A RG: group or semi-group?

Demystifier

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Summary
Is renormalization group a true group for renormalizable theories?
It is often said that the renormalization group (RG) is not a true group but only a semi-group, because the RG transformation is not invertible. But for renormalizable theories, the renormalized Hamiltonian has the same form as the original Hamiltonian, only with some different values of the finite number of parameters (coupling constants). Denoting the set of these "coupling constants" as ##g=(g_1,\ldots,g_n)##, for a renormalizable theory we have a set of coupled equations of the form
$$\frac{dg_i(t)}{dt}=\beta_i(g(t))$$
where ##t## is the scale parameter. Such equations can be solved uniquely in both forward and backward directions of ##t##, so in this sense the RG transformation is invertible. Does it mean that RG is a true group for renormalizable theories?
 
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Demystifier

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Near the bottom of p42
I see, thanks! The relevant quote is this:
"... the RG procedure is sometimes referred to as a semi-group. The term applies
to the action of RG on the space of configurations: each magnetization profile is mapped uniquely to one at larger scale, but the inverse process is non-unique as some short scale information is lost in the coarse graining. (There is in fact no problem with inverting the transformation in the space of the parameters of the Hamiltonian.)"


But in particle physics the RG is usually formulated without any coarse graining, using only the requirement that certain physical quantities do not depend on the scale ##\mu## appearing in dimensional regularization. It looks as if in such a formulation there is no reason to say that RG is only a semi-group.
 
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Demystifier

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I have just checked the book Pascual and Tarrach, QCD: Renormalization for the Practitioner, which indeed is a particle physics book based entirely on dimensional regularization. In Sec. IV The Renormalization Group, they state explicitly that a composition of renormalizations has the full group structure whenever the composition is defined, but note that composition is not defined for arbitrary two elements of the group, in which case it has only a grupoid structure. (A grupoid is even more primitive than a semi-group in the sense that the product in a grupoid does not need to be associative.)
 
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A. Neumaier

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There are two different concepts of RG, The earlier by Stückelberg is a group, the later by Wilson is only a semigroup.
Stückelberg's RG describes a change in parameterization of a single QFT, which makes it a group. On the other hand, Wilson's RG describes a parameterized family of effective QFTs involving more and more coarsening it in the process, whence it is not invertible.
 

Demystifier

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Stückelberg's RG describes a change in parameterization of a single QFT, which makes it a group. On the other hand, Wilson's RG describes a parameterized family of effective QFTs involving more and more coarsening it in the process, whence it is not invertible.
But when the theory is renormalizable, the members of a Wilson family differ only by the values of the parameters (coupling constants), which makes it effectively the same as the Stückelberg's change of parameterization. Sure, the Wilson coarsening is not invertible, but its effect - the change of effective parameters - is invertible.

To make an analogy, this is somewhat like computation of an invertible function on a physical computer. The work of a physical computer is not invertible (it produces heat), but the result of computation is invertible.

I think it would not be wrong to say that Wilson's RG describes what is really going on physically, while Stückelberg's RG is a convenient recipe to get the final results of practical interest. Roughly speaking, something like QM with and without "hidden variables". The final measurable results are the same, but the philosophies are very different.
 
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Demystifier

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No. The large momentum behavior is altered!
But from the Wilson point of view that's irrelevant, because it's the small momentum behavior (small compared to the Wilson cutoff in the momentum space) that one wants to describe with an effective theory.
 

A. Neumaier

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But from the Wilson point of view that's irrelevant, because it's the small momentum behavior (small compared to the Wilson cutoff in the momentum space) that one wants to describe with an effective theory.
It is irrelevant for the low energy behavior. But it makes mathematically the essential difference between a group and a semigroup. For a group one needs invertible mappings, but from a low energy effective theory one cannot recover the higher energy effective theory.
 

Demystifier

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but from a low energy effective theory one cannot recover the higher energy effective theory
It depends on what one means by "theory". If the theory is just the Hamiltonian (or Lagrangian), and not one particular solution associated with that Hamiltonian, then one can recover it for renormalizable theories. That's indeed what Kardar said (post #5) in different words, by saying that RG is not invertible in the space of configurations, but is invertible in the space of parameters.
 

vanhees71

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Effective theories are renormalizable order by order in the momentum expansion. Some are non-renormalizable in the sense of "Dyson-renormalizable", which latter means that the theory is renormalizable order by order in the expansion of powers of ##\hbar## (or order by order in powers of the coupling constant, where however one has to be careful with gauge invariance).
 

