Geometric proof of rotational invariance of Laplacian I am just looking for a more intuitive proof that the $\nabla^2$ operator is rotationally invariant in Euclidean space.
That is:
If $u(x)$ solves $\nabla ^2 u=0$, then $v(x)=u(Rx)$ solves it too, where $R$ is the usual rotation matrix ($RR^T=Id$). Here $x\in \mathbb R^n$.
I was able to prove it using brute force index gymnastics, but I am sure there something prettier out there.
 A: To expand on Daniel Fischer's comment, the solutions to the equation $\nabla^2u=0$ are precisely the integrable functions $u$ satisfying
$$u(x)=\frac{1}{V(B(x,r))}\int_{B(x,r)} u(y)dy$$
for any $x$ and any $r>0$, where $B(x,r)$ is the ball of radius $r$ around $x$ and $V$ denotes volume. Let $R$ be a rotation about $0$. Then we have 
$$\begin{align*}
u(Rx) &=\frac{1}{V(B(Rx,r))}\int_{B(Rx,r)} u(y)dy\;\; \text{because $u$ satisfies MVP}\\
&=\frac{1}{V(B(Rx,r))}\int_{RB(x,r)} u(y)dy\;\; \text{because}\; RB(x,r)=B(Rx,r)\\
&=\frac{1}{V(B(Rx,r))}\int_{B(x,r)} u(Ry)dy
\end{align*}$$
as desired.
A: Just treat the rotation like a function: $v(x) = (u \circ R)(x)$.  Then use the chain rule.  Let $a$ be an arbitrary vector.  The chain rule is then:
$$a \cdot \nabla v|_x = [(a \cdot \nabla) R(x)] \cdot \nabla u|_{R(x)}$$
$R$ is a linear function, and as such, $(a \cdot \nabla) R(x) = R(a)$.  For brevity, let $R(x) = x'$, and we get
$$a \cdot \nabla v|_x = R(a) \cdot \nabla u|_{x'} \implies \nabla (u \circ R)|_x = R^T(\nabla u)|_{x'}$$
(You may be thinking, "What on earth does it mean for a rotation matrix to act on $\nabla$?  Well, you just treat the partial derivatives as if they were components of a vector.)
A similar line of logic would show that, for a vector field $F$,
$$\nabla \cdot (F \circ R)|_x = R^T(\nabla) \cdot F |_{x'}$$
So the Laplacian would take the form
$$\nabla^2 (u \circ R)|_x = R^T(\nabla) \cdot R^T( \nabla) u|_{x'}$$
But the right-hand side reduces to $R^T R(\nabla) \cdot \nabla u|_{x'}$, and we know that $R^T R = 1$, leaving us with $\nabla^2 u|_{x'} = \nabla^2 v |_x$.  If $\nabla^2 u = 0$ everywhere, then $\nabla^2 v = 0$ everywhere as well.
