Using index notation to prove vector identities

Question

I am asked to use index notation to prove that $\nabla\cdot(\overrightarrow{u}\times\overrightarrow{v})=(\nabla\times\overrightarrow{u})\cdot\overrightarrow{v}-\overrightarrow{u}\cdot(\nabla\times\overrightarrow{v})$. Below is so far I have worked:

By index notation, there is \begin{equation*} \overrightarrow{u}\times\overrightarrow{v}=u_{i}v_{j}\epsilon_{ijk}\hat{e}_{k} \end{equation*} \begin{equation*} \Downarrow \end{equation*} \begin{equation*} \nabla\cdot(\overrightarrow{u}\times\overrightarrow{v})=\partial_{l}u_{i}v_{j}\epsilon_{ijk}\delta_{kl} \end{equation*} Then we notice \begin{equation*} (\nabla\times\overrightarrow{u})\cdot\overrightarrow{v}=\partial_{l}u_{i}\epsilon_{lik}\hat{e}_{k}\cdot v_{j}\hat{e}_{j}=\partial_{l}u_{i}v_{j}\epsilon_{lik}\delta_{kj} \end{equation*} \begin{equation*} \overrightarrow{u}\cdot(\nabla\times\overrightarrow{v})=u_{i}\hat{e}_{i}\cdot\partial_{l}v_{j}\epsilon_{ljk}\hat{e}_{k}=\partial_{l}u_{i}v_{j}\epsilon_{ljk}\delta_{ik} \end{equation*}

Then I am stuck here. It seems that we have to show $\epsilon_{lik}\delta_{kj}-\epsilon_{ljk}\delta_{ik}=\epsilon_{ijk}\delta_{kl}$, yet I have no clue of doing so. Any hint please?

Answer 1

Yes, you are on the right track. As already said in the comments, you can use the fact that $\epsilon_{lik}\delta_{kj}=\epsilon_{lij}$ and the product rule to get the desired result. However, it is worth pointing out that you do not need to explicitly use the Kronecker delta and the basis vectors $\hat e_k$ to arrive at the result. Note that

\begin{align} \nabla\cdot(\vec u\times\vec v)&=\partial_i(\vec u\times\vec v)_i\\ &=\epsilon_{ijk}\partial_i(u_jv_k)\\ &=\epsilon_{ijk}\partial_i(u_j)v_k+\epsilon_{ijk}u_j\partial_iv_k\\ &=\epsilon_{kij}\partial_i(u_j)v_k-\epsilon_{jik}u_j\partial_iv_k\\ &=(\nabla\times\vec u)_kv_k-u_j(\nabla\times\vec v)_j\\&=(\nabla\times\vec u)\cdot\vec v-\vec u\cdot(\nabla\times\vec v).\end{align}