A. Neumaier

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Effective theories are renormalizable order by order in the momentum expansion. Some are non-renormalizable in the sense of "Dyson-renormalizable", which latter means that the theory is renormalizable order by order in the expansion of powers of ##\hbar## (or order by order in powers of the coupling constant, where however one has to be careful with gauge invariance).
I n this sense all QFTs are renormalizable, and the term used in this sence has lost its discriminating meaning.
 

vanhees71

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Well, the distinguishing property is that Dyson-renormalizable theories have a finite number of coupling constants, while the effective theories have infinitely many since in any higher loop (##\hbar##) order (i.e., also higher powers of momentum) you need new counterterms.
 

Demystifier

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No. The Wilson RG is a map between effective theories, and these are all nonrenormalizable.
No, effective theories can also be renormalizable. For example, the Standard Model is also likely to be an effective theory (emerging perhaps from string theory, or something else).
 

A. Neumaier

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No, effective theories can also be renormalizable.
The Wilson RG is a map between effective theories, and these are almost all nonrenormalizable.
The few exceptions don't change the argument.

For example, the Standard Model is also likely to be an effective theory (emerging perhaps from string theory, or something else).
The standard model may be an approximation to an underlying theory but it need not be regarded as an effective theory. Even when one does, one cannot recover from it the underlying true theory, since it lacks the more fundamental degrees of freedom. This gives an explicit counterexample to your claim that
If the theory is just the Hamiltonian (or Lagrangian), and not one particular solution associated with that Hamiltonian, then one can recover it for renormalizable theories.
 

Demystifier

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Even when one does, one cannot recover from it the underlying true theory, since it lacks the more fundamental degrees of freedom.
I agree and I never claimed the opposite. Just because RG gives well defined non-infinite results at small distances doesn't mean that it is right at small distances. And the same is true for large distances too.
 
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vanhees71

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The Wilson RG is a map between effective theories, and these are almost all nonrenormalizable.
The few exceptions don't change the argument.


The standard model may be an approximation to an underlying theory but it need not be regarded as an effective theory. Even when one does, one cannot recover from it the underlying true theory, since it lacks the more fundamental degrees of freedom. This gives an explicit counterexample to your claim that
Do you think the standard model really has no problems with Landau poles? Isn't at least QED as part of the SM likely to have a Landau pole? If it has one, I think it's validity is constrained to energy-momentum scales not too close to the Landau pole, or is there a way to avoid the trouble with the Landau pole somehow and make it a theory valid on all energy scales?
 

vanhees71

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I agree and I never claimed the opposite. Just because RG gives well defined non-infinite results at small distances doesn't mean that it is right at small distances. And the same is true for large distances too.
Indeed, and that's good. After all the experimentalists should also have some work to do in ruling out theorists' phantasies ;-))).
 

A. Neumaier

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Do you think the standard model really has no problems with Landau poles? Isn't at least QED as part of the SM likely to have a Landau pole? If it has one, I think it's validity is constrained to energy-momentum scales not too close to the Landau pole, or is there a way to avoid the trouble with the Landau pole somehow and make it a theory valid on all energy scales?
The Gross-Neveu model in 1+1D has a Landau pole but has been rigorously constructed.
QCD, which is asymptotically free, has, due to infrared effects, a Landau pole (and unlike QED) at physically realizable energies. Nevertheless it seems to be a sound theory.

Therefore, in contrast to the original beliefs, Landau poles say nothing about the existence or nonexistence of a theory. They are just an artifact of incorrectly summing up pieces of the perturbative expansion. We discussed this here in another thread.
 

vanhees71

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What do you mean by "rigorously constructed"? Is it well-defined in the mathematical sense, but has a Landau pole? But then doesn't it break down at this pole as a physical theory (i.e., one that can in principle be interpreted as describing observable entities (which you may call "particles"?) at any energy scale)?

The 2nd paragraph seems to contradict the first. You say the GN model in 1+1 dimensions has been rigorously constructed and has a Landau pole. On the other hand you say, the Landau pole is an artifact of incorrectly summing up pieces of the pert. series. Shouldn't this imply that a completely constructed 1+1-dim. GN model shouldn't have a Landau pole since there you don't use a pert. expansion to begin with?

The other thread makes more sense to me, but it also tells you that perturbative QED doesn't make sense close to the Landau pole. Since QED is well-defined in the low-energy region, it's thus an effective theory with the upper energy scale defined by the Landau pole. I think it's pretty clear that the problem of a Landau pole doesn't depend on the specific regularization you use to define the renormalized theory. I think it's also there in BPHZ, where you just substract integrands without introducing an explicit regularization at all.

Concerning pQCD it's pretty similar: The perturbative expansion makes sense at large scales. If there's a Landau pole at lower scales, you cannot go close to it either, but that makes perfect sense, since in the low-energy limit your perturbative expansion becomes ill-defined. FAPP you simply use the wrong effective degrees of freedom. The low-energy effective theory of QCD is some hadronic field theory, using some symmetry principles like approximate chiral symmetry in the light-quark sector, heavy-quark effective theory in the heavy-quark sector etc.
 

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