are submersions (we can take $I = [0,1]$ to ensure that everything is a Frechet manifold at every stage of the construction here). I believe that this is the case. My "proof" is that I think that one can choose the local additions used in the definition of the manifold structure on mapping spaces in a manner that "respects" the submersion $X_1 \to X_0$. Indeed, my conjecture is the following: **Conjecture**: Let $M, N$ be smooth finite dimensional manifolds (without boundary) and $S$ a suitably compact suitable smooth space. Let $f \colon M \to N$ be a regular mapping. Then $C^\infty(S,f) \colon C^\infty(f,M) \to C^\infty(f,N)$ is a regular mapping. I have the sketch of the proof, and fully intend to fill in the details (just to stake my claim!). The sketch is: **Ingredients** 1. $M$ a smooth finite dimensional manifold, 1. $S$ a suitably compact smooth space, 1. $\gamma \colon S \to M$ a smooth map, 1. For each $s \in S$, a $k$-dimensional ($k$ fixed) submanifold near $f(s)$ that "varies smoothly" over $S$ in that for each $s \in S$ there is an open neighbourhood $U_s$ of $s$ (in a suitable topology) and a chart $\psi_s \colon V_s \to \mathbb{R}^m$ of $f(s)$ with the property that for $s' \in U_s$ then the submanifold at $f(s')$ is precisely $\mathbb{R}^k \subseteq \mathbb{R}^m$. **Recipe** 1. From the above data, we build a [[local addition]] for $M$ that varies smoothly with $s \in S$ and which respects the submanifolds. This will probably involve choosing suitable sections of $f^* T M$, extending them to vector fields on $M$, and applying the exponential mapping to get diffeomorphisms of $M$ which are then evaluated to give a suitable map $f^* T M \to S \times M$. The inverse function theorem will then say that this is a diffeomorphism in a neighbourhood of the zero section (which will map to $(s,f(s))$). 1. Then applying the standard mapping-space construction, we get charts for $C^\infty(S,M)$ in which the structure of the submanifolds is seen as $C^\infty(S,\mathbb{R}^k)$ sitting inside $C^\infty(S,\mathbb{R}^m)$. This will need extending the mapping-space construction to these varying local additions (shouldn't be a problem). **Baking** 1. To apply this to the case in hand, $f \colon M \to N$, we apply it to the level sets of $f$ in $M$ and the image of $f$ in $N$. If done properly, we will even be get that for these charts, $f$ is simply the composition $$C^\infty(S,\mathbb{R}^m) \to C^\infty(S,\mathbb{R}^k) \to C^\infty(S,\mathbb{R}^n)$$ as desired. The main detail is constructing the local additions. As hinted above, I think that this will go along the lines of [[propagating flows]] whereby we build suitable vector fields and apply the exponential function. Using vector fields gives us more control over the resulting diffeomorphisms and so makes it easier to show that they satisfy suitable properties that means that the resulting local addition has the right properties.

   This is a partial continuation of a conversation between me and David Roberts about path spaces of Lie groupoids. The conversation has so far taken place by email and MathOverflow comments, but was getting unwieldy for either so we’re moving it here. As always with nForum discussions, anyone else is welcome to join in.

   There was a MathOverflow question: http://mathoverflow.net/questions/85960/induced-map-on-path-manifolds-is-it-a-submersion which really had two parts. One was about defining a manifold of suitable paths that aren’t quite the usual paths, the other was about regular maps.

   Let me try to outline the setup - this will test to see if I’ve truly understood it.

   We start with a Lie groupoid $X$. So that should be a groupoid internal to the category of (finite dimensional) smooth manifolds, so my instinct says:

   1. Smooth manifold of objects, $X_0$
   2. Smooth manifold of morphisms (total space of all morphisms), $X_1$
   3. All maps are regular, more specifically the source and target maps should be submersions and the identity map an embedding.

   Let’s check: Lie groupoid is a bit confused (the Idea section isn’t even a sentence!), but the above seems a reasonable starting point.

   Okay, so now we want a “path space” (or loop space) of the above. The idea is to take a smooth path in the set of objects, but allow for modification by morphisms. However this construction ends up, a step along the way is the following construction. Take a finite open cover of $S^1$, let’s say by intervals (not sure if that’s necessary or just nice), $\{I_1,\dots,I_k\}$, and let’s assume that these are such that the intersections, $I_{ij}$, are empty for non-successive intervals. Each interval is diffeomorphic to $(0,1)$ so we can consider the smooth spaces (not manifold) $C^\infty(I_j, X_0)$ and take their product. On overlaps, $I_{i j}$, we want the paths to match by a path of morphisms. So for a “path” $(\alpha_j)$” there should be a smooth map $\beta_{i j} \colon I_{i j} \to X_1$ with $\beta_{i j} \alpha_j = \alpha_i$ on $I_{i j}$. We consider a smooth path in $X$ to be a collection $\{\alpha_i, \beta_{i j}\}$ satisfying these conditions.

   I’m pretty confident that this is a Frechet manifold. The difficulty with open paths isn’t present here because the openness of the intervals is a Chimera. All you really need is for the intervals to non-trivially overlap so you could use a closed cover of $S^1$ with the property that the intersections were either empty or had non-void interior. This, I think, is the clearest way to see that the issue with open intervals is a non-issue.

   Then there is a question about submersions. The maps $s, t \colon X_1 \to X_0$ are submersions. The question is whether or not the induces mappings $C^\infty(I, X_1) \to C^\infty(I,X_0)$ are submersions (we can take $I = [0,1]$ to ensure that everything is a Frechet manifold at every stage of the construction here). I believe that this is the case. My “proof” is that I think that one can choose the local additions used in the definition of the manifold structure on mapping spaces in a manner that “respects” the submersion $X_1 \to X_0$. Indeed, my conjecture is the following:

   Conjecture: Let $M, N$ be smooth finite dimensional manifolds (without boundary) and $S$ a suitably compact suitable smooth space. Let $f \colon M \to N$ be a regular mapping. Then $C^\infty(S,f) \colon C^\infty(f,M) \to C^\infty(f,N)$ is a regular mapping.

