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.


  • $\begingroup$ You could use the formula $\Delta = \nabla^* \nabla$, where $\nabla$ is the Levi-Civita connection. The Levi-Civita connection can in turn be represented via the second fundamental form. $\endgroup$ – Matthias Ludewig Oct 21 '12 at 17:57
  • $\begingroup$ Thanks Kofi. The tricky part there is that you then need an expression for the second fundamental form that satisfies the criteria above. In other words, imagine that you start with just $f$, $N$, and the differential. How can you build up all the objects you need to define $\Delta$ without appealing to coordinates or exterior calculus? (Also note that something like "$\Delta = \mathrm{div} \circ \mathrm{grad}$" doesn't satisfy the first criterion above.) $\endgroup$ – TerronaBell Oct 21 '12 at 18:03
  • 2
    $\begingroup$ @fuzzytron: Actually, the second fundamental form is easy in these terms: For any pair of tangent vectors $X,Y$, one has $$ II(X,Y) = =-\tfrac12\bigl(df(X)\cdot dN(Y) + df(Y)\cdot dN(X)\bigr) $$ @Kofi: How you are going to use $II$ to get the Levi-Civita connection? While I know how to do it, I don't think it's straightforward. $\endgroup$ – Robert Bryant Oct 22 '12 at 1:19
  • $\begingroup$ Interestingly enough, (df(X) \cdot dN(Y)) is already symmetric -- i.e., you don't even need to symmetrize. $\endgroup$ – TerronaBell Oct 22 '12 at 21:44
  • $\begingroup$ I think you could extend $\phi$ and $N$ to a neighbourhood of $f(M)$; take the ordinary 3D gradient: $v=\nabla\phi$; project it onto the surface: $w=v-(v\cdot N)N$; take the 3D divergence: $\Delta\phi=\nabla\cdot w$; and restrict this to $f(M)$. $\endgroup$ – mr_e_man Dec 16 '19 at 21:47

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.

  • $\begingroup$ I love it, actually -- thanks Robert. I'm not completely clear on how to "translate" this guy back into the usual $\star d \star d$, though. Looks like $df \wedge d(N \times \nabla \phi) = -\star \Delta \phi$, and $N \times$ looks like a quarter rotation / application of the 1-form Hodge star. Not sure where to go from there. $\endgroup$ – TerronaBell Oct 22 '12 at 19:55
  • $\begingroup$ @fuzzytron: I'm glad you found it useful. Yes, the point is that crossing with $N$ is the substitute for the Hodge star, once you convert everything into vector fields. In my opinion, though, not allowing the students either local coordinates or differential forms is depriving them of valuable tools for doing calculations. It's often difficult to see what is really going on without one or the other. The problem with your 'formula' above is that you are wedging two vector-valued $1$-forms and getting a real-valued $2$-form. You need to get the dot product into the formula; use transpose? $\endgroup$ – Robert Bryant Oct 25 '12 at 9:00
  • $\begingroup$ @Robert -- I absolutely agree; we're starting with exterior calculus next week. Just wanted to see how far we could get without it (always good to understand which objects are truly necessary and which are there just for convenience). Thanks again! $\endgroup$ – TerronaBell Oct 25 '12 at 20:07

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.