Inconsistent statements about Laplacian versus Laplace-Beltrami 
*

*The Laplace-Beltrami operator is said to be intrisic: it can be defined in terms of the metric and without reference to the "ambient" coordinate system.

*Not so for the Laplacian, it is defined in terms of the ambient coordinate system, e.g.
$\nabla = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ where $x,y$ 
are coordinates of the ambient space.
Statement 1 should be true for whichever metric is involved. But if the metric is identity (a perfectly valid metric), then the Laplace-Beltrami reduces to the Laplacian, and statement 1 is contradicted.
What is the flaw?
 A: Let $(M,g)$ be a Riemannian manifold with Levi-Civita connection $\nabla$.  In Do Carmo's Riemannian geometry book, we find the following definitions.


*

*$\operatorname{div} X:M\rightarrow \mathbb{R}$ is the map $p\mapsto $ the trace of linear mapping $Y(p)\mapsto \nabla_Y X(p)$.


*$\operatorname{grad} f$ is the vector field defined by $g(\operatorname{grad} f(p), Y(p)) = d_pf \, Y(p)$.


*$\triangle f = \operatorname{div}\operatorname{grad} f$.

None of these definitions uses coordinates, so your statement 1. is true.
Now, let's assume $(M,g) = (\mathbb{R}^n, g_0)$, where $g_0$ is the usual dot product on $\mathbb{R}^n$.  Let's compute $\triangle f$ using the above definition.
First, we claim that $\operatorname{grad}f$ is the usual gradient of $f$, that is, a vector of partial derivatives of $f$.  At a fixed point $p\in \mathbb{R}^n$, consider the usual basis $e_1,..., e_n$ of $T_p\mathbb{R}^n\cong \mathbb{R}^n$.  Then $\operatorname{grad}f(p) = \sum a_i e_i$ for some real numbers $a_i \in \mathbb{R}$.  Let $Y(p) = e_j$.  Then \begin{align*} a_j &= g_0\left(\sum a_i e_i, e_j\right)\\ &= g_0(\operatorname{grad}(f)(p), Y(p)) \\ &= d_p f\, e_j.\end{align*}
The way we compute $d_p f\, e_j$ is by picking a curve $\gamma$ with $\gamma(0) = p$, $\gamma'(0) = e_j$, and then computing $\frac{d}{dt}|_{t=0} f(\gamma(t))$.  Let's pick $\gamma(t) = p + te_j$.  Then $$ \frac{d}{dt}|_{t=0} f(\gamma(t)) = \lim_{h\rightarrow 0} \frac{f(p + te_j) - f(p)}{h}$$ which is the definition of $\frac{\partial f}{\partial e_j}$.  Hence, $a_j = \frac{\partial f}{\partial e_j}$, as claimed.
Now, for a vector field $X$, what is $\operatorname{div} X$?  We claim it's the usual divergence.  To see this, fix $p\in \mathbb{R}^n$.  Let's write $X = \sum x_i e_i$, where the $x_i$ are functions on $\mathbb{R}^n$.
Let $Y(p) = e_j$.  Then $\nabla_Y X =  e_j(x_i)e_i = \frac{\partial x_i}{\partial e_j} e_i$, so the trace of the map $Y\mapsto \nabla_Y X$ is $\sum \frac{\partial x_j}{\partial e_j}$.
Finally, let's compute $\triangle f$ by combining these two ideas.  We already saw that $\operatorname{grad} f = \sum \frac{\partial f_i}{\partial e_i} e_i$.  Then $\operatorname{div} \operatorname{grad} f = \operatorname{div} \sum \frac{\partial f_i}{\partial e_i} e_i = \sum \frac{partial^2 f}{\partial e_i^2}$.  In other words, $\triangle = \sum \frac{\partial^2}{\partial e_j^2}$ in $(\mathbb{R}^n, g_0)$.
