Question about irred rep of SU(2) that factor by the covering map

[marcus]

Let φ SU(2) ---> SO(3) be the double covering

Any irreducible representation of SO(3) pulls back by φ
to provide an irreducible representation of SU(2) on the
same finite dimensional Hilbert space.

this seems clear, almost not worth saying:
the pullback is obviously irred. and has the same dimension.

I have a question about the other direction----suppose an
irred. rep. of SU(2) factors thru φ

then it obviously gives a representation of SO(3), same space
same dimension and all, but

under what circumstances is the representation irreducible?

 

[jeff]

Originally posted by marcus
Let φ SU(2) ---> SO(3) be the double covering

Any irreducible representation of SO(3) pulls back by φ
to provide an irreducible representation of SU(2) on the
same finite dimensional Hilbert space.

this seems clear, almost not worth saying:
the pullback is obviously irred. and has the same dimension.

I have a question about the other direction----suppose an
irred. rep. of SU(2) factors thru φ

then it obviously gives a representation of SO(3), same space
same dimension and all, but

under what circumstances is the representation irreducible?

A simpler way to think about this is to note that the natural isomorphism SU(2)/Z₂ ≅ SO(3) trivially induces an isomorphism between representations, but doesn't determine which ones. That, and in particular whether they're irreducible, is usually determined by the choice of space of objects on which they operate.

One interesting fact is that despite the way that switching corresponding reps will usually produce equivalent theories on the classical level, it will in general produce different theories on the quantum level. For example, they may differ in the spectra of their observables.

 

[marcus]

Originally posted by jeff
A simpler way to think about this is to note that the natural isomorphism SU(2)/Z₂ ≅ SO(3) trivially induces an isomorphism between representations, but doesn't determine which ones. That, and in particular whether they're irreducible, is usually determined by the choice of space of objects on which they operate.

One interesting fact is that despite the way that switching corresponding reps will usually produce equivalent theories on the classical level, it will in general produce different theories on the quantum level. For example, they may differ in the spectra of their observables.

Jeff has not answered the question, which still stands as asked, namely when, under what conditions, is the representation obtained in this way irreducible?

I believe I have figured out the answer, but for the time being will leave the problem up, in case any mathematically inclined person here wants to think about it.

To restate it: Suppose there is an irreducible representation ψ of SU(2) which factors thru the covering map φ

That is, there exists a map σ such that the representation ψ = σφ

Then σ is a clearly a representation of SO(3) in the operators on the same Hilbert space, and the question is-----is σ irreducible?

If it is not always, what under what conditions will it be?

One possibility is that σ is always irreducible---this would require a proof. Another possibility is that it depends on the dimension of the Hilbert space.

Here's the original post for reference:

********************
Let φ SU(2) ---> SO(3) be the double covering

Any irreducible representation of SO(3) pulls back by φ
to provide an irreducible representation of SU(2) on the
same finite dimensional Hilbert space.

this seems clear, almost not worth saying:
the pullback is obviously irred. and has the same dimension.

I have a question about the other direction----suppose an
irred. rep. of SU(2) factors thru φ

then it obviously gives a representation of SO(3), same space
same dimension and all, but

under what circumstances is the representation irreducible?

 

[jeff]

Originally posted by marcus
Jeff has not answered the question

Respectfully, I have, my point being that because representations may be trivially viewed as "identity functions" on their respective groups, and the special relation between SU(2) and SO(3), there's no real need to involve pullbacks etc. So yes, you can set up the same correspondence in both directions with a surjection in the one and an injection in the other, which is made obvious by the isomorphism I defined. But anything more about the corresponding reps is a matter of choice that depends on what you're doing with them. They're simply is no hard and fast rule about the nature of the reps that follows directly from the correspondence between their respective group manifolds.

I would appreciate very much being addressed personally.

This is not you're personal forum and you should know that I'm not going to be giving you any sort of free pass.

I don't know if anything will come of it, but you should also know that I've reported some of your posts to the administrator and will continue to do so until you stop reacting in such a manifestly inappropriate and unjustifiably personal and insulting way.

