Proof of the Laplace-Beltrami operator Consider the local representation of the Laplace-Beltrami operator on a 2-dimesnsional manifold (immersed in $\mathbf R^3$) with Riemannian metric $(g_{ij})$.  Please, I want help in showing that:
\begin{equation}
\Delta f = \frac{1}{\sqrt{\text{det}(g)}} \sum_{i,j } \frac{\partial}{\partial x_i} (g^{ij} \sqrt{\text{det}(g)}\frac{\partial}{\partial x_j}f) 
\end{equation}
I have looked on the web and found this but the proof used terms I'm not familiar with. I have some background in vector analysis though. I hope someone would be kind in enough to take me through the proof.
 A: The following is well known and can be found in multiple stanard reference books (Michael Taylor, Chris Sogge, Jurgen Jost, etc). Here is one of many ways to derive it. Recall that
$$
(\nabla\cdot X)\omega=\mathcal{L}_{X}\omega=d\omega\rfloor X+d(\omega \rfloor X)
$$
where $\rfloor$ is the contraction operator. Now since $\omega$ is a top degree form on $M$, we have $d\omega=0$. Thus we have
$$
(\nabla\cdot X)\omega=d(\omega \rfloor X)
$$
While we know a volume form can always be written as
$$
\omega=\sqrt{\det(G)}dx_{1}\wedge \cdots \wedge dx_{n}
$$
Thus we have
$$
\omega \rfloor X=\sum (-1)^{j+1}X_{j}\sqrt{\det(G)}dx_{1}\wedge\cdots \wedge\widehat{dx_{j}}\wedge\cdots \wedge dx_{n}
$$
and taking the exterior derivative we have 
$$
d(\omega \rfloor X)=\sum \frac{\partial (X_{j}\sqrt{\det(G)})}{\partial x_{j}}dx_{1}\wedge\cdots \wedge dx_{n}
$$
Hence comparing original equation $(\nabla\cdot X)\omega=d(\omega \rfloor X)$ we have
$$
\nabla\cdot X=(\det(G))^{-1/2}\frac{\partial (X_{j}\sqrt{\det(G)})}{\partial x_{j}}=g^{-1/2}\partial x_{j}(g^{1/2}X_{j})
$$
On the other hand we have
$$
\nabla f=(df)^{*}=(\sum \frac{\partial f}{\partial x_{i}}dx_{i})^{*}=\sum (\frac{\partial f}{\partial x_{i}}g^{ij})\partial x_{j}
$$
where the $(*)$ operator is the isomorphism between tangent and cotangent bundle given by the metric tensor. Since we go from the cotangent to tangent bundle, we use the inverse (of the transpose). Thus substitute $\Delta f=\nabla\cdot (\nabla f)$ you get the the desired formula. 