   I have the sketch of the proof, and fully intend to fill in the details (just to stake my claim!). The sketch is:

   Ingredients

   1. $M$ a smooth finite dimensional manifold,
   2. $S$ a suitably compact smooth space,
   3. $\gamma \colon S \to M$ a smooth map,
   4. For each $s \in S$, a $k$-dimensional ($k$ fixed) submanifold near $f(s)$ that “varies smoothly” over $S$ in that for each $s \in S$ there is an open neighbourhood $U_s$ of $s$ (in a suitable topology) and a chart $\psi_s \colon V_s \to \mathbb{R}^m$ of $f(s)$ with the property that for $s' \in U_s$ then the submanifold at $f(s')$ is precisely $\mathbb{R}^k \subseteq \mathbb{R}^m$.

   Recipe

   1. From the above data, we build a local addition for $M$ that varies smoothly with $s \in S$ and which respects the submanifolds. This will probably involve choosing suitable sections of $f^* T M$, extending them to vector fields on $M$, and applying the exponential mapping to get diffeomorphisms of $M$ which are then evaluated to give a suitable map $f^* T M \to S \times M$. The inverse function theorem will then say that this is a diffeomorphism in a neighbourhood of the zero section (which will map to $(s,f(s))$).
   2. Then applying the standard mapping-space construction, we get charts for $C^\infty(S,M)$ in which the structure of the submanifolds is seen as $C^\infty(S,\mathbb{R}^k)$ sitting inside $C^\infty(S,\mathbb{R}^m)$. This will need extending the mapping-space construction to these varying local additions (shouldn’t be a problem).

   Baking

   1. To apply this to the case in hand, $f \colon M \to N$, we apply it to the level sets of $f$ in $M$ and the image of $f$ in $N$. If done properly, we will even be get that for these charts, $f$ is simply the composition $$C^\infty(S,\mathbb{R}^m) \to C^\infty(S,\mathbb{R}^k) \to C^\infty(S,\mathbb{R}^n)$$ as desired.

   The main detail is constructing the local additions. As hinted above, I think that this will go along the lines of propagating flows whereby we build suitable vector fields and apply the exponential function. Using vector fields gives us more control over the resulting diffeomorphisms and so makes it easier to show that they satisfy suitable properties that means that the resulting local addition has the right properties.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJan 30th 2012
   • (edited Jan 30th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> the _Idea_ section isn't even a sentence! I have added the missing period.

   the Idea section isn’t even a sentence!

   I have added the missing period.

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 30th 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexLie groupoid - check 'Paths/loops' - check >All you really need is for the intervals to non-trivially overlap so you could use a closed cover of $S^1$ with the property that the intersections were either empty or had non-void interior. This, I think, is the clearest way to see that the issue with open intervals is a non-issue. ....almost. There are further requirements which are irrelevant to the construction of the Frechet manifold of functors, but which preclude the use of closed intervals. For instance, I need the intersection of two covers (of the interval or circle) of whatever sort we are considering to also be a cover. This is tricky to ensure for closed intervals, even the ones you mention, e.g. the intersection of the closed covers $[0,1/2] \coprod [1/3,1]$ and $[0,2/3]\coprod [1/2,1]$. I'm not saying it's not possible, I'm just looking further down the track. If such withholding of information is slowing progress, I will gladly supply, but I didn't want to overwhelm with new constructions and details all at once. >I have the sketch of the proof, and fully intend to fill in the details (just to stake my claim!) Oh good!

   Lie groupoid - check

   ’Paths/loops’ - check

   All you really need is for the intervals to non-trivially overlap so you could use a closed cover of $S^1$ with the property that the intersections were either empty or had non-void interior. This, I think, is the clearest way to see that the issue with open intervals is a non-issue.

   ….almost. There are further requirements which are irrelevant to the construction of the Frechet manifold of functors, but which preclude the use of closed intervals. For instance, I need the intersection of two covers (of the interval or circle) of whatever sort we are considering to also be a cover. This is tricky to ensure for closed intervals, even the ones you mention, e.g. the intersection of the closed covers $[0,1/2] \coprod [1/3,1]$ and $[0,2/3]\coprod [1/2,1]$.

   I’m not saying it’s not possible, I’m just looking further down the track. If such withholding of information is slowing progress, I will gladly supply, but I didn’t want to overwhelm with new constructions and details all at once.

   I have the sketch of the proof, and fully intend to fill in the details (just to stake my claim!)

   Oh good!

4. • CommentRowNumber4.
   • CommentAuthorAndrew Stacey
   • CommentTimeJan 31st 2012
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexIt would help to have a couple of illustrative examples. If going with closed covers, I think you'd have to build in something where when intersecting two covers, you actually took the closures of the intersections of the interiors. However, the important point there is that I can think of several ways to "fix" this issue, and the "right" one will probably only become apparent further down the line so arguing small details now is probably not productive (if you feel like adding the details now so that we can try to decide which one then that's fine, but if you'd rather wait, that's also fine). All I wanted to say with that is that it is *fixable*. To try to put it in context, one can regard just an ordinary smooth path in a manifold in a similar fashion. Certainly there, openness of the cover doesn't cause problems because the "ragged edge" of one interval is well contained in the interior of the next. *That's* the real fix. The only piece where this might be important is in the paths in the morphism spaces. As these are only defined on intersections, their "ragged edges" will be exposed so there some care might be needed, but it might be that in the specific cases you are considering then this is a non-issue. For example, if there is some condition which means that once the source and target are specified then the morphisms are discrete (or that the allowable morphisms are discrete, for example if they are horizontal with respect to some connection somewhere). Otherwise, one could have a path in the morphism space that looked perfectly respectable when projected down to the objects but which wiggled uncontrollably in the vertical direction. Urs: Yes, point taken! I was skimming the page in "reader mode", but I should at least have put the full stop in. I *did* find the rest of the page a little confusing. I'll try reorganising it to see if I can make it clearer (though I don't really know anything about the subject).

   It would help to have a couple of illustrative examples.

   If going with closed covers, I think you’d have to build in something where when intersecting two covers, you actually took the closures of the intersections of the interiors. However, the important point there is that I can think of several ways to “fix” this issue, and the “right” one will probably only become apparent further down the line so arguing small details now is probably not productive (if you feel like adding the details now so that we can try to decide which one then that’s fine, but if you’d rather wait, that’s also fine). All I wanted to say with that is that it is fixable.

