Warning: one might be tempted to write R|_{T₂} = R^{E_k}, that is: the representation restricted to T₂ is the subrepresentation that leaves E_k invariant. But on dimensionality grounds, this cannot be right: the LHS (R|_{T₂}) is still a 3-dimensional representation, while the RHS is now 2-dimensional.
Ok so I do guess it's kind of obvious: In our study of the representation spaces V of an arbitrary representation R of SU(2), we first only consider the action of the subgroup U(1) of SU(2), yielding a subrepresentation, and allowing us to write V = ⊕_k U_k. The reason we can still do this for SU(3), I think, is that _we didn't use anything about the complement SU(2) \ U(1)_ to arrive at the conclusion (V = ⊕_k U_k). So again, we won't, at first, have to bother with the complement SU(3) \ U(1) to again produce the conclusion that V = ⊕_k U_k (with the appropriate U_k, of course). But now it gets interesting, because in SU(3), we can now do this _twice_! With two "distinct" U(1)'s. Even more so, even as they are distinct, _they dó_ portray a very interesting interplay! ("The subrep of one fixes the other" or something to that tune.) I'll have to work out the details, but I'll talk to my supervisor on Wednesday, so then I'm sure I will have all this very clear. This has been a great help, thanks again!
It seems you answered your own question before I got chance to. You're right: what we used in both cases was the fact that there is a subgroup isomorphic to U(1). Then we restrict the representation to that subgroup and apply the classification theory for U(1) representations. It doesn't matter whether that U(1) is living in SU(2), SU(3) or something completely different. And you're also right that with SU(3) we have two essentially different ways of doing this which (importantly) commute with one another. In SU(2) we could also take two different U(1) subgroups, but they wouldn't commute.
@@jonathanevans27 Oh wow that last part is insightful, I would never have considered another copy of U(1) in SU(2). Or that it mattered whether they'd commute. Before you go on: you don't have to respond, you owe me nothing, I'm not your student, and I will talk to my own supervisor about it tomorrow. I'm just putting my own thoughts in order. That said... :p Another thing is that, yeah, I 'answered' the question, but I didn't fully gét it. Maybe I do now: the fact that restricting our attention to U(1), and immediately obtaining the desired decomposition for an irrep of SU(2) we were just, in a sense, lucky..? i.e., the rep. space U = sp{v, Yv, Y² v, ..., Yⁿv} you constructed sort of _happened to be_ irreducible? It just feels wrong. Where is the rest of SU(2)? Why didn't we have to appeal to more information about the whole of SU(2)? Since it is larger group, the fact that U is invariant under U(1) certainly doesn't mean it also will be under SU(2). Then of course the whole Lie algebra argument ensued, where things get kind of messy for an layman as I, we showed that U was invariant under the Lie _algebra_ *sl*(2, ℂ), and therefore certainly under *su*(2, ℂ), upon which Lie's Homomorphism Theorem -- quite miraculously -- ensures that U will in fact be an invariant subspace of V under the action of _SU(2)._ So maybe that's why it felt wrong to me. A lot of heavy lifting here is done by Lie's Theorem, and happens behind the scenes, in some sense. Ahhh I think I may have something here, thanks :)
Of course, more detail can be added. We didn't _just_ decompose our rep. space V, we decomposed it into summands which were eigenspaces! Which eigenspaces? Well, the eigenspaces of R^ℂ_*(H) at eigenvalues -n + 2k, k ∈ {0, 1, ..., n}! Now we jump to SU(3). Again, we first look at U(1), probably (correct me if I'm wrong) applying the entire Lie algebra argument again¹ (maybe now with *sl*(3, ℂ)), and then, _again by Lie's Homomorphism Theorem??,_ show that we have an invariant subspace that decomposes as the eigenspaces of R(D(θ₁, 0)) at eigenvalues e^(ikθ₁)? How they [these new eigenspaces], in turn, can be decomposed as eigenspaces 'of the other torus' will be a topic for tomorrow's thought :) ¹This may have gone a little less under the radar next time.. :p
