In my general relativity course, we are discussing infinitesimal diffeomorphisms defined by $x^{\mu}\rightarrow y^{\mu}(x) = x^{\mu} + \xi^{\mu}(x)$. We have been examining how different objects transform under this. For example, the metric transformation $g_{\mu\nu}(x)dx^{\mu}dx^{\nu} \rightarrow g_{\mu\nu}(y) dy^\mu dy^\nu$ can be found via

$$g_{\mu\nu}(y) dy^\mu dy^\nu = g_{\mu\nu}(x+\xi)\frac{\partial(x+\xi)^\mu}{\partial x^\alpha} \frac{\partial (x +\xi)^\nu}{\partial x^\beta} dx^\alpha dx^\beta \tag{1} $$

From here, it is my understanding that we Taylor expand the expressions to get:

$$ = \left( g_{\mu\nu}(x) + \xi^\alpha \partial_\alpha g_{\mu\nu}(x)\,+\, ... \right)(\delta^\mu_\alpha + \partial_\alpha \xi^\mu ) (\delta^\nu_\beta + \partial_\beta \xi^\nu\,) \mathrm d{x^\alpha} \mathrm dx^\beta \tag{2}$$ $$= \left( g_{\mu\nu}(x) +\, \xi^\sigma \partial_\sigma g_{\mu\nu}(x) +\, g_{\mu\sigma} \partial_\nu \xi^\sigma +\, g_{\nu\sigma} \partial_\mu \xi^\sigma\,+\,... \right) \mathrm dx^\mu \mathrm dx^\nu \tag{ 3}$$

where the dots are just higher order terms that we can ignore. From here, one can show that the metric varies with the Lie derivative in the direction of $\xi^{\mu}$.

I am really struggling to comprehend the mathematical calculation and simplification in this problem. My specific questions are:

Q1) From (1) to (2), how does one taylor expand an expression like $\partial(x+\xi)^\mu / \partial x^\alpha$? I am having trouble understanding how one gets to the resulting expression.

Q2) From (2) to (3), I am completely lost on what was done. It seems like the kronecker delta's were applied to $\mathrm dx^\alpha \mathrm dx^\beta$. I assume they would need to be factored out of the binomial expressions, but how does that affect the other terms?

Q3) I am often confused when new indices are relabeled. What is the motivation for introducing this sigma index?

Q4) On a more conceptual note, I understand that we want to use this result to show that the action is invariant under infinitesimal diffeomorphisms. But why look at infinitesimal diffeomorphisms rather than diffeomorphisms in general?

I apologize if much of this question is trivially mathematical. I debated putting it in the Math forum but figured I'd rather get a physicist's answer.


1 Answer 1


Q1) The derivative is a linear operator hence \begin{equation} \partial_\alpha (x+ \xi)^\mu = \partial_\alpha x^\mu + \partial_\alpha \xi^\mu = \delta^\mu_\alpha + \partial_\alpha \xi^\mu \end{equation} The statement that those 2 factors get Taylor expanded is actually wrong, only the first factor gets Taylor expanded.

Q2) To first order in the derivatives we have something like \begin{equation} (g+A)(\delta+B)(\delta+C) = g \delta \delta + A\delta \delta + gB\delta + gC\delta \end{equation} A bit cryptic, but it gives you the four terms considered.

Q3) There isn't really a motivation, just know that you can always relabel whenever there are contraction $\xi^\alpha \xi_\alpha = \xi^\beta \xi_\beta$. Presumably they got rid of the different $\alpha$'s and $\beta$'s and replaced it with $\sigma$'s.

Q4) For most groups used in physics the general symmetry transformation can be obtained from the infinitesimal transformation, called generators. Therefore, one looks at the transformation under these infinitesimal transformations and want them to be of a certain form which results in an equation to solve for these infinitesimal generators.
I think that the only necessary condition is that the group should be connected to the identity such that for generators $\xi^A$ and parameters $\epsilon_A$: \begin{equation} g = e^{\epsilon^A \xi_A} \end{equation} $A$ is not a spacetime label, but an internal index of the symmetry group. $g$ is the finite group element.

  • 1
    $\begingroup$ Okay I have worked out the problem with your comments and I got it to work out! Thanks! Also, I am not familiar with groups/generators. Do you know any good resources for an introduction to them? $\endgroup$
    – ZacharyC
    Commented Oct 30, 2020 at 17:44
  • 2
    $\begingroup$ No problem, glad I could help! About references, most books on QFT have a part about Lie groups and their generators when introducing non-Abelian gauge theories (like Pesking & Schroeder eg.) Other references may be: Group Theory in a Nutshell for Physicists - A. Zee and Symmetries, Lie Algebras and Representations - J. Fuchs (more mathematically inclined) $\endgroup$
    – Guliano
    Commented Oct 30, 2020 at 18:03
  • $\begingroup$ Just a comment but $\xi_A$ are not the generators, $\xi^A \partial_A$ is the generator. You can think of diffeos as position dependent translations,so the $\xi_A$ are not the generators themselves but spacetime dependent paramenters, exactly like in a gauge transformation. $\endgroup$ Commented Sep 1, 2023 at 2:09
  • $\begingroup$ if you tried to apply your operator g to $x^\mu$ you would get $g x^\mu = x^\mu + \epsilon^A \xi_A x^\mu + \dots$ which is not how diffeos are supposed to act on coordinates. $\endgroup$ Commented Sep 1, 2023 at 2:10
  • $\begingroup$ Well, $A$ here is read as an internal index of the symmetry group, not necessarily a spacetime index. But yes, I believe you could write $\xi^{A \mu} \partial_\mu$ if you want to be more specific in this case instead of seeing $\xi^A$ as a generator. $\endgroup$
    – Guliano
    Commented Sep 2, 2023 at 3:18

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