$$\nabla\cdot(\mathbf{u}\times\mathbf{v}) = (\nabla\times\mathbf{u})\cdot\mathbf{v} - (\nabla\times\mathbf{v})\cdot\mathbf{u}$$
Hi, the above is a vector equation, where u and v are vectors. I am trying to prove this identity using index notation. I am able to get the first term of the right-hand side, but I don't see where the second term with the minus in front comes from. Any help? Thanks!
EDIT: I don't think this is a duplicate of that thread since the question in that thread has a scalar f and a vector v, whereas my problem has two vectors u and v, and no scalars.
Well, here is my answer:
the left-hand term is read in components as
$$ \nabla \cdot (u\times v) = \sum_{i,j,k} \frac{\partial}{\partial x_i} (\epsilon_{ijk} u_j v_k) = \sum_{i,j,k} \epsilon_{ijk}\left(v_k\frac{\partial u_j}{\partial x_i} + u_j\frac{\partial v_k}{\partial x_i} \right), $$
whereas $$ (\nabla \times u)\cdot v - (\nabla\times v)\cdot u = \sum_{i,j,k}v_i\epsilon_{ijk} \frac{\partial u_k}{\partial x_j} - \sum_{i,j,k}u_i\epsilon_{ijk} \frac{\partial v_k}{\partial x_j}. $$
Now observe:
$$ \sum_{i,j,k}v_i\epsilon_{ijk} \frac{\partial u_k}{\partial x_j} = - \sum_{i,j,k}v_i\epsilon_{jik} \frac{\partial u_k}{\partial x_j} = (-1)^2 \sum_{i,j,k}v_i\epsilon_{jki} \frac{\partial u_k}{\partial x_j} \equiv \sum_{i,j,k}v_k\epsilon_{ijk} \frac{\partial u_j}{\partial x_i}, $$
where I have just reindexed in the last equality. Similarly:
$$ - \sum_{i,j,k}u_i\epsilon_{ijk} \frac{\partial v_k}{\partial x_j} = (-1)^2 \sum_{i,j,k}u_i\epsilon_{jik} \frac{\partial v_k}{\partial x_j} \equiv \sum_{i,j,k}u_j\epsilon_{ijk} \frac{\partial v_k}{\partial x_i}. $$
Hence:
$$ (\nabla \times u)\cdot v - (\nabla\times v)\cdot u = \sum_{i,j,k}u_j\epsilon_{ijk} \frac{\partial v_k}{\partial x_i} + \sum_{i,j,k}v_k\epsilon_{ijk} \frac{\partial u_j}{\partial x_i} = \sum_{i,j,k}\epsilon_{ijk}\left( v_k\frac{\partial u_j}{\partial x_i} + u_j\frac{\partial v_k}{\partial x_i}\right) = \nabla \cdot (u\times v), $$
such as we want.
Remember: in the first equation I have used the Leibniz rule to achive two terms. The second trick is a game with indices and properties of the Levi-Civita symbol $\epsilon_{ijk}$.
I hope this is the answer you were looking for.