SU(2) a double cover for Lorentz group????

strangerep

Quote by friend

Yes, I've looked into Clifford Algebra a bit. It makes use of hypercomplex numbers like quaternions and octonions. As I recall, complex numbers are a representation of U(1), quaternions are a representation of SU(2), and octonions are a representation of SU(3), right?

Octonions are nonassociative. But this is drifting outside my competence areas.

I wonder if your copy of Greiner covers the CM theorem, maybe in the last chapter?

I don't think so. Greiner's textbooks are introductory. I studied the CM theorem from the original paper and from an early chapter of Weinberg vol 3.

Quote by friend

"the Coleman-Mandula theorem says is that if special relativity is true, you're very restricted in your choice of internal group"... (namely U(1), SU(2), and SU(3) I assume).

And,

The Coleman-Mandula theorem states that the most general symmetries that a quantum field theory (QFT) can possess are Lorentz invariance (special relativity) and gauge symmetries like conservation of charge, lepton number, etc. (whose generators belong to Lie Algebras [33]).

And that's very close to what I'm trying to prove.

None of those statements about the CM thm, nor the more extensive quotes from the originating urls, are reliable. Too often, people try to "interpret" the CM no-go result, applying their own spin to it, but don't state the actual theorem. I get a bit tired of that sort of thing, actually.

There is a reasonable statement of the CM theorem in this paper:
http://arxiv.org/abs/hep-th/9605147

Since there is no coordinate independent form of QFT, the CM thm must start with a Lorentz metric, right?

The fields in QFT are constructed as "unitary irreducible representations of the Poincare group". There is a large amount of math to understand inside these quotation marks, and until you become able to explain what my quoted phrase means in full mathematical detail, your chances of achieving anything worthwhile in this direction are exactly zero. You're trying to bake a cake before you've learned what flour is.

And then given the QFT formulation, the only allowed symmetries are U(1), SU(2), and SU(3).

No, the CM thm doesn't say that.

Except I'm trying to go the other way, given U(1), SU(2), and SU(3) does the Lorentz Group or metric follow from that?

No. One starts from Poincare invariance and constructs local gauge fields and interactions that are compatible with Poincare invariance. The gauge groups are chosen by essentially phenomenological means, i.e., "these choices give theories compatible with experiment".

Some statements of the CM thm state for QFT you can only have both the Lorentz and these Lie groups.

That's not what the CM thm says.

Does that mean given QFT and the Lie groups you must have the Lorentz group?

No, that's backwards. One constructs QFT by combining quantum principles with special relativity. That's part of the (broad) meaning of "unitary irreducible representations of the Poincare group".

[...] And I may have both QM and SR from logic.
Thank you for helping realize that.

No, you're hearing what you want to hear, not what I'm actually trying tell you.

friend

Quote by strangerep

There is a reasonable statement of the CM theorem in this paper:
http://arxiv.org/abs/hep-th/9605147

Yea, I tried reading that. I feel like I can understand some of the definitions separately, but I'm at a total lost when it comes to why they would define things that way or what the implication of those statements mean. It's not complete gibberish to me. But it's too involved to have any value for me. I could have used a little more "interpretation" in the conclusion section.

Quote by strangerep

The fields in QFT are constructed as "unitary irreducible representations of the Poincare group". There is a large amount of math to understand inside these quotation marks, and until you become able to explain what my quoted phrase means in full mathematical detail, your chances of achieving anything worthwhile in this direction are exactly zero. You're trying to bake a cake before you've learned what flour is.

At this point I'm only trying to learn with an appreciation of the potential it may hold for my efforts. Does your quote mean the generators of the internal symmetries commutes with the elements of the Poincare group?

Quote by strangerep

One starts from Poincare invariance and constructs local gauge fields and interactions that are compatible with Poincare invariance. The gauge groups are chosen by essentially phenomenological means, i.e., "these choices give theories compatible with experiment".

because there is no other formulism to give internal symmetries other than by experiment. Instead of curve fitting, we're symmetry fitting. Wouldn't it be nice if we had a means of providing the internal symmetries on principle alone?

Quote by strangerep

That's not what the CM thm says.
No, that's backwards. One constructs QFT by combining quantum principles with special relativity. That's part of the (broad) meaning of "unitary irreducible representations of the Poincare group".

