Surface Laplace-Beltrami without coordinates, exterior calculus?

Question

Let $f: M \rightarrow \mathbb{R}^3$ be an immersion of a surface $M$. For pedagogical purposes (i.e., I'm teaching a class!) I am looking for an expression for the scalar Laplace-Beltrami operator $\Delta$ applied to a real function $\phi$ on $f(M)$ that:

1. explicitly depends on the immersion $f$,
2. does not rely on local coordinates, and
3. does not use exterior calculus.

A standard coordinate expression is

$$\Delta \phi = \frac{1}{\sqrt{g}} \partial_i (\sqrt{|g|} g^{ij} \partial_j \phi),$$

and a standard expression using exterior calculus is

$$\Delta\phi = \star d \star d \phi.$$

However, the students do not have exposure to exterior calculus, and I am discouraging the use of coordinates whenever possible (and have so far been able to get by without them).

To give a concrete example of the "style" of expression I'm looking for, consider the normal curvature in a direction $X \in TM$, which can be expressed as

$$\kappa_n(X) = -\frac{dN(X) \cdot df(X)}{|df(X)|^2},$$

where $N: M \rightarrow S^2 \subset \mathbb{R}^3$ is the Gauss map and $\cdot$ denotes the usual Euclidean inner product. This expression uses the differential $d$ of a function, but it does not use the exterior derivative on $k$-forms (at least, not for $k>0$), nor does it use the Hodge star, nor does it rely on a coordinate system.

In English, $\Delta$ is not hard to describe: take the sum of second derivatives along orthogonal directions in the ambient space. But after a lot of digging, I'm surprised to find there isn't a more suggestive algebraic description.

Thanks!

Answer 1

You probably will disallow this, but the following recipe does work:

First, let $\nabla\phi:M\to\mathbb{R}^3$ be the (unique) vector-valued function that satisfies $$ d\phi(X) = \nabla\phi\cdot df(X)\qquad\text{and}\qquad \nabla\phi\cdot N = 0. $$ for all vector fields $X$ on $M$. Then $\Delta\phi:M\to\mathbb{R}$ is the function that satisfies $$ df(X)\cdot d(N\times \nabla\phi)(Y)-df(Y)\cdot d(N\times \nabla\phi)(X) = -\Delta \phi\ \ N\cdot\bigl(df(X)\times df(Y)\bigr). $$ for all vector fields $X$ and $Y$ on $M$.

This only uses $d$ on functions. The thing you may not like is the use of 'arbitrary' vector fields $X$ and $Y$ on $M$, which, essentially, replaces the use of differential forms.

NB: I introduced the minus sign so that it now matches your convention for $\Delta$ as you gave it in the question; your Laplacian is the opposite of the usual geometer's Laplacian.

Answer 2

This is just a riff on Robert Bryant's answer but thought I would throw it out there -- its the way I think about this stuff at least....

Suppose $\mathbf{H}: M\to \mathbb{R}^3$ is the mean curvature vector (i.e. locally $\mathbf{H}=-H\mathbf{n}$ were $\mathbf{n}$ is a unit normal vector field to $f(M)$ and $H=tr A$ is the mean curvature -- this is well defined even if $M$ is unoriented). This of course depends on the immersion.

If $\phi$ is a function on $\mathbb{R}^3$ which restricts to $f(M)$ as the given function $\phi$ then we have that $$ \Delta_{f(M)} \phi =\Delta_{\mathbb{R}^3} \phi -\nabla^2_{\mathbb{R}^3} \phi (\mathbf{n}, \mathbf{n})+\mathbf{H}\cdot \nabla_{\mathbb{R}^3} \phi $$

Note that $\nabla^2_{\mathbb{R}^3} \phi (\mathbf{n}, \mathbf{n})$ also does not depend on choice of $\mathbf{n}$ so this is also well defined on unoriented surfaces.