Ok, new perspective. And this time perhaps an actually well-thought-out question. ^^ You have made a claim that is either entirely obvious, or entirely unwarranted (in current form). You write V = ⊕ U_k with U_k = [eigenspace of R(D(θ₁, 0)) with eigenvalue e^(ikθ₁)]. This is either 1) entirely obvious because Lie's Theorem immediately translates [eigenspace of R(D(θ₁, 0)) with eigenvalue e^(ikθ₁)] to [eigenspace of R^ℂ_*(D(θ₁, 0)) with eigenvalue k], so we actually still end up with U_k being the eigenspace of a linear operator with integer eigenvalues, _ór_ 2) this 'translation' is indeed true, but entirely non-obvious: we must this time inspect the Lie algebra *su*(3, ℂ), a completely different beast, that is generated not by (i times) the Pauli matrices, but by the (3 × 3) Gell-Mann matrices! Would we not have to prove again that we have a lemma like before, that says that R^ℂ_*(H) acts on W_m like mI (m times the identity matrix)?? At any rate, not looking at *su*(3, ℂ) seems to me like letting Lie do heavier lifting than it can handle all by itself, if you receive my meaning ;). So which is true? 1, 2, 1 ∧ 2 or ¬(1 ∨ 2)? :)
Hi Jos, Sorry for the late reply! The point is that the subgroup of SU(3) given by diagonal matrices e^{i\theta},1,e^{-i\theta} is isomorphic to U(1), so if V is a rep of SU(3) then it is also a rep of this subgroup (just by restriction). Now we apply the earlier result about U(1)-reps (which in particular gives integrality of the weights). The point is to find a subgroup isomorphic to U(1). If you try and do something like this with a Lie group like the Heisenberg group (upper triangular matrices with 1s on the diagonal) then you can't find a subgroup isomorphic to U(1) so this fails, and indeed its rep theory is completely different (its Lie algebra is nilpotent).
@@jonathanevans27 Thanks for any reply at all! But I do now feel that I have since grasped the concepts, so unless you really want to, don't bother to answer all of my other questions! They're certainly not all useful, I fear.. :p
@@jonathanevans27 Yes, I remember now, for some reason I was struggling to see how this new eigenspace (of R(D(0, θ₂)), I think), could play the part of _V_ from our study of general representation spaces of SU(2).. If you're into it, btw, I _dó_ still have a question posed in video 32, which I haven't yet answered.. ^^
Well done Jonathan. Thanks
Warning: one might be tempted to write R|_{T₂} = R^{E_k}, that is: the representation restricted to T₂ is the subrepresentation that leaves E_k invariant. But on dimensionality grounds, this cannot be right: the LHS (R|_{T₂}) is still a 3-dimensional representation, while the RHS is now 2-dimensional.
Ok so I do guess it's kind of obvious: In our study of the representation spaces V of an arbitrary representation R of SU(2), we first only consider the action of the subgroup U(1) of SU(2), yielding a subrepresentation, and allowing us to write V = ⊕_k U_k.
The reason we can still do this for SU(3), I think, is that _we didn't use anything about the complement SU(2) \ U(1)_ to arrive at the conclusion (V = ⊕_k U_k).
So again, we won't, at first, have to bother with the complement SU(3) \ U(1) to again produce the conclusion that V = ⊕_k U_k (with the appropriate U_k, of course). But now it gets interesting, because in SU(3), we can now do this _twice_! With two "distinct" U(1)'s. Even more so, even as they are distinct, _they dó_ portray a very interesting interplay! ("The subrep of one fixes the other" or something to that tune.)
I'll have to work out the details, but I'll talk to my supervisor on Wednesday, so then I'm sure I will have all this very clear.
This has been a great help, thanks again!
It seems you answered your own question before I got chance to. You're right: what we used in both cases was the fact that there is a subgroup isomorphic to U(1). Then we restrict the representation to that subgroup and apply the classification theory for U(1) representations. It doesn't matter whether that U(1) is living in SU(2), SU(3) or something completely different. And you're also right that with SU(3) we have two essentially different ways of doing this which (importantly) commute with one another. In SU(2) we could also take two different U(1) subgroups, but they wouldn't commute.
@@jonathanevans27 Oh wow that last part is insightful, I would never have considered another copy of U(1) in SU(2). Or that it mattered whether they'd commute.
Before you go on: you don't have to respond, you owe me nothing, I'm not your student, and I will talk to my own supervisor about it tomorrow. I'm just putting my own thoughts in order.
