# Lie Derivatives and Symmetry

• I
I'm trying to better understand how people refer to symmetry in Physics and Differential Geometry. In "Exterior Differential Systems and Euler Lagrange Partial Differential Equations," by Bryant, Griffiths and Grossman, it seems a vector field is a symmetry of a Lagrangian if the Lie derivative of the Lagrangian with respect to the vector field vanishes. I don't see how this connects to the notion of a symmetry as a conserved quantity arising from invariance of a Lagrangian under a group of transformations as these seem to be disjoint conceptually; what is the link I wonder?

jambaugh
Gold Member
This is my understanding from a foundational Lie group/algebra perspective.

A Lie derivative is applied to a function over a differentiable manifold. You can't have a symmetry of the function if you are not already working with a symmetry of the manifold itself. The Lie group of continuously generated symmetries of a differentiable manifold (with no structure other than topology) is the diffeomorphism group and the generators of diffeomorphism are the Lie derivatives. You identify a vector field with a Lie derivativ by treating the vectors as rates of point flows for the one parameter diffeomorphism subgroup generated by a Lie derivative.

SIDENOTE:I think there's still something unsettled here but only in so far as this defines vector and thence tensor fields in toto without actually defining vectors at a given point per se. But this usually is resolved once a metric or similar structure is imposed so it never comes up in applications.

So if you are looking for a symmetry of a field you are looking for a subgroup of the diffeomorphism group which will map the field to itself, and this means the generators of that subgroup must, as infinitesimal actors, map the field to zero. For scalar fields:
$$\mathcal{D}_\epsilon \phi = \exp(\epsilon \mathcal{L}) \phi \approx ( \boldsymbol{1}+\epsilon \mathcal{L} ) \phi \to \phi$$
for small $\epsilon$.

Now to transform a vector field you need a bit more... but by identifying vector fields with the Lie derivatives you have the natural adjoint action of the Lie derivative a.k.a. vector fields, on themselves in the form of the Lie (commutator) product. But the gist of it is that we can thereby extend the actions of Lie derivatives and thus diffeomorphisms to vector and thence tensor fields.

Given a tensor field $\mathbf{A}$ and a vector field $X$ with corresponding Lie derivative $\mathcal{L}_X$, the generated 1-parameter Lie subgroup: $\{ \Gamma(\theta)=\exp(\theta\mathcal{L}_X)\}_{\theta \in \mathbb{R}}$ is a symmetry of the field if each element of the group leaves the field invariant:
$$\Gamma(\theta)\mathbf{A} = \mathbf{A}\to$$
$$e^{\theta\mathcal{L}_X}\mathbf{A}=\left[\boldsymbol{1}+\theta \mathcal{L}_X + \theta^2 \mathcal{L}_X^2 +\ldots\right]\mathbf{A} = \mathbf{A}+\boldsymbol{0}\to$$
$$\mathcal{L}_X \mathbf{A} = \boldsymbol{0}$$