$SO(N)$ model and Noether current This is a very simple calculation that, for some reason, I cannot find the solution to. I want to show that along the equation of motion for the following $SO(N)$ model $$\mathcal{L} = \frac{1}{2}\partial_\mu\phi_i\partial^\mu\phi_i-\mu^2\phi_i\phi_i-g(\phi_i\phi_i)^2$$ the Nöether current $$J_\mu^a = \partial_\mu\phi_i\lambda^a_{ij}\phi_k$$ is conserved $$\partial^\mu J_\mu^a=0.$$
The progress I've made, which is a bit embarrassing to be honest, is up to this point: I found the EOM $$(\square+2\mu^2) \phi_i =-4g(\phi_j\phi_j) \phi_i$$ and that $$\partial^\mu J_\mu^a = \square\phi_i\lambda^a_{ij}\phi_j+\partial^\mu\phi_i\lambda^a_{ij}\partial_\mu\phi_j$$ Now I don't know where to go. Surely there's some trickery tied to the exchange of some partial derivatives throwing away boundary terms and so on, but I cannot see how.
 A: I'll give a more general approach. Recall the basic setup of Noether's theorem. If $\delta\phi_i$ is any variation of the fields of your theory you have $$\delta{\cal L}=\dfrac{\partial\cal L}{\partial \phi_i}\delta\phi_i+\dfrac{\partial\cal L}{\partial(\partial_\mu\phi_i)}\partial_\mu\delta\phi_i+\cdots+\dfrac{\partial \cal L}{\partial (\partial_{\mu_1}\cdots\partial_{\mu_k}\phi_i)}\partial_{\mu_1}\cdots\partial_{\mu_k}\delta \phi_i\tag{1}.$$
Now you can use Liebnitz rule several times on (1) to remove derivatives from $\delta \phi_i$ and extract terms proportional to $\delta \phi_i$ that combine with the first term in (1). Doing this reveals the general structure of $\delta{\cal L}$:$$\delta{\cal L}=E_i\delta\phi_i+\partial_\mu\Theta^\mu\tag{2},$$
where $E_i$ are the equations of motion and $\Theta^\mu[\phi_i;\delta\phi_i]$ depends on the fields and the specific variation. We call $\Theta^\mu$ the pre-sympletic potential current density and it takes one important role in the Covariant Phase Space formalism.
Now suppose that $\delta \phi_i$ is a (quasi-)symmetry of your Lagrangian. This means that if $\phi_i$ is a field configuration (not necessarily on-shell) then $\delta{\cal L}=\partial_\mu k^\mu$. If $k^\mu=0$ we have a real symmetry (not a quasi-symmetry), but Noether's theorem applies to both of them. In either case applying (2) to this specific variation you find that $$E_i\delta\phi_i+\partial_\mu\Theta^\mu=\partial_\mu k^\mu\Longrightarrow \partial_\mu(\Theta^\mu-k^\mu)=-E_i\delta \phi_i\tag{3}$$
and therefore setting $j^\mu = \Theta^\mu-k^\mu$ you find that if the configuration about which you have taken the variation obeys the equations of motion $E_i$ vanishes and so does $\partial_\mu j^\mu$ giving current conservation. Equality modulo equations of motion is often written as $\approx$ so that you might say that $\partial_\mu j^\mu \approx 0$.
This is Noether's first theorem. Now apply this to your Lagrangian. First find $\Theta^\mu$ by taking $\delta {\cal L}$. If you do so you will recognize that $\Theta^\mu[\phi_i;\delta\phi_i]:=\partial^\mu \phi_i \delta \phi_i$.
Now I believe you are OK in verifying that ${\cal L}$ is invariant under ${\rm SO}(N)$, in which case that Noether's theorem applies. Now consider the infinitesimal ${\delta}_{{\rm SO}(N)}\phi_i$ transformation. If $\phi_i$ is a multiplet transforming in a representation with generators $T^a$ we can write ${\delta}_{{\rm SO}(N)}\phi_i=\omega^a (T^a)_{ij}\phi_j$. In that case we have $$\Theta^\mu[\phi_i;\delta_{{\rm SO}(N)} \phi_i]=\partial^\mu \phi_i \delta_{{\rm SO}(N)}\phi_i=\omega^a\partial^\mu \phi_i (T^a)_{ij}\phi_j\tag{4}.$$
Noether's theorem implies that (4) is conserved on-shell for all $\omega^a$. In particular taking $\omega^a = \delta^{ab}$ for each value of $b$, one at a time, we obtain $$J_\mu^{a}=\partial_\mu \phi_i (T^a)_{ij}\phi_j\tag{5},$$
and you have found through Noether's theorem the statement of the conservation of $J_\mu^{a}$, modulo equations of motion of course, $\partial^\mu J_\mu^a\approx 0$. Assuming $\lambda^a_{ij}$ in your question are the matrix elements of the generators in the representation $\phi_i$ lives this is exactly the current you want to prove to be conserved.