   To try to put it in context, one can regard just an ordinary smooth path in a manifold in a similar fashion. Certainly there, openness of the cover doesn’t cause problems because the “ragged edge” of one interval is well contained in the interior of the next. That’s the real fix. The only piece where this might be important is in the paths in the morphism spaces. As these are only defined on intersections, their “ragged edges” will be exposed so there some care might be needed, but it might be that in the specific cases you are considering then this is a non-issue. For example, if there is some condition which means that once the source and target are specified then the morphisms are discrete (or that the allowable morphisms are discrete, for example if they are horizontal with respect to some connection somewhere). Otherwise, one could have a path in the morphism space that looked perfectly respectable when projected down to the objects but which wiggled uncontrollably in the vertical direction.

   Urs: Yes, point taken! I was skimming the page in “reader mode”, but I should at least have put the full stop in. I did find the rest of the page a little confusing. I’ll try reorganising it to see if I can make it clearer (though I don’t really know anything about the subject).

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeFeb 1st 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexOK, here we start getting down to the nitty-gritty. >It would help to have a couple of illustrative examples. There are several main ones: 1. A manifold, as you have pointed out. The way things are defined (as opposed to my previous attempts) we recover the usual set of smooth paths. 1. A Cech groupoid - given a surjective submersion $Y \to M$, consider the Lie groupoid with object manifold $Y$ and arrow manifold $Y\times_M Y$, and projections for source and target. Denote this by $C(Y)$. 1. A Lie group - here the object manifold is the point, and the arrow manifold is the Lie group itself. [Actually this example raises all sorts of questions, see below] Such a groupoid is denoted $\mathbf{B}G$. 1. An action groupoid - this generalises the previous point. Consider an action of a Lie group $G$ on a manifold $M$. [we might want this action to be proper] The object manifold is $M$, the arrow manifold is $G\times M$, the source is projection on second factor, the target is the action map. 1. Consider a Cech groupoid $C(Y)$, and then consider another groupoid also with object manifold $Y$, but with arrow manifold $E$ such that the canonical map $E \to Y\times_M Y$ is a principal $U(1)$-bundle. This is basically a bundle gerbe in different language. The example of $\mathbf{B}G$ leads me to think I really do need some conditions, because a 'path' in $\mathbf{B}G$ is a finite collection of maps $(s,t) \to G$ for various subintervals of $[0,1]$. The space of such is not very pretty. One condition on the Lie groupoid that will be present in all relevant examples is _properness_ **Definition:** a Lie groupoid $X$ is _proper_ if the map $(s,t):X_1 \to X_0 \times X_0$ is proper. I don't know if this helps, but we can assume it if we want. It is one of the assumptions used in <http://arxiv.org/pdf/0712.3857v2.pdf>, section 5.2, so it is probably a necessary assumption. >The only piece where this might be important is in the paths in the morphism spaces. As these are only defined on intersections, their "ragged edges" will be exposed so there some care might be needed, .... Otherwise, one could have a path in the morphism space that looked perfectly respectable when projected down to the objects but which wiggled uncontrollably in the vertical direction. I thought I had this sorted, using the following logic: * a 'path' in $X$ (i.e. a functor with codomain $X$) defines a 'path' in $X^\mathbf{2}$, the groupoid with object manifold the arrow manifold of $X$, and an irrelevant manifold of arrows (albeit the 'path' has a different domain), such that the object component of this new functor is precisely the arrow component of the original functor. Thus if we can show that paths in the object manifold of an arbitrary Lie groupoid behave, we immediately get the result that paths in the arrow manifold behave. But the example of $\mathbf{B}G$ seems to undo this reasoning, even if we assume that $G$ is compact (perhaps you can disabuse me of a misconception?) In practice, we (Ray Vozzo and I) are looking at Cech groupoids and those groupoids underlying bundle gerbes, but a general theory is desirable, because of links to the theory of differentiable stacks.

   OK, here we start getting down to the nitty-gritty.

   It would help to have a couple of illustrative examples.

   There are several main ones:

   1. A manifold, as you have pointed out. The way things are defined (as opposed to my previous attempts) we recover the usual set of smooth paths.

   2. A Cech groupoid - given a surjective submersion $Y \to M$, consider the Lie groupoid with object manifold $Y$ and arrow manifold $Y\times_M Y$, and projections for source and target. Denote this by $C(Y)$.

   3. A Lie group - here the object manifold is the point, and the arrow manifold is the Lie group itself. [Actually this example raises all sorts of questions, see below] Such a groupoid is denoted $\mathbf{B}G$.

   4. An action groupoid - this generalises the previous point. Consider an action of a Lie group $G$ on a manifold $M$. [we might want this action to be proper] The object manifold is $M$, the arrow manifold is $G\times M$, the source is projection on second factor, the target is the action map.

   5. Consider a Cech groupoid $C(Y)$, and then consider another groupoid also with object manifold $Y$, but with arrow manifold $E$ such that the canonical map $E \to Y\times_M Y$ is a principal $U(1)$-bundle. This is basically a bundle gerbe in different language.

   The example of $\mathbf{B}G$ leads me to think I really do need some conditions, because a ’path’ in $\mathbf{B}G$ is a finite collection of maps $(s,t) \to G$ for various subintervals of $[0,1]$. The space of such is not very pretty.

   One condition on the Lie groupoid that will be present in all relevant examples is properness

   Definition: a Lie groupoid $X$ is proper if the map $(s,t):X_1 \to X_0 \times X_0$ is proper.

   I don’t know if this helps, but we can assume it if we want. It is one of the assumptions used in http://arxiv.org/pdf/0712.3857v2.pdf, section 5.2, so it is probably a necessary assumption.

   The only piece where this might be important is in the paths in the morphism spaces. As these are only defined on intersections, their “ragged edges” will be exposed so there some care might be needed, …. Otherwise, one could have a path in the morphism space that looked perfectly respectable when projected down to the objects but which wiggled uncontrollably in the vertical direction.

   I thought I had this sorted, using the following logic:

   • a ’path’ in $X$ (i.e. a functor with codomain $X$) defines a ’path’ in $X^\mathbf{2}$, the groupoid with object manifold the arrow manifold of $X$, and an irrelevant manifold of arrows (albeit the ’path’ has a different domain), such that the object component of this new functor is precisely the arrow component of the original functor. Thus if we can show that paths in the object manifold of an arbitrary Lie groupoid behave, we immediately get the result that paths in the arrow manifold behave.

