# Why we only consider ''group'' symmetry but not general symmetry?

1. Nov 22, 2011

### ndung200790

Why do we only consider symmetry group(Lie group and Lie algebras) but not general symmetry(the transform that keeps Lagrangian invarian) in QFT?Is it because the symmetry group is more simple and more beautiful and in reality the forces of Nature obey the symmetry U(1)xSU(2)xSU(3)?
Thank you very much for your kind helping.

2. Nov 22, 2011

### strangerep

Which "transform that keeps Lagrangian invariant" do you mean? Can you give an example? Many of these transformations do indeed form a group. Perhaps you're thinking of dynamical groups instead of just symmetry groups?

Or are you asking why we use groups and algebras based on commutators rather than anticommutators?

Sometimes people use semigroups (group without the requirement for an inverse).
An example is the theory of time-asymmetric phenomena such as formation and decay of resonances.

3. Nov 22, 2011

### Fredrik

Staff Emeritus
He might also be asking why we spend so much time studying Lagrangians with "groups of symmetries" rather than just "symmetries". ndung200790, you may have to explain what you meant.

One reason why groups of symmetries are interesting in quantum theories is that Stone's theorem says (roughly) that for any group homomorphism U from a 1-parameter commutative group into the group of unitary operators on a Hilbert space, there's a self-adjoint operator A such that $U(t)=e^{iAt}$ for all t. (Click the link if you want to see a more exact statement of the theorem). The operator A (or is it -A?) is called the generator of this representation.

Pretty much all the interesting observables in quantum theories are generators of a representation of some symmetry group.

4. Nov 23, 2011

### dextercioby

You have to define what a symmetry is and apparently the only way to do it consistently and generally leads to the semigroup condition for the transformations you're looking for.

5. Nov 23, 2011

### ndung200790

I think there would be a transformation that the generators do not form a group,meaning that if we combine two generators,the result is not coresponded to any generator we considering.So can we demontrate that any transformation which remains Lagrangian invariant is a group(e.g.satisfying the closed characteristic of a group)?

6. Nov 23, 2011

### DrDu

In the case of discrete symmetry operations like charge conjugation, parity or time reversal you don't have any generator at all which does not mean that the operations don't form a group.

7. Nov 23, 2011

### naima

8. Nov 23, 2011

### strangerep

What do you mean by "combine two generators"?

If the commutator of two generators always yields another generator then you've probably got a Lie algebra, which can (usually, but not always, iiuc) be exponentiated to form a Lie group.

OTOH, if a commutator of two generators [A,B] yields something (call it C) which is not in your set of generators, then you must calculate higher commutators like [A,C] and [B,C] and try to determine whether this higher order commutator algebra eventually closes. If it doesn't then you've got an infinite-dimensional Lie algebra (probably better thought of as a noncommutative Poisson algebra).

[Edit: I wish Arnold Neumaier wasn't so busy with other things right now. He could give a much better answer to this.]

Last edited: Nov 23, 2011
9. Nov 24, 2011

### ndung200790

The combination of two generators means the product of two elements in group theory.

10. Nov 24, 2011

### tom.stoer

ndung200790,

everything you said so far can be understood as a symmetry operation represented by a (finite or infinite or even continuous) group.

In some cases it makes sense to discuss generators (in case of Lie groups and algebras), in some other cases (like C, P, T or crystallographic groups) not.

There are rather general cases of symmetry structures like the diffeomorphism or mapping class group for GR, infinite dimensional Kac-Moody algebras (as generalizations of finite dimensional Lie algebras) with central extension in string theory, supersymmetry / supergravity / graded algebras, quantum deformations of U(1), SU(2), dynamical symmetries (ordinay symmetry groups like SU(n) for the n-dim. harmonic oscillator) and perhaps many more which I am not aware of. I haven't seen anything else that does not belong to such a (generalized) symmetry structure.

The reason is rather simple.

Consider you have a Lagrangian L[x]; now you do something with it and get L[x'] where x has been transformed using something called 'g', but L is invariant (b/c it's a symmetry ;-). No you do something else with it called 'h' and get L[x'']. You can now write formally

x'= g*x
x''= h*x' = hg*x

which automatically results in a group structure!