That said... :p
Another thing is that, yeah, I 'answered' the question, but I didn't fully gét it. Maybe I do now: the fact that restricting our attention to U(1), and immediately obtaining the desired decomposition for an irrep of SU(2) we were just, in a sense, lucky..? i.e., the rep. space U = sp{v, Yv, Y² v, ..., Yⁿv} you constructed sort of _happened to be_ irreducible? It just feels wrong. Where is the rest of SU(2)? Why didn't we have to appeal to more information about the whole of SU(2)? Since it is larger group, the fact that U is invariant under U(1) certainly doesn't mean it also will be under SU(2). Then of course the whole Lie algebra argument ensued, where things get kind of messy for an layman as I, we showed that U was invariant under the Lie _algebra_ *sl*(2, ℂ), and therefore certainly under *su*(2, ℂ), upon which Lie's Homomorphism Theorem -- quite miraculously -- ensures that U will in fact be an invariant subspace of V under the action of _SU(2)._
So maybe that's why it felt wrong to me. A lot of heavy lifting here is done by Lie's Theorem, and happens behind the scenes, in some sense. Ahhh I think I may have something here, thanks :)
Of course, more detail can be added. We didn't _just_ decompose our rep. space V, we decomposed it into summands which were eigenspaces! Which eigenspaces? Well, the eigenspaces of R^ℂ_*(H) at eigenvalues -n + 2k, k ∈ {0, 1, ..., n}!
Now we jump to SU(3). Again, we first look at U(1), probably (correct me if I'm wrong) applying the entire Lie algebra argument again¹ (maybe now with *sl*(3, ℂ)), and then, _again by Lie's Homomorphism Theorem??,_ show that we have an invariant subspace that decomposes as the eigenspaces of R(D(θ₁, 0)) at eigenvalues e^(ikθ₁)?
How they [these new eigenspaces], in turn, can be decomposed as eigenspaces 'of the other torus' will be a topic for tomorrow's thought :)
¹This may have gone a little less under the radar next time.. :p
Ok, new perspective. And this time perhaps an actually well-thought-out question. ^^
You have made a claim that is either entirely obvious, or entirely unwarranted (in current form).
You write V = ⊕ U_k with U_k = [eigenspace of R(D(θ₁, 0)) with eigenvalue e^(ikθ₁)]. This is either
1) entirely obvious because Lie's Theorem immediately translates [eigenspace of R(D(θ₁, 0)) with eigenvalue e^(ikθ₁)] to [eigenspace of R^ℂ_*(D(θ₁, 0)) with eigenvalue k], so we actually still end up with U_k being the eigenspace of a linear operator with integer eigenvalues, _ór_
2) this 'translation' is indeed true, but entirely non-obvious: we must this time inspect the Lie algebra *su*(3, ℂ), a completely different beast, that is generated not by (i times) the Pauli matrices, but by the (3 × 3) Gell-Mann matrices! Would we not have to prove again that we have a lemma like before, that says that R^ℂ_*(H) acts on W_m like mI (m times the identity matrix)?? At any rate, not looking at *su*(3, ℂ) seems to me like letting Lie do heavier lifting than it can handle all by itself, if you receive my meaning ;).
So which is true? 1, 2, 1 ∧ 2 or ¬(1 ∨ 2)? :)
Hi Jos, Sorry for the late reply! The point is that the subgroup of SU(3) given by diagonal matrices e^{i\theta},1,e^{-i\theta} is isomorphic to U(1), so if V is a rep of SU(3) then it is also a rep of this subgroup (just by restriction). Now we apply the earlier result about U(1)-reps (which in particular gives integrality of the weights). The point is to find a subgroup isomorphic to U(1). If you try and do something like this with a Lie group like the Heisenberg group (upper triangular matrices with 1s on the diagonal) then you can't find a subgroup isomorphic to U(1) so this fails, and indeed its rep theory is completely different (its Lie algebra is nilpotent).
@@jonathanevans27 Thanks for any reply at all! But I do now feel that I have since grasped the concepts, so unless you really want to, don't bother to answer all of my other questions! They're certainly not all useful, I fear.. :p
@@jonathanevans27 Yes, I remember now, for some reason I was struggling to see how this new eigenspace (of R(D(0, θ₂)), I think), could play the part of _V_ from our study of general representation spaces of SU(2)..
If you're into it, btw, I _dó_ still have a question posed in video 32, which I haven't yet answered.. ^^