Simplify $\ \nabla\cdot (\vec{G}\times\vec{H})$
I have very little experience using Einstein Notation, but I believe it is the quickest way to solve this problem.
Attempt: $$\ \nabla\cdot (\vec{G}\times\vec{H})=\sum_{l=1}^4e_{ijkl}\partial_i(G_jH_k)\mathbf{e}_l=\sum_{l=1}^4e_{ijkl}(G_j\partial_i H_k+H_k\partial_i G_j)\mathbf{e}_l$$ Now I'm unsure if what I have done is correct thus far. Furthermore, I'm sure of how to proceed after this point.
The vector product is defined as: $$\vec{G}\times\vec{H}=\sum_{i,k,l}\varepsilon_{ikl}G_kH_l\vec{e}_i\tag{1}$$ And the divergence is defined: $$\textrm{div}(\vec{v})=\sum_{i}\frac{\partial v_i}{\partial x_i}\tag{2}$$ The component $i$ of $(1)$ is (taking the scalar product with $\vec{e}_j$ and seting $j=i$: $$(\vec{G}\times\vec{H})_i=\sum_{k,l}\varepsilon_{ikl}G_kH_l$$ Seting $v_i=(\vec{G}\times\vec{H})_i$ in $(2)$ we finaly have: $$\textrm{div}(\vec{G}\times\vec{H})=\sum_{i,k,l}\epsilon_{ikl}\frac{\partial G_k H_l}{\partial x_i}$$
If you want to go further you can say that: $$\textrm{div}(\vec{G}\times\vec{H})=\sum_{i,k,l}\epsilon_{ikl}\frac{\partial G_k }{\partial x_i}H_l+\sum_{i,k,l}\epsilon_{ikl}\frac{\partial H_l }{\partial x_i}G_k\tag{3}$$ We first recognise that an even permutation of the symbol $\epsilon_{ikl}=\epsilon_{lik}=\epsilon_{kli}=1$, and the odd permutations $\epsilon_{ilk}=\epsilon_{lki}=\epsilon_{kil}=-1$, therefore the symbols $\epsilon_{ikl}=\epsilon_{lik}=-\epsilon_{kil}$ are equivalent. Applying this result to $(3)$ one has: $$\textrm{div}(\vec{G}\times\vec{H})=\sum_{i,k,l}H_l\epsilon_{lik}\frac{\partial G_k }{\partial x_i}-\sum_{i,k,l}G_k\epsilon_{kil}\frac{\partial H_l }{\partial x_i}\tag{4}$$
Finally the scalar product of $\vec{v}\cdot\vec{w}$ may be written: $$\vec{v}\cdot\vec{w}=\sum_{i}v_iw_i\tag{5}$$ You can recognise that every summand of $(4)$ is a scalar product between $\vec{H}$ and $\vec{\textrm{curl}}(\vec{G})$ and $\vec{G}$ and $\vec{\textrm{curl}}(\vec{H})$ respectively: $$\textrm{div}(\vec{G}\times\vec{H})=\vec{H}\cdot\vec{\textrm{curl}}(\vec{G})- \vec{G}\cdot \vec{\textrm{curl}}(\vec{H})$$