Seeing that the second fundamental form is the orthogonal component of the Laplacian

I have come across the statement a few times that, for a mapping $u:M\to N$ between a Riemannian manifold $(M,g)$ and a submanifold $N$ of Euclidean space $\mathbb{R}^n$, the part of the Laplacian of M orthogonal to the tangent plane of $N$ is given by the second fundamental form $II$ of $N$ in $\mathbb{R}^n$. $$(\Delta_gu)^\perp=g^{ij}II(u)(\partial_i u ,\partial_j u)$$ I can't find a proof of this fact or see how to demonstrate it myself. Would anyone be able to offer a proof, or a sketch of a proof?

Several months on and I finally got round to understanding this and since noone was able to post an answer I thought I would share my proof here :)

Let $\bf{n}$ be a vector which is normal to $N$ at $u \in \mathbb{R}^n$ (with respect to the Euclidean inner product $\langle,\rangle$).

Firstly, the Laplacian is $$\Delta_gu= \frac{1}{\sqrt g} \partial_i(\sqrt{g}g^{ij}\partial_ju)=g^{ij}\partial_{i}\partial_ju + \frac{1}{\sqrt g} \partial_i(\sqrt g g^{ij})\partial_ju$$ and since the vectors $\partial_j u$ of the second term are in the tangent plane to $N$, it follows $$\langle\Delta_gu,{\bf{n}}\rangle=\langle g^{ij}\partial_i\partial_ju,{\bf n}\rangle.$$

Secondly, if $\bar{\nabla}$ denotes the flat connection on $\mathbb{R}^n$ and $\nabla$ denotes the induced connection on $N$, by definition the second fundamental form $II$ of $N$ in $\mathbb{R}^n$ is given by $$II(\partial_i u, \partial_ju) = \bar{\nabla}_{\partial_i u} \partial_j u - \nabla_{\partial_i u } \partial_ju = \partial_i\partial_j u-\nabla_{\partial_i u } \partial_ju$$ and again since the second term is tangential to $N$ we have that $$\langle II(\partial_iu,\partial_ju),{\bf n}\rangle= \langle \partial_i \partial_ju,{\bf n} \rangle$$ and thus $$\langle g^{ij} II(\partial_iu,\partial_ju),{\bf n}\rangle= \langle g^{ij}\partial_i \partial_ju,{\bf n} \rangle.$$

Since this holds for an arbitrary normal vector $\bf n$ and since the second fundamental form is by definition orthogonal to $N$ (i.e. $II^\perp=II$) we arrive at $$(\Delta_gu)^\perp = g^{ij}II(\partial_iu,\partial_ju)$$

• I found you post to be very useful, thank you for sharing. By the way, there's one part of your proof I don't quite follow. Would you mind explain why is it the case that $$\overline{\nabla}_{\partial_i u} \partial_j u = \partial_i\partial_j u$$? Commented Mar 30, 2018 at 16:53
• Actually I got inspired by this post of yours and made a question based on this. If you don't mind, could you elaborate it in a bit more detailed there and I'll gladly accept it as an answer. Commented Mar 30, 2018 at 21:53
• Commented Mar 30, 2018 at 21:53
• OK i'll take the discussion over there :) Commented Mar 30, 2018 at 22:02