Understanding the Double Coset H\backslash G / K

[Oxymoron]

I was reading an introduction to something called Mackey Theory in a Representation Theory pdf and I came across the following statement:

Definition

The double coset $H\backslash G / K$ is the set of $H \times K$-orbits on $G$, for the left action of $H$ and the right action of $K$.

Does this mean that the double coset $H\backslash G / K$ can be understood to be a set of orbits? That is, a set of orbits of the left action of the product group $H \times K$ on $G$?

What I don't get is this: I've also read that the double coset is the set of orbits for the left action of $H$ on the coset $G/K$ induced by the action $(h,g)\mapsto hg$ of $H$ on $G$.

So which definition of the double coset is correct? Or are they the same?

 

[matt grime]

Ok, let's sort that out.

1. H\G/K is not a double coset. It is the set of double cosets. A double coset is usually written HgK and it is the set { hgk : h in H and k in K}.

2. Each double coset is an orbit under the action of HxK (this is not a left action of HxK, it is a left-right action, H has a left action, and K has a right action, in terms of the orits, though this is the same as the left action of HxK on G by (h,k)g = hgk^{-1}). So H\G/K is a set of orbits.

3. Why don't you try to prove or disprove the equivalence of your two definitions. What I wrote in 1. should make it trivial.

 

[Oxymoron]

Posted by Matt Grime:

1. H\G/K is not a double coset. It is the set of double cosets. A double coset is usually written HgK and it is the set { hgk : h in H and k in K}.

Ah yes of course.

Posted by Matt Grime:

2. Each double coset is an orbit under the action of HxK (this is not a left action of HxK, it is a left-right action, H has a left action, and K has a right action, in terms of the orits, though this is the same as the left action of HxK on G by (h,k)g = hgk^{-1}). So H\G/K is a set of orbits.

May I ask how you know that each double coset is an orbit under the action of $H \times K$? I mean, as far as I know (which isn't very far) the orbit (under the action of $H\times K$) of an element g in G is a set. It is a set of all elements of G to which g can be mapped to by the elements of $H \times K$. That is, you have to take an element of $H\times K$, say $(h,k)$ and an element of $G$, say $g$, and then see where $((h,k),g)$ gets mapped to. Then the set of all elements to which g gets mapped to is the orbit of g. But then elements of H\G/K look like HgK - how does this mean that HgK are orbits?

Also, you said the action on G is given by $(h,k)g = hgk^{-1}$. Why not $(h,k)g = hgk$?

So are you saying that you can think of the set of double cosets, H\G/K, as a set of orbits!? In two ways then?

1. The LEFT action of H on G/K by $(h,g)\mapsto hg$
2. The RIGHT action of K on H\G by $(k,g)\mapsto gk^{-1}$

 

[Oxymoron]

"Each double coset is an orbit of $g \inG$ under the action of $H \times K$"

The action of $H$ on $G$ is a function. It takes an element of $H$ and an element of $G$ and maps it to a product of those two elements in the following way:

$$G \times H \rightarrow H$$
$$(g,h) \mapsto gh$$

which satisfies

$$eh = h \quad \forall\,h \in H$$
$$(gh)g' = g(hg') \quad \forall\,g,\,g' \in G \mbox{ and } \forall\,h \in H$$


So following from this basic definition, the action of $H \times K$ on $G$ is a function

$$G \times (H \times K) \rightarrow (H\times K)$$
$$(g,(h,k)) \mapsto g(hk)$$

what is wrong with this map?

 

[matt grime]

May I ask how you know that each double coset is an orbit under the action of $H \times K$?

because of the definition of orbit.

I mean, as far as I know (which isn't very far) the orbit (under the action of $H\times K$) of an element g in G is a set. It is a set of all elements of G to which g can be mapped to by the elements of $H \times K$. That is, you have to take an element of $H\times K$, say $(h,k)$ and an element of $G$, say $g$, and then see where $((h,k),g)$ gets mapped to. Then the set of all elements to which g gets mapped to is the orbit of g. But then elements of H\G/K look like HgK - how does this mean that HgK are orbits?

Look how I defined HgK. It was precisely the orbit of g under the action of HxK.

Also, you said the action on G is given by $(h,k)g = hgk^{-1}$. Why not $(h,k)g = hgk$?

Actually I said the precise opposite of that. The action of HxK is such that (h,g) sends g to hgk. It is neither a left nor a right action.

 

[matt grime]

what is wrong with this map?

It is ass-backward, that is what is wrong with it. Your definition of the action of H on G is at best an action of G on H, and I have no idea what the part immediately following 'which satisfies' comes from.

 

[Oxymoron]

Posted by Matt Grime:

because of the definition of orbit.

But the definition of orbit is "a the set of elements to which a group element can be mapped to under a group action". How then is HgK one of these?

Posted by Matt Grime:

Look how I defined HgK. It was precisely the orbit of g under the action of HxK.

Yeah, but I define HgK as equivalence classes. And I also define H\G/K as a set of equivalence classes of the relation ~ where

$$x\sim y \Leftrightarrow \exists h \in H, k \in K \mbox{ such that }y = hgk$$

This relation is an equivalence relation whose classes are of the form HgK where g is in G. Then I write H\G/K as the set of classes of the relation ~.

So forgive me if I am having trouble understanding the apparently trivial idea that H\G/K is the set of orbits, because I am very much used to thinking of H\G/K as the set of equivalence classes of ~.

So back to the quote - let me restate my question: You define HgK (not as an equivalence class) but precisely as the orbit of g under the action of HxK. How does this relate to thinking of HgK as equivalence classes [g]?

 

[matt grime]

The equivalence relation is 'x~y if they are in the same orbit'. If you read what you have written yourself it should be clear that these are all the same way of saying the same thing.

