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Three observations:

  1. A topological group does not have a (nontrivial) Lie algebra structure. A Lie group has a (generically) nontrivial Lie algebra structure.
  2. Noether's first theorem is usually proved and stated in terms of "generators" of a group, i.e., a basis of the Lie algebra corresponding to the (necessarily Lie by 1.) group.
  3. Confusingly (to me) Noether's theorem is always stated in terms of continuous symmetries (which has to do with topology and not smoothness).

Leading to the question:

Does Noether's first theorem refer to Lie groups or is it a stronger statement about topological groups?

Valter Moretti
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Silly Goose
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3 Answers3

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First of all, it is necessary to stress that a topological group which is also a topological manifold (no smothness is required) is automatically a Lie group. So, not only it is smooth, but it is also analytic: it is an analytic manifold and the group operations are analytic with respect analytic coordinates. The Lie group structure is uniquely determined by the topological group together with the topological manifold structures (assumed to be compatible).

In practice one starts with a topological space which is locally identified with $\mathbb{R}^n$ through continuous maps with continuous inverse. One also assume that on that space there is a group structure whose operations are continuous with respect the underlying topology.

The mentioned result states that, among the various (continuous) local coordinate systems, there exists a uniquely determined subfamily of local charts whose elelements are connected by means of smooth (actually analytic) coordinate transformations. Using this subfamily of local charts, the group turns out to be a Lie group. Hence there is in particular a Lie algebra in the tangent space to the unit element.

This is the content of the celebrated Gleason-Montgomery-Zippin theorem, that solved Hilbert's fifth problem.

Regarding the Noether theorem, actually one needs less. In fact the theorem deals with one-parameter Lie groups, that is $1$-dimensional Lie groups which satisfy some further depending on the type of formulation (e.g. one parameter groups of canonical transformations, in case if a Hamiltonian formulation).

According to the GMZ theorem, it is sufficient a group of elements one-to-one labelled on $\mathbb{R}$ (or also $S^1$) such that the group operations are continuous on $\mathbb{R}$. With a change of coordinates on $\mathbb{R}$ it becomes a one-dimensional Lie group.

As a final comment I stress that, in QM, the Noether group does not even require a topological structure because it is valid also for discrete simmetries unitarily dynamical represented, when the unitaries are also selfadjoint (think of the parity operator).

Valter Moretti
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I have only ever seen it apply to Lie groups. I have never seen an example in physics of a topological group which was not a Lie group. Indeed, in the standard textbook proof of Noether's theorem, one uses "generators" of the symmetry infinitesimally close to the origin in order to prove the theorem. The requirement that the group be a Lie group can be thought of as a (usually unstated) assumption of Noether's theorem.

user1379857
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Noether's theorem requires a smooth structure for its statement. Thus it only applies to symmetries that are Lie groups and not topological groups.

Mozibur Ullah
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