I have no idea how to talk about a symmetries or transformations which do not form to a group.

http://en.wikipedia.org/wiki/Group_(mathematics)#Definition

Which property can be relaxed?

Closure: you have a symmetry transformation, a second symmetry transformation, but both transformation together are not a symmetry ???
Associativity: OK, perhaps one could play around with Octonions, non-associative algebras ...
Identity element: having no identity element would mean that it's not possible to do nothing!
Inverse element: having no inverse means that certain transformations cannot be undone; I have no idea how this would look like

Please give me a hint what you have in mind

11. Nov 25, 2011

### ndung200790

I owe all your helping very much.As Mr Tom.Stoer pointed,now I understand under transformations the Lagrangian to be invariant,the transformation ''group'' has closure characteristic(that I had wondered).

12. Nov 25, 2011

### Fredrik

Staff Emeritus
How would you define a symmetry? A reasonable definition is that a symmetry transformation of the Lagrangian is a transformation that doesn't change the equations of motion. With this definition, to do nothing to the Lagrangian is a symmetry transformation. Let's denote it by 1. If S is a symmetry transformation, then it must be invertible and S-1 must be a symmetry transformation. So {1,S,S-1} is closed under composition of functions, and is therefore a group.

13. Nov 25, 2011

### tom.stoer

Exactly!

14. Nov 25, 2011

### haael

Indeed. Every symmetry transformation set is a group.

15. Nov 25, 2011

### ndung200790

Does the equation of 4-divergence of Noether conserved current equaling zero be a ''type of motion equation''(and the equation does not change under the symmetry(Frederik's definition)?

16. Nov 26, 2011

### ndung200790

By the way,please teach me what is ''dynamical group''?

17. Nov 26, 2011

### tom.stoer

Afaik it's a symmetry group acting in phase space, not in position space.

Consider the 1-dim. harmonic oscillator. In position space there is no obvious continuous symmetry, but in phase space you can define a SO(2) rotation in the (p,x) plane via a canonical transformation which leaves the Hamiltonian H ~ p² + x², i.e. the length of the vector (p,x) invariant.

For the N-dim. harmonic oscillator you get the SU(N) symmetry in phase space.

18. Nov 27, 2011

### strangerep

The generators in a "symmetry algebra" commute with the Hamiltonian.

The generators in a "dynamical algebra" do not necessarily commute with the Hamiltonian, however such a commutator preserves the set of such generators.

More explicitly, suppose the set $\{ g_1, g_2, g_3 \}$ spans a Lie algebra ${\mathbb g}$. Then if $[g_i,H] = 0$, the $g_i$ are said to span a "symmetry algebra". However, if the commutator between the Hamiltonian H and $\mathbb{g}$ stays in $\mathbb{g}$, i.e., if
$$[g_i,H] = \dots ~\mbox{linear combination of the}~g_i \dots$$
then the $g_i$ are said to span a "dynamical algebra". In other words, the action of the Hamiltonian transforms any element of the dynamical algebra into some other element of the dynamical algebra.

Thus, if one can find the maximal dynamical algebra for a given Hamiltonian (and a convenient representation thereof), then one has almost solved the whole problem since elements of the dynamical algebra evolve in time only amongst themselves.

BTW, in classical mechanics, one has Poisson brackets. We then try to "quantize" the system by modifying this classical dynamical algebra and representing it somehow as operators on a Hilbert space.

Last edited: Nov 27, 2011
19. Nov 27, 2011

### tom.stoer

Then my example of the N-dim. harmonic oscillator is not the most general case of a dynamical group b/b H is the the generator of the trivial U(1) factor and commutes with all SU(N) generators, i.e. we have U(N) = U(1)*SU(N).

Is there a simple example where one can see explicitly [H, gi] = f(gi)?

(I am thinking about the constraint algebra in canonical LQG - but this is definitly not a simple example)

20. Nov 27, 2011

### ndung200790

By the way,now I have known the very difference between Lie group and Lie algebra.(The confusing leads me to this question)