Showing a rotational symmetry of the Navier-Stokes / Euler equation I'm going through Majda and Bertozzi's "Vorticity and Incompressible Flow" and am having a hard time verifying what looks like an easy equality; I suspect its due to my very poor understanding/familiarity of multivariable calculus.
That is, I am trying to prove the following. Recall the Navier Stokes equations (NSE) on $ℝ^n$ with viscosity $\nu$, where $n=2,3$:
$$ \begin{align}\frac{Dv}{Dt} &= -∇ p + \nu \Delta v \\ \text{div }v &= 0 \\ v|_{t=0} &= v_0\end{align}  $$
Where the material derivative is $$\frac{D}{Dt} := \frac{\partial}{\partial t} + \sum_{i=1}^n v^i\frac{\partial}{\partial x^i} = \frac{\partial}{\partial t} + (v· ∇)$$
Suppose $(v,p)$ is a solution of the NSE, and $Q$ is a rotation i.e. $QQ^t = I$. Then $(v_Q,p_Q)$ is another solution of the NSE, where
\begin{align}
v_Q(x,t)&:=Q^t v(Qx,t)\\
p_Q(x,t)&:=p(Qx,t)
\end{align}

So I plan to compute $\frac{Dv_Q}{Dt}$, $∇p_Q$, $Δ v_Q$, $\text{div }v_Q$.  I can compute $∇ p_Q$,
$$ ⟨ D_x[ p(Qx,t)],h⟩  = ⟨ D_xp(Qx,t) \circ Q, h \rangle = ⟨ D_xp(Qx,t), Qh⟩ = ⟨[\text{adj }Q] D_x p(Qx,t),h⟩ =Q^t∇_xp(Qx,t)h$$
so $∇ p_Q(x,t) = Q^t ∇ p(Qx,t)$. Also, $\partial_t v_Q = Q^t \partial_t v(Qx,t)$. Apart from these, I am at a loss; trying to apply chain rule leaves me in a mess. Can someone help me?

*

*Majda, Andrew J.; Bertozzi, Andrea L., Vorticity and incompressible flow, Cambridge Texts in Applied Mathematics. Cambridge: Cambridge University Press (ISBN 0-521-63057-6/hbk; 0-521-63948-4/pbk). xii, 545 p. (2002). ZBL0983.76001.

 A: So i managed to do it, with the magic of Einstein summation and a much more pedestrian approach than finding the Riesz representative.
$$ \let\del\partial \del_j (v^i(Qx)) = \del_j [v^i(q_{kl}x^l\mathbf{e}_k)] =[\del_k v^i](Qx)\del_j[q_{kl}x^l] = \del_k v^i(Qx) q_{kj} = q^T_{jk} \del_k v^i(Qx),\tag{$*$}$$
so $$\nabla (v\circ Q)(x) = Q^T \nabla v(Qx).$$Taking another derivative and summing,
$$\del_j\del_j [v^i(Qx)] = q^T_{jk}\del_j[\del_k v^i(Qx)] = q^T_{jk} [\del_m\del_kv^i](Qx)\del_j[q_{mn}x^n]=q_{mj}q^T_{jk} \del_m\del_kv^i(Qx) = \del_k\del_k v^i(Qx),$$
where we used that $q_{mj}q^T_{jk} = \delta_{mk}$. Hence
$$ \Delta(v\circ Q)(x) = \Delta v(Qx),$$
so that by linearity $\Delta v_Q(x) = Q^T \Delta v(Qx)$.
Also, $$\nabla\cdot v_Q (x)= \partial_j(Q^tv(Q(x)))_j=\partial_jq_{ij}(v^i( Qx))\overset{(*)}=q_{ij}q^T_{jk} \del_k v^i(Qx)=\delta_{ik}\del_k v^i(Qx)=\partial_i v^i(Qx)=0.$$
Similarly for $[(v\circ Q)\cdot \nabla] (v\circ Q)(x)$,
\begin{align} [[(v\circ Q)\cdot \nabla] (v\circ Q)(x)]^i
& = v^{\ j}(Qx)\del_j [v^i(Qx)]\\ 
& = v^{\ j}(Qx)q_{kj} \del_k v^i(Qx) \\
& = [(Qv)_k \del_k] v^i(Qx) \\
& = [(Qv)· \nabla]v^i(Qx)
\end{align}
So again by linearity we have
$$[v_Q· \nabla]v_Q(x) = Q^T [(v· \nabla)v](Qx)$$
So plugging into Navier-Stokes or Euler, we see that the $Q^T$s magically cancel and we have a new solution.
