I have this question related to the the Noether's Theorems. I want to know a rigorous enough enunciation of this theorem, the context is Classical Field Theory without fancy geometrical structures but the usual stuff you need to know to do QFT and the use of Lie Groups(without being too abstract, I need a sensible connection with particle physics).

For what I read around in standard classical mechanics texts, Peskin, Brading and Brown and the thesis of one those authors,it is not really clear to me what a symmetry transformation and Noether's theorems are if viewed with a group theoretic perspective and the knowledge constraints mentioned above.

For what I read in other posts on the site, the group that acts on the Lagrangian and gives the conservation of currents(conservation law) and Noether's theorems is the group of transformations on the space of fields $ \mathfrak{F(\mathcal{M})}$ that is $\mathcal{G} = \{ \Lambda: \mathcal{M} \rightarrow G \} $ where $\mathcal{M}$ is a manifold(for now let just say Minkowski space or Euclidean ), $\Lambda(x)$ is the transformation and $G$ is a group usually compact as $SU(N)$ or Poincare/Galileo. But the problem is that there is a big distance in understanding between this fact and what I have read about Noether's theorems using the literature mentioned above .

Following what they do in the paper, lets define the total variation of the action as $\hat{\delta}S\equiv S(\phi'(x'), \partial_{\mu},\phi(x'),x') -S(\phi(x),\partial_{\mu}\phi(x),x) $. They also define a generic transformation of the action as $\Delta S \equiv \tilde{S}(\phi'(x'), \partial_{\mu},\phi(x'),x') -S(\phi(x),\partial_{\mu}\phi(x),x) $. The transformations that give both variations are "infinitesimal"(what are they more rigorously?). Question a) Are these elements of $\mathcal{G}$? I think they are.

In the thesis they define a symmetry as the transformations (I suppose elements of $\mathcal{G}$) that give $\hat{\delta} S = 0$. Here I think they are again talking about the action of some infinitesimal transformations of $\mathcal{G}$. Then they proceed to derive the so called Noether relations without imposing the Lagrange Euler conditions. Those are: \begin{equation} \sum_{i=1}^{N} \left( \partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_i)} - \frac{\partial \mathcal{L}}{\partial \phi_i} \right)\delta \phi_i= \sum_{i=1}^N \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_i)}\delta \phi_i + \mathcal{L}\delta x^{\mu} \right) \end{equation} Then they do following. To enunciate Noether's First Theorem they restrict to "finite dimensional continuous group of transformations depending smoothly on $\rho$ independent parameters $\omega_{i}, (i= 1, \cdots, \rho)$ " that give $\hat{\delta}S = 0$. Then they proceed to expand $\delta \phi$ around the parameters and impose Lagrange Euler equations and get the conservation theorem. For what I understand is that they restrict to the "infinitesimal group action" of some group of transformations that depends on finite parameters. Q b) This is not a Lie group, right? What is the relation with the usual physicist defnition of global symmetries as "the infinitesimal action of a finite dimensional Lie group that leaves $S$ invarant"

Question c) Are they talking about a subgroup of $\mathcal{G}$ that is finite? how is that if $\mathcal{G}$ is infinite dimensional? Are they talking about $G$? or is that the subgroup of $\mathcal{G}$ is somehow isomorphic to G? They seem to act the same way.

In the paper they just refer to transformation that do not act on coordinates ("they defined them as gauge transformations") but in the thesis the same approach is done with one that changes coordinates.

For Noether's second theorem they consider the infinite dimensional group of transformations with finite parameters that depend on x (i.e. functions). I really don't understand this. How is that having the parameters depending explicitly on spacetime changes your the dimension of the group of transformations. How this Second Theorem relates to usual local symmetries as defined in textbooks of physics is even muddier at least for me.

Thanks in advance.


1 Answer 1


This question (v1) asks many questions. Let us here make some general remarks, which OP hopefully will find useful.

  1. Noether's theorem only needs infinitesimal transformations to work. Hence the important object is not the set $G$ of finite transformations, but rather the set $\mathfrak{g}$ of infinitesimal transformations.

