When introducing the vielbein formalism in general relativity, I came across the use of an infinitesimal general transformation, or Einstein transformation. The latter term seems to not be covered on standard references, so my asking here.

I quote the lectures in which I found this term: Horatiu Nastase - Introduction to Supergravity, p.12, right above (2.2).

On both the metric and the vielbein we have also general coordinate transformations. We can check that an infinitesimal general coordinate transformation ("Einstein" transformation) $\delta x^\mu = \xi^\mu$ acting on the metric gives $$ \tag{2.2} (\delta_\xi g)_{\mu\nu} (x) = (\xi^\rho \partial_\rho) g_{\mu\nu} + (\partial_\mu \xi^\rho) g_{\rho\nu} + (\partial_\nu \xi^\rho) g_{\rho \nu} $$ where the first term corresponds to a translation (the linear term in the Fourier expansion of a field), but there are extra terms. Thus the general coordinate transformations are the general relativity version, i.e. the local version of the (global) $P_\mu$ translations in special relativity (in special relativity we have a global parameter $\xi^\mu$, but now we have a local $\xi^\mu(x)$).

  1. What is meant by an Einstein transformation?
  2. What is meant with $\delta_\xi$, and how is the relation (2.2) obtained?
  3. The following transformation for the vielbein is given: $$ \tag{2.3} (\delta_\xi e)_\mu^a(x) = (\xi^\rho \partial_\rho) e_\mu^a + ( \partial_\mu \xi^\rho) e_\rho^a. $$ Why is this relation different from (2.2)? Why is the last term missing?
  • 2
    $\begingroup$ I'm pretty sure $\delta_\xi$ is just the Lie derivative $\mathcal{L}_\xi$ in strange notation. $\endgroup$
    – Ryan Unger
    Jan 16, 2015 at 10:13
  • 1
    $\begingroup$ @0celo7 and how can I write it in coordinates notation? A simple $$ g_{\mu\nu}(x) \rightarrow g_{\mu\nu}(x+\xi(x))$$ only gives the first term in (2.2) $\endgroup$
    – glS
    Jan 16, 2015 at 10:16
  • 1
    $\begingroup$ the metric at the point $x+\xi(x)$ is, in general, different from the metric at $x$, so perhaps you have to transform that as well through $\xi^\mu_{,\nu}$ $\endgroup$
    – Phoenix87
    Jan 16, 2015 at 10:19
  • $\begingroup$ I guess its just the Lie derivative wrt a vector field $$\xi$$.. $\endgroup$
    – GRrocks
    Jan 16, 2015 at 14:29

1 Answer 1


Comments to the question (v3):

  1. $\delta_\xi\equiv {\cal L}_\xi$ is a Lie-derivative wrt. the vector field $\xi$ in the active picture, or a general coordinate transformation in the passive picture.

  2. The terminology Einstein transformation yields (at the time of writing this answer) less than a 1000 Google hits, so we conclude that it is not widely used. It seems to be just another word for the before-mentioned Lie-derivative/coordinate transformation.

  3. Concerning the vielbein in eq. (2.3), the before-mentioned Lie-derivative/coordinate transformation does not "act" on a flat index "$a$", so to speak. See also e.g. this and this Phys.SE posts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.