Divergence of this expression I am currently struggeling to calculate the following expression. Let $O$ be an orthogonal matrix and $u: \mathbb{R}^n \rightarrow \mathbb{R}$
$$div( O^T (\nabla (u)(Ox))).$$
So I want to calculate the divergence of the vector we get from the matrix multiplication of the transpose of an orthogonal matrix and the gradient of $u$ evaluated at $Ox.$
 A: We note that the $i$'th component of $Ox$ is given by $\sum_jO_{ij}x_j$.  Then, we have
$$\begin{align}
\nabla u(Ox) &=\sum_k \hat x_k\frac{\partial u(Ox)}{\partial x_k}\\\\
&=\sum_k \hat x_k\sum_i u_iO_{ik}\tag 1
\end{align}$$
where $u_i$ designates the partial derivative of $u$ with respect to the $i$'th argument.
Now, taking the divergence of $(1)$ gives
$$\begin{align}
\nabla \cdot \left(\sum_k \hat x_k\sum_i u_iO_{ik}\right)&=\sum_{l} \hat x_l\frac{\partial }{\partial x_l} \cdot \left(\sum_k \hat x_k\sum_i u_iO_{ik}\right)\\\\
&=\sum_{l}\sum_{k}\sum_i (\hat x_l\cdot \hat x_k) \frac{\partial u_i}{\partial x_l}O_{ik}\\\\
&= \sum_{l}\sum_{k}\sum_i \sum_{m}(\hat x_l\cdot \hat x_k) u_{im}O_{ml}O_{ik} \tag 2\\\\
&=\sum_{l}\sum_{k}\sum_i \sum_{m}\delta_{lk} u_{im}O_{ml}O_{ik} \tag 3\\\\
&= \sum_{k}\sum_i \sum_{m} u_{im}O_{mk}O_{ik} \tag 4\\\\
&=\sum_i \sum_{m} u_{im}\sum_{k}\,O_{mk}O_{ik} \tag 5\\\\
&=\sum_i \sum_{m} u_{im}\delta_{im}\tag 6\\\\
&=\sum_i  u_{ii}\tag 7\\\\
&=\nabla^2 u
\end{align}$$

NOTES:
In going from $(2)$ to $(3)$, we used the orthogonality of $\hat x_l$ and $\hat x_k$.  The term $\delta_{kl}$ is the Knonecker Delta and is equal to $1$ when $k=l$ and $0$ otherwise.
In going from $(3)$ to $(4)$, we used the sifting property of the Knonecker Delta.
In going from $(4)$ to $(5)$, we simply interchanged the order of summation.
In going from $(5)$ to $(6)$, we made use of the fact the $O$ is an orthogonal matrix.
In going from $(6)$ to $(7)$, we used the sifting property of the Kronecker Delta once again.
