Automorphisms of which structure form a Lie groupoidWhat is the local structure of a Lie groupoid?Is there a “geometric” language that describes the equivalence groupoid of a foliated manifold?What are the possible symplectic structures on a given Lie groupoid?When are $k$-sectors of a Lie groupoid a manifold?Integrating representations of Lie algebroidsIsotropy group of a Lie groupoid is a Lie groupCondition on a Lie groupoid to be represented by manifold/group or an action groupoidStack associated to Lie group and manifoldLie groupoid cohomologyIsotropy subgroupoid of a regular Lie groupoid

Automorphisms of which structure form a Lie groupoid


What is the local structure of a Lie groupoid?Is there a “geometric” language that describes the equivalence groupoid of a foliated manifold?What are the possible symplectic structures on a given Lie groupoid?When are $k$-sectors of a Lie groupoid a manifold?Integrating representations of Lie algebroidsIsotropy group of a Lie groupoid is a Lie groupCondition on a Lie groupoid to be represented by manifold/group or an action groupoidStack associated to Lie group and manifoldLie groupoid cohomologyIsotropy subgroupoid of a regular Lie groupoid













4












$begingroup$


Given a manifold $M$, the collection of all automorphisms of $M$, denoted by $textAut(M)$ forms a Lie group.



Do we have similar setting in case of Lie groupoid?



Is there "a structure" whose "automorphisms" forms a Lie groupoid?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:29










  • $begingroup$
    what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
    $endgroup$
    – user140765
    May 24 at 6:51










  • $begingroup$
    Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
    $endgroup$
    – Adrien
    May 24 at 6:53










  • $begingroup$
    @kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:57






  • 2




    $begingroup$
    Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
    $endgroup$
    – YCor
    May 24 at 7:03















4












$begingroup$


Given a manifold $M$, the collection of all automorphisms of $M$, denoted by $textAut(M)$ forms a Lie group.



Do we have similar setting in case of Lie groupoid?



Is there "a structure" whose "automorphisms" forms a Lie groupoid?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:29










  • $begingroup$
    what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
    $endgroup$
    – user140765
    May 24 at 6:51










  • $begingroup$
    Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
    $endgroup$
    – Adrien
    May 24 at 6:53










  • $begingroup$
    @kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:57






  • 2




    $begingroup$
    Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
    $endgroup$
    – YCor
    May 24 at 7:03













4












4








4





$begingroup$


Given a manifold $M$, the collection of all automorphisms of $M$, denoted by $textAut(M)$ forms a Lie group.



Do we have similar setting in case of Lie groupoid?



Is there "a structure" whose "automorphisms" forms a Lie groupoid?










share|cite|improve this question











$endgroup$




Given a manifold $M$, the collection of all automorphisms of $M$, denoted by $textAut(M)$ forms a Lie group.



Do we have similar setting in case of Lie groupoid?



Is there "a structure" whose "automorphisms" forms a Lie groupoid?







dg.differential-geometry lie-groupoids






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 24 at 15:19







Praphulla Koushik

















asked May 24 at 6:28









Praphulla KoushikPraphulla Koushik

1,2241626




1,2241626











  • $begingroup$
    I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:29










  • $begingroup$
    what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
    $endgroup$
    – user140765
    May 24 at 6:51










  • $begingroup$
    Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
    $endgroup$
    – Adrien
    May 24 at 6:53










  • $begingroup$
    @kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:57






  • 2




    $begingroup$
    Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
    $endgroup$
    – YCor
    May 24 at 7:03
















  • $begingroup$
    I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:29










  • $begingroup$
    what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
    $endgroup$
    – user140765
    May 24 at 6:51










  • $begingroup$
    Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
    $endgroup$
    – Adrien
    May 24 at 6:53










  • $begingroup$
    @kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
    $endgroup$
    – Praphulla Koushik
    May 24 at 6:57






  • 2




    $begingroup$
    Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
    $endgroup$
    – YCor
    May 24 at 7:03















