Laplacian is the only 2nd order operator translation and rotation invariant How to show that the Laplacian is the only 2nd order operator that is translation and rotation invariant such that $L0=0$?
I have shown that it is rotation and translation invariant but I could not show the uniqueness. 
 A: As I was suspecting, this question is not clear. 
The question can be written more precisely as: Let $L[u](x)=\sum_{ij} a_{ij}(x)D_{ij}u(x)+\sum_i b_i(x)D_iu(x)+c(x)u(x)$ be a translation and rotation invariant operator, show that $L[u]=\lambda\Delta u+cu$ whre $\lambda$ and $c$ are constants. 
I have just shown that $b=0$ and $c=k=constant$ but I couldn't prove yet that $a_{ij}=\lambda \delta_{ij}$. First of all, being translation and rotation invariant means that 
$$
L[u\circ T](x)=Lu\circ T(x)\;(*)
$$
for all translation or rotation $T$. Taking $u\equiv k=constant$ and using the hypothesis (*) we get $c(x)\cdot k=c(T(x))\cdot k$, for any traslation $T$ so $c$ is constant. Now taking $u(x)=x_k$, $k=1,...,n$ and $T$ to be the translation $T_v(x)=x+v$ we obtain $u\circ T(x)=u(x+v)=x_k+v_k$ whence $D_k(u\circ T)=1$ and, as before, $b(x)=b(T(x))$, so $b$ is constant. Now to conclude that $b_k=0$ by applying the hypothesis to the rotation $T(x)=-x$ and $u(x)=x_k$ again.
It remains to show that $a_{ij}=\lambda \delta_{ij}$. 
The idea is is to take $u$ as somtehing with degree two, let say, $u(x)=x_k\cdot x_l$ or sometring like this and the rotation as being $T(x)=(x_1,...,-x_k,...,x_n)$, that is, the reflection about the $k$-th axis. 
I still did not figure out how to conclude, could someone please give me some help?
A: I guess you figured out the answer by now, but if someone still wants to know you might proceed by showing something like that if R is a rotation matrix acting on x, then for all functions u, $L[u\circ R]=\sum_{ij} a_{ij}(x)D_{ij}u(Rx)=\sum_{ij}a_{ij}\sum_{kl}R_{ik}R_{jl}D_{kl}u(x)=\sum_{kl}\left(\sum_{ij}R_{ik}R_{jl}a_{ij}\right)D_{kl}u(x)$; substituting some nice test functions with specified quadratic behavior at some point $x_0$ then ought to give $a_{kl}=\left(\sum_{ij}R_{ik}R_{jl}a_{ij}\right)$ so that the result follows rotational invariance of $a_{ij}$.
