# Mackey Theory

1. Oct 17, 2006

### Oxymoron

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

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 dont 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?

Last edited: Oct 17, 2006
2. Oct 17, 2006

### 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.

Last edited: Oct 17, 2006
3. Oct 17, 2006

### Oxymoron

Ah yes of course.

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

Last edited: Oct 17, 2006
4. Oct 17, 2006

### 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?

5. Oct 17, 2006

### matt grime

because of the definition of orbit.

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

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.

6. Oct 17, 2006

### matt grime

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.

7. Oct 17, 2006

### Oxymoron

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?

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]?

8. Oct 17, 2006

### 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.

9. Oct 17, 2006

### matt grime

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...

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

and they are the same thing.

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

10. Oct 21, 2006

### Oxymoron

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

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).

Last edited: Oct 21, 2006
11. Oct 21, 2006

### 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.

12. Oct 21, 2006

### matt grime

a pointless restriction but what the heck...

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.

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.

13. Oct 21, 2006

### Oxymoron

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 Im 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.

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.

Last edited: Oct 21, 2006
14. Oct 21, 2006

### 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}.