 

[marcus]

the answer is ALWAYS

I guess I've left this question up on the board long enough for anyone who wanted to try it to reply. No one has given the answer (although an attempt was made). So here is the
answer: ALWAYS.

If a irreducible representation of SU(2) factors thru the covering
the resulting map defined on SO(3)

(i) is a representation of SO(3) on the same vector space, and
(ii) is irreducible.

Originally posted by marcus
Jeff has not answered the question, which still stands as asked, namely when, under what conditions, is the representation obtained in this way irreducible?

I believe I have figured out the answer, but for the time being will leave the problem up, in case any mathematically inclined person here wants to think about it.

To restate it: Suppose there is an irreducible representation ψ of SU(2) which factors thru the covering map φ

That is, there exists a map σ such that the representation ψ = σφ

Then σ is a clearly a representation of SO(3) in the operators on the same Hilbert space, and the question is-----is σ irreducible?

If it is not always, what under what conditions will it be?

One possibility is that σ is always irreducible---this would require a proof. Another possibility is that it depends on the dimension of the Hilbert space.

 

[jeff]

Originally posted by marcus
I guess I've left this question up on the board long enough for anyone who wanted to try it to reply. No one has given the answer (although an attempt was made). So here is the
answer: ALWAYS.

If a irreducible representation of SU(2) factors thru the covering
the resulting map defined on SO(3)

(i) is a representation of SO(3) on the same vector space, and
(ii) is irreducible.

Yes, that's true. In this case, we chose an SO(3) vector space and the action of an IR upon it. But these choices were not determined by the mapping between the two lie groups.

 

[marcus]

When does an irred rep of SU(2) factor thru the covering?

We seem to be on a roll here. Here is another question related to representations which kind of follows up on the previous one.

When does an irredicible representation of SU(2) factor through the covering map φ?

***************

Hints, discussion etc: As before the answer might be "always" or it could depend on something, like the dimension of the vector space of the representation.

To spell the question out in more detailed fashion, if
ψ is an irreducible representation of SU(2) on a vector space V of dimension n, then under what circumstances can one say that there exists an irreducible representation σ of SO(3)
such that

ψ = σ φ

where φ denotes the double covering.

 

[marcus]

Originally posted by marcus
We seem to be on a roll here. Here is another question related to representations which kind of follows up on the previous one.

When does an irredicible representation of SU(2) factor through the covering map φ?

***************

Hints, discussion etc: As before the answer might be "always" or it could depend on something, like the dimension of the vector space of the representation.

To spell the question out in more detailed fashion, if
ψ is an irreducible representation of SU(2) on a vector space V of dimension n, then under what circumstances can one say that there exists an irreducible representation σ of SO(3)
such that

ψ = σ φ

where φ denotes the double covering.

The answer is NOT ALWAYS but only when the dimension of the vector space is odd.

(I'm answering because this has been up for a day or so, just for closure.)

 

[jeff]

Again, we agree. I took your question to be about conditions on the map between the two manifold groups. My answer was that it doesn't matter.

The reason for the mix up is that the answer to the question that you actually asked was so mind-numbingly obvious, it didn't occur to me that it was a question anyone who understands group theory would ever ask.

 

[marcus]

Originally posted by jeff
Again, we agree. I took your question to be about conditions on the map between the two manifold groups. My answer was that it doesn't matter.

The reason for the mix up is that the answer to the question that you actually asked was so mind-numbingly obvious, it didn't occur to me that it was a question anyone who understands group theory would ever ask.

Your question on 6-13 in the SO(3)+LQG thread:

Originally posted by jeff on 6-13
Here's a question inspired by various posts above.

In LQG spin networks, are equal integer spin IRs (irreducible representations) of SU(2) and SO(3) interchangeable?

 

[jeff]

Originally posted by marcus
Your question on 6-13 in the SO(3)+LQG thread:

The point of that question isn't mathematical, it's physical. My question was whether there is a physical reason that such exchanges of representations may not be allowed in LQG. Such exchanges are usually allowed on the classical level, but not on the quantum level. For example, the spectra of observables could change, suggesting that in this case the area operator should be scrutinized.