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.

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    $\begingroup$ What do you mean by $\Delta \mathbb n$? Specifically, what do you mean by the Laplace-Beltrami of a vector field? $\endgroup$ – Giuseppe Negro May 5 at 0:56
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    $\begingroup$ It looks to me that you could join your efforts with math.stackexchange.com/q/3213841/2002 $\endgroup$ – 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 $$ D_i D_j \mathbf n = - \gamma_{ij} \mathbf n\,. $$


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