TITLE: Diagonalization of Riemannian Metric and the Laplace Beltrami Operator
QUESTION [8 upvotes]: Consider the local representation of the Laplace Beltrami operator on a Riemannian n - dimensional manifold $(M,g)$:
\begin{equation}
\triangle_g = \frac{1}{\sqrt{\text{det}(g)}} \sum^n_{i,j = 1} \frac{\partial}{\partial x^i} g^{ij} \sqrt{\text{det}(g)}\frac{\partial}{\partial x^j} 
\end{equation}
Some time ago I read in some textbook that one can, locally at any point $p \in M$, choose a small enough neigborhood $U \subset M$ so that by a linear transformation of the coordinates the matrix representation of $g$ in $U$ is $g_{ij} = \delta_{ij}$.
But wouldn't this imply that the above expression always simplifies to the expression
\begin{equation}
 \triangle_g = \sum^n_{i = 1} \frac{\partial^2}{\partial (x^i)^2} \quad ?
\end{equation}
Alternatively I would think that even though the metric evaluates to the Kronecker delta that doesn't mean that its derivatives are zero. So do we actually then have
\begin{equation}
\triangle_g =  \sum^n_{j = 1} \frac{\partial^2}{\partial (x^j)^2} + \sum_{j = 1}^n ( \frac{\partial}{\partial x^j} \sqrt{\text{det}(g)} + \sum_{i = 1}^n \frac{\partial}{\partial x^i} g^{ij})\frac{\partial}{\partial x^j} 
\end{equation}
I am sure at least one my impression is false, if anybody could help pointing out what I got wrong, or refer to a reference where I could read about diagonalization of metrics that would be so helpful, many thanks !

REPLY [3 votes]: This question is partly about the metric and how to diagonalize it, so let me focus on this.  I think some amphiboly is present here: one must be careful to distinguish between coordinates and coordinates.
By "normal coordinates", one means really, choose an orthonormal basis for the tangent bundle over a neighborhood about $p$.  As Zhen Lin points out, the obstruction to integrating this basis to coordinates is curvature.  Such an orthonormal basis is also called a tetrad or a verbein.
Contrast these normal "coordinates" to a coordinate system about $p$ in the sense of a smooth manifold, a homeomorphism between a neighborhood about $p$ to $\mathbb{R}^n$. Coordinate vector fields correspond to the images of the coordinate vector fields in $\mathbb{R}^n$ under this map.  Such fields are of course a local basis for $TM$, but not all bases for $TM$ can be integrated to coordinates.
So "coordinates" means two different things: "local basis for $TM$" if you're talking about tensor quantities, or "local homeomorphism to $\mathbb{R}^n$" otherwise.