$begingroup$
I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
$endgroup$
– Praphulla Koushik
May 24 at 6:29




$begingroup$
I have checked examples from some books and some online notes.. I could not find any thing related to "automorphisms" of "a structure" forming a Lie groupoid...
$endgroup$
– Praphulla Koushik
May 24 at 6:29












$begingroup$
what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
$endgroup$
– user140765
May 24 at 6:51




$begingroup$
what is a manifold, and an automorphism of a manifold? Do you mean specifically smooth manifolds and self-diffeomorphisms?
$endgroup$
– user140765
May 24 at 6:51












$begingroup$
Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
$endgroup$
– Adrien
May 24 at 6:53




$begingroup$
Well, automorphisms are by definition self-maps, i.e. maps from something to itself, while somehow the notion of groupoid is precisely designed to encodes isomorphisms between different things as well. So you might want to look e.g. at a collection of manifolds instead.
$endgroup$
– Adrien
May 24 at 6:53












$begingroup$
@kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
$endgroup$
– Praphulla Koushik
May 24 at 6:57




$begingroup$
@kartop_man I think Adrien answers about self maps.. I mean smooth manifold...
$endgroup$
– Praphulla Koushik
May 24 at 6:57




2




2




$begingroup$
Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
$endgroup$
– YCor
May 24 at 7:03




$begingroup$
Automorphisms will always form a group. To get a groupoid, you should consider isomorphisms between various objects and not a single structure.
$endgroup$
– YCor
May 24 at 7:03










2 Answers
2






active

oldest

votes


















5












$begingroup$

Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books.



Take a principal $G$-bundle, $picolon Pto M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m_1$ to $m_2$ is a triple $(m_1,m_2,phi)$ where $phi$ is an isomorphism $phicolon P_m_1 stackrelsimeqto P_m_2$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m_1,m_2,phi) mapsto m_i$ are surjective submersions. Composition is smooth, and inversion is $(m_1,m_2,phi)mapsto (m_2,m_1,phi^-1)$. This gives a Lie groupoid.



In slightly more detail, the collection of all isomorphisms $P_m_1 stackrelsimeqto P_m_2$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m_1$ and $m_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $Ptimes P$ by the diagonal action of $G$, namely $(p_1,p_2) mapsto (p_1g,p_2g)$. The source and target maps are given by $[p_1,p_2]mapsto pi(p_i)$. (Edited to confirm and correct this definition).




The automorphisms $Xstackrelsimeqto X$ of a single object $X$ will never give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:04










  • $begingroup$
    You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
    $endgroup$
    – David Roberts
    May 25 at 0:20










  • $begingroup$
    Thanks thanks... I will see that also..
    $endgroup$
    – Praphulla Koushik
    May 25 at 5:58


















2












$begingroup$

Just some quick addenda to David's comments:



  1. You can see that the automorphisms of a Lie groupoid do not form a richer (groupoid) structure in some more specialised examples. It is well known that proper etale Lie groupoids are a Lie groupoid formulation of orbifolds (they are often called orbifold groupoids). Now their automorphisms corresponds to the group of orbifold diffeomorphisms and these can be turned into an infinite dimensional Lie group (this was the whole point of my PhD thesis, see The diffeomorphism group of a non-compact orbifold). The point is that these objects really form a group and not some higher structure like a groupoid or a $2$-Lie group.


  2. In the literature people have considered groups of smooth maps on the arrow space of a Lie groupoid. Basically you look at diffeomorphisms $operatornameDiff(G)$ which preserve source and target fibres. Again this yields groups not groupoids and has the advantage that the bisection group of the Lie groupoid identifies a subgroup of the automorphism group, called the inner automorphisms (by sending a bisection to the associated left translation). To my knowledge no smooth structure has yet been constructed on this automorphism group in general (though it seems to be an interesting question if there is one).
    For more references let me sneak in my recent paper The Lie group of vertical bisections of a regular Lie groupoid. The references can be found in the introduction together with an explanation of how this is related to the automorphism group.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks.. I will see the paper :)
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:10











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books.



