I have been studying Lie derivatives and some applications. While searching the web I found a reference with the following statement:

For a general Riemannian manifold $M$, take a tangent vector field $k=k^\mu \partial _\mu$ and consider the infinitesimal coordinate transformation, $$ x^\mu \to x^\mu + \alpha k^\mu~, $$ where $|\alpha| \ll 1$. Then it is possible to find that the metric components $g_{\mu \nu}$ transform as $$ g_{\mu \nu} \to g_{\mu \nu} + \alpha(\partial_\mu k_\nu + \partial _\nu k_\mu + k^\sigma \partial _\sigma g_{\mu\nu} ) +O(\alpha^2)~. $$

Now, how does the author get this? Does the author just uses the usual transformation law for a $(0,2)$-type tensor? This does not seem to be the case, since he finds so many terms. Moreover, I do not know the inverse transformation, so I wouldn't be able to apply it. Can somebody help me?


1 Answer 1


The metric being a rank $(0,2)$ tensor transforms under general coordinate transformations $x^\mu \to x'^\mu(x)$ as $$ g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x) $$

Now set $x'^\mu (x) = x^\mu + \alpha k^\mu(x)$ in the above expression and take a limit of small $\alpha$. You should then get $$ g'_{\mu\nu}(x) = g_{\mu\nu}(x) - \alpha ( g_{\mu\rho} \partial_\nu k^\rho + g_{\nu\rho} \partial_\mu k^\rho + k^\rho \partial_\rho g_{\mu\nu} ) + {\cal O} (\alpha^2) $$

The above is the correct version of the transformation. I believe you have gotten it wrong in the question (v1)

  • $\begingroup$ This must be a really stupid question but what do you mean substitute in the expression?Make a Taylor expansion? $\endgroup$
    – PML
    Jan 29, 2016 at 16:17
  • 1
    $\begingroup$ BTW, to be clear, the inverse transformation is $x^\mu = x'^\mu - a k^\mu + O(a^2)$. $\endgroup$ Jan 29, 2016 at 16:20
  • $\begingroup$ @PML - I mean, substitute the expression for $x'^\mu$ that I've written down into the first equation and then do a Taylor expansion in $\alpha$. $\endgroup$
    – Prahar
    Jan 29, 2016 at 16:23
  • $\begingroup$ @Prahar why do you have $\partial_\rho g_{\mu\nu}$ instead of $\partial_\rho g_{\mu\nu}'$ when you Taylor expand? $\endgroup$
    – AnOrAn
    Feb 6, 2020 at 16:51
  • $\begingroup$ I write $\partial_\rho g'_{\mu\nu} = \partial_\rho g_{\mu\nu} + {\cal O}( \alpha)$. $\endgroup$
    – Prahar
    Feb 6, 2020 at 16:52

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