SU(2) a double cover for Lorentz group?

[friend]

SU(2) a double cover for Lorentz group?

I'm presently reading the new book, "Symmetry and the Standard Model", by Matthew Robinson. On page 120, he writes, "the Lorentz group (SO(1,3), pg 117) is actually made up of two copies of SU(2). We want to reiterate that this is only true in 1+3 spacetime dimensions." I suppose this is not a surprise since they started with the 1+3 version of special relativistic QFT to derive the SU(2) group to begin with.

But suppose we had some formulism that necessitated the internal SU(2) structure before imposing spacetime coordinates. If we found "two copies of SU(2)" in this formulism, would that necessitate the existence of a SO(1,3) Lorentz group on any underlying spacetime? So I guess I'm asking about internal verses external symmetries, about local verses global symmetries, and about how to connect internal SU(2) symmetry with spacetime SO(1,3) symmetries. At this point I don't really know how "two copies of SU(2)" "cover" SO(1,3). If anyone had a summary or a broad outline of how this all works, I'd appreciate it. If I can understand this, then perhaps I could understand how that one formulism could derive both QM and GR. Thanks.

 

[dextercioby]

It looks like a bad book.

1. SU(2)xSU(2) is not a cover for SO(1,3), but a double cover for SO(4).

2. SU(2) is a gauge symmetry group, and the gauge symmetry cannot precede the spacetime symmetry.

3. There's a famous theorem by Coleman and Mandula about the ways a global symmetry and a gauge symmetry are related. This should cover the Standard Model.

 

[Bill_K]

At this point I don't really know how "two copies of SU(2)" "cover" SO(1,3).

SU(2) x SU(2) is isomorphic to SO(3,1). Take the Lorentz group generators L_μν, and form M_i = ε_ijkL_jk and N_i = L_0i.

Then J_i = ½(M_i + N_i) and K_i = ½(M_i - N_i) generate SU(2) x SU(2).

 

[friend]

SU(2) x SU(2) is isomorphic to SO(3,1). Take the Lorentz group generators L_μν, and form M_i = ε_ijkL_jk and N_i = L_0i.

Then J_i = ½(M_i + N_i) and K_i = ½(M_i - N_i) generate SU(2) x SU(2).

So does "isomorphic to" mean that SU(2)XSU(2) covers SO(3,1) uniquely? Or could SU(2)XSU(2) cover other symmetry groups as well?

Another question is whether SO(3,1) applies to both special relativity and general relativity. Or does it only apply to special relativity?

dextercioby: I am seeing this "SU(2) is a double cover to SO(3,1)" on a number of places on the web, even in the arXiv. Sometimes I see it as SU(2) is a double cover of SO(3). I don't know if that's shorthand for SO(3,1), however.

 

[fzero]

So does "isomorphic to" mean that SU(2)XSU(2) covers SO(3,1) uniquely? Or could SU(2)XSU(2) cover other symmetry groups as well?

"Isomorphic" in this case means that there is an isomorphism between the Lie algebras of the two groups. There are other Lie groups that are isomorphic to SU(2)XSU(2) in this sense. However, a Lie group has more structure than the algebra, so for instance, while SO(3,1) and SO(4) have equivalent algebras, SO(4) is compact, but SO(3,1) is not. Compactness has a specific mathematical meaning, but in part it means that the volume is finite. The notion of a cover (or covering space) involves the global structure of the group.

Another question is whether SO(3,1) applies to both special relativity and general relativity. Or does it only apply to special relativity?

Special relativity involves invariance under the SO(3,1) invariance of the Minkowski metric. General relativity does not assume SO(3,1) invariance. In fact, GR does not assume any fixed metric, so SO(3,1) is not singled out in GR.

dextercioby: I am seeing this "SU(2) is a double cover to SO(3,1)" on a number of places on the web, even in the arXiv. Sometimes I see it as SU(2) is a double cover of SO(3). I don't know if that's shorthand for SO(3,1), however.

SU(2) and SO(3) have the same dimension (number of Lie algebra generators), in this case 3, while SO(3,1) is 6-dimensional. SU(2)XSU(2) is isomorphic (as a group, not just algebra) to Spin(4), which is the double cover of SO(4). The double cover of SO(3,1) is called Spin(3,1) which is isomorphic to SL(2,C). All of these 6-dimensional examples have isomorphic algebras.