Take a principal $G$-bundle, $picolon Pto M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m_1$ to $m_2$ is a triple $(m_1,m_2,phi)$ where $phi$ is an isomorphism $phicolon P_m_1 stackrelsimeqto P_m_2$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m_1,m_2,phi) mapsto m_i$ are surjective submersions. Composition is smooth, and inversion is $(m_1,m_2,phi)mapsto (m_2,m_1,phi^-1)$. This gives a Lie groupoid.



In slightly more detail, the collection of all isomorphisms $P_m_1 stackrelsimeqto P_m_2$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m_1$ and $m_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $Ptimes P$ by the diagonal action of $G$, namely $(p_1,p_2) mapsto (p_1g,p_2g)$. The source and target maps are given by $[p_1,p_2]mapsto pi(p_i)$. (Edited to confirm and correct this definition).




The automorphisms $Xstackrelsimeqto X$ of a single object $X$ will never give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:04










  • $begingroup$
    You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
    $endgroup$
    – David Roberts
    May 25 at 0:20










  • $begingroup$
    Thanks thanks... I will see that also..
    $endgroup$
    – Praphulla Koushik
    May 25 at 5:58















5












$begingroup$

Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books.



Take a principal $G$-bundle, $picolon Pto M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m_1$ to $m_2$ is a triple $(m_1,m_2,phi)$ where $phi$ is an isomorphism $phicolon P_m_1 stackrelsimeqto P_m_2$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m_1,m_2,phi) mapsto m_i$ are surjective submersions. Composition is smooth, and inversion is $(m_1,m_2,phi)mapsto (m_2,m_1,phi^-1)$. This gives a Lie groupoid.



In slightly more detail, the collection of all isomorphisms $P_m_1 stackrelsimeqto P_m_2$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m_1$ and $m_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $Ptimes P$ by the diagonal action of $G$, namely $(p_1,p_2) mapsto (p_1g,p_2g)$. The source and target maps are given by $[p_1,p_2]mapsto pi(p_i)$. (Edited to confirm and correct this definition).




The automorphisms $Xstackrelsimeqto X$ of a single object $X$ will never give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:04










  • $begingroup$
    You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
    $endgroup$
    – David Roberts
    May 25 at 0:20










  • $begingroup$
    Thanks thanks... I will see that also..
    $endgroup$
    – Praphulla Koushik
    May 25 at 5:58













5












5








5





$begingroup$

Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books.



Take a principal $G$-bundle, $picolon Pto M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m_1$ to $m_2$ is a triple $(m_1,m_2,phi)$ where $phi$ is an isomorphism $phicolon P_m_1 stackrelsimeqto P_m_2$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m_1,m_2,phi) mapsto m_i$ are surjective submersions. Composition is smooth, and inversion is $(m_1,m_2,phi)mapsto (m_2,m_1,phi^-1)$. This gives a Lie groupoid.



In slightly more detail, the collection of all isomorphisms $P_m_1 stackrelsimeqto P_m_2$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m_1$ and $m_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $Ptimes P$ by the diagonal action of $G$, namely $(p_1,p_2) mapsto (p_1g,p_2g)$. The source and target maps are given by $[p_1,p_2]mapsto pi(p_i)$. (Edited to confirm and correct this definition).




The automorphisms $Xstackrelsimeqto X$ of a single object $X$ will never give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.






share|cite|improve this answer











$endgroup$



Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books.



Take a principal $G$-bundle, $picolon Pto M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m_1$ to $m_2$ is a triple $(m_1,m_2,phi)$ where $phi$ is an isomorphism $phicolon P_m_1 stackrelsimeqto P_m_2$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m_1,m_2,phi) mapsto m_i$ are surjective submersions. Composition is smooth, and inversion is $(m_1,m_2,phi)mapsto (m_2,m_1,phi^-1)$. This gives a Lie groupoid.