  2. In general, the set $\mathfrak{g}$ does not have to constitute a Lie algebra or even a Lie algebroid. The "Lie bracket" of two infinitesimal transformations might only close on-shell, i.e. modulo Euler-Lagrange equations. (This is known as an open algebra.)

  3. A horizontal infinitesimal transformation $\delta x^{\mu}$ changes the spacetime point $x^{\mu}$, while a vertical infinitesimal transformation $\delta_0 \phi^{\alpha}(x)$ changes the fields $\phi^{\alpha}(x)$ without moving the spacetime point $x$. A general infinitesimal transformation is a combination of horizontal and vertical infinitesimal transformations.

  4. A vertical infinitesimal transformation is typically of the form $$\tag{1} \delta_0 \phi^{\alpha}(x) ~=~\varepsilon^a(x) ~Y^{\alpha}_a(\phi(x),\partial\phi(x),x) + d_{\mu}\varepsilon^a(x)~ Y^{\alpha, \mu}_a(\phi(x),\partial\phi(x),x),$$ where $\varepsilon^a(x)$ are infinitesimal transformation parameters, which are coordinates of a section $\varepsilon(x)$ in a vector bundle $E$ over spacetime.

  5. To apply Noether's first theorem for a finite subspace of global$^1$ infinitesimal transformations, one identifies a finite-dimensional subspace of sections $\varepsilon_{(1)}(x)$,$\ldots,$ $\varepsilon_{(m)}(x)$, in $E$. Thus the global infinitesimal transformations are of the form $$\tag{2} \varepsilon(x)~=~ \sum_{r=1}^m \omega^{(r)}~\varepsilon_{(r)}(x), $$ where the parameters $\omega^{(1)}$, $\ldots$, $\omega^{(m)}$, are $x$-independent. In coordinates, $$\tag{3} \varepsilon^a(x)~=~ \sum_{r=1}^m \omega^{(r)}~\varepsilon^a_{(r)}(x). $$


$^1$ A global (local) transformation refers in this physics context to an $x$-independent ($x$-dependent) transformation, respectively. What are $x$-independent are here really the $\omega^{(r)}$ parameters, not necessarily the basis elements $\varepsilon_{(r)}(x)$. Thus the notion of global transformations depends in principle on the choice of section basis $\varepsilon_{(1)}(x)$,$\ldots,$ $\varepsilon_{(m)}(x)$. [Local and global transformation in physics should not be confused with the mathematical notion of locally and globally defined objects. All transformations in this answer (local as well as global) are assumed to be globally defined on the entire spacetime. Locally defined transformations take us to the realm of gerbes.]

  • $\begingroup$ Thanks for the answer. I'm understanding a bit more but not enough. For example in the case of $U(1)$ symmetry of classical electromagnetism. The structure group of the fiber bundle is the same but if we want to apply a Noether theorem we have to select the first for transformations that conserve "global symmetry" and the second for "local symmetry". This is because the "group of transformations" (what group?)is finite or infinite respectively. I understand points 1-4 above but 5 not too much. For example in U(1) what is a finite subspace of sections? Any references about these topics? $\endgroup$
    – Mitor
    Oct 28, 2013 at 4:04
  • $\begingroup$ In the case of EM, the finite-dimensional subgroup $H\subseteq G$ is typically just one-dimensional. Here $H$ and $G=\{\mathbb{R^4}\to U(1)\}$ are the groups of global and local gauge transformations, respectively. $H$ is conventionally the set of $x$-independent sections $\mathbb{R^4}\to U(1)$. $\endgroup$
    – Qmechanic
    Oct 28, 2013 at 14:57
  • $\begingroup$ OK. $G$ is infinite dimensional since is a space of functions (right?) So how do I say that $H$ is one dimensional? because it is isomorphic to $U(1)$? $\endgroup$
    – Mitor
    Oct 29, 2013 at 2:39
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Oct 29, 2013 at 2:44

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