My question refers to example theory introduced in the book "Supergravity" from D.Z.Freedman & A. van Proeyen p.80. Its Lagrangian is given by

$${\cal L}(Z,F) =-\frac{1}{4}(Im Z)F_{\mu\nu}F^{\mu\nu} -\frac{1}{8}(Re Z)\epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}F_{\rho\sigma}=-\frac{1}{2}Im(Z F_{\mu\nu}^- F^{\mu\nu-})$$

where $F_{\mu\nu}$ is the field tensor of a $U(1)$ gauge field and $Z$ is a complex scalar field. The field tensor has a dual

$$F^\tilde{\mu\nu} =-\frac{1}{2}i\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$$

and (apart from a possible sign) self-dual combinations can be defined:

$$F^{\mu\nu\pm} = \frac{1}{2}(F^{\mu\nu} \pm F^\tilde{\mu\nu})$$

The negative linear combination is used in the definition of the Lagrangian. This theory gives rise to the following field equations (FEs):

$$\partial_\mu F^\tilde{\mu\nu}=0 \quad\quad\text{and}\quad\quad \partial_\mu[(ImZ) F^{\mu\nu} + i(ReZ) F^\tilde{\mu\nu}]=0 $$

which by using the definition

$$ G^{\mu\nu} = \epsilon^{\mu\nu\rho\sigma}\frac{\delta S}{\delta F^{\rho\sigma}} = -i(ImZ) F^\tilde{\mu\nu} + (ReZ) F^{\mu\nu} $$

can be also written in the following form:

$$\partial_\mu Im F^{\mu\nu-} =0\quad\quad \text{and}\quad\quad \partial_\mu Im G^{\mu\nu-} =0$$

where the same definition of the self dual combination was applied on $G^{\mu\nu}$. $G$ also fulfils:

$$G^{\mu\nu-} = Z F^{\mu\nu-}$$.

Now the authors claim that the field equations are invariant to the following transformation:

$$\left(\begin{array}{c} F'^-\\ G'^-\end{array}\right) = {\cal S} \left(\begin{array}{c} F^-\\ G^-\end{array}\right)$$

with ${\cal S} \in SL(2,\mathbb{R})$, i.e.

$${\cal S} = \left(\begin{array}{cc} d & c \\ b & a \end{array} \right) \quad\quad \text{with} \quad\quad ad-cb=1$$

Invariance is supposed to be that both $F'^-$ and $G'^-$ fulfill the same FEs than $F^-$ & $G^-$ as well as that $Z$ transforms like:

$$ Z' = \frac{aZ+b }{cZ+d }$$ where $Z'$ is defined as:

$$ G'^{\mu\nu-} = Z' F'^{\mu\nu-}$$

I checked the claimed invariance and it is indeed realised. The curious thing is that on the next page the authors claim that the Lagrangian is not invariant to the same transformation:

$${\cal L}(Z',F') = -\frac{1}{2} Im(Z(1+cZ)F_{\mu\nu}^- F^{\mu\nu-})\neq {\cal L}(Z,F)$$

I was quite surprised by this result. Would it be mean that a (more subtle) duality as presented here leaves the field equations invariant, but the Lagrangian not? I always thought that an invariance found on the FEs corresponds one-to-one to an invariance of the Lagrangian. Is this not the case for a duality? Any help is appreciated.

  • $\begingroup$ "I always thought that an invariance found on the FEs corresponds one-to-one to an invariance of the Lagrangian" that's not true, even for ordinary symmetries: e.g. galilean boosts for the nonrelativistic particle (where the lagrangian changes by a total derivative). For a duality in the sense here this is even worse because $SL(2;\mathbb R)$ does not act on the space of field configurations, but only on the field equations, so you can't even prove invariance up to total derivatives. I'll see if I can recall a paper where this is discussed... $\endgroup$
    – user21299
    Commented Aug 14, 2020 at 18:45
  • $\begingroup$ @alexarvanitakis: may be you can expand this to a full answer. In particular what do you mean with "$SL(2,\mathbf{R})$ does not act on the space of field configurations, but only on the field equations", And actually why is duality so much appreciated respectively discussed if it does not leave the action (more general than the Lagrangian) invariant. $\endgroup$ Commented Aug 14, 2020 at 18:50
  • $\begingroup$ @alexarvanitakis: still curiously waiting for a post from you using the paper you mentioned. $\endgroup$ Commented Aug 21, 2020 at 8:35
  • $\begingroup$ sorry, haven't found time yet... $\endgroup$
    – user21299
    Commented Aug 21, 2020 at 17:44
  • $\begingroup$ How do you get that Lagrangian you wrote in the second line of your question?. Freedman doesn't have an explanation for that. $\endgroup$ Commented Jun 2, 2023 at 2:02

