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Consider the Laplacian operator $\nabla^{2}=\frac{1}{\sqrt{g}}\frac{\partial}{\partial q^{i}}\left(\sqrt{g}g^{ij}\frac{\partial}{\partial q^{j}}\right)$.

How does the Laplacian operator transform under an infinitesimal change $x^{i}(x)=x^{i}+\varepsilon\xi^{i}(x)$?

I know how to work out the transformation rules for the metric and the derivatives, but I don't undertand how $\sqrt{g}$ should transform.

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closed as off-topic by Spencer, max_zorn, Cesareo, mrtaurho, Javi Apr 30 at 10:45

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  • $\begingroup$ What have you tried? $\endgroup$ – Spencer Apr 30 at 4:48
  • $\begingroup$ I know how the metric and the derivatives changes and I tried to plug everything together, but I'm struggling with all the expansions. $\endgroup$ – AndresB Apr 30 at 4:55
  • $\begingroup$ It is certainly a very involved calculation. Are you clear on how $\sqrt{g}$ transforms? Are you neglecting terms of order $\epsilon^2$? $\endgroup$ – Spencer Apr 30 at 4:57
  • $\begingroup$ No, I'm not certain how it transforms, and keeping things to first order is a pain so far. $\endgroup$ – AndresB Apr 30 at 5:00
  • $\begingroup$ You aren't certain about $\sqrt{g}$? $\endgroup$ – Spencer Apr 30 at 5:00
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The symbol $g$ without indices refers to the determinant of the matrix of components of the metric tensor $g_{ij}$. Lets refer to this matrix as $\mathbb{G}=[g_{ij}]$.

Now lets see how $g=\mathrm{det} \mathbb{G}$ transforms when $g_{ij} \rightarrow g_{ij}+\delta g_{ij}.$

$$ \mathrm{det}\mathbb{G} \rightarrow \mathrm{det}(\mathbb{G} + \delta \mathbb{G}) $$

$$=\mathrm{det}\Big(\mathbb{G}( \mathbb{I}+ \mathbb{G}^{-1}\delta \mathbb{G})\Big) $$ $$=\mathrm{det}\Big(\mathbb{G}\Big) \mathrm{det}\Big( \mathbb{I}+ \mathbb{G}^{-1}\delta \mathbb{G})\Big) $$ $$=\mathrm{det}\Big(\mathbb{G}\Big) \mathrm{det}\Big( \mathbb{I}+ \mathbb{G}^{-1}\delta \mathbb{G})\Big) $$

Now note that $\det(\mathbb{I}+\epsilon A) \approx 1 + \epsilon Tr(A)$

$$=\mathrm{det}\Big(\mathbb{G}\Big) \Big( 1+ Tr(\mathbb{G}^{-1}\delta \mathbb{G})\Big) $$

$$=\mathrm{det}(\mathbb{G})+ \mathrm{det}(\mathbb{G}) \sum_{i,k} \mathbb{G}^{-1}_{ik} \delta \mathbb{G}_{ki} $$

$$=g+ g \ g^{\mu\nu} \delta g_{\mu\nu} $$

So we have,

$$g \rightarrow g+ g \ g^{\mu\nu} \delta g_{\mu\nu}, $$

or

$$\delta g = g \ g^{\mu\nu} \delta g_{\mu\nu},$$

from this it is easy to use derivatives to show that,

$$ \sqrt{g} \rightarrow \sqrt{g+\delta g} $$

$$ = \sqrt{g(1 +\delta g/g)} $$

$$ = \sqrt{g}\sqrt{1 +\delta g/g} $$

$$ = \sqrt{g}\Big(1 +\frac12 \frac{\delta g}{g}\Big) $$

$$ = \sqrt{g} +\frac12 \frac{\delta g}{\sqrt{g}} $$

$$ = \sqrt{g} +\frac12 \frac{ g\ g^{\mu\nu} \delta g_{\mu\nu}}{\sqrt{g}} $$

$$ = \sqrt{g} +\frac12 \sqrt{g} \ g^{\mu\nu} \delta g_{\mu\nu} $$

so we have,

$$ \sqrt{g} \rightarrow \sqrt{g} +\frac12 \sqrt{g} \ g^{\mu\nu} \delta g_{\mu\nu} $$

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  • $\begingroup$ This is the extent to which I am willing to work the problem You indicated in the comments that you don't know how $\sqrt{g}$ transforms so I have worked it out here. For the rest you are either on your own, or you will need someone willing to slog through the rest of the calculation. $\endgroup$ – Spencer Apr 30 at 5:17
  • $\begingroup$ Thanks, how do you get the identity $\det(\mathbb{I}+\epsilon A) \approx 1 + \epsilon Tr(A)$? $\endgroup$ – AndresB Apr 30 at 14:12
  • $\begingroup$ It is a well known identity. You can see my derivation of it here math.stackexchange.com/questions/1174639/… . $\endgroup$ – Spencer Apr 30 at 14:34
  • $\begingroup$ You can also search something like "derivative of determinant near identity". I think Terrance Tao has an article with a derivation. $\endgroup$ – Spencer Apr 30 at 14:34
  • $\begingroup$ This is the Terrance Tao article I was thinking of. The identity is in equation (4). He doesn't derive it, but this might give you some search terms for further investigation. google.com/amp/s/terrytao.wordpress.com/2013/01/13/… $\endgroup$ – Spencer Apr 30 at 14:39

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