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In my example, the isotropy group is $A_n$, not $S_n$.
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André Henriques
André Henriques
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One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$$A_n$ at point -1" with "$S_n$$A_n$ at point +1" via any group element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$$A_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(S_n)$$Out(A_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$A_n$ at point -1" with "$A_n$ at point +1" via any element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $A_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(A_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

deleted 2 characters in body; deleted 3 characters in body
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André Henriques
André Henriques
  • 43.2k
  • 5
  • 130
  • 264

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is the element ofits class in $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $M$$m\in M$ mapping to $x$. A morphisAn arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is the element of $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $M$ mapping to $x$. A morphis from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

added 2000 characters in body; deleted 15 characters in body; deleted 4 characters in body; added 1 characters in body; edited body; added 12 characters in body
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André Henriques
André Henriques
  • 43.2k
  • 5
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  • 264

One topicsituation in which it is essential to use groupoids is the study of orbifolds.

Slogan: the set of points of an orbifold is a groupoid.The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Here's a concrete classification problem thatQuestion: What is illuminated by the language"monodromy" of groupoids.going around that stratum?

Suppose I haveAt first glance, one might guess that it's an orbifoldelement of $M$ and a singuar stratum$Aut(G)$.
That's wrong! The monodromy is an element of $X$ that$Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is isomorphica group associated to every point of $S^1$$X$. Yet, and whose isotropythe monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is the element of $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the symmetric groupconfusion.
It all becomes more clear once you realize that the thing that is associated to a point of $S_n$$X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $M$ mapping to $x$. A morphis from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

[Sorry, I pressedThe monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to quickly onfix any of the "post" button.objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices. I'm still writing up my answer]

One topic in which it is essential to use groupoids is the study of orbifolds.

Slogan: the set of points of an orbifold is a groupoid.

Here's a concrete classification problem that is illuminated by the language of groupoids.

Suppose I have an orbifold $M$ and a singuar stratum $X$ that is isomorphic to $S^1$, and whose isotropy group is the symmetric group $S_n$.

[Sorry, I pressed to quickly on the "post" button... I'm still writing up my answer]

One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.

Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is the element of $Out(S_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $M$ mapping to $x$. A morphis from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

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André Henriques
André Henriques
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