   But the example of $\mathbf{B}G$ seems to undo this reasoning, even if we assume that $G$ is compact (perhaps you can disabuse me of a misconception?)

   In practice, we (Ray Vozzo and I) are looking at Cech groupoids and those groupoids underlying bundle gerbes, but a general theory is desirable, because of links to the theory of differentiable stacks.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeFeb 1st 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItex> **Definition:** a Lie groupoid $X$ is _proper_ if the map $(s,t):X_1 \to X_0 \times X_0$ is proper. Just for the record: there is an entry _[[proper Lie groupoid]]_.

   Definition: a Lie groupoid $X$ is proper if the map $(s,t):X_1 \to X_0 \times X_0$ is proper.

   Just for the record: there is an entry proper Lie groupoid.

7. • CommentRowNumber7.
   • CommentAuthorzskoda
   • CommentTimeFeb 1st 2012
   • (edited Feb 1st 2012)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexIs it standard to define the same way proper topological groupoids ? It looks to me a more natural generality for this definition.

   Is it standard to define the same way proper topological groupoids ? It looks to me a more natural generality for this definition.

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeFeb 1st 2012
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexI'm not sure I understand you, Zoran. This is pretty much the standard defintion, and for Lie groupoids only depends on the underlying topological groupoid, though the fact manifolds are locally compact may come into play here.

   I’m not sure I understand you, Zoran. This is pretty much the standard defintion, and for Lie groupoids only depends on the underlying topological groupoid, though the fact manifolds are locally compact may come into play here.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeFeb 2nd 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexwe have _[[proper topological groupoid]]_

   we have proper topological groupoid

10. • CommentRowNumber10.
    • CommentAuthorAndrew Stacey
    • CommentTimeFeb 2nd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexDavid, > a 'path' in $X$ (i.e. a functor with codomain $X$) defines a 'path' in $X^\mathbf{2}$, How?

    David,

    a ’path’ in $X$ (i.e. a functor with codomain $X$) defines a ’path’ in $X^\mathbf{2}$,

    How?

11. • CommentRowNumber11.
    • CommentAuthorAndrew Stacey
    • CommentTimeFeb 2nd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItex(Sorry, that was a bit terse! I was intending on writing a longer comment about the regular map thing, but then it got gnarly so I left it at the above.)

    (Sorry, that was a bit terse! I was intending on writing a longer comment about the regular map thing, but then it got gnarly so I left it at the above.)

12. • CommentRowNumber12.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 2nd 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@Andrew - actually it doesn't. This where my purported proof falls down. I didn't check all the functor conditions at the time, but I rethought it yesterday to see why the example of $\mathbf{B}G$ broke my 'theorem'.

    @Andrew - actually it doesn’t. This where my purported proof falls down. I didn’t check all the functor conditions at the time, but I rethought it yesterday to see why the example of $\mathbf{B}G$ broke my ’theorem’.

13. • CommentRowNumber13.
    • CommentAuthorAndrew Stacey
    • CommentTimeFeb 3rd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexOkay, leave that one then. Here's the next question. How am I meant to think of a path in a Lie groupoid? I remember thinking about this a few years ago, not necessarily with Lie groupoids but more general "category objects in Diff" (if that makes sense). There, my thinking was that I really wanted a path in the object space, but for some reason every now and then I'd reach the "edge" of something and I wouldn't be able to continue my path in the object space. Fortunately, I'd have some instructions that gave me a way to leap from one patch of the object space to another via arrows. So as I got near the edge, I'd find another patch and continue my path in that one. To know that I could do this, I'd have to have a path in the arrows that connected the two on the overlap. In the situation I was considering, that path of arrows wasn't really important - it was just the instructions on how to match paths. All I really cared about was that such a path *existed*, not what it was.

    Okay, leave that one then.

    Here’s the next question. How am I meant to think of a path in a Lie groupoid? I remember thinking about this a few years ago, not necessarily with Lie groupoids but more general “category objects in Diff” (if that makes sense). There, my thinking was that I really wanted a path in the object space, but for some reason every now and then I’d reach the “edge” of something and I wouldn’t be able to continue my path in the object space. Fortunately, I’d have some instructions that gave me a way to leap from one patch of the object space to another via arrows. So as I got near the edge, I’d find another patch and continue my path in that one. To know that I could do this, I’d have to have a path in the arrows that connected the two on the overlap. In the situation I was considering, that path of arrows wasn’t really important - it was just the instructions on how to match paths. All I really cared about was that such a path existed, not what it was.

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2012
    • (edited Feb 3rd 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> How am I meant to think of a path in a Lie groupoid? I am not sure what exactly you may be after. But just for the record, notice that there is a canonical notion: We have the [[path groupoid]] functor $$ P_1(-) : SmoothManifolds \to SmoothGroupoids; $$ and we have the 2-Yoneda embedding $$ y : SmoothManifolds \to SmoothGroupoids \,. $$ Therefore there is canonically the 2-[[Yoneda extension]] of $P_1$ to smooth groupoids $$ P_1(-) : SmoothGroupoids \to SmoothGroupoids \,. $$ Some comments on this that happen to have made their way to the $n$Lab are [here](http://ncatlab.org/nlab/show/smooth%20Lorentzian%20space#PathnCategory) in the section "The path 2-groupoid of an orbifold". As I said, this may not be what you are after. But this is sort of the canonical notion of paths in a smooth groupoid (or smooth higher groupoid).

    How am I meant to think of a path in a Lie groupoid?

    I am not sure what exactly you may be after. But just for the record, notice that there is a canonical notion:

    We have the path groupoid functor

    $$P_1(-) : SmoothManifolds \to SmoothGroupoids;$$

    and we have the 2-Yoneda embedding

    $$y : SmoothManifolds \to SmoothGroupoids \,.$$

    Therefore there is canonically the 2-Yoneda extension of $P_1$ to smooth groupoids

    $$P_1(-) : SmoothGroupoids \to SmoothGroupoids \,.$$

    Some comments on this that happen to have made their way to the $n$Lab are here in the section “The path 2-groupoid of an orbifold”.

    As I said, this may not be what you are after. But this is sort of the canonical notion of paths in a smooth groupoid (or smooth higher groupoid).

