Inconsistent statements about Laplacian versus Laplace-Beltrami

Question

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?

Answer 1

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. $\operatorname{div} X:M\rightarrow \mathbb{R}$ is the map $p\mapsto $ 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 $p\in \mathbb{R}^n$, 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 $(\mathbb{R}^n, g_0)$.