Laplace operator and tensor calculus: I'm studying Tensor calculus and I found this interesting problem:

Show that:
$$ \Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)$$

Here's some attempts, hope it helps, even I find them useless!

Well, we know that: $$\Delta F=\nabla\cdot \nabla F $$
And : $$\nabla \cdot \mathbf{V}=\nabla_iv ^i$$Using it : $$\Delta F=\nabla_i (g^{ik}\partial_kF)$$

That's the only advance I've made till now, I'm thinking about a property but I'm not that much certain about its validity here.

$$\Delta F=g^{ik}\nabla_i(\partial_k F)$$

Being true or false I think it's not useful to derive this formula.
 A: Not from the first principles, but based on physical intuition proof looks as follows. Consider the action for scalar field:
$$
S = \int d^D x \ \sqrt{g} g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi 
$$
This is the only viable GR covariant expression for the action of scalar field without free indices, and $d^D x \sqrt{g}$ is an invariant volume element. Integrating this expression by parts, one gets:
$$
S = -\int d^D x \ \phi \partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = -
\int d^D x \ \sqrt{g} \phi \frac{1}{\sqrt{g}}\partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = -\int d^D x \ \phi \Delta \phi
$$
Where we have assumed that boundary terms vanis, and recovered in the last equality the invariant volume element.
A: Well, the Laplace operator is a composite operator:
$$ \Delta F = div\ grad\ F = \nabla\cdot\nabla F $$
and as you wrote
$$ (grad\ F)^r = (\nabla F)^r = \frac{\partial F}{\partial x^k}\,g^{rk} = V^r $$
You obtain the divergence by contraction of the derivation operator $\nabla$
and we emphasize that the contraction has to be performed on the covariant derivative:
$$ div\ \boldsymbol{V} = \nabla_iV^i =
  V^i_{\phantom{i};\,i}= 
        \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir}  $$
By use of a property of the levi-Civita connection coefficients
$$
\Gamma^i_{ki} = \frac{1}{2} g^{ij} \frac{\partial g_{ij}}{\partial x^k}
              = \frac{1}{2g} \frac{\partial g}{\partial x^k}
              = \frac{\partial \,log \sqrt{|g|}}{\partial x^k}
$$
you can write further
$$ div\ \boldsymbol{V} = \nabla_iV^i =
  V^i_{\phantom{i};\,i}= 
        \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir} =
        \frac{\partial V^r}{\partial x^r} + V^r\;
                \frac{\partial \,log \sqrt{|g|}}{\partial x^r} =
        \frac{1}{\sqrt{|g|}}\; 
        \frac{\partial}{\partial x^r} (\sqrt{|g|}\; V^r) $$
Finally, substituting $V^r$ gives the desired result:
$$ \Delta F = div\ grad\ F = \frac{1}{\sqrt{|g|}}\; 
        \frac{\partial}{\partial x^r} (\sqrt{|g|}\;
\frac{\partial F}{\partial x^k}\,g^{rk} ) $$
A: Here's a quick derivation of this problem:
As you said:
$$\Delta F= \nabla\ .\nabla F$$
And always using your steps:
$$\nabla \ .\ F=\nabla_iv^i$$
And for those who don't know why he involved the "($g^{ik}\partial_k F$)" well that's the contravariant components of the gradient operator.
\begin{align}
\Delta F&=  \nabla_iv^i\\
&=\nabla_i\left(g^{ik}\partial_kF\right)\\
&=g^{ik}\nabla_i\left(\partial_k F\right)
\end{align}

Recall:
$$\nabla_i(\partial_k F)=\partial_{ik}F-\Gamma_{ik}^l\partial_lF$$

Hence:
$$\Delta F=g^{ik}(\partial_{ik}F-\Gamma_{ik}^l\partial_lF)$$

Another recall: :)
$$\nabla\ .\ \mathbf{V}=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(v^i \sqrt{\vert g\vert}\right)\quad{(1)}$$

Involving the contravariant components of $\mathbf{grad}F$ in $(1)$ we got the following:
$$\bbox[silver,5px,border:2px solid teal] {\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)}$$
and that's true, because when $g^{ik}=\delta^{ik}$ we get the classic expression of the Laplacien operator:
$$\Delta \ F=\partial_{kk}F.$$
