Ytukyg

1. 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.




2. 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?

Let $(M,g)$ be a Riemannian manifold with Levi-Civita connection $nabla$. In Do Carmo's Riemannian geometry book, we find the following definitions.

1. $operatorname{div} X:Mrightarrow mathbb{R}$ is the map $pmapsto $ the trace of linear mapping $Y(p)mapsto nabla_Y X(p)$.




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




3. $triangle f = operatorname{div}operatorname{grad} f$.

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 $pin mathbb{R}^n$, consider the usual basis $e_1,..., e_n$ of $T_pmathbb{R}^ncong mathbb{R}^n$. 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_0left(sum a_i e_i, e_jright)\ &= 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 $$ frac{d}{dt}|_{t=0} f(gamma(t)) = lim_{hrightarrow 0} frac{f(p + te_j) - f(p)}{h}$$ 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 $pin mathbb{R}^n$. 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 $Ymapsto nabla_Y X$ 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 $(mathbb{R}^n, g_0)$.