There are many interesting partial differential equations which can be expressed as the Euler-Lagrange equations corresponding to some Lagrangian. Thus they are equivalent to the condition that the action defined by the Lagrangian is stationary under all variations. Sometimes we want to study solutions of the equations which are invariant under some symmetry group. Starting from the original equations, it is possible to calculate the symmetry-reduced equations. This is what I and many others usually do, without worrying about a Lagrangian formulation. Suppose that in some particular case the task of doing a symmetry reduction of the Lagrangian is significantly easier than the corresponding task for the differential equations. Then it is tempting to take the Euler-Lagrange equations corresponding to the symmetry-reduced action and hope that for symmetric solutions they are equivalent to the Euler-Lagrange equations without symmetry. But is this always true? The Euler-Lagrange equations without symmetry are equivalent to stationarity under all variations while the Euler-Lagrange equations for the symmetry-reduced action are equivalent to stationarity under symmetric perturbations. The second property is a priori weaker than the first. This procedure is often implicit in physics papers, where the variational formulation is more at the centre of interest than the equations of motion.

The potential problem just discussed is rarely if ever mentioned in the physics literature. Fortunately this question has been examined a long time ago by Richard Palais in a paper entitled ‘The principle of symmetric criticality’ (Commun. Math. Phys. 69, 19). I have known of the existence of this paper for many years but I never took the trouble to look at it seriously. Now I have finally done so. Palais shows that the principle is true if the group is compact or if the action is by isometries on a Riemannian manifold. Here the manifold is allowed to be an infinite-dimensional Hilbert manifold, so that examples of relevance to field theories in physics are included. The proof in the Riemannian case is conceptually simple and so I will give it here. Suppose that is a Riemannian manifold and a function on . Let a group act smoothly on leaving and invariant. Let be a critical point of the restriction of to the set of fixed points of the group action. It can be shown that is a smooth totally geodesic submanifold. (In fact in more generality a key question is whether the fixed point set is a submanifold. If this is not the case even the definition of the principle may be problematic.) The gradient of at is orthogonal to . Now consider the geodesic starting at with initial tangent vector equal to the gradient of . It is evidently invariant under the group action since all the objects entering into its definition are. It follows that this geodesic consists of fixed points of the action of and so must be tangent to . Hence the gradient of vanishes.

When does the principle fail? Perhaps the simplest example is given by the action of the real numbers on the plane generated by the vector field and the function . This has no critical points but its restriction to the fixed point set, which is the -axis, has critical points everywhere.

March 9, 2011 at 12:44 pm |

Hello Alan. You may not remember, but we’ve met once or twice over the years. I have also had some interest in the Principle of Symmetric Criticality, which has a fairly long history in the context of general relativity. In fact, Ian Anderson, Mark Fels and I did some relatively recent work on it. We found that if one restricts attention to local Lagrangian field theories, one can give a rather more detailed description of the obstructions to the principle. If you are still interested in this stuff, have a look here:

http://arXiv.org/pdf/1011.3429

http://dx.doi.org/10.1088/0264-9381/19/4/303

Sincerely,

Charles Torre

March 9, 2011 at 6:58 pm |

Hi,

Thanks for the references,

Alan