It is well-known that if $O$ is an orthogonal map, then $\Delta u(Ox) = \Delta u$ where $\Delta$ is the Laplacian.


Now, let $A$ be a constant invertible matrix, then we define the weighted Laplacian


$$\Delta\_A = \langle \nabla, A \nabla \rangle.$$


My question is: Does there exist a function $\varphi\_O$ such that $Gf(x):=e^{\varphi\_O(x)}f(Ox)$ satisfies for all $f$ but fixed orthogonal $O$


$$\Delta\_A Gf(x) = G(\Delta\_A f)(x)?$$



Orthogonal invariance of (weighted) Laplacian
$\newcommand{\De}{\Delta}$The answer is: not in general.


Indeed, suppose the contrary. Let
\begin{equation}
 A:=\begin{pmatrix}
 0&1/2\\
 1/2&0
\end{pmatrix},
\end{equation}
so that
\begin{equation}
 (\De\_A f)(s,t)=\frac{\partial^2 f(s,t)}{\partial s\,\partial t}=f^{(1,1)}(s,t). 
\end{equation}
Let
\begin{equation}
O:=\begin{pmatrix}
 0&-1\\
 1&0
\end{pmatrix}
\end{equation}
and
\begin{equation}
 g:=\varphi\_O, 
\end{equation}
so that
\begin{equation}
 (Gf)(s,t):=e^{g(s,t)}f(-t,s), 
\end{equation}
and the difference between the right-hand side and left-hand side of the equality in question
divided by $e^{g(s,t)}$ is
\begin{equation}
\begin{aligned}
 d\_f(s,t)&:=\frac{(G(\De\_A f))(s,t)-(\De\_A(Gf))(s,t)}{e^{g(s,t)}} \\ 
 &=\frac{e^{g(s,t)}f^{(1,1)}(-t,s)
 -\dfrac{\partial^2 f(s,t)}{\partial s\,\partial t}
 \big(e^{g(s,t)}f(-t,s)\big)}{e^{g(s,t)}} \\ 
 &=-f^{(0,1)}(-t,s) g^{(0,1)}(s,t)+f^{(1,0)}(-t,s) g^{(1,0)}(s,t) 
 +2 f^{(1,1)}(-t,s) \\ 
 &-f(-t,s)
 \left(g^{(0,1)}(s,t) g^{(1,0)}(s,t)+g^{(1,1)}(s,t)\right)=0 
\end{aligned}
\end{equation}
for all $f\in C^1$ and all real $s,t$.


Letting now $f\_1(s,t):=s$, $f\_2(s,t):=t$, and $f\_4(s,t):=t^2$, we get
\begin{equation}
 0=s d\_{f\_2}(s,t)-d\_{f\_4}(s,t)=s g^{(0,1)}(s,t),
\end{equation}
so that $g^{(0,1)}=0$ and hence $g^{(1,1)}=0$, which implies $0=d\_{f\_1}=g^{(1,0)}$.
So, $g$ is a constant, and hence
$0=d\_f(s,t)=2f^{(1,1)}(-t,s)$ for all $f\in C^1$ and all real $s,t$, which is absurd. $\quad\Box$