To learn about SU(2) as a double cover of SO(3) (as well as basics about lie groups), I'd suggest looking at a geometry book, like Nakahara, Geometry Topology and Physics. For the Lorentz group, a basic discussion appears in any decent QFT text, but the one in Weinberg Vol 1 is probably the most complete, if not the easiest to digest.

 

[friend]

To learn about SU(2) as a double cover of SO(3) (as well as basics about lie groups), I'd suggest looking at a geometry book, like Nakahara, Geometry Topology and Physics. For the Lorentz group, a basic discussion appears in any decent QFT text, but the one in Weinberg Vol 1 is probably the most complete, if not the easiest to digest.

Brilliant! Thank you, fzero, for that contribution. Yes, I have Nakahara's book. I actually read some of it years ago. Now that these issues are becoming relevant, I suppose I should go back and read it again.

Another question: when they say SU(2)XSU(2), can that mean one copy of the SU(2) group could be the hermitian conjugate of the other copy of the SU(2) group? I'm thinking in terms of inner products to form probabilities, and wondering if the metric, or even gravity, might be a manifestation of the probability side of things. Yes, I know this sounds vague, but I'm grasping at straws here. I'm still not sure how to connect internal symmetries to spacetime symmetries.

 

[dextercioby]

[...]dextercioby: I am seeing this "SU(2) is a double cover to SO(3,1)" on a number of places on the web, even in the arXiv. [...]

It's wrong.

[...]Sometimes I see it as SU(2) is a double cover of SO(3) [...]

It's right.

[...]I don't know if that's shorthand for SO(3,1), however.

It can't be, SO(3) and SO(3,1)/S0(1,3) are totally different groups.

Fzero reccomended Weinberg Vol.1, I say Sexl and Urbantke.

 

[fzero]

Another question: when they say SU(2)XSU(2), can that mean one copy of the SU(2) group could be the hermitian conjugate of the other copy of the SU(2) group? I'm thinking in terms of inner products to form probabilities, and wondering if the metric, or even gravity, might be a manifestation of the probability side of things. Yes, I know this sounds vague, but I'm grasping at straws here. I'm still not sure how to connect internal symmetries to spacetime symmetries.

Let me first say that, in general, we do not try to connect internal symmetries to spacetime symmetries. But perhaps there is a confusion of nomenclature. Usually what we mean by internal symmetries are the ones like gauge symmetries or global flavor symmetries. There is a theorem called the Coleman-Mandula theorem that states that there are no non-trivial combinations of these types of internal symmetries with spacetime ones.

Now, I suspect that by "internal symmetry" you might mean something different, which would ordinarily be called the representation of the Lorentz group. See, the Lorentz group is defined to be the group of symmetries of spacetime that preserve the Minkowski metric. The action of the Lorentz group must also extend to quantum states and fields and the terminology we use in this regard is to say that fields lie in representations of the Lorentz group. Mathematically a representation is a vector space, upon which the group generators can be associated to specific matrix-valued generators.

For example, we can represent the action of SO(3) rotations on the 2-sphere with differential operators $J^i = \epsilon^{ijk} x_j \partial_k$. We can also consider a 3-vector representation, along with the usual action of 3x3 orthogonal matrices. In quantum mechanics you learned this as the $j=1$ representation. The 3-vector can be described using the basis of states $|j,m_j\rangle$, where $m_j = -1,0,+1$. The $j=1/2$ states are a different representation, called the spinor representation, where now we have a 2x2 matrix action. I suspect that you are calling this action on the quantum states an "internal symmetry," but that is not the usual terminology.

Now to get back to your question about the two SU(2)s in the algebra of the 3d Lorentz group, I would say, no, you cannot consider one as the hermitian conjugate of the other. BillK, in post #3, gave a description of how the SO(3,1) generators are written in terms of two sets of generators of SU(2). One set correspond to rotations, while the other set are boosts. They are not conjugate to one another.

