Lie derivatives and the tetrad formalism

Question

I have been trying to learn about the tetrad formalism in general relativity and I understand the basic idea, but there is one issue that I can't seem to figure out: Is there a definition of a Lie derivative that gives covariant results when applied to a tensor with both coordinate and Lorentz indices?

Example: letting $\mu,\nu\dots$ represent coordinate indices and $a,b,\dots$ represent Lorentz indices, suppose I have a vector field $v^\mu$ and a mixed tensor $t_\mu^a$. Naïvely taking the Lie derivative of $t_\mu^a$ along $v^\mu$ gives $$ \mathcal L_v t_\mu^a = v^\nu \partial_\nu t_\mu^a + \partial_\mu v^\nu t_\nu^a. $$ While such an object transforms nicely under spacetime diffs, it does not transform nicely under local Lorentz transformations.

Question: Is there a way to modify the Lie derivative so that the result is Lorentz-covariant? Or are Lie derivatives just not so useful when using the tetrad formalism?

P.S. I understand that this is more of a math than physics question, but if I ask it on a Math forum, the answer is likely to be in terms of math symbols that I will find incomprehensible. So I would appreciate any answer in terms of index notation as that will be far easier for me to understand.

Answer 1

The Lie derivative of a vector can be written as $$\mathcal{L}_v A^\mu = A^{\mu }{}_{,\gamma} v^\gamma - A^\gamma{} v^\mu{}_{,\gamma}$$ where the commas stand for partial derivatives. This formulas should hold even if we substitute $A^\gamma = A^a e_a^\mu$, where $e_a^\mu, a=0,...,3$ is the tetrad. Now we use the usual Lie derivative formula for $e_a^\mu$ and obtain that it must hold that $$\mathcal{L}_v A^a = A^a{}_{,\gamma} v^\gamma$$ That is, in the most straightforward definition, tetrad components behave like scalars, exactly as you have said. However, it is possible to generalize this formula by requiring that the gradient above is replaced by the Lorentz-covariant nabla that annihilates the tetrad $\nabla_\gamma e^\mu_a =0$ or $$\nabla_\gamma A^a = A^a{}_{;\gamma} + e^a_\mu e^\mu_{b;\gamma} A^b$$ where the semi-colon involves coordinate-covariant derivatives (with the Christoffel symbols and all), and $e^a_\mu$ is the dual tetrad. This yields a modified definition of the action of the Lie derivative $\tilde{\mathcal{L}}_v$ that is Lorentz-covariant. (An alternative way of deriving it would be to require that $\tilde{\mathcal{L}}_v e_\mu^a =0$ and work from there as in the first few paragraphs.)

Answer 2

You can use the covariant Lie derivative. For instance, see https://arxiv.org/abs/2012.12094. Unfortunately I think there's no good reference, but the idea is simple as I outline below.

The Lie derivative can be defined in an "operational" way from 1) the Leibniz rule and 2) the Cartan magic formula which is $$\mathcal{L}_X \alpha = d \iota_X \alpha + \iota_X d \alpha$$ for a differential form $\alpha$. Then we define the gauge-covariant Lie derivative w.r.t. any gauge connection $\nabla$ on some vector bundle by demanding that $$\mathcal{L}^\nabla_X \alpha = d^\nabla \iota_X \alpha + \iota_X d^\nabla \alpha \,,$$ where $d^\nabla$ is the gauge-covariant exterior derivative. And now, by construction, gauge covariance is guaranteed for any "bundle-valued" tensor fields.

About your last point: It turns out that this covariant Lie derivative can be a super useful tool in some theoretical physics research (at least I find so). For instance you can use it for a manifestly gauge covariant formalism describing a particle (or a string (!)) moving in a Yang-Mills or gravity background. Or also for field theoryㅡlike the Ashtekar paper I attached. You can also go more mathematical and investigate the algebra of this gauge-covariant Lie derivatives and ponder on its physical significance. Or, you may utilize it for studying the covariant phase space of Yang-Mills, Einstein-Cartan theory, etc. Plus, I believe that we can also find nice examples in classical general relativity that can be efficiently described with covariant Lie derivative.