Ok to recap the structure of what you have done up to this point: - Symⁿ(V) is a subrepresentation of the tensor product representation V ⊗ V ⊗ ... ⊗ V (of..? SU(2)? sl(2, ℂ)?); - Any _irreducible_ representation V of sl(2, ℂ)(?) has this basis {v, Yv, ..., Yⁿv}, and is therefore unique up to isomorphism. So the second part shows that _íf_ we would have an irrep, then it must be of that form (span{v, Yv, ..., Yⁿv}). But do we in fact have these? Part one shows that there is a representation Symⁿ(V) for each n, but those are not necessarily irreducible. By complete reducibility, we can look for irreps within Synⁿ(V) I suppose, but then we'd lose sight of what they will look like, and I remember from earlier videos that the irreps are in fact precisely the reps Symⁿ(ℂ²). You said these will turn out to be irreducible, but if you've shown that I must have missed it.
I'm not sure what the question is, but here's a summary. We start with a representation V of SU(2). We get a rep of the complexified lie algebra (sl(2,C)). We pick a highest weight vector v and show that v,Yv,Y^2v,... span a subrep U for sl(2,C) (and hence for SU(2)). Moreover, we know exactly how the generators of the Lie algebra act, and this is determined by the highest weight. So we get a list of representations (one for each nonnegative integer) at least one of which must occur inside any rep, so this list contains all the irreps. In fact they are precisely the Sym^n(C^2): to see this, pick a highest weight vector in Sym^n(C^2): this generates a subrep U on our list and you can see by comparing weight spaces that U has to be the whole of Sym^n(C^2). Finally, we want to know that these are all irreducible. I think this was an exercise in my course, so I don't want to post the answer here! (Hint: suppose there's a subrep, pick a highest weight vector in this subrep and think about how X acts on this weight space)
@@jonathanevans27 Please don't take my silence as disinterest! I'm very grateful but also still trying to work through it. I first want to talk to someone in my institute to help me with this, because I'm still a bit lost on some points, but this is certainly a great help (again)!
You show that U is invariant under the aciton of sp{X, Y, H}, but this is sl(2, ℂ), isn't it? So in fact we now have a subrep of sl(2, ℂ), while we wanted one of SU(2). Now, any rep of a Lie group G induces one of its Lie algebra g, but not necessarily the other way around, is it? So I feel I can't quite tie a bow on it yet..
@Jos: In fact, because SU(2) is simply connected, every lie algebra representation of su(2) comes from a representation of the group. But this doesn't matter for the proof: we're supposing that we have a representation of the group to start with, so if we find a subrep for the action of the Lie algebra it will come from a representation of the group.
Ok to recap the structure of what you have done up to this point:
- Symⁿ(V) is a subrepresentation of the tensor product representation V ⊗ V ⊗ ... ⊗ V (of..? SU(2)? sl(2, ℂ)?);
- Any _irreducible_ representation V of sl(2, ℂ)(?) has this basis {v, Yv, ..., Yⁿv}, and is therefore unique up to isomorphism.
So the second part shows that _íf_ we would have an irrep, then it must be of that form (span{v, Yv, ..., Yⁿv}). But do we in fact have these? Part one shows that there is a representation Symⁿ(V) for each n, but those are not necessarily irreducible. By complete reducibility, we can look for irreps within Synⁿ(V) I suppose, but then we'd lose sight of what they will look like, and I remember from earlier videos that the irreps are in fact precisely the reps Symⁿ(ℂ²). You said these will turn out to be irreducible, but if you've shown that I must have missed it.
I'm not sure what the question is, but here's a summary. We start with a representation V of SU(2). We get a rep of the complexified lie algebra (sl(2,C)). We pick a highest weight vector v and show that v,Yv,Y^2v,... span a subrep U for sl(2,C) (and hence for SU(2)). Moreover, we know exactly how the generators of the Lie algebra act, and this is determined by the highest weight. So we get a list of representations (one for each nonnegative integer) at least one of which must occur inside any rep, so this list contains all the irreps. In fact they are precisely the Sym^n(C^2): to see this, pick a highest weight vector in Sym^n(C^2): this generates a subrep U on our list and you can see by comparing weight spaces that U has to be the whole of Sym^n(C^2). Finally, we want to know that these are all irreducible. I think this was an exercise in my course, so I don't want to post the answer here! (Hint: suppose there's a subrep, pick a highest weight vector in this subrep and think about how X acts on this weight space)
@@jonathanevans27 Please don't take my silence as disinterest! I'm very grateful but also still trying to work through it. I first want to talk to someone in my institute to help me with this, because I'm still a bit lost on some points, but this is certainly a great help (again)!
Is the irrep that is the direct sum of weight space is the general irrep of su(2) or it is still just an irrep of subspace u(1)? Thanks
You show that U is invariant under the aciton of sp{X, Y, H}, but this is sl(2, ℂ), isn't it? So in fact we now have a subrep of sl(2, ℂ), while we wanted one of SU(2). Now, any rep of a Lie group G induces one of its Lie algebra g, but not necessarily the other way around, is it? So I feel I can't quite tie a bow on it yet..
@Jos: In fact, because SU(2) is simply connected, every lie algebra representation of su(2) comes from a representation of the group. But this doesn't matter for the proof: we're supposing that we have a representation of the group to start with, so if we find a subrep for the action of the Lie algebra it will come from a representation of the group.