We can go a bit further to clarify the general case. Suppose the quantum states $|\psi_i\rangle$ are a basis set for a definite representation (call it $R$) of some symmetry group. Then the group acts on $|\psi_i\rangle$ by

$$|\psi_i'\rangle =\sum_j M_{ij} |\psi_j\rangle,$$

where $M$ is some matrix. The conjugate representation is obtained by taking the Hermitian conjugate of this,

$$\langle \psi_i'| =\sum_j \langle \psi_j |M^*_{ji} .$$

The conjugate states $\langle \psi_j |$ are said to be in $R^*$, the dual representation to $R$. Orthogonality of the states will lead to the conclusion that the representation must be unitary, i.e., $M$ must be a unitary matrix to conserve probabilities.

 

[friend]

Let me first say that, in general, we do not try to connect internal symmetries to spacetime symmetries. But perhaps there is a confusion of nomenclature. Usually what we mean by internal symmetries are the ones like gauge symmetries or global flavor symmetries. There is a theorem called the Coleman-Mandula theorem that states that there are no non-trivial combinations of these types of internal symmetries with spacetime ones.

Just a moment. The author in post 1 writes, "the Lorentz group (SO(1,3), pg 117) is actually made up of two copies of SU(2). We want to reiterate that this is only true in 1+3 spacetime dimensions."... "in 1+3 spacetime dimensions"? Maybe he's just plain wrong, as dextercioby suggests. Or maybe if there does exist 2 copies of SU(2), then there does exist SO(1,3) [or is it SO(3,1)?]. If so, then is this SO(1,3) not about the underlying spacetime? If it's not about spacetime, then doesn't there still need to exist a spacetime that's 1+3? If so, then the two copies of SU(2) at least specify a background of 1+3. Is that the Lorentz signature? Your responses are very much appreciated. Thanks.

 

[strangerep]

A reasonable textbook (in this case, the one by Robinson) can seem incomprehensible if read by someone who does not have the necessary prerequisites.

The author in post 1 writes, "the Lorentz group (SO(1,3), pg 117) is actually made up of two copies of SU(2). We want to reiterate that this is only true in 1+3 spacetime dimensions."... "in 1+3 spacetime dimensions"? Maybe he's just plain wrong, as dextercioby suggests.

No, what Robinson actually said is ok. (It helps to read the pages in the book properly before passing judgement.)

On p120, Robinson is explaining essentially the same thing that Bill_K briefly explained in his post #3. I.e., one can re-express the generators of the Lorentz algebra as a sum of 2 separate su(2) algebras.

The Lorentz algebra so(1,3) [or so(3,1) depending on one's choice of metric signature] is indeed the direct sum of two separate copies of su(2). Hence the Lorentz group is a product of 2 copies of SU(2).

This is quite different from the notion of "double cover". SU(2) is a double-cover for SO(3) [the ordinary 3D rotation group] in this sense: the Lie algebras su(2) and so(3) coincide,
but when exponentiated as groups we find that for every element of SO(3) there are 2 elements of SU(2). That's what "double-cover" means.

Glancing through the earlier parts of Robinson's book, I see that he covers a fair bit of this but it's at a level you haven't yet reached. Try another book, such as Greiner's "Quantum Mechanics -- Symmetries" for an introduction to group theory in physics which might be at a more suitable level.

 

[friend]

A reasonable textbook (in this case, the one by Robinson) can seem incomprehensible if read by someone who does not have the necessary prerequisites.

No, what Robinson actually said is ok. (It helps to read the pages in the book properly before passing judgement.)

I must admit, I found myself wishing he had put in some exercises so I could practice the algebraic manipulations he refers to. He does recommend finishing the rest of the generators after giving an example. Maybe I'll try doing that on my second reading.

On p120, Robinson is explaining essentially the same thing that Bill_K briefly explained in his post #3. I.e., one can re-express the generators of the Lorentz algebra as a sum of 2 separate su(2) algebras.

The Lorentz algebra so(1,3) [or so(3,1) depending on one's choice of metric signature] is indeed the direct sum of two separate copies of su(2). Hence the Lorentz group is a product of 2 copies of SU(2).

For a minute there I thought I learned something when you said "direct sum of", but then you said "is a product of". I thought there was a difference between sums and products of spaces and groups.

This is quite different from the notion of "double cover". SU(2) is a double-cover for SO(3) [the ordinary 3D rotation group] in this sense: the Lie algebras su(2) and so(3) coincide,
but when exponentiated as groups we find that for every element of SO(3) there are 2 elements of SU(2). That's what "double-cover" means.

