# Have I shown $\nabla^2$ is invariant under rotation?

In a homework problem I was asked to show that Laplacian operator (3D) is invariant under rotation. The below is my reasoning, i don't know whether it is correct.

For the 3D rotation R, I can represent it as an orthogonal matrix $D(R)$ which satisfies $D^{T}(R)D(R)=1$, since it is an element of $SO(3)$ group. For a function $f(x_1, x_2, x_3)$ that I apply $R$ upon, from $Rf(x_1,x_2,x_3)= f(R^{-1}(x_1,x_2,x_3))$, I can denote the function after rotation by $f(x_1^{'}, x_2^{'}, x_3^{'})$, and the basis just transform as $$(x_1^{'}, x_2^{'}, x_3^{'})=(x_1, x_2,x_3)D(R^{-1})$$ which means $$x_{j'}= \sum D_{ij'}(R^{-1})x_i$$ and so we have $$\frac{\partial {x_{j'}}}{\partial {x_i}}=D_{ij'}(R^{-1})$$ After setting all this up, we can calculate how Laplacian $\nabla^2= \partial^{2}_x+\partial^{2}_y+\partial^2_z$ will act on the new function

Since $$\frac{\partial }{\partial x_k}f(x_1^{'},x_2^{'},x_3^{'})=\sum_{j^{'}} \frac{\partial{f}}{\partial x_{j'}} \frac {\partial x_{j'}}{\partial x_k}=\sum_{j'} \frac{\partial f}{\partial x_{j'}}D_{kj'}(R^{-1})$$ So, we have $$\frac{\partial^2}{\partial {x_{k}}^2}f(x_1^{'},x_2^{'},x_3^{'})=\sum_{j', l'} \frac{\partial^2 f}{\partial x_{j'} \partial x_{l'}}D_{kj'}(R^{-1})D_{kl'}(R^{-1})$$ Since Laplacian can be denoted as $\sum_{k} \frac{\partial^2}{\partial {x_{k}}^2}$, and we know that the matrix $D(R)$ is orthogonal, so that $$\sum_{k} D_{kj'}(R^{-1})D_{kl'}(R^{-1})= \delta_{j',l'}$$ So finally, we will arrive at $$\nabla^2 f(x_1^{'},x_2^{'},x_3^{'})=\sum _{j',l'}\frac{\partial^2 f}{\partial x_{j'} \partial x_{l'}} \delta_{j',l'}= \sum _{j'} \frac {\partial^2 f}{\partial x_{j'}^2}$$

I think I'm done here for showing that Laplacian is invariant under rotation, but the result is somewhat not that satisfying to me.

So have i shown Laplacian is invariant?