I am very confused by these two points of view. Consider a theory whose space of fields is $V$ and that has an action $S$. Thinking of a gauge symmetry as a redundancy is your description means that one should identify certain fields in $V$ as physically equivalent. In other words, the space of physically distinguishable fields is some quotient of $V$. In most examples this happens by the existence of a gauge Lie algebra $\mathfrak{g}$ which acts on $V$ through a Lie algebra homomorphism that assigns to every element $X\in\mathfrak{g}$ a vector field $\int d^dxd^dyR^i_a(x,y)X^a(y)\frac{\delta}{\delta\phi^i(x)}$ on $V$.

In the second point of view, gauge symmetries are understood as constraints in the equations of motion. These usually take the form $\int d^dx F^i(x)\frac{\partial S}{\partial\phi^i(x)}=0$.

Both ideas are connected by noticing that, if the action is symmetric under a Lie group, understood in the first sense, then the equations of motion are constrained by a Noether identity $\int d^dxd^dyR^i_a(x,y)X^a(y)\frac{\delta S}{\delta\phi^i(x)}=0$. One may thus define the Gauge algebra as simply the set of all vector fields on $V$ under which the action is invariant (modulus trivial vector fields of the form $\int d^dxd^dy \frac{\partial S}{\partial\phi^i(x)}F^{ij}(x,y)\frac{\delta }{\delta\phi^i(y)}$ for $F^{ij}(x,y)=-F^{ji}(y,x)$).

  1. Isn't the Gauge algebra then simply the set of all local symmetries?
  2. Are gauge transformations equivalent to local symmetries?
  3. In case they aren't, is there any QFT with a local symmetry which is not a gauge transformation?

With respect to the third question, I have been given finite dimensional examples. On the other hand, I thought that maybe diffeomorphisms in Chern-Simons would be a valid example. However, I then read that these are equivalent to the usual gauge transformations of Chern-Simons modulus trivial local symmetries. Maybe Weyl transformations in electrodynamics?

By the way, most of the concepts that I am using are from Henneaux and Teitelboim, Quantization of Gauge Systems and Gomis, París and Samuel, Antibracket, Antifields and Gauge-Theory Quantization

EDIT: I was informed of an example that clearly shows that not all vector fields leaving the action invariant are gauge symmetries. Namely, consider a scalar field $\phi$ with action $S(\phi)=\int d^dx\partial_\mu\phi(x)\partial^\mu(\phi)$. Then the vector field corresponding to a translation $a^\mu$ is $\int d^dx\partial_\mu\phi a^\mu\frac{\partial}{\partial\phi(x)}$. This vector field acting on the action yields $$2\int d^dx d^dy \partial_\mu\phi(x) a^\mu\partial_\nu\delta(y-x)\partial^\nu\phi(y)=-2\int d^dx\partial_\mu\phi(x) a^\mu\square\phi(x).$$ One might think that this action vanishes only when the equations of motion are satisfied (which is true for any vector field by the very definition of the equations of motion). However, we have $$\int d^dx \partial_\mu\phi(x) a^\mu\square\phi(x)=\int d^dx \partial_\mu\square\phi(x) a^\mu\phi(x).$$ Therefore $$\int d^dx\partial_\mu\phi a^\mu\frac{\partial S}{\partial\phi(x)}=-\int d^dx\partial_\mu\phi(x) a^\mu\square\phi(x)-\int d^dx\partial_\mu\square\phi(x) a^\mu\phi(x)=-\int d^dxa^\mu\partial_\mu(\phi\square\phi)(x),$$ which is a total derivative, vanishing if we ask our fields to vanish at infinity. However, no one would gauge this symmetry. Maybe the problem is that this symmetry is not local?

  • $\begingroup$ Read where? Which page? Which finite dimensional examples? $\endgroup$ – Qmechanic May 4 '20 at 11:36
  • $\begingroup$ As for the finite dimensional example, I don't feel comfortable with sharing here since I didn't come up with it. However, this case is trivial because local symmetries are the same as global symmetries and there are clearly theories with global symmetries that are not gauge symmetries. As for the reference for Chern-Simons I'll look it up and update this comment as soon as possible. $\endgroup$ – Iván Mauricio Burbano May 4 '20 at 11:59
  • $\begingroup$ The statement regarding Chern-Simons is found in Exercise 1 of sciencedirect.com/science/article/pii/092056329090647D $\endgroup$ – Iván Mauricio Burbano May 4 '20 at 13:25
  1. Yes, the gauge algebra as a set refers to the infinitesimal gauge transformations/vector fields. The gauge algebra as an algebra encodes (at the infinitesimal level) the fact that the composition of gauge transformations is again a gauge transformations (possibly modulo EOM).

  2. Yes, according to Refs. 1-3, gauge transformations are local transformations that leave the Boltzmann-factor $e^{\frac{i}{\hbar}S}$ invariant up to boundary terms. So e.g. Yang-Mills/Chern-Simons-type gauge transformations and diffeomorphisms are both examples of gauge transformations. However, strictly speaking, that's a matter of terminology & conventions, and other authors may disagree.

OP mentions constraints. This sounds a bit like the Dirac conjecture that in the Hamiltonian formalism the first class constraints are generators of gauge symmetry. In this context it is natural to ask the following question.

Why can the Lagrangian gauge algebra be open (i.e. only close modulo EOM), while the Hamiltonian gauge algebra is always closed?

That's a good question. The only answer I know is that that's what the Lagrangian BRST (aka. Batalin-Vilkovisky) formalism and the Hamiltonian BRST (aka. Batalin-Fradkin-Vilkovisky) formalism can accommodate without violating ghost number conservation, etc., respectively (at least in their original form).


  1. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

  2. M. Henneaux, Lectures on the antifield-BRST formalism for gauge theories, Nucl. Phys. B Proc. Suppl. 18 (1990) 47.

  3. J. Gomis, J. Paris & S. Samuel, Antibracket, Antifields and Gauge-Theory Quantization, arXiv:hep-th/9412228.

  • $\begingroup$ Thank you very much for your reply. I don't know if this is really what I am asking since the I am not very familiar with the BFV formalism. I will modify my question to make it clearer. $\endgroup$ – Iván Mauricio Burbano May 4 '20 at 11:09
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic May 5 '20 at 9:25

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