1 Answer 1


In Maxwell theory, duality transformation reads infinitesimally $\delta F_{\mu\nu}=\tilde F_{\mu\nu}$ or $\delta \vec E= \vec B,\; \delta \vec B= -\vec E$. Using this in the action with Lagrangian ${\cal L}=-1/4 F_{\mu\nu}F^{\mu\nu}$ we find \begin{align} \delta S&=-1/2 \int F^{\mu\nu} \tilde F_{\mu\nu}=-1/2 \int \epsilon^{\mu\nu\alpha\beta}F_{\mu\nu} F_{\alpha\beta}\\ &=-1/2\oint \partial_\alpha \left(\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}A_\beta\right) \end{align} However, this can be regarded as a true symmetry of the Lagrangian if it is represented as a transformation on the dynamical field $A_\mu$. It is shown by Deser and Teitelboim here that this is possible, but the duality transformation $\delta A$ is non-local, see their eq. 2.12. Therefore duality transformation is a symmetry of the action as it changes the action by a boundary (Chern-Simons) term.

However, more generally it is not true that every symmetry of the field equations can be realized in the Lagrangian.

There is a class of symmetries called hidden symmetries defined as those transformations that can be realized on the (Hamiltonian) phase space, but not on the configuration space. More explicitly on the phase space $(x,p)$ a symmetry transformation is one that preserves the symplectic form $dx\wedge dp$ and the Hamiltonian $H$. Here, the transformation of $x,p$ are independent of each other. However, the Lagrangian formulation is based on the tangent bundle of the configuration space (parametrized by $q$) and therefore the symmetries are written as diffeomorphisms on the configuration space leaving the action invariant (up to boundary terms). The transformation on $\dot q$ is implied by the transformation on $q$. Therefore there are some symmetries in the Hamiltonian formulation that cannot be realized in the Lagrangian formulation. The simplest example is the Runge-Lenz vector in the Kepler problem, see e.g. this paper by Cariglia.

  • $\begingroup$ you mean a duality transformation is a transformation in the phase space that has no equivalent on the configuration space ? Actually I am not completely convinced since on wikipedia it is shown that the conservation of the Runge-Lenz vector can be associated with a transformation in configuration space up to a total time derivative of the Lagrangian. I did not exclude that possibility of a total t-derivative of $L$ in my post, but anyway a duality transformation does not leave Lagrangian invariant up to a total t-derivative. $\endgroup$ Commented Aug 21, 2020 at 8:33
  • $\begingroup$ I updated my answer to clarify. $\endgroup$
    – Ali Seraj
    Commented Aug 21, 2020 at 10:32
  • 1
    $\begingroup$ @AliSeraj: nice find! (on the paper by Deser and Teitelboim) $\endgroup$
    – user21299
    Commented Aug 21, 2020 at 17:39
  • $\begingroup$ @Ali Seraj the infinitesimal duality transformation you mention seems to change the action (or the Lagrangian) only by a boundary term what is not the case for the more general one mentioned in the book of Freedman and van Proeyen, a fact that still makes me wonder. So I will consider it as a hidden symmetry. I also will assign the bounty to you. Unfortunately now (weekend) I don't have access to the paper Deser & Teitelboim, but yesterday I could have a look on it, and I will have more thorough look on it soon. $\endgroup$ Commented Aug 22, 2020 at 11:43
  • $\begingroup$ @Frederic Thomas: Can you show that finite translation is a symmetry of a Poincare symmetric action? In that case also one only requires that an infinitesimal translation changes the action by a boundary term. I don't think this is true for a finite translation as well. For internal symmetries where the Lagrangian is invariant under symmetry transformations, then the finite transformation also leaves the action invariant, but this is a special case $\endgroup$
    – Ali Seraj
    Commented Aug 23, 2020 at 5:36

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