Laplacian Invariance by Rotations 
Problem (Laplacian Invariance by Rotations). Let $U \subset \mathbb{R}^{m}$ be an open set. For any function $f: U \to \mathbb{R}$ twice differentiable, the Laplacian of $f$ is the function $\triangle f: U \to \mathbb{R}$, defined by
  $$\displaystyle \triangle f = \frac{\partial^{2} f}{\partial x^{2}_{1}} + \cdots + \frac{\partial^{2} f}{\partial x^{2}_{m}}.$$
  Prove that if $T: \mathbb{R}^{m} \to \mathbb{R}^{n}$ is an orthogonal linear transformation, then
  $$\triangle (f \circ T) = (\triangle f)\circ T: V \to \mathbb{R},$$
  where $V = T^{-1}(U)$. 

I tried to solve this using the Hessian matrice and writing the partial derivatives as inner product, but I didn't get something promising. I appreciate any hint!
 A: Let $e_1, \ldots, e_m$ be the standard $\mathbb R^m$ basis. Then we have $$Te_j = \sum_{i}a_{ij}e_i$$
where $a_{ij} = \langle Te_j , e_i \rangle$. Let $T_i(x)$ be the $i$-th coordinate of $T(x)$.
So we have $$T_i(x) = \sum_{j} a_{i,j}x_j.$$
Lets denote $\partial / \partial x_i$ by $\partial_i$. We have
$$\partial_i [f(T(x))] = \sum_j \partial_i T_j(x)\partial_j(f)(T(x)) $$
and since $ \partial_i T_j(x) = a_{ij}$ we obtain
$$\partial_i [f(T(x))] = \sum_j a_{ij}\partial_j(f)(T(x)).$$
Let's now compute the second derivative:
$$\partial_i^2 [f(T(x))] = \sum_j a_{ij}\partial_i[\partial_j(f)(T(x))].$$
By the above calculation, putting $\partial_j(f)(T(x))$ in the place of $f(T(x))$, we obtain
$$\partial_i[\partial_j(f)(T(x))] = \sum_k a_{ik}\partial_k\partial_j(f)(T(x))$$ 
and therefore
$$\partial_i^2 [f(T(x))] = \sum_{j,k} a_{ij}a_{ik}\partial_k\partial_j(f)(T(x)).$$
Since $T$ is ortonormal we have $$\sum_{i}a_{ij}a_{ik} =\langle Te_j,Te_k\rangle = \langle e_j, e_k \rangle = \delta_{jk}.$$
Therefore
\begin{align*}
\Delta[f\circ T(x)] &= \sum_{i}\partial_i^2 [f(T(x))]  =\sum_i\sum_{j,k}a_{ij}a_{ik}\partial_k\partial_j(f)(T(x)) 
\\&= \sum_{j,k} \partial_k\partial_j(f)(T(x)) \sum_i  a_{ij}a_{ik} = \sum_{j,k} \partial_k\partial_j(f)(T(x)) \delta_{jk} 
\\&= \sum_k \partial_k^2(f)(T(x)) = \Delta(f) \circ T(x). 
\end{align*}
And that ends the proof.