15. • CommentRowNumber15.
    • CommentAuthorAndrew Stacey
    • CommentTimeFeb 3rd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexUrs, looks interesting but from my understanding it looks as if everything is one level too high. That is, in David's construction then the paths are the objects, not the morphisms. Nonetheless, worth keeping in mind. David, I missed something that may make the groupoid structure much, much simpler. If you restrict to situations where the source and target maps are *proper* then they sort-of become fibre bundles - at least, modulo number of components. Then it's clear that the mapping space of a fibre bundle is again a fibre bundle and you can use that to get that the projection map is again a submersion.

    Urs, looks interesting but from my understanding it looks as if everything is one level too high. That is, in David’s construction then the paths are the objects, not the morphisms. Nonetheless, worth keeping in mind.

    David, I missed something that may make the groupoid structure much, much simpler. If you restrict to situations where the source and target maps are proper then they sort-of become fibre bundles - at least, modulo number of components. Then it’s clear that the mapping space of a fibre bundle is again a fibre bundle and you can use that to get that the projection map is again a submersion.

16. • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItex> the paths are the objects, not the morphisms That's what you get by passing to the corresponding [[arrow category]] (as a smooth groupoid).

    the paths are the objects, not the morphisms

    That’s what you get by passing to the corresponding arrow category (as a smooth groupoid).

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2012
    • (edited Feb 3rd 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexYou can also look at the internal hom smooth groupoid $$ \mathcal{G}^I := [I, \mathcal{G}] $$ where $\mathcal{G}$ is the given Lie groupoid and $I = [0,1]$ is the standard interval, regarded as a smooth space.

    You can also look at the internal hom smooth groupoid

    $$\mathcal{G}^I := [I, \mathcal{G}]$$

    where $\mathcal{G}$ is the given Lie groupoid and $I = [0,1]$ is the standard interval, regarded as a smooth space.

18. • CommentRowNumber18.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 4th 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@Andrew #13 'paths' in a Lie groupoid $X$ (or topological groupoid, or even internal categories in Diff or Top) are functors with domain a Cech groupoid for an open cover of the unit interval $I$. As such they are anafunctors in $Diff$ (or $Top$ etc) $I \to X$. This implies several things 1. they 'are' maps of stacks on $Diff$ ($Top$ etc, I'll stop now) 1. they give rise to (pretty much canonical) paths on the geometric realisation of the nerve. (any choices are irrelevant up to homotopy) So in a sense, these 'paths' see more geometry/topology than just the object space and the arrow space on their own. There is a construction by Behrend, Ginot, Noohi and Xu of the free loop stack (and path stack) of a differentiable stack (given by a Lie groupoid). It is essentially what I'm envisaging here, except only using the compact-open topology on the space of (continuous) 'loops'. There is no subtlety as in our case, but they only get a topological groupoid. (I'm not sure I entirely believe their construction, or at least they are very brief and don't get what I expect should be the case for one of their constructions. That doesn't mean I don't believe it works.) Anyway, to get back to our case, you can think of 'paths' as Lie-groupoid valued Cech cocycles on $I$, and arrows between them as coboundaries. There is no difficulty in what we are doing if we pass to smooth spaces (Chen or whatever), but I believe we should be able to get an honest Frechet Lie structure on the ordinary groupoid of these. One thing I forgot to say is that eventually I'm only interested in restricting to finite (open?) covers of $I$ where the endpoints are dyadic rationals, so that we have a countable collection of covers. Or more generally I want to be able to only use covers from a countable family which is closed under intersections. I'm sorry to reveal details in a such a slow way, some seem irrelevant to start with, and then turn out to be necessary for the longer perspective.

    @Andrew #13

    ’paths’ in a Lie groupoid $X$ (or topological groupoid, or even internal categories in Diff or Top) are functors with domain a Cech groupoid for an open cover of the unit interval $I$. As such they are anafunctors in $Diff$ (or $Top$ etc) $I \to X$. This implies several things

    1. they ’are’ maps of stacks on $Diff$ ($Top$ etc, I’ll stop now)
    2. they give rise to (pretty much canonical) paths on the geometric realisation of the nerve. (any choices are irrelevant up to homotopy)

    So in a sense, these ’paths’ see more geometry/topology than just the object space and the arrow space on their own.

    There is a construction by Behrend, Ginot, Noohi and Xu of the free loop stack (and path stack) of a differentiable stack (given by a Lie groupoid). It is essentially what I’m envisaging here, except only using the compact-open topology on the space of (continuous) ’loops’. There is no subtlety as in our case, but they only get a topological groupoid.

    (I’m not sure I entirely believe their construction, or at least they are very brief and don’t get what I expect should be the case for one of their constructions. That doesn’t mean I don’t believe it works.)

    Anyway, to get back to our case, you can think of ’paths’ as Lie-groupoid valued Cech cocycles on $I$, and arrows between them as coboundaries. There is no difficulty in what we are doing if we pass to smooth spaces (Chen or whatever), but I believe we should be able to get an honest Frechet Lie structure on the ordinary groupoid of these.

    One thing I forgot to say is that eventually I’m only interested in restricting to finite (open?) covers of $I$ where the endpoints are dyadic rationals, so that we have a countable collection of covers. Or more generally I want to be able to only use covers from a countable family which is closed under intersections.

    I’m sorry to reveal details in a such a slow way, some seem irrelevant to start with, and then turn out to be necessary for the longer perspective.

19. • CommentRowNumber19.
    • CommentAuthorjim_stasheff
    • CommentTimeFeb 4th 2012
    • PermaLink
    Author: jim_stasheff
    Format: Text@ David: the endpoints are dyadic rationals curious or significant coincidence? these are exactly the endpoints that occur in proving the Ainfty structure on the space of paths from [0,1]

    @ David: the endpoints are dyadic rationals
    
    curious or significant coincidence? these are exactly the endpoints that occur
    in proving the Ainfty structure on the space of paths from [0,1]
20. • CommentRowNumber20.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 6th 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexHi Jim, possibly a coincidence, possibly not. I just need a countable family of covers of [0,1] closed under intersections. Rational endpoints would do, but dyadic rationals are in a sense less dense; and as you say, arise as the 'join points' when one concatenates copies of [0,1] according to some bracketing a la your polytope. In a _continuous_ version of what I am trying to do the analogy is much closer, because then you take covers by closed intervals corresponding exactly to these iterated concatenations. Would be nice to display some sort of A_oo structure on the Lie groupoid (or possibly topological groupoid, if we don't get that far) of these 'loops'. Might be a weakened version appropriate to stacks etc.