In slightly more detail, the collection of all isomorphisms $P_m_1 stackrelsimeqto P_m_2$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m_1$ and $m_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $Ptimes P$ by the diagonal action of $G$, namely $(p_1,p_2) mapsto (p_1g,p_2g)$. The source and target maps are given by $[p_1,p_2]mapsto pi(p_i)$. (Edited to confirm and correct this definition).




The automorphisms $Xstackrelsimeqto X$ of a single object $X$ will never give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 24 at 7:50

























answered May 24 at 7:21









David RobertsDavid Roberts

18k464184




18k464184











  • $begingroup$
    Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:04










  • $begingroup$
    You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
    $endgroup$
    – David Roberts
    May 25 at 0:20










  • $begingroup$
    Thanks thanks... I will see that also..
    $endgroup$
    – Praphulla Koushik
    May 25 at 5:58
















  • $begingroup$
    Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:04










  • $begingroup$
    You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
    $endgroup$
    – David Roberts
    May 25 at 0:20










  • $begingroup$
    Thanks thanks... I will see that also..
    $endgroup$
    – Praphulla Koushik
    May 25 at 5:58















$begingroup$
Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
$endgroup$
– Praphulla Koushik
May 24 at 8:04




$begingroup$
Thanks. So, you are not considering the collection of automorphisms of the fibre bundle.. You are considering the collection of isomorphism of fibres $E_xxrightarrow cong E_y$.. This is not what I expected but I am ok with this... This seems to be reasonable version of automorphisms
$endgroup$
– Praphulla Koushik
May 24 at 8:04












$begingroup$
You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
$endgroup$
– David Roberts
May 25 at 0:20




$begingroup$
You might also wish to look at the Lie groupoids of germes of semigroups of local diffeomorphisms, say arising from foliations, eg math.toronto.edu/mein/teaching/MAT1341_LieGroupoids/…
$endgroup$
– David Roberts
May 25 at 0:20












$begingroup$
Thanks thanks... I will see that also..
$endgroup$
– Praphulla Koushik
May 25 at 5:58




$begingroup$
Thanks thanks... I will see that also..
$endgroup$
– Praphulla Koushik
May 25 at 5:58











2












$begingroup$

Just some quick addenda to David's comments:



  1. You can see that the automorphisms of a Lie groupoid do not form a richer (groupoid) structure in some more specialised examples. It is well known that proper etale Lie groupoids are a Lie groupoid formulation of orbifolds (they are often called orbifold groupoids). Now their automorphisms corresponds to the group of orbifold diffeomorphisms and these can be turned into an infinite dimensional Lie group (this was the whole point of my PhD thesis, see The diffeomorphism group of a non-compact orbifold). The point is that these objects really form a group and not some higher structure like a groupoid or a $2$-Lie group.


  2. In the literature people have considered groups of smooth maps on the arrow space of a Lie groupoid. Basically you look at diffeomorphisms $operatornameDiff(G)$ which preserve source and target fibres. Again this yields groups not groupoids and has the advantage that the bisection group of the Lie groupoid identifies a subgroup of the automorphism group, called the inner automorphisms (by sending a bisection to the associated left translation). To my knowledge no smooth structure has yet been constructed on this automorphism group in general (though it seems to be an interesting question if there is one).
    For more references let me sneak in my recent paper The Lie group of vertical bisections of a regular Lie groupoid. The references can be found in the introduction together with an explanation of how this is related to the automorphism group.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks.. I will see the paper :)
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:10















2












$begingroup$

Just some quick addenda to David's comments:



  1. You can see that the automorphisms of a Lie groupoid do not form a richer (groupoid) structure in some more specialised examples. It is well known that proper etale Lie groupoids are a Lie groupoid formulation of orbifolds (they are often called orbifold groupoids). Now their automorphisms corresponds to the group of orbifold diffeomorphisms and these can be turned into an infinite dimensional Lie group (this was the whole point of my PhD thesis, see The diffeomorphism group of a non-compact orbifold). The point is that these objects really form a group and not some higher structure like a groupoid or a $2$-Lie group.


