# CPT and Diffeomorphism Invariance

1. Oct 28, 2013

### Rena Cray

I suspect this is somewhat off the beaten track here, but there may be some few that could give it a go.

Einstein called his concept of coordinate independent physical theory General Covariant. The mathematicians call coordinate independent differential topology, diffeomorphism invariant.

In any case, I don’t think that CPT and diffeomorphism invariance are logically compatible.

Am I right about this??? Deciding this correctly is an embarrassing mental block on my part. And I do hope someone can help me overcome this failing. At the crux of this is general relativity itself. Perhaps G_d prevents me. I am at your mercy.

-Rená

2. Oct 28, 2013

### Staff: Mentor

Hi Rena, do you understand how CPT symmetry and Lorentz invariance are related (or more precisely CPT symmetry and Poincare symmetry)?

Basically, CPT symmetry implies Poincare symmetry and Poincare symmetry is a local symmetry of diffeomorphism invariance in a (-+++) manifold. So I am trying to understand if your question is related to the connection between CPT and Poincare or between Poincare and diffeomorphism.

EDIT: note, I have the implication backwards here, please see below

Last edited: Oct 28, 2013
3. Oct 28, 2013

### vanhees71

That's an interesting statement! Is it really true that CPT symmetry implies already Poincare symmetry? The other way is standard QFT but already a pretty tedious proof in QFT. Is there a proof for the CPT->Poincare statement in the literature?

4. Oct 28, 2013

### Staff: Mentor

Oops, you are correct, I have the implication backwards. If you have Poincare symmetry and a Hermitian Hamiltonian for a local QFT then that implies CPT symmetry.

5. Oct 28, 2013

### vanhees71

and the Hamiltonian should also be bounded from below. Then it's the famous Pauli-Lüders theorem. I'd be surprised if one could prove the opposite direction, i.e., CPT->Poincare, which on my feeling would be much stronger, because it would prove a large continuous symmetry (Poincare group) from a single discrete symmetry (CPT).

6. Oct 29, 2013

### Rena Cray

Thank you both, vanhees and Dale. I'm happy to see I am being helped by two of the big guns of this forum.

Dale and vanhees: I am, in particular, concerned with CPT and diffeomorphism invariance rather than Poincare, though I will look into it as well.

My question was was intended to be within the realm of relativity theory--classical. However, Vanhees, I will certainly search and learn what I can of Pauli-Lüders' theorem as it may still apply (Does it?) as well as Dale's CPT-Poincare connection. The references are much welcomed.

Symmetries are the core of physics in my way of thinking, and I do appreciate the references. Symmetries (and the odd non-symmetries) could be a wonderful topic of discussion.

7. Oct 29, 2013

### atyy

"A CPT invariant theory whose spectrum contains a state of helicity +λ must therefore also contain a state with helicity −λ. For instance, a photon has two polarizations (with helicities ±1), the same is true for a graviton (with helicities ±2), and a left-handed Weyl spinor always comes with its right-handed conjugate."

Last edited by a moderator: May 6, 2017
8. Oct 29, 2013

### Staff: Mentor

OK, then you will have to explain the connection you see between the two. Other than their mutual connection to Poincare symmetry I don't see any relationship. In particular, I don't see what would lead you to say:

9. Oct 29, 2013

### atyy

Could the answer depend on whether one sees GR with a curved spacetime metric, so there is no global inertial frame, and so no global CPT? OTOH, if one sees GR as spin 2 on flat spacetime, then there is a global inertial frame, and so there is still global CPT?

10. Nov 1, 2013

### K^2

Symmetries are the standard now, anyhow. And that's where Dale is coming from on this. The typical way of introducing GR is via the topological interpretation, which is how you arrive at diffeomorphism invariance. The connection to Poincare group is that it's the very group of continuous transformations under which the theory is invariant.

Which leads to an alternate approach to GR. You start by requiring that your field theory is invariant under continuous local Poincare transformations. This immediately leads you to introducing a covariant derivative and then you get all of the differential topology as a fallout.

At any rate, this is where Pauli-Lüders theorem comes in. Poincare invariance leads to CPT, and since Poincare local symmetry is your diffeomorphism invariance, there can't be an incompatibility.

11. Nov 16, 2013

### Rena Cray

Whether or not we include improper rotations O(3,1) or not is really the crux of this isn't it?

As I understand it, the definition of CPT invariance says:

1) The laws of physics are unchanged under inversion of all three of C, P and T performed simultaneously.

2) E.g.: the laws of physics are invariant under (q, x, y, z, t) --->(-q, -x, y, z, -t).

3) Trivially, (q, x, y, z, t) --->(q, x, y, z, t).

4) Implicitly, the laws of physics are not invariant under an inversion of a single space or time dimension.

5) E.g.: the laws of physics are not invariant under (q, x, y, z) --> (q, -x, y, z, t), for instance.