    Hi Jim,

    possibly a coincidence, possibly not. I just need a countable family of covers of [0,1] closed under intersections. Rational endpoints would do, but dyadic rationals are in a sense less dense; and as you say, arise as the ’join points’ when one concatenates copies of [0,1] according to some bracketing a la your polytope. In a continuous version of what I am trying to do the analogy is much closer, because then you take covers by closed intervals corresponding exactly to these iterated concatenations.

    Would be nice to display some sort of A_oo structure on the Lie groupoid (or possibly topological groupoid, if we don’t get that far) of these ’loops’. Might be a weakened version appropriate to stacks etc.

21. • CommentRowNumber21.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 23rd 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexAndrew, here is a question which I have no intuition for: Let $G$ be a compact Lie group. Can we give the set $G^{(0,1)}$ the structure of a manifold? Or do we have to put restrictions on the derivatives of the maps $(0,1) \to G$? This gives us a chance at figuring out what the path groupoid is for proper, one-object Lie groupoids, because it is built up from copies of products of $G^{(0,1)}$.

    Andrew,

    here is a question which I have no intuition for: Let $G$ be a compact Lie group. Can we give the set $G^{(0,1)}$ the structure of a manifold? Or do we have to put restrictions on the derivatives of the maps $(0,1) \to G$?

    This gives us a chance at figuring out what the path groupoid is for proper, one-object Lie groupoids, because it is built up from copies of products of $G^{(0,1)}$.

22. • CommentRowNumber22.
    • CommentAuthorAndrew Stacey
    • CommentTimeFeb 23rd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexYes, you need more conditions. The set $G^{(0,1)}$ is diffeomorphic (in whatever category) to $G^{\mathbb{R}}$. The problem is that without extra conditions, the ends have to be "free" as derivatives can be arbitrarily unbounded. Once the ends are free, they can wander all over the target space meaning that "nearby" paths can be arbitrarily far apart. But that's made me ponder what the absolute weakest conditions could be. The problem is with the curve itself, not the derivatives. Once you have the curve tamed then the derivatives can be wild because they take values in a linear space and so are naturally contained in charts. The key requirement is that if $U$ is an open subset of $G$ then $U^{(0,1)}$ must be an open subset of $G^{(0,1)}$. This isn't true if you simply take the compact open topology. But if you arrange for this, then I don't think that you really need any conditions on the higher derivatives. But do you really need this? I thought that the use of $(0,1)$ was somewhat artificial and that when you took a path in total, although it restricted to a family of elements in these spaces then because of the overlaps, the "edge effects" didn't actually occur.

    Yes, you need more conditions. The set $G^{(0,1)}$ is diffeomorphic (in whatever category) to $G^{\mathbb{R}}$.

    The problem is that without extra conditions, the ends have to be “free” as derivatives can be arbitrarily unbounded. Once the ends are free, they can wander all over the target space meaning that “nearby” paths can be arbitrarily far apart.

    But that’s made me ponder what the absolute weakest conditions could be. The problem is with the curve itself, not the derivatives. Once you have the curve tamed then the derivatives can be wild because they take values in a linear space and so are naturally contained in charts. The key requirement is that if $U$ is an open subset of $G$ then $U^{(0,1)}$ must be an open subset of $G^{(0,1)}$. This isn’t true if you simply take the compact open topology. But if you arrange for this, then I don’t think that you really need any conditions on the higher derivatives.

    But do you really need this? I thought that the use of $(0,1)$ was somewhat artificial and that when you took a path in total, although it restricted to a family of elements in these spaces then because of the overlaps, the “edge effects” didn’t actually occur.

23. • CommentRowNumber23.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 23rd 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex>But do you really need this? I thought that the use of (0,1) was somewhat artificial and that when you took a path in total, although it restricted to a family of elements in these spaces then because of the overlaps, the "edge effects" didn't actually occur. this is true, but when the Lie groupoid we are considering has one object, the overlaps become useless in constraining the edge effects. But this _may_ not occur in practice (one-object groupoids that is).

    But do you really need this? I thought that the use of (0,1) was somewhat artificial and that when you took a path in total, although it restricted to a family of elements in these spaces then because of the overlaps, the “edge effects” didn’t actually occur.

    this is true, but when the Lie groupoid we are considering has one object, the overlaps become useless in constraining the edge effects. But this may not occur in practice (one-object groupoids that is).

24. • CommentRowNumber24.
    • CommentAuthorAndrew Stacey
    • CommentTimeMar 2nd 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexLet me see if I can work through this example from your latest email. Ingredients: 1. $\mathcal{O}$ the set of open covers of $I = [0,1]$, ordered by $U_1 \lt U_2$ if there is a map $U_1 \to U_2$ over $I$. Now, I'm to think of an open cover $U$ as a disjoint union of intervals, so presumably the map should be continuous and so what this amounts to is a choice for each $I_j \in U_1$ of some $I_k \in U_2$ with $I_j \subseteq I_k$. As a topologist, I'd then say that $U_1$ *refines* $U_2$ (do we have a page [[refinement]], I wonder). 1. To an open cover, $U$, I assign the [[Čech groupoid]] (let's lab-link like loonies), $C(U)$. So $C(U)_0 = \{(t,I_i) : t \in I_i \in U,\}$ and $C(U)_1 = \{(t,I_i,I_j) : t \in I_i \cap I_j, I_i, I_j \in U\}$. 1. Next we take the set $Ana(I,X)_0$ of [[anafunctors]] $I \to X$, which means functors $C(U) \to X$ for $U \in O$. Now, I've not worked with anafunctors much before (though I've hung around here enough to understand roughly what they are). Do you really want the set of *all* anafunctors, or is there some equivalence notion whereby two are considered equivalence if one factors through or lifts the other (not sure which way round it would be). 1. Now we do the same with some other family to get morphisms, but I'll skip that part for now. Couple of questions so far: 1. The equivalence bit above. 1. Are your covers finite? Otherwise things like: > So consider a cofinal countable $C \subset O$ which is also filtered. For example the set of open covers with rational endpoints, or dyadic rational endpoints. don't seem so sure. Take a subset of the set of dyadic rationals, pair up its points, and take the corresponding open cover with those points as endpoints (maybe add some more to get a cover). Seems an injection from an uncountable set to me. Right, so to the case where $X = G$. Then given an open cover $U$, we look at functors $C(U) \to \underline{G}$. This means functions $\alpha_j \colon I_j \to \{\star\}$ and $\alpha_{j k} \colon I_{j k} \to G$ (smooth). So it does look like $G^{I_{j k}}$ which, under these assumptions, is Bad News. But only if you really are going to take the disjoint union of all of these things. Because if I restrict this particular anafunctor to some shrinkage of $U$, then life is very much easier (amnesia's to thank) because if we shrink each $I_j$ slightly, say to $J_j$, with the property that $\overline{J}_j \subseteq I_j$, then we get that $\alpha_{j k}$ restricted to $J_{j k} = J_j \cap J_k$ does extend to its closure, being a subset of $\overline{J}_j \cap \overline{J}_k \subseteq I_{j k}$. Does that help at all?

