Register to reply 
Mackey Theory 
Share this thread: 
#1
Oct1706, 03:50 AM

P: 867

I was reading an introduction to something called Mackey Theory in a Representation Theory pdf and I came across the following statement:
What I dont get is this: I've also read that the double coset is the set of orbits for the left action of [itex]H[/itex] on the coset [itex]G/K[/itex] induced by the action [itex](h,g)\mapsto hg[/itex] of [itex]H[/itex] on [itex]G[/itex]. So which definition of the double coset is correct? Or are they the same? 


#2
Oct1706, 06:10 AM

Sci Advisor
HW Helper
P: 9,397

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 leftright 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. 


#3
Oct1706, 10:03 AM

P: 867

Also, you said the action on G is given by [itex](h,k)g = hgk^{1}[/itex]. Why not [itex](h,k)g = hgk[/itex]? 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 [itex](h,g)\mapsto hg[/itex] 2. The RIGHT action of K on H\G by [itex](k,g)\mapsto gk^{1}[/itex] 


#4
Oct1706, 10:24 AM

P: 867

Mackey Theory
"Each double coset is an orbit of [itex]g \inG[/itex] under the action of [itex]H \times K[/itex]"
The action of [itex]H[/itex] on [itex]G[/itex] is a function. It takes an element of [itex]H[/itex] and an element of [itex]G[/itex] and maps it to a product of those two elements in the following way: [tex]G \times H \rightarrow H[/tex] [tex](g,h) \mapsto gh[/tex] which satisfies [tex]eh = h \quad \forall\,h \in H[/tex] [tex](gh)g' = g(hg') \quad \forall\,g,\,g' \in G \mbox{ and } \forall\,h \in H[/tex] So following from this basic definition, the action of [itex]H \times K[/itex] on [itex]G[/itex] is a function [tex]G \times (H \times K) \rightarrow (H\times K)[/tex] [tex](g,(h,k)) \mapsto g(hk)[/tex] what is wrong with this map? 


#5
Oct1706, 03:44 PM

Sci Advisor
HW Helper
P: 9,397




#6
Oct1706, 03:47 PM

Sci Advisor
HW Helper
P: 9,397

It is ***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
Oct1706, 06:24 PM

P: 867

[tex]x\sim y \Leftrightarrow \exists h \in H, k \in K \mbox{ such that }y = hgk[/tex] 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
Oct1706, 06:42 PM

Sci Advisor
HW Helper
P: 9,397

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
Oct1706, 06:47 PM

Sci Advisor
HW Helper
P: 9,397




#10
Oct2106, 10:17 AM

P: 867

Let's see if I can get it right this time...
[tex]H\backslash G/K = \{HgK\,:\,g \in G\}[/tex] That is, H\G/K is the set of all double cosets [tex]HgK = \{hgk\,:\,h \in H\,\,k\in K\}[/tex] Now, the orbit of an element [itex]g \in G[/itex] (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 [itex]H\times K[/itex], where [tex](h,k) \in H \times K[/tex] Hence we obtain a left action of [itex]H \times K[/itex] on G which is simply a function [tex](H\times K) \times G \rightarrow G[/tex] given by [tex]((h,k),g) \mapsto hgk^{1}[/tex] 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 [itex]H \times K[/itex] 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 [itex]H \times K[/itex] 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 [itex]H \times K[/itex] 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 [itex]k^{1}[/itex]. i.e. [itex]hgk^{1}[/itex]. Now let me introduce [itex]G/K[/itex] 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: [tex](h,g) \mapsto hg[/tex] This works simply because elements of [itex]G/K[/itex] are of the form gK where K represents an element from the group K. Therefore, all elements of [itex]G/K[/itex] are already orbits! Therefore, if we take an element [tex]h \in H[/tex] and [tex]g \in G/K[/tex] then the left action of H on [itex]G/K[/itex] is simply [tex](h,g)\mapsto hg[/tex] and since g is an element of G/K this takes care of everything (I think). 


#11
Oct2106, 10:37 AM

P: 867

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



#12
Oct2106, 10:52 AM

Sci Advisor
HW Helper
P: 9,397

a pointless restriction but what the heck... { 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
Oct2106, 11:00 AM

P: 867




#14
Oct2106, 01:36 PM

Sci Advisor
HW Helper
P: 9,397

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}.



Register to reply 
Related Discussions  
Algorithmic information theory and the existence of a Unified Field Theory  General Physics  5  
String theory ~ the theory of physical theory?  Beyond the Standard Model  32  
Applied mathematics of Game Theory overlooked as a representaiton in string theory?  Beyond the Standard Model  0  
Simple contradiction/flaw in many worlds theory/theory explanation.  General Discussion  31  
Matrix string theory, contact terms, and superstring field theory  Beyond the Standard Model  2 