Show that $( \nabla \times {\bf G} ) {\bf v} = \nabla \times ({\bf G}^T {\bf v})$ (curl formula) The effect of the curl of a (second order) tensor field $\bf G$ on a constant vector $v$ is defined by the relation
$
( \nabla \times {\bf G} ) {\bf v} = \nabla \times ({\bf G}^T {\bf v})
$
I'd like to derive the RHS of the above relation from the LHS. 
Update
This is how far I have gotten following the example in section B.3.5 of this thesis. The only problem now is the inconsistency in the index of the partial derivatives in the results of $( \nabla \times  G ) \, {\bf v}$ and  $\nabla \times  ( G^T \, {\bf v})$ i.e. $\frac{\partial G_{ij}}{\partial x_k}$ versus $\frac{\partial G_{ji}}{\partial x_k}$.
$$
\begin{equation}\begin{aligned}
( \nabla \times  G ) \, {\bf v}& = \left[({\bf e}_k \partial_k) \times (G_{ij} {\bf e}_i \otimes {\bf e}_j)\right] v_l {\bf e}_l \\
& = \frac{\partial G_{ij}}{\partial x_k} v_l \left[ {\bf e}_k \times ({\bf e}_i \otimes {\bf e}_j)\right] {\bf e}_l \\
& = \frac{\partial G_{ij}}{\partial x_k} v_l \left[ ({\bf e}_k \times {\bf e}_i ) \otimes {\bf e}_j \right] {\bf e}_l \\
& = \frac{\partial G_{ij}}{\partial x_k} v_l \, \epsilon_{mki} \,  ({\bf e}_m \times {\bf e}_j ) \, {\bf e}_l \\
& = \frac{\partial G_{ij}}{\partial x_k} v_l \, \epsilon_{mki} \, \delta_{jl} \, {\bf e}_m \\
& = \frac{\partial G_{ij}}{\partial x_k} v_j \, \epsilon_{mki} \, {\bf e}_m \\
\end{aligned}\end{equation}
$$
whereas because $G^T = G_{ji} ({\bf e}_i \otimes {\bf e}_j)$ 
$$
\begin{equation}\begin{aligned}
\nabla \times  ( G^T \, {\bf v}) & = \left[({\bf e}_k \partial_k) \times (G_{ji} {\bf e}_i \otimes {\bf e}_j)\right] v_l {\bf e}_l \\
& = \frac{\partial G_{ji}}{\partial x_k} v_l {\bf e}_k \times \left[   ({\bf e}_i \otimes {\bf e}_j) \, {\bf e}_l \right] \\
& = \frac{\partial G_{ji}}{\partial x_k} v_l {\bf e}_k \times \left[   {\bf e}_i \,  ( {\bf e}_j . {\bf e}_l ) \right] \\
& = \frac{\partial G_{ji}}{\partial x_k} v_l {\bf e}_k \times ({\bf e}_i \,  \delta_{jl}) \\
& = \frac{\partial G_{ji}}{\partial x_k} v_j \, ( {\bf e}_k \times {\bf e}_i ) \\
& = \frac{\partial G_{ji}}{\partial x_k} v_j \, \epsilon_{mki} \, {\bf e}_m \\
\end{aligned}\end{equation}
$$
 A: One thing needs to be clear: $\nabla$ is not a vector. I repeat: nabla is not a vector. In particular, taking the curl is not "doing cross product with $\nabla$".
In the thesis you referred to, the cross product of a vector $a$ on the left of a (2,0)-tensor $B$ is taken as
$$(a\times B)^{ij}=\epsilon^i{}_{kl}a^kB^{lj}$$
and the dot product of a vector $c$ on the right of a (2,0)-tensor $B$ is defined as
$$(B\cdot c)^i = B^{ij}\delta_{jr}c^r.$$
Now you want to take the curl of a (2,0)-tensor. This is not a standard thing to do, so there may be various definitions depending on the author.

Option 1 (This is what you did, and what I would do, too).
Looking at formula for the curl of a vector $(\nabla\times v)^i = \epsilon^{ik}{}_l\partial_kv^l$, one is tempted to define the curl of a (2,0)-tensor as
$$(\nabla\times G)^{ij} = \epsilon^{ik}{}_l\partial_kG^{lj}\text.$$
With this definition we obtain
$$\begin{align*}
((\nabla\times G)\cdot v)^i
&= (\nabla\times G)^{ij}\delta_{jr}v^r \\
&= \epsilon^{ik}{}_l\partial_kG^{lj}\delta_{jr}v^r \\
&= \epsilon^{ik}{}_l\partial_k(G^{lj}\delta_{jr}v^r) && \text{(because $\partial_kv^r=0$)} \\
&= \epsilon^{ik}{}_l\partial_k(G\cdot v)^l \\
&= (\nabla\times(G\cdot v))^i
\end{align*}$$
Since you made this coice, this is the result you got to and, as you already noted, there are no transposes at sight.

Option 2.
Another possible choice (kinda weird if you ask me), is the definition made in the slides you also gave me. There the author deines the curl of a (2,0)-tensor as
$$(\nabla\times G)^{ij} = \epsilon^{ik}{}_l\partial_kG^{jl}\text.$$
(the contraction with the Levi-Civita symbol is on the second index of $G$, rather than the first). With that definition, one gets
$$\begin{align*}
((\nabla\times G)\cdot v)^i
&= (\nabla\times G)^{ij}\delta_{jr}v^r \\
&= \epsilon^{ik}{}_l\partial_kG^{jl}\delta_{jr}v^r \\
&= \epsilon^{ik}{}_l\partial_k(G^{jl}\delta_{jr}v^r) && \text{(because $\partial_kv^r=0$)} \\
&= \epsilon^{ik}{}_l\partial_k((G^T)^{lj}\delta_{jr}v^r) \\
&= \epsilon^{ik}{}_l\partial_k(G^T\cdot v)^l \\
&= (\nabla\times(G^T\cdot v))^i
\end{align*}$$
as you wanted.
In short, you couldn't get to the identity that appears in those slides because the author and you have different definitions for the curl of a $(2,0)$-tensor.