    Let me see if I can work through this example from your latest email.

    Ingredients:

    1. $\mathcal{O}$ the set of open covers of $I = [0,1]$, ordered by $U_1 \lt U_2$ if there is a map $U_1 \to U_2$ over $I$. Now, I’m to think of an open cover $U$ as a disjoint union of intervals, so presumably the map should be continuous and so what this amounts to is a choice for each $I_j \in U_1$ of some $I_k \in U_2$ with $I_j \subseteq I_k$. As a topologist, I’d then say that $U_1$ refines $U_2$ (do we have a page refinement, I wonder).

    2. To an open cover, $U$, I assign the Čech groupoid (let’s lab-link like loonies), $C(U)$. So $C(U)_0 = \{(t,I_i) : t \in I_i \in U,\}$ and $C(U)_1 = \{(t,I_i,I_j) : t \in I_i \cap I_j, I_i, I_j \in U\}$.

    3. Next we take the set $Ana(I,X)_0$ of anafunctors $I \to X$, which means functors $C(U) \to X$ for $U \in O$. Now, I’ve not worked with anafunctors much before (though I’ve hung around here enough to understand roughly what they are). Do you really want the set of all anafunctors, or is there some equivalence notion whereby two are considered equivalence if one factors through or lifts the other (not sure which way round it would be).

    4. Now we do the same with some other family to get morphisms, but I’ll skip that part for now.

    Couple of questions so far:

    1. The equivalence bit above.

    2. Are your covers finite? Otherwise things like:

       So consider a cofinal countable $C \subset O$ which is also filtered. For example the set of open covers with rational endpoints, or dyadic rational endpoints.

       don’t seem so sure. Take a subset of the set of dyadic rationals, pair up its points, and take the corresponding open cover with those points as endpoints (maybe add some more to get a cover). Seems an injection from an uncountable set to me.

    Right, so to the case where $X = G$. Then given an open cover $U$, we look at functors $C(U) \to \underline{G}$. This means functions $\alpha_j \colon I_j \to \{\star\}$ and $\alpha_{j k} \colon I_{j k} \to G$ (smooth). So it does look like $G^{I_{j k}}$ which, under these assumptions, is Bad News. But only if you really are going to take the disjoint union of all of these things. Because if I restrict this particular anafunctor to some shrinkage of $U$, then life is very much easier (amnesia’s to thank) because if we shrink each $I_j$ slightly, say to $J_j$, with the property that $\overline{J}_j \subseteq I_j$, then we get that $\alpha_{j k}$ restricted to $J_{j k} = J_j \cap J_k$ does extend to its closure, being a subset of $\overline{J}_j \cap \overline{J}_k \subseteq I_{j k}$.

    Does that help at all?

25. • CommentRowNumber25.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 4th 2012
    • (edited Mar 4th 2012)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexAh, this - the last paragraph - looks something like exactly what we'll need! Thanks for persevering with this, I'll get more details to you soon (got to ride to work v. soon).

    Ah, this - the last paragraph - looks something like exactly what we’ll need! Thanks for persevering with this, I’ll get more details to you soon (got to ride to work v. soon).

26. • CommentRowNumber26.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 5th 2012
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexOk, a little bit of time now to expand on what I said in #25. Consider a compact manifold $K$ and a Lie groupoid $X$. A priori, the set of objects of the (bare) groupoid of anafunctors $LieGpd_{ana}(K,X)$ is a coproduct indexed by the set $O$ of open covers of $K$, which is too big. We can restrict to the (equivalent!) full subgroupoid taking the subset indexed by the set $F$ of finite open covers. This is still too big, but better. Now assume that we have a countable subset $C \subset F$ which is closed under intersections (and contains the trivial cover $K\to K$, but we can add this if not) and is cofinal in $F$ under refinement. Then taking the full subgroupoid indexed by $C$ we again get a groupoid equivalent to $LieGpd_{ana}(K,X)$. I know I can do this last step for $K = I$, and presumably for some other simple manifolds like $I^n \times (S^1)^m$, but I think it may be possible to extend the naive techniques I'm thinking of to just use second countability of $K$ together with the fact finite covers are cofinal in all covers, so that there is a countable family as we need. Now the property you describe, that each open cover $U$ has a refinement $V$ such that the closure of $V$ (i.e. the closed cover $\widebar{V}$ given by the closure of each open set in $V$) still refines $U$, looks familiar but I can't quite remember what it is called (something like having a halo or halo-extension or similar). This means we can restrict from _all_ anafunctors $C(U) \to X$ indexed by $C$ to those that extend to the closure of $U$, that is those that extend along $C(U) \to C(\widebar{U})$. Note that we ask that the whole functor extends, not just object or arrow components (the latter is what you were describing in the example). If we take the full subgroupoid on these anafunctors, then we again get an equivalent groupoid. _Now_ we consider the disjoint union indexed by $C$, but restricting to these 'extendible' anafunctors. For each $U \in C$, we (well, you!) should be able to get a Frechet manifold structure on $LieGpd_{ext}(C(U),X)$, which I think is in bijection with $LieGpd(C(\widebar{U}),X)$. Phew! The next stage involves stacks and we need to consider the interplay between stacks on the site of finite-dim manifolds and on the site of Frechet manifolds. But we'll get to that...

