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 !
Best Answer
In fact, the following is true: At any point $p$, there exists a neighborhood $U$ of $p$ such that $g_{ij}=\delta_{ij}$, the Kronecker delta, and $\displaystyle\frac{\partial g_{ij}}{\partial x_k}$ at $p$ is zero. (It's important to keep in mind that $\displaystyle\frac{\partial g_{ij}}{\partial x_k}$ only vanishes at $p$, not in a neighborhood; otherwise, it has constant curvature.) If I remember correctly, this is called normal coordinates. If you look at the book of Riemannian Geometry of Do Carmo, you need to prove the existence of normal coordinates in an exercise. Maybe in other books they have the proof.
Anyway, in the normal coordinates, as you have calculated, we 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} Because $\displaystyle\frac{\partial g_{ij}}{\partial x_k}=0$ at $p$, the second term vanishes. So at the point $p$, the Laplace Beltrami operator is given by \begin{equation} \triangle_g = \sum^n_{i = 1} \frac{\partial^2}{\partial (x^i)^2}. \end{equation}