What is the change of Laplacian operator under small coordinate transformation? 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. 
 A: 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, 
$$ \sqrt{g} \rightarrow \sqrt{g} +\frac12 \sqrt{g} \  g^{\mu\nu} \delta g_{\mu\nu} $$