    Ok, a little bit of time now to expand on what I said in #25.

    Consider a compact manifold $K$ and a Lie groupoid $X$. A priori, the set of objects of the (bare) groupoid of anafunctors $LieGpd_{ana}(K,X)$ is a coproduct indexed by the set $O$ of open covers of $K$, which is too big. We can restrict to the (equivalent!) full subgroupoid taking the subset indexed by the set $F$ of finite open covers. This is still too big, but better. Now assume that we have a countable subset $C \subset F$ which is closed under intersections (and contains the trivial cover $K\to K$, but we can add this if not) and is cofinal in $F$ under refinement. Then taking the full subgroupoid indexed by $C$ we again get a groupoid equivalent to $LieGpd_{ana}(K,X)$. I know I can do this last step for $K = I$, and presumably for some other simple manifolds like $I^n \times (S^1)^m$, but I think it may be possible to extend the naive techniques I’m thinking of to just use second countability of $K$ together with the fact finite covers are cofinal in all covers, so that there is a countable family as we need.

    Now the property you describe, that each open cover $U$ has a refinement $V$ such that the closure of $V$ (i.e. the closed cover $\widebar{V}$ given by the closure of each open set in $V$) still refines $U$, looks familiar but I can’t quite remember what it is called (something like having a halo or halo-extension or similar). This means we can restrict from all anafunctors $C(U) \to X$ indexed by $C$ to those that extend to the closure of $U$, that is those that extend along $C(U) \to C(\widebar{U})$. Note that we ask that the whole functor extends, not just object or arrow components (the latter is what you were describing in the example).

    If we take the full subgroupoid on these anafunctors, then we again get an equivalent groupoid. Now we consider the disjoint union indexed by $C$, but restricting to these ’extendible’ anafunctors. For each $U \in C$, we (well, you!) should be able to get a Frechet manifold structure on $LieGpd_{ext}(C(U),X)$, which I think is in bijection with $LieGpd(C(\widebar{U}),X)$. Phew!

    The next stage involves stacks and we need to consider the interplay between stacks on the site of finite-dim manifolds and on the site of Frechet manifolds. But we’ll get to that…

27. • CommentRowNumber27.
    • CommentAuthorAndrew Stacey
    • CommentTimeMar 5th 2012
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexSo then we're back with one of my earlier suggestions. When looking at covers of $I$, we use *closed* covers, but all the important properties depend on the interiors. There's probably a proper name for this. Here's the conditions that I would like: Work with $U \subseteq \wp(I)$ where: 1. Elements of $U$ are (unions of?) closed intervals, 1. The interiors of the elements of $U$ cover $I$. If we allow elements of $U$ to be unions of closed intervals, then we should probably insist that each be the closure of its interior. Or we could simply ban singleton sets. I'd rather work with this directly than with open intervals and extensions. The reason being that it is easier to keep track of the topology. The topology (and other structure) in use will depend not just on being extendible but on the actual extension. Although these are equivalent (since the extension is unique), I find it easier to keep track of things this way.

    So then we’re back with one of my earlier suggestions. When looking at covers of $I$, we use closed covers, but all the important properties depend on the interiors. There’s probably a proper name for this. Here’s the conditions that I would like:

    Work with $U \subseteq \wp(I)$ where:

    1. Elements of $U$ are (unions of?) closed intervals,
    2. The interiors of the elements of $U$ cover $I$.

    If we allow elements of $U$ to be unions of closed intervals, then we should probably insist that each be the closure of its interior. Or we could simply ban singleton sets.

    I’d rather work with this directly than with open intervals and extensions. The reason being that it is easier to keep track of the topology. The topology (and other structure) in use will depend not just on being extendible but on the actual extension. Although these are equivalent (since the extension is unique), I find it easier to keep track of things this way.

28. • CommentRowNumber28.
    • CommentAuthorDavidRoberts
    • CommentTimeMar 5th 2012
    • (edited Mar 5th 2012)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexI think of a cover $U \to I$, as we are considering it, as a finite disjoint union of closed intervals such that the interiors cover $I$. Since we don't care so much about the intervals individually, only their disjoint union as a space of $I$, then we don't think of the cover as consisting of a collection $\{U_a\}$ of subspaces of $I$. I suppose more correctly, being a category theorist, I should say that any space over $I$ isomorphic to one of this form is also a cover, but we can be a little sloppy without risk. I think if the interiors form a cover, then we don't have to worry too much about singletons turning up, as the behaviour of the functor on them should be tightly controlled - if the interiors cover $I$, then the intersections of adjacent intervals have non-empty interior. Oh, one other thing, we can restrict to covers by intervals such that only adjacent intervals have the interior of their intersection non-empty. So, where does this leave us with respect to path groupoids? Assuming that a submersion $M \to N$ of (finite dimensional) manifolds induces a submersion $M^I \to N^I$ of Frechet manifolds, we should have little trouble forming the Frechet Lie groupoid $Hom(C(U),X)$ (there is a little matter of the map $M^I \to N^J$ induced by an inclusion of closed intervals $I \to J$).

    I think of a cover $U \to I$, as we are considering it, as a finite disjoint union of closed intervals such that the interiors cover $I$. Since we don’t care so much about the intervals individually, only their disjoint union as a space of $I$, then we don’t think of the cover as consisting of a collection $\{U_a\}$ of subspaces of $I$. I suppose more correctly, being a category theorist, I should say that any space over $I$ isomorphic to one of this form is also a cover, but we can be a little sloppy without risk.

    I think if the interiors form a cover, then we don’t have to worry too much about singletons turning up, as the behaviour of the functor on them should be tightly controlled - if the interiors cover $I$, then the intersections of adjacent intervals have non-empty interior.

    Oh, one other thing, we can restrict to covers by intervals such that only adjacent intervals have the interior of their intersection non-empty.

    So, where does this leave us with respect to path groupoids? Assuming that a submersion $M \to N$ of (finite dimensional) manifolds induces a submersion $M^I \to N^I$ of Frechet manifolds, we should have little trouble forming the Frechet Lie groupoid $Hom(C(U),X)$ (there is a little matter of the map $M^I \to N^J$ induced by an inclusion of closed intervals $I \to J$).