# a good introduction to Laplace Beltrami operator over differential manifolds?

I'd like to have a good reference to understand how the Laplacian operator get generalized over differential manifolds.

More concretely, I want to understand and prove the equation : $$\Delta Id_{\mathbf{X}}=H.\mathbf{N}$$ Where $\mathbf{X}$ is a smooth surface in $\mathbb{R}^3$, $Id_\mathbf{X}$ the identity function defined on $\mathbf{X}$, $H$ its mean curvature and $N$ the normal vector to the surface $\mathbf{X}$ (ie its Gaussian map).

I have read Andrew Pressley's book Elementary Differential Geometry. I'm used to the classical Laplacian of differential calculus that operates on scalar fields, but I dont really understand what the Laplacian of a vector field represents, how it relates to the scalar version or how you define a Laplacian over manifolds.

I have not finished Pressley, but it does not seem to cover this topic.

But trying to look into it, I also saw lot of references to things like tensors, connection form, volum form, that I dont know.

If you have also good recommandation regarding those topics, I would be grateful. Basically, I would like to understand how to do multivariable calculus over surfaces. Thank you.

• Could you try to clarify the notation? For example, what does $\Delta X$ mean for $X$ a smooth surface in $\mathbb{R}^n$? Commented Mar 22, 2014 at 23:31
• I'm afraid you do have to learn "things like tensors, connection form, volume form". The Laplace operator and, especially, Laplace--Beltrami operators are parts of what is called Hodge theory. Probably, one can define $\Delta$ of a function on a surface in $\mathbb{R}^3$ in terms of coordinates (I never tried this), but such a definition would be pretty much meaningless (as any definition in terms of coordinates). The true definition $\Delta=dd^*+d^*d$ uses differential forms a lot, and it cannot be explained in just a few words.
– Alex Degtyarev
Commented Mar 22, 2014 at 23:41
• Yan, I can't make sense of your equation as it is written. The $\Delta$ operator acts on functions (or differential forms), not on manifolds. Even assuming the LHS was $\Delta_X$ in the sense of "the Laplace-Beltrami operator on the manifold $X$", the RHS is not a differential operator but a vector field on $X$. So, please, clarify. Commented Mar 23, 2014 at 0:56
• Or, maybe, $X$ is a map $X: \mathbb{R}^2 \to\mathbb{R}^3$ and $\Delta X$ means "take the Euclidean Laplacian of each component of $X$ ", and the equation says you obtain the components of a vector field on the image of $X$ which is proportional to the normal vector field? Commented Mar 23, 2014 at 1:00
• You are right, I was abusing notations.
– Yan
Commented Mar 23, 2014 at 1:37

This is rather standard notation which admittedly could be a bit confusing if you've never seen it before. In this equation $\Delta X$ has nothing to do with differential forms, all you need to know is the definition of the Laplacian for a submanifold. FYI, I think that my sign convention is different than yours.

Write $X = (x_1,x_2,x_3)$ and the normal vector $N=(n_1,n_2,n_3)$. All this equation claims is that for $\Sigma^2\hookrightarrow \mathbb{R}^3$,

$$\Delta_\Sigma x_i = -H n_i$$

for $i=1,2,3$. Here, we just mean that $x_i$ are the Euclidean coordinates, restricted to the surface. In particular, a often used consequence of this equation is that $\Sigma$ is a minimal surface if and only if $\Delta_\Sigma x_i=0$ for $i=1,2,3$.

To prove this, you need to know about the second fundamental form, and how it relates the Levi-Civita connection on $\Sigma$ with the Euclidean connection. Remember that $$(D^\Sigma)_X Y = (D^{\mathbb{R}^3})_X Y + II(X,Y) N$$ (I don't know exactly what notation you're used to, so feel free to ask for clarification). Also, everyone has different sign conventions here, so be careful.

Then, if $e_1,e_2$ is an orthonormal basis for $T_p\Sigma$, we can extend it to a neighborhood of $p$ in $\mathbb{R}^3$ so that it is parallel with respect to $D^{\mathbb{R}^3}$. Then, we compute: \begin{align*} \Delta_\Sigma x_i & = tr_{g_\Sigma}(D^\Sigma)^2 x_i\\ & = \sum_{j=1}^2 ((D^\Sigma)_{e_j}d x_i)(e_j) \\ & = \sum_{j=1}^2 ((D^\Sigma)_{e_j}(d x_i(e_j)) - dx_i((D^\Sigma)_{e_j}e_j)\\ & = \sum_{j=1}^2 ((D^{\mathbb{R}^3})_{e_j} dx_i)(e_j) - dx_i((D^{\mathbb{R}^3})_{e_j}e_j) - II(e_j,e_j)dx_i(N)\\ & =- Hn_i \end{align*}

• Thanks a lot for the very detailled answer. I know the 1st and 2nd fundamental forms and what they represent. The problem is that I don't know what is the Levi Civita connection. I have seen the term several times, but it isn't cover by the book I'm reading. When reading the wikipedia article, I see a lot of terme I am not familiar with (connection ? metric connection ? fiber/vector bundle ?) That's why I am looking for a more formal introduction to the subject. But there is a huge number of books on the subject of differential geometry and I dont really know where to look.
– Yan
Commented Mar 23, 2014 at 2:46
• A good textbook on differential geometry is what you need. There are several, and you should consult a few of them. Commented Mar 23, 2014 at 3:22
• @Yan, one of my favorite is Lee "Riemannian Manifolds: Introduction to Curvature" amazon.com/….
– Otis
Commented Mar 23, 2014 at 4:52
• @Otis Lee's book states already in the introduction that it will not touch upon Laplace-Beltrami operator.
– DeM
Commented Jun 21, 2017 at 17:22

Another neat proof I found recently: given $v \in \Bbb R^3$, define $h: \Sigma \to \Bbb R$ by $h(p) = \langle p,v\rangle$. On one hand, $\triangle h = \langle \triangle\iota, v\rangle$, where $\iota\colon \Sigma \hookrightarrow \Bbb R^3$ is the inclusion, and on the other hand ${\rm Hess}\,h(X,Y) = \langle {\rm II}(X,Y),v\rangle$. Since $\triangle = \rm tr\; {\rm Hess}$, we obtain $\triangle h = \langle 2H, v\rangle$, where $H = {\rm tr}({\rm II})/2$ is the mean curvature vector field. Comparing, we obtain $\triangle \iota = 2H$.

This generalizes to isometric immersions $f:\Sigma^n \hookrightarrow \Bbb R^{n+m}_\nu$ giving $\triangle f = n H$, even when $\Sigma$ is a pseudo-Riemannian manifold.