# $\Delta \mathbf n = -2 \mathbf n$ on the Euclidean sphere

Let us consider the Euclidean two-sphere, defined by the embedding in the three dimensional Euclidean space as $$\mathbf n \cdot \mathbf n = 1\,,$$ where $$\cdot$$ denotes the standard scalar product. The metric on the sphere, in some coordinates $$x^i$$, is expressed as $$\gamma_{ij}=\mathbf e_i \cdot \mathbf e_j\,,$$ where $$\mathbf e_i=\partial_i\mathbf n$$. For instance, in the standard spherical coordinates $$\mathbf n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$ and $$\gamma_{\theta\theta}=1\,,\qquad \gamma_{\phi\phi}=\sin^2\theta\,,\qquad \gamma_{\theta\phi}=0\,.$$ We define the Laplace-Beltrami operator on the sphere by $$\Delta = \gamma^{ij} D_iD_i$$, where $$\gamma^{ij}$$ is its inverse and $$D_i$$ is the associated Levi-Civita connection.

I would like to prove that $$\Delta \mathbf n = -2 \mathbf n$$ and that in higher dimensions the same holds with $$2$$ replaced by the dimension of the sphere. I came to believe that this is true by an explicit check in spherical coordinates in dimensions $$3$$, $$4$$ and $$6$$.

Considering that parallel transport of a given tangent vector $$\mathbf v$$ defined at the point $$x+dx$$ to the point $$x$$ is defined by keeping constant its embedding components and then projecting it on the sphere at the point $$x$$, we have $$\mathbf v_{\parallel}(x+dx,x)=\mathbf v(x+dx)-\mathbf v(x+dx)\cdot \mathbf n (x)\, \mathbf n(x)$$ hence $$D_i\mathbf v\, dx^i = \mathbf v_{\parallel}(x+dx,x)- \mathbf v (x)= (\partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n)dx^i$$ where we have used $$\mathbf n \cdot \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v=0$$ and $$D_i\mathbf v = \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n\,.$$ Applying this to the basis vectors $$\mathbf e_j =\partial_j\mathbf n$$ affords $$D_i\mathbf e_j = \partial_i \mathbf e_j+\gamma_{ij}\mathbf n\,.$$ But unfortunataly I am not able to go further.

• What do you mean by $\Delta \mathbb n$? Specifically, what do you mean by the Laplace-Beltrami of a vector field? – Giuseppe Negro May 5 at 0:56
• It looks to me that you could join your efforts with math.stackexchange.com/q/3213841/2002 – Yuri Vyatkin May 5 at 4:03

There is a more general formula. For every immersion $$\mathbf x \colon M^n \to \mathbb{E}^m$$ of an $$n$$-dimensional manifold $$M$$ into $$\mathbb{E}^m$$ that $$\Delta \mathbf{x} = n H$$ where $$H$$ is the mean curvature vector of $$\mathbf{x}$$. This is sometimes called the formula of Beltrami. For $$M=S^n$$ the mean curvature is $$1$$, so one obtains $$\Delta \mathbf{x} = n \mathbf{n}.$$

Note that this formula differs from yours by a minus sign. I can think of two reasons.

1. Some authors put a minus sign in their definition of the Laplacian, some don't. Your definition doesn't have a minus sign, so in the calculation below there doesn't appear a minus sign.
2. Replacing the normal vector $$\mathbf{n}$$ by $$-\mathbf{n}$$ also gives a change of sign.

Proof. Let $$v$$ be an arbitrary vector in $$\mathbb{E}^m$$ and $$p\in M$$. If $$\{e_1,\ldots, e_n\}$$ is an orthonormal basis of $$T_p M$$, we can extend $$e_1,\ldots, e_n$$ to an orthonormal frame $$E_1,\ldots, E_n$$ such that $$D_{E_i} E_j = 0 \quad \text{at p for i,j=1,\ldots,n,}$$ where $$D$$ is the Levi-Civita connection of $$M$$. Then at $$p$$ we have \begin{align*} (\Delta \langle \mathbf{x},v\rangle)_p &= \sum_{i=1}^n e_i\langle E_i,v\rangle = \sum_{i=1}^n \langle \bar D_{e_i}E_i,v\rangle \\ &= \sum_{i=1}^n \langle h(e_i,e_i),v\rangle = n \langle H,v\rangle(p). \end{align*} Here $$\bar D$$ stands for the Levi-Civita connection on $$\mathbb{E}^n$$. Since both $$\Delta x$$ and $$H$$ are independent of the choice of local basis, we have $$\langle\Delta x, v\rangle = n \langle H,v \rangle$$ for any $$v$$. Since $$v$$ was arbitrary and the inner product is non-degenerate, the formula of Beltrami follows.

Reference: Pseudo-Riemannian Geometry and Delta Invariants by B.-Y. Chen.

As Yuri Vyatkin pointed out in the comments, your question is related to this question, which was also recently asked.

A way to prove the above is the following (this is probably a special case of the more general answer given by @Ernie060, but I still need to fill in a few details).

In three-dimensional Euclidean space $$\mathbb R^3$$, the metric in Cartesian coordinates is $$\delta_{IJ}=\mathrm{diag}(1,1,1)$$ and in spherical coordinates, defined by $$\mathbf x = r\,\mathbf n(x^i)$$, reads $$g_{rr}=1$$ and $$g_{ij}=r^2\gamma_{ij}$$, with $$\gamma_{ij}=\partial_i\mathbf n\cdot\partial_j\mathbf n$$. Comparing the two expressions for the Laplacian in the given coordinate systems, we have $$0=\Delta_{\mathbb R^{3}}\mathbf x=\frac{1}{r^2}\partial_r(r^2 \mathbf n)+\frac{1}{r}\Delta_{S^2}\mathbf n$$ which yields precisely $$\Delta_{S^2}\mathbf n = -2\mathbf n\,.$$ This is actually a consequence of the fact that any second covariant derivative of $$\mathbf x$$ vanishes (since it vanishes in the Cartesian coordiante frame), hence in particular $$0=\nabla_i \nabla_j \mathbf x =r D_i D_j\mathbf n-\Gamma_{ij}^r\mathbf n\,,$$ but explicit calculation affords $$\Gamma_{ij}^r=-r\gamma_{ij}$$ and hence $$D_i D_j \mathbf n = - \gamma_{ij} \mathbf n\,.$$