As I understand diffeomophism invariance:

1) It is an arbitrary smooth coordinate transformation.

2) This includes coordinate transforms having Jacobian matrices of negative determinate.

3) To be sure, this includes the full Lorentz group.

4) This includes coordinate transforms such as (q, x, y, z) --> (q, -x, y, z, t) in contradiction to number 5), above.

From this I seem to conclude CPT and diffeomorphism invariance are incompatible. Did I error in this?

Last edited: Nov 16, 2013
12. Nov 17, 2013

### Bill_K

Rena,

What you're missing, and what none of the comments so far have pointed out, is that in a general coordinate system, P and T are NOT reflections of the coordinates. The above relation holds only in Minkowski coordinates in a flat spacetime.

The CPT theorem applies to physics locally at a point. And to describe local physics you need to use not only a system of coordinates but a basis of 4-vectors as well, i.e. a tetrad. All spacetimes, even ones with curvature, are locally Minkowskian, and there always exists such a set of unit vectors at every point. In fact there are many such sets at every point, and they can be obtained from one another by a Lorentz transformation. (Independently, at every point.) Again, these Lorentz transformations affect only the basis, not the coordinates, which may be cylindrical, or spherical, or anything you like. They remain unchanged. The tetrad basis is especially needed to describe spinor fields.

The P and T operations in the CPT theorem are reflections of the tetrad basis.

13. Nov 17, 2013

### Staff: Mentor

In addition to the excellent point by Bill_K, your number 4 and 5 are NOT implied by CPT symmetry. The presence of CPT symmetry does not imply the absence of other symmetries. Numbers 4 and 5 are correct, but not in any way implied by CPT symmetry.

14. Nov 26, 2013

### Rena Cray

Thank you for your kindly response. I am familiar with vielbein. My (perhaps overly) schematic notation referred to coordinate basis rather than tetrads. I don't suspect it is not necessary to examine this beyond coordinate bases.

I placed in bold your reference to Lorentz transformations. What no one has yet made clear is whether they are writing of Lorentz transforms that include improper rotations or those of continuous transformations. As this is so central, it seems to be an oversight.

Please, if you will, do you see any logical errors in the analysis of my previous post?

Last edited: Nov 26, 2013
15. Nov 27, 2013

### Rena Cray

I wrote:
"I don't suspect it is not necessary to examine this beyond coordinate bases", is what I wrote.

Forgive me, It was my intent to write:
"I don't suspect it is necessary to examine this beyond coordinate bases," sans the negation.

---------------------------------------------------------------------------------------

I will skip the platitudes:-

CPT violates parity conservation, in the sense we take charge as a coordinate.

But more strongly, it is not necessary to pretend that charge is a coordinate to violate space-time parity conservation under CPT:

Inverting one spatial dimension and also time is parity invariant, but violates CPT invariance.

Does anyone else see a problem or an issue with this??:

1) CPT is odd on a space-time-charge manifold, and is not precluded from being odd on a space-time manifold.

2) A mobius strip would be topologically equivalent to a normal rubber band. It is not.

This would demand that space-time has no oriention sense, wouldn't it? Would this be the case, models involving operands such as cross products would be verboten, no?

-------------------------------------------------------------

This is all very convoluted to the initiate, I'm sure. Few venture into manifold orientation. I could very well have been in error at some point.

I am not at all confident of my analysis.

I am not "baiting." I have real doubts and confusion.

Last edited: Nov 27, 2013
16. Nov 27, 2013

### Staff: Mentor

I don't know what else can be done for you. I have pointed out that far from being incompatible Poincare symmetry (with some conditions on the Hamiltonian) implies CPT. Personally, I think that does it. Bill_K further pointed out that diffeomorphism invariance of the coordinates is irrelevant to the discussion, all that needs to be considered is local Lorentz symmetry of the tetrad.

Given that, I don't see what you are still struggling with. Topology and orientation are global properties of the manifold, this is a local symmetry of the tetrad, so I don't see the relevance.

Sorry I cannot help more, but my knowledge is not terribly deep here. I just don't know enough to be able to figure out what is troubling you here.

17. Nov 28, 2013

### Rena Cray

No problem. I have what I needed.

18. Dec 8, 2013

### Rena Cray

I'm coming back to this because it bothers me that I seem to be in general disagreement.

No one has yet defined what they mean when writing of Lorentz transforms; if they imply improper rotations inclusive or not. This qualification is pivotal.

I think everything we have to know is embodied in my post #11, without appeal to other authors.

Am I to assume all, except Dale have dropped their arguments on examining the evidence?

I have a great deal of respect for those responding here, so it continues to bother me greatly to find myself in disagreement.

19. Dec 8, 2013

### K^2

I finally see what you mean. No, in this context, because we are talking about subgroup of Poincare group, Lorentz group is only the SO(1, 3) subgroup. (Proper Lorentz group). No improper rotations.