 

[matt grime]

But the definition of orbit is "a the set of elements to which a group element can be mapped to under a group action". How then is HgK one of these?

that is not quite the definition of orbit. You must forget that G is a group. It is not important. It is just a set that H and K act on, albeit by virtue of the fact that they are subgroups. What did I define HgK to be? The set of elements of the form hgk for all h,k in H and K resp. this is precisely an equivalence class of...

Yeah, but I define HgK as equivalence classes. And I also define H\G/K as a set of equivalence classes of the relation ~ where

$$x\sim y \Leftrightarrow \exists h \in H, k \in K \mbox{ such that }y = hgk$$

Aye, so they're in the same orbit if and only if they are related by ~.

This relation is an equivalence relation whose classes are of the form HgK where g is in G. Then I write H\G/K as the set of classes of the relation ~.

So forgive me if I am having trouble understanding the apparently trivial idea that H\G/K is the set of orbits, because I am very much used to thinking of H\G/K as the set of equivalence classes of ~.

and they are the same thing.

So back to the quote - let me restate my question: You define HgK (not as an equivalence class) but precisely as the orbit of g under the action of HxK. How does this relate to thinking of HgK as equivalence classes [g]?

Being in the same orbit is an equivalence relation. It is precisely ~. The orbits are the equivalence classes.

 

[Oxymoron]

Let's see if I can get it right this time...

Posted by Matt Grime:

1. H\G/K is not a double coset. It is the set of double cosets. A double coset is usually written HgK and it is the set { hgk : h in H and k in K}.

So

$$H\backslash G/K = \{HgK\,:\,g \in G\}$$

That is, H\G/K is the set of all double cosets

$$HgK = \{hgk\,:\,h \in H\,\,k\in K\}$$

Now, the orbit of an element $g \in G$ (where, as you pointed out, G can be treated simply as a set) is merely a set of elements of G to which g can be mapped to under some left action of the group element. (Note: I only want to consider left actions!)

Suppose the left action on g is by the set $H\times K$, where

$$(h,k) \in H \times K$$

Hence we obtain a left action of $H \times K$ on G which is simply a function

$$(H\times K) \times G \rightarrow G$$

given by

$$((h,k),g) \mapsto hgk^{-1}$$

QUESTION: Why is there an inverse on the k? I reckon it is because I am only considering the left action? See below.

Now, suppose we act on g by $H \times K$ and see where it gets mapped to. The collection of all these points is called the orbit of g. Suppose we act on every element g in G by $H \times K$ and see where they all get mapped to, and then collect up all those elements. Then we have a set of orbits.

QUESTION: Does this "set of orbits" correspond to the set of double cosets, H\G/K?

Well, since H\G/K is defined to be the set of all double cosets, and each double coset is of the form HgK, then each of them is a union of right cosets, Hg and left cosets gK (But, by definition, Hg and gK are orbits themselves!) - That is, each double coset is a union of orbits for the group action of $H \times K$ on G with H acting by left multiplication hg and K acting by right multiplication gk. But I made the clear indication that I wanted left actions. So, I believe that in order to have K act on the left by multiplication, I must act on g by $k^{-1}$. i.e. $hgk^{-1}$.Now let me introduce $G/K$ as the group G modulo K. In this setting I believe that H\G/K can be understood to be simply the set of orbits of the left action of H on G/K! - where the action is by left multiplication in the following way:

$$(h,g) \mapsto hg$$

This works simply because elements of $G/K$ are of the form gK where K represents an element from the group K. Therefore, all elements of $G/K$ are already orbits! Therefore, if we take an element

$$h \in H$$

and

$$g \in G/K$$

then the left action of H on $G/K$ is simply

$$(h,g)\mapsto hg$$

and since g is an element of G/K this takes care of everything (I think).

 

[Oxymoron]

Furthermore, H\G/K can be viewed as the set of orbits of the left action of K on $H \backslash G$ by the action $(k,g)\mapsto gk^{-1}$ where the inverse is because K is forced to act on the left.

 

[matt grime]

(Note: I only want to consider left actions!)

a pointless restriction but what the heck...

QUESTION: Why is there an inverse on the k? I reckon it is because I am only considering the left action? See below.

correct. Surely you're aware that the action g -->gk is not a left action, but it is a right action. The distinction is largely irrelevant here.

QUESTION: Does this "set of orbits" correspond to the set of double cosets, H\G/K?

Since an orbit is precisely the same thing as a double coset what do you think? You do realize that the following sets are the same:

{ hgk : h in H, k in K}
{hgk^{-1} : h in H k in K}

don't you? The first is a double coset the second an orbit.

 

[Oxymoron]

Posted by Matt Grime:a pointless restriction but what the heck...

Not really!...well maybe...If I consider only left actions (in all cases) my job at understanding H\G/K as the set of orbits of the (left) action of K on H\G induced by the action $(h,g)\mapsto gk^{-1}$ of K on G by right multiplication becomes more consistent and nicer looking - without having to keep interchanging which side I am acting on. Alternatively I could reverse everything and consider right actions, but it would be a shame to have to keep track of the chirality - why not simply stick with one.

Posted by Matt Grime:

Since an orbit is precisely the same thing as a double coset what do you think? You do realize that the following sets are the same:

{ hgk : h in H, k in K}
{hgk^{-1} : h in H k in K}

don't you?

No I dont. This is probably the reason why I haven't immediately sprung to realize the synonymity between the set of orbits and the set of double cosets.

 

[matt grime]

Ok. If you did understand that it would make it all very obvious. I thought that since inverting elements is a bijection on a group you'd've realized this. The reason not to bother keeping track of what side the action on is because it doesn't matter: Every left action of G is a right action and every right action of G using the inversion k-->k^{-1}.