Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a surface $S$ is endowed with a metric given by the matrix $$M=\begin{pmatrix} E&F\\F&G\end{pmatrix}$$

And $f,g$ are scalar functions defined on the surface. What then is the (geometric) significance of the scalar function given by ${1\over \sqrt{\det(M)}}{\partial \over \partial x_i}\left(f\sqrt{\det(M)} (M^{-1})_{ij} {\partial \over \partial x_j} g\right)$?

I have been told that if we set $f=1$, we get an operator equivalent to the Laplacian acting on the function $g$. Why does the Laplacian become this form? Is there an intuitive geometric explanation of what is going on?

Thank you.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Yes, there is a more intuitive geometric explanation, though it is kind of difficult to see if you just get to see this formula. In differential (Riemannian) geometry one looks at curved (in contrast to flat, like Euclidean space) surfaces or higher dimensional manifolds. The metric you are looking at (more precisely: the pointwise norm associated with the pointwise scalar product it defines) is kind of an infinitesimal measure for distances in these surfaces.

It turns out that in this context you can set up differential calculus, the basic operation of which, when applied to vector fields, is the so called covariant differentiation. If you are looking at surfaces embedded in Euclidean 3 space this is basically a differentiation in that surrounding space followed by an orthogonal projection onto the tangent plane to the surface, but one can define this in an abstract manner, too (without an ambient manifold). If you then take a function and it's gradient (a concept which is also to be defined and depends on the metric) and take the covariant derivative of this object, the trace of this object (wrt the metric) is the Laplacian of the function (as is in Euclidean space, the Laplacian is the trace of the Hessian). In local coordinates this happens to look like the object you wrote down (when $f=1$). While in this form it looks a bit arbitrary it turns out to have some interesting properties. In particular it is invariant under coordinate changes, that is, it is well defined as a differential operator on the surface.

If you want to learn more about this you should fetch some basic textbook on Riemannian Geometry, like do Carmo.

share|improve this answer
    
Thank you, Thomas. I will look for a Riemannian Geometry textbook. If you don;t mind, could you explain a little about the covariant derivative, or perhaps provide a link to a more visual explanation for that? Thanks again! –  Brio Jun 30 '12 at 13:29
    
@Brio I added a link (in the answer) to a wikipedia page about covariant differentiation. –  user20266 Jun 30 '12 at 13:41
    
Thank you, Thomas! –  Brio Jun 30 '12 at 16:36

Thomas has already given a good answer. However it might be helpful to mention that the operator in the question(v1) with $f=1$ is in fact the 2D Laplace-Beltrami operator.

share|improve this answer
    
Thank you, Qmechanic! –  Brio Jun 30 '12 at 16:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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