Thanks

Glancing through the earlier parts of Robinson's book, I see that he covers a fair bit of this but it's at a level you haven't yet reached. Try another book, such as Greiner's "Quantum Mechanics -- Symmetries" for an introduction to group theory in physics which might be at a more suitable level.

Is there any difference between the $124, 2004 version of the book and the $24, 2008 version? See here.

 

[strangerep]

I must admit, I found myself wishing he had put in some exercises so I could practice the algebraic manipulations he refers to.

This is where Greiner's series of textbooks are great for self-learning. There's lots of worked examples, etc, and in the main text he doesn't skip steps (which can often leave students feeling bamboozled).

For a minute there I thought I learned something when you said "direct sum of", but then you said "is a product of". I thought there was a difference between sums and products of spaces and groups.

Indeed there is. (To understand this in depth you'll need to acquire more knowledge of Lie groups.) A Lie algebra is also a vector space, so "sum" is the right word there. As a (very) rough analogy, think about the sum of 2 numbers $a$ and $b$ and their exponentiations $e^a$ and $e^b$. The sum $a+b$ corresponds to the product $e^a \, e^b = e^{a+b}$, hence we talk about the "sum" of the numbers, but the "product" of their exponentials.

Is there any difference between the $124, 2004 version of the book and the $24, 2008 version?

I don't know, but probably. It is good that Greiner keeps updating his extensive series of textbooks, and that they seem to have come down a long way in price from when I bought my (much older) editions a long time ago. You'd probably benefit from others in his series, such as "QM -- An Introduction" and "QM - Wave Equations" (not sure if I got the title right -- it's an introduction to relativistic QM).

(As an introduction to non-rel QM, I tend to like Ballentine more, but he demands a higher level of math from his readers than Greiner does.)

 

[friend]

This is where Greiner's series of textbooks are great for self-learning. There's lots of worked examples, etc, and in the main text he doesn't skip steps (which can often leave students feeling bamboozled).

I intend to buy the book. Can I count on you to answer some questions about it, if I don't abuse the privledge?

Indeed there is. (To understand this in depth you'll need to acquire more knowledge of Lie groups.) A Lie algebra is also a vector space, so "sum" is the right word there. As a (very) rough analogy, think about the sum of 2 numbers $a$ and $b$ and their exponentiations $e^a$ and $e^b$. The sum $a+b$ corresponds to the product $e^a \, e^b = e^{a+b}$, hence we talk about the "sum" of the numbers, but the "product" of their exponentials.

Thank you. That's pretty clear.

I don't know, but probably. It is good that Greiner keeps updating his extensive series of textbooks, and that they seem to have come down a long way in price from when I bought my (much older) editions a long time ago. You'd probably benefit from others in his series, such as "QM -- An Introduction" and "QM - Wave Equations" (not sure if I got the title right -- it's an introduction to relativistic QM).

(As an introduction to non-rel QM, I tend to like Ballentine more, but he demands a higher level of math from his readers than Greiner does.)

You know, it's been some years since I studied these things in school. And I don't really work with it. But recently I've been forced to reconsider these things for philosophical reasons. So I've been doing some self-study as time and insight allow. I've stummbled on a formulism that promises to derive physics from logic alone. (Incredible!) It seems to have shown the foundations of quantum theory, where QM comes from, how QFT is iterated from that, and where the U(1)SU(2)SU(3) symmetries come from. You'd be surprised how easy it is. So now I'm wondering if it can also be used to derive special and/or general relativity. Thus the series of questions I've asked in this thread.

Note to moderator: I did not explain any detail at all about this theory. Therefore, I'm not advocating speculative theories. I'm only explaining my motivation for the questions I ask in this thread. If anyone is interested in details they can send me a private message.

 

[strangerep]

I intend to buy the book. Can I count on you to answer some questions about it, if I don't abuse the priviledge?

Depends how much spare time I have, and the nature of the questions. But there are lots of other people here who are more than capable of answering questions at that level, so it doesn't necessarily have to be me.

And they're probably more likely to help you if you park that other philosophy stuff outside the PF door, and pick it up when you leave...

 

[friend]

The Lorentz algebra so(1,3) [or so(3,1) depending on one's choice of metric signature] is indeed the direct sum of two separate copies of su(2). Hence the Lorentz group is a product of 2 copies of SU(2).

