I've had a brief look through similar threads on this topic to see if my question has already been answered, but I didn't find quite what I was looking for, perhaps it is because I'm finding it hard to put words on my question and I hope that you will be able to help me ask it clearly.

I'm trying to link what I know from mathematics to what we are writing in physics for Noether's theorem. If I understand correctly we are looking at the symmetries of the action i.e. under which symmetry groups is it invariant. Noether's theorem allows us to calculate a conserved current in the case of a continuous symmetry (Lie groups), by means of so-called infinitesimal symmetries which I believe to be elements of the Lie algebra (so the tangent space to the neutral element) of the Lie group.

I'm guessing that if the action is invariant under the symmetry then it's "variation" should be 0 when we vary the system (space-time coordinate, or field) using this symmetry; it is exactly this step that I would like to understand better, how can I formalize this step correctly mathematically? How should I understand this variation, and how does it's calculation give rise to the elements of the Lie algebra?


1 Answer 1

  1. Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws.

  2. Lie's third theorem guarantees that a finite-dimensional Lie algebra can be exponentiated into a Lie group, cf. e.g. Wikipedia & n-Lab.

  3. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws often together form a representation of the Lie algebra $L$), then one may focus on a 1-dimensional Abelian Lie subalgebra $u(1)\cong \mathbb{R}$.

  4. In the context of field theory, there should be Lie algebra homomorphisms from the Lie algebra $L$ to the Lie algebra of vector fields on the field configuration space (so-called vertical transformations) and to the Lie algebra the vector fields of spacetime (so-called horizontal transformations).

  5. That the action functional $S[\phi]$ possesses a symmetry (quasisymmetry) means that the appropriate Lie derivatives of $S$ wrt. the above vector fields should vanish (be a boundary term), respectively.

  6. Note that the Noether currents & charges do not always form a representation of the Lie algebra $L$. There could e.g. appear central extensions, cf. this and this Phys.SE posts.

  • $\begingroup$ 1. So if I choose as in 3), to consider a 1 dimensional sub -algebra, I therefore have 1 generator and a parameter (say an angle) that I can differentiate with respect to: I would look at the change of the Lagrangian with respect to the change of the angle, which should be zero? 2. Another thought that comes to mind (probably an easy question), does invariance with respect to the "real" transformation (by that I mean element of Lie group), imply invariance w.r.t. the "infinitesimal" symmetry? $\endgroup$
    – Jack
    Feb 6, 2015 at 16:44
  • $\begingroup$ 1. Apart from the issue of Lagrangian vs action, then yes. 2. Yes. $\endgroup$
    – Qmechanic
    Feb 6, 2015 at 16:54
  • $\begingroup$ For question 2. How can I see this? Maybe this is the link that I'm missing as the tangent vectors aren't necessarily elements of the group itself... $\endgroup$
    – Jack
    Feb 6, 2015 at 18:11
  • 1
    $\begingroup$ 1. Firstly, there should be a Lie group homomorphism $\Phi$ from the Lie group $G$ to the group of diffeomorphisms/flows on the appropriate space. 2. Secondly, the action functional $S$ should be invariant when composing the fields with the above flows. 3. The derivative $\Phi_{\ast}$ of this map becomes a Lie algebra homomorphisms from the Lie algebra $L$ to the Lie algebra of vector fields. 4. Now combine these facts to show that $S$ must be invariant under the corresponding vector fields. $\endgroup$
    – Qmechanic
    Feb 6, 2015 at 18:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.