I also found this on the web, which might prove to be a little more of a gentle introduction and still be useful:

This a bit of a sketch;

The S-matrix acts on to shift the state or momentum state of a particle. A state with two particle states |p,p′> is acted upon by the S-matrix through the T matrix

S = 1 – i(2π)⁴δ⁴(p – p′)T

So that T|p,p′> ≠ 0 . For zero mass plane waves scatter at almost all energy. The Hilbert space is then an infinite product of n-particle subspaces H = ⊗_n Hⁿ . As with all Hilbert spaces there exists a unitary operator U , often U = exp(iHt) , which transforms the states S acts upon. U transforms n-particle states into n-particle states as tensor products. The unitary operator commutes with the S-matrix

SUS⁻¹ = [1–i(2π)⁴δ⁴(p – p′)T]U[1 + i(2π)⁴δ⁴(p – p′)T^†]

=U + i(2π)⁴δ⁴(p – p′)[TU – UT^†] + [(2π)⁴δ⁴(p – p′)]² (TUT^†).

By Hermitian properties and unitarity it is not difficult to show the last two terms are zero and that the S-matrix commutes with the unitary matrix. The Lorentz group then defines operator p_μ and M_μν for momentum boosts and rotations. The S-matrix defines changes in momentum eigenstates, while the unitary operator is generated by a internal symmetries A_a, where the index a is within some internal space (the circle in the complex plane for example, and we then have with some

[A_a , p_μ ] = [A_a , M_μν ] = 0.

This is a sketch of the infamous “no-go” theorem of Coleman and Mundula. This is what prevents one from being able to place internal and external generators or symmetries on the same footing.

What seems key to me is the last commutator relation where the internal symmetries, A_a commute with the elements of the Lorentz group p_μ and M_μν.

Now as I understand you, this means that given the p_μ and M_μν, these commutator relations restrict the kind of internal symmetries that can exist.

But I wonder if it can also be read, given the internal symmetries these commutators restrict the spacetime symmetries to the Lorentz group? In other words, once these commutators have been proven to be TRUE, does it matter how they were proven, that they were proven with the assumption of the Lorentz group? Can we not just take these commutators to be inherently true and use them as I've suggested?

Quote by strangerep

No, you're hearing what you want to hear, not what I'm actually trying tell you.

Not really. I feel like I'm narrowing the questions and starting to learn something. Thank you. Try not to get frustrated with me. I do appreciate your efforts.

strangerep

Quote by friend

Does your quote mean the generators of the internal symmetries commutes with the elements of the Poincare group?

No, but that's an outcome of the CM thm, as you already found elsewhere.

Wouldn't it be nice if we had a means of providing the internal symmetries on principle alone?

Only if it predicts unexpected new physics, subsequently confirmed by experiment.

What seems key to me is the last commutator relation where the internal symmetries, A_a commute with the elements of the Lorentz group p_μ and M_μν.

Yes, but...

Now as I understand you, this means that given the p_μ and M_μν, these commutator relations restrict the kind of internal symmetries that can exist.

Not really. If a generator commutes with another, a better way to think of them is that they're pretty much independent -- in the sense that performing a transformation corresponding to one of the generators has no effect on a transformation corresponding to the other. The transformations can be performed in either order, yielding the same outcome either way. They don't affect/interfere with each other.

But I wonder if it can also be read, given the internal symmetries these commutators restrict the spacetime symmetries to the Lorentz group?

No, for the reasons I explained in the previous paragraph. And recall what I said earlier about how the gauge transformations are unphysical, in the sense that all observable quantities are gauge-invariant. A gauge transformation cannot affect the numbers you get from any experiment. But Poincare transformations can, and are therefore physical.

I feel like I'm narrowing the questions and starting to learn something. Thank you. Try not to get frustrated with me. I do appreciate your efforts.

Well, you'd better stop trying to subtly maneuver the discussion too far away from mainstream physics towards your own theories or this thread will end abruptly.

friend

Quote by strangerep

Yes, but...
Not really. If a generator commutes with another, a better way to think of them is that they're pretty much independent -- in the sense that performing a transformation corresponding to one of the generators has no effect on a transformation corresponding to the other. The transformations can be performed in either order, yielding the same outcome either way. They don't affect/interfere with each other.

