Vorticity equation in index notation (curl of Navier-Stokes equation) I am trying to derive the vorticity equation and I got stuck when trying to prove the following relation using index notation:
$$
{\rm curl}((\textbf{u}\cdot\nabla)\mathbf{u}) = (\mathbf{u}\cdot\nabla)\pmb\omega - ( \pmb\omega \cdot\nabla)\mathbf{u}
$$
considering that the fluid is incompressible $\nabla\cdot\mathbf{u} = 0 $, $\pmb \omega = {\rm curl}(\mathbf{u})$ and that $\nabla \cdot \pmb \omega = 0.$
Here follows what I've done so far:
$$
(\textbf{u}\cdot\nabla) \mathbf{u} = u_m\frac{\partial u_i}{\partial x_m} \mathbf{e}_i  = a_i  \mathbf{e}_i \\
{\rm curl}(\mathbf{a}) = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \mathbf{e}_i = \epsilon_{ijk} \frac{\partial}{\partial x_j}\left( u_m\frac{\partial u_k}{\partial x_m} \right) \mathbf{e}_i = \\
= \epsilon_{ijk}\frac{\partial u_m}{\partial x_j}\frac{\partial u_k}{\partial x_m}  \mathbf{e}_i + \epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i    \\
$$
the second term $\epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i$ seems to be the first term "$(\mathbf{u}\cdot\nabla)\pmb\omega$" from the forementioned identity. Does anyone have an idea  how to get the second term? 
 A: The trick is the following:
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_m}{\partial x_k} = 0 $$
by antisymmetry. 
So you can rewrite
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_k}{\partial x_m} = \epsilon_{ijk} \frac{\partial u_m}{\partial x_j}\left( \frac{\partial u_k}{\partial x_m} - \frac{\partial u_m}{\partial x_k} \right) $$
Note that the term in the parentheses is something like $\pm\epsilon_{kml} \omega_l$
Lastly use the product property for Levi-Civita symbols
$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$
A: Not a direct answer to the original post's question, but I have done a full write-up on deriving the vorticity transport equations. I go into more detail in my post, but I've copied the general gist of the derivation below:
Derivation
Incompressible conservation of momentum equations:
$$ \partial_t u_i + u_j \partial_j u_i = - \tfrac{1}{\rho} \partial_i p +
\nu \partial_j^2 u_i $$
To get vorticity evolution, we can take the curl of the momentum transport equations:
$$ \varepsilon_{k\ell i} \partial_\ell [\partial_t u_i + u_j
\partial_j u_i = - \tfrac{1}{\rho} \partial_i p + \nu \partial_j^2 u_i ]$$
Distributing this across the terms, we get:
$$ \begin{align}
 \underbrace{\varepsilon_{k\ell i} \partial_\ell
        \partial_t u_i}_\text{Temporal Term} +
  \underbrace{\varepsilon_{k\ell i} \partial_\ell
       u_j \partial_j u_i}_\text{Advection Term} & = 
  \underbrace{- \varepsilon_{k\ell i} \partial_\ell
       \tfrac{1}{\rho} \partial_i p}_\text{Pressure Term} +
  \underbrace{\varepsilon_{k\ell i} \partial_\ell
        \nu \partial_j^2 u_i}_\text{Viscous Term} \\\\
\Rightarrow \quad \mathbb{T} + \mathbb{A} & = \mathbb{P} + \mathbb{V}
\end{align}$$
Temporal Term $\mathbb{T}$
$$\mathbb{T} = \varepsilon_{k\ell i} \partial_\ell \partial_t u_i \Rightarrow \ \partial_t \varepsilon_{k\ell i} \partial_\ell u_i \Rightarrow \ \partial_t \omega_k $$
Pressure Term $\mathbb{P}$
Since the curl of the gradient of a scalar is 0, $\mathbb{P} = 0$.
Viscous Term $\mathbb{V}$
$$ \mathbb{V} = \varepsilon_{k\ell i} \partial_\ell \nu \partial_j^2 u_i  \Rightarrow \quad \nu \partial_j^2 \varepsilon_{k\ell i} \partial_\ell u_i  \Rightarrow \quad \mathbb{V} = \nu \partial_j^2 \omega_k $$
Advection Term $\mathbb{A}$
$$ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell u_j \partial_j u_i $$
Using:
$$ u_j \partial_j u_i = \partial_i (\tfrac{1}{2} u_j u_j ) +
\varepsilon_{ijq} u_q (\underbrace{\varepsilon_{jmn} \partial_m u_n}_{\omega_j})  $$
get:
$$ \mathbb{A} = \varepsilon_{k\ell i} (\partial_i (\tfrac{1}{2} u_j u_j ) +
\varepsilon_{ijq} u_q \omega_j $$
For the lefthand term, note that $u_j u_j$ is just a scalar. Therefore, the
left expression can be surmised as the curl of the gradient of a scalar and it
is then equal to zero. This leaves us with:
$$\Rightarrow \ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell (
\varepsilon_{ijq} \omega_j u_q ) 
\Rightarrow \ \varepsilon_{ik\ell} \varepsilon_{ijq}
\partial_\ell \omega_j u_q$$
Plug:
$$ \varepsilon_{ik\ell} \varepsilon_{ijq} = \delta_{kj}\delta_{\ell q} -
\delta_{kq}\delta_{\ell j} $$
into previous expression:
$$ \mathbb{A} = (\delta_{kj}\delta_{\ell q} - \delta_{kq}\delta_{\ell j} )
\partial_\ell \omega_j u_q $$
$$ \Rightarrow \ \mathbb{A} = \partial_q \omega_k u_q - \partial_j \omega_j u_k \Rightarrow \ (u_q \partial_q \omega_k + \omega_k \partial_q  u_q) -
( u_k \partial_j \omega_j  + \omega_j \partial_j  u_k)$$
By incompressibility $\partial_q u_q =0$. Also, $\partial_j \omega_j$ also equals zero:
$$ \Rightarrow \  \mathbb{A} = \underbrace{u_q \partial_q
\omega_k}_\text{Vorticity Advection}  - \underbrace{\omega_j \partial_j
u_k}_\text{Vorticity Stretching} $$
Putting It All Together
$$
\underbrace{\partial_t \omega_k}_\mathbb{T}  +
\underbrace{u_q \partial_q \omega_k  - \omega_j \partial_j u_k}_\mathbb{A} =
\underbrace{0}_\mathbb{P} +
\underbrace{\nu \partial_j^2 \omega_k}_\mathbb{V}
$$