  2. In the literature people have considered groups of smooth maps on the arrow space of a Lie groupoid. Basically you look at diffeomorphisms $operatornameDiff(G)$ which preserve source and target fibres. Again this yields groups not groupoids and has the advantage that the bisection group of the Lie groupoid identifies a subgroup of the automorphism group, called the inner automorphisms (by sending a bisection to the associated left translation). To my knowledge no smooth structure has yet been constructed on this automorphism group in general (though it seems to be an interesting question if there is one).
    For more references let me sneak in my recent paper The Lie group of vertical bisections of a regular Lie groupoid. The references can be found in the introduction together with an explanation of how this is related to the automorphism group.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks.. I will see the paper :)
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:10













2












2








2





$begingroup$

Just some quick addenda to David's comments:



  1. You can see that the automorphisms of a Lie groupoid do not form a richer (groupoid) structure in some more specialised examples. It is well known that proper etale Lie groupoids are a Lie groupoid formulation of orbifolds (they are often called orbifold groupoids). Now their automorphisms corresponds to the group of orbifold diffeomorphisms and these can be turned into an infinite dimensional Lie group (this was the whole point of my PhD thesis, see The diffeomorphism group of a non-compact orbifold). The point is that these objects really form a group and not some higher structure like a groupoid or a $2$-Lie group.


  2. In the literature people have considered groups of smooth maps on the arrow space of a Lie groupoid. Basically you look at diffeomorphisms $operatornameDiff(G)$ which preserve source and target fibres. Again this yields groups not groupoids and has the advantage that the bisection group of the Lie groupoid identifies a subgroup of the automorphism group, called the inner automorphisms (by sending a bisection to the associated left translation). To my knowledge no smooth structure has yet been constructed on this automorphism group in general (though it seems to be an interesting question if there is one).
    For more references let me sneak in my recent paper The Lie group of vertical bisections of a regular Lie groupoid. The references can be found in the introduction together with an explanation of how this is related to the automorphism group.






share|cite|improve this answer











$endgroup$



Just some quick addenda to David's comments:



  1. You can see that the automorphisms of a Lie groupoid do not form a richer (groupoid) structure in some more specialised examples. It is well known that proper etale Lie groupoids are a Lie groupoid formulation of orbifolds (they are often called orbifold groupoids). Now their automorphisms corresponds to the group of orbifold diffeomorphisms and these can be turned into an infinite dimensional Lie group (this was the whole point of my PhD thesis, see The diffeomorphism group of a non-compact orbifold). The point is that these objects really form a group and not some higher structure like a groupoid or a $2$-Lie group.


  2. In the literature people have considered groups of smooth maps on the arrow space of a Lie groupoid. Basically you look at diffeomorphisms $operatornameDiff(G)$ which preserve source and target fibres. Again this yields groups not groupoids and has the advantage that the bisection group of the Lie groupoid identifies a subgroup of the automorphism group, called the inner automorphisms (by sending a bisection to the associated left translation). To my knowledge no smooth structure has yet been constructed on this automorphism group in general (though it seems to be an interesting question if there is one).
    For more references let me sneak in my recent paper The Lie group of vertical bisections of a regular Lie groupoid. The references can be found in the introduction together with an explanation of how this is related to the automorphism group.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 24 at 15:55









LSpice

3,06322630




3,06322630










answered May 24 at 7:57









Alexander SchmedingAlexander Schmeding

1,083710




1,083710











  • $begingroup$
    Thanks.. I will see the paper :)
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:10
















  • $begingroup$
    Thanks.. I will see the paper :)
    $endgroup$
    – Praphulla Koushik
    May 24 at 8:10















$begingroup$
Thanks.. I will see the paper :)
$endgroup$
– Praphulla Koushik
May 24 at 8:10




$begingroup$
Thanks.. I will see the paper :)
$endgroup$
– Praphulla Koushik
May 24 at 8:10

















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