Surface Laplace-Beltrami without coordinates, exterior calculus? 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:


*

*explicitly depends on the immersion $f$,

*does not rely on local coordinates, and

*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!
 A: 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.
A: 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.