I beg your indulgence. Let me try this one more time. You say "depending on one's choice of metric signature",... "metric signature"? So here it seems there is a reference to the underlying spacetime metric, which is what I'm going for. So when the author writes, "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). Or do you still think I need to address the Coleman-Mandula theorem?

 

[strangerep]

You say "depending on one's choice of metric signature",... "metric signature"?

Some authors prefer to write the line element as
$$
ds^2 ~=~ c^2dt^2 - dx^2 - dy^2 - dz^2 ~.
$$This tends to be preferred in particle physics. But one can adopt a different convention:
$$
ds^2 ~=~ -c^2dt^2 + dx^2 + dy^2 + dz^2 ~,
$$ which tends to be preferred by relativists because the spatial part is the same as in the usual 3D Euclidean geometry, and it can be more convenient when taking a nonrelativistic limit.

There is a significant body of mathematics called "Clifford Algebras" which (among other things) delves into the interrelationships between such groups, metric signatures, generalized Dirac gamma matrices, and so. But it's easy to become hypnotized by such elegant maths unless one stays grounded through physics.

[...] Or do you still think I need to address the Coleman-Mandula theorem?

There's an important difference between physical transformation groups (such the Lorentz, the rotation group, and SU(2) as the double-cover of the latter). By themselves, these having nothing to do with the CM thm.

The gauge groups such as $SU(2)_L \times U(1)$ in the electroweak interaction indicate degrees of freedom arising from the way we build interaction terms in the Hamiltonian out of free fields. In that sense, gauge freedoms are unphysical since all observable quantities are necessarily gauge-independent. The gauge transformations are thus merely an aspect of our abstract mathematical modelling of physical situations.

However, the enormous success of gauge theories of interactions fosters a belief that the gauge transformations must indeed be somehow physical, hence people tried for a long time to unify gauge transformations with the physical Poincare transformations. Alas, the CM Thm showed that this can only be done in a rather trivial way, much to everyone's disappointment. Then followed attempts to evade the CM Thm by constructing theories in more general frameworks not covered by CM -- such as supersymmetry (SUSY) in which the algebras involve anticommutators as well as commutators. This led to a vast body of research in SUSY extensions to the Standard Model -- but now the increasingly-inconvenient (non-)results from the LHC leave many people red-faced on that score.

As to whether you should "address the CM thm", well, the proof is challenging even for people well-versed in group theory and QFT. So the standard advice of "learn to walk before you run" remains applicable.

 

[friend]

There is a significant body of mathematics called "Clifford Algebras" which (among other things) delves into the interrelationships between such groups, metric signatures, generalized Dirac gamma matrices, and so. But it's easy to become hypnotized by such elegant maths unless one stays grounded through physics.

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?

As to whether you should "address the CM thm", well, the proof is challenging even for people well-versed in group theory and QFT. So the standard advice of "learn to walk before you run" remains applicable.

That's for sure. I wonder if your copy of Greiner covers the CM theorem, maybe in the last chapter? I bought the book. It looks like a good reference if nothing else.

This Coleman-Mandula theorem poses the biggest threat to my efforts so far. Or maybe it's the biggest help. I'm not sure yet. So I think that's where I'm headed in my studies.

So I'm searching around the web a bit to see if I can get a clearer interpretation. And it seems I'm really not asking anything more than what the theorem proposes. What I read is this:

"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). I got this http://boards.straightdope.com/sdmb/archive/index.php/t-166128.html.

And,

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

And that's very close to what I'm trying to prove. Since there is no coordinate independent form of QFT, the CM thm must start with a Lorentz metric, right? And then given the QFT formulation, the only allowed symmetries are U(1), SU(2), and SU(3). 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? Some statements of the CM thm state for QFT you can only have both the Lorentz and these Lie groups. Does that mean given QFT and the Lie groups you must have the Lorentz group? I hope so. If so, that's a God sent. And I may have both QM and SR from logic.

Thank you for helping realize that.

 

[strangerep]

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.

"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 possesses 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]

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.

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?

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?

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 [STRIKE]on[/STRIKE] 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?

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]

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]

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]

[...] 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 $\mathrm{O}(1,3)^{\uparrow}$ is $\mathrm{SL}(2,\mathbb{C})$.

 

[friend]

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

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]

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 $so(1,3) = su(2) \oplus su(2)$. 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]

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.