No, for the reasons I explained in the previous paragraph. And recall what I said earlier about how the gauge transformations are unphysical, in the sense that all observable quantities are gauge-invariant. A gauge transformation cannot affect the numbers you get from any experiment. But Poincare transformations can, and are therefore physical.

Forgive me, but this leaves me a little confused. I thought these commutator relations were the very thing they were trying to prove. And I thought commutators form a restriction on the kinds of symmetries allowed. Are these assumptions not right?

But perhaps something more is needed for these commutator relations to be meaningful. Is it that they must be a commutation relation between Lorentz group and internal symmetries? In other words, it's not a commutation relation between the Lorentz group and arbitrary groups; it must be with internal symmetrys. And it's not a commutation relation between internal symmetries and arbitrary spacetime symmetries; it must be with Lorentz group, right? Is it the QFT formulism that connects the two symmetries in the commutator, that requires the symmetries to be of the Lorentz group and the internal symmetries? In other words, it's only valid within the framework of a relativistic QFT that has internal symmetries? Or can it involve a QFT that has internal symmetries but is not necessarily Lorentz invariant, and visa versa, if such can exist?

strangerep

Quote by friend

[...] I thought commutators form a restriction on the kinds of symmetries allowed. Are these assumptions not right?

Your assumptions are distorted. A commutator algebra defines a particular symmetry algebra, nothing more.

Just try being a student and study Greiner's books properly.

I have already answered the other things in your last post, and I have other things to do.

vanhees71

I haven't read through the whole thread (due to lack of time, sorry). I just want to mention that the universal covering group of [itex]\mathrm{O}(1,3)^{\uparrow}[/itex] is [itex]\mathrm{SL}(2,\mathbb{C})[/itex].

friend

I got a little side-tracked on the Coleman-Mandula theorem. But I never really got a good answer to the following:

Quote by friend

So when the author (Matthew Robinson, see post 1) writes (on page 120), "this is only true in 1+3 spacetime dimensions", I'm still seeing a connection between SU(2) and the need for a Lorentz metric (1,3).

The author is referring to how 2 copies of SU(2) cover SO(1,3), but only in 1 +3 dimensions. So my question is if one can find that some formalism has 2 copies of SU(2), in the way the author describes, does that mean that spacetime has an underlying 1+3 signature? Or is SO(1,3) synonymous with the Lorentz metric in 1+3 dimensions.

fzero

Quote by friend

I got a little side-tracked on the Coleman-Mandula theorem. But I never really got a good answer to the following:

The author is referring to how 2 copies of SU(2) cover SO(1,3), but only in 1 +3 dimensions. So my question is if one can find that some formalism has 2 copies of SU(2), in the way the author describes, does that mean that spacetime has an underlying 1+3 signature? Or is SO(1,3) synonymous with the Lorentz metric in 1+3 dimensions.

It is not true that 2 copies of SU(2) cover SO(1,3) (it is the cover of SO(4)!). It is true that the algebra of SO(1,3) contains two copies of the algebra of SU(2). It is common as strangerep has been doing to denote the algebra with lowercase letters. Then we can write [itex]so(1,3) = su(2) \oplus su(2)[/itex]. But this is not what we mean by covering space (for which you should really look at the references given earlier in the thread).

The appearance of 2 copies of the algebra su(2) here is not because of the signature, but is a property of any so(p,q) algebra for which p+q=4. This is the sense in which the relationship is special to four dimensions.

friend

Quote by fzero

The appearance of 2 copies of the algebra su(2) here is not because of the signature, but is a property of any so(p,q) algebra for which p+q=4. This is the sense in which the relationship is special to four dimensions.

I thank all of you for your help in this. I've got some resouces to consider. And I now know what to be mindful of as I study.

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vanhees71 Recognitions: Science Advisor	I haven't read through the whole thread (due to lack of time, sorry). I just want to mention that the universal covering group of [itex]\mathrm{O}(1,3)^{\uparrow}[/itex] is [itex]\mathrm{SL}(2,\mathbb{C})[/itex].