Edited after @J. Martel's comment: Let us consider the sphere $S^n$ (embedded in $\mathbb{R}^{n+1}$). We know that if $X\_i$ represent the vector fields on $S^2$ giving the rotation about the $x\_i$-axis, then the Laplacian on $S^2$ is given by $X^2\_1 + X^2\_2 + X^2\_3$. My question is:


Can we have a similar expression for higher dimensional spheres? Specifically, if $Y\_i$ generate the Lie algebra of $SO(n+1)$, can we conclude how the Laplacian would look like on $S^n$ from the fact that one can write $S^n$ as a coset space $SO(n+1)/SO(n)$?


Any help will be greatly appreciated. If there is a more general way of answering this question for coset spaces like $G/K$, I would be glad to know about it. Thanks...



Laplacian on coset spaces
I'll give an answer in a few parts:


First, for the $n$-sphere $S^n\subset\mathbb{R}^{n+1}$: The obvious thing to do is to consider the vector fields
$$
X\_{ij} = x\_i\frac{\partial }{\partial x\_j} - x\_j\frac{\partial }{\partial x\_i}\ ,
\qquad 0\le i < j\le n.
$$
Then one easily computes that the operator
$$
L = \sum\_{0\le i < j\le n} {X\_{ij}}^2
$$
is equal to the Laplacian for the induced metric on $S^n$. 


Second, notice that one doesn't always have uniqueness of this representation. For example, when $n=3$, consider the vector fields
$$
Y\_1^\pm = X\_{01}\pm X\_{23},\quad Y\_2^\pm = X\_{02}\pm X\_{31},\quad Y\_3^\pm = X\_{03}\pm X\_{12}.
$$
Then, for functions on $S^3$, one has that the Laplacian equals
$$
 L = (Y\_1^+)^2 + (Y\_2^+)^2 + (Y\_3^+)^2 = (Y\_1^-)^2 + (Y\_2^-)^2 + (Y\_3^-)^2,
$$


Third, in the general case of a Riemannian homogeneous space $M=G/K$, where $G$ acts effectively on $M$ and fixes a metric $g$ on $M$, I gather that the question is whether there always exists a quadratic polynomial $\lambda\in S^2({\frak{g}})$ (where $\frak{g}$ is the Lie algebra of $G$, thought of as $g$-Killing vector fields on $M$) such that $\lambda$, when regarded as a self-adjoint second-order differential operator on $M$, is equal to the Laplacian of the metric $g$. 


This certainly is true in a large number of cases. For example, if $G$ is semi-simple and $G/K$ is a Riemannian symmetric space, this is true. Note, though, that $\lambda$ may not be 'positive definite' in the sense that it may not be the sum of squares of a basis of $\frak{g}$. For example, see the $S^3$ case above. As another example, if $G=\mathrm{SO}(2,1)$ and $K = \mathrm{SO}(2)$, then $G/K$ is the Poincaré disk, and $\lambda$ in this case turns out to be of the form ${X\_1}^2 + {X\_2}^2 - {X\_3}^2$ for an appropriate basis $X\_i$ of ${\frak{so}}(2,1)$.


Whether it is true in all cases (and how unique the representation is) is not clear to me at first glance, but I'll think about it if I get the chance. It seems that it probably is true when $M$ is compact, or when $M$ is isotropy irreducible, but I should check to be sure before attempting a definitive answer.


