I'm having some trouble with proving that the curl of gradient of a vector quantity is zero using index notation: $\nabla\times(\nabla\vec{a}) = \vec{0}$.
In index notation, I have $\nabla\times a_{i,j}$, where $a_{i,j}$ is a two-tensor. But is this correct? If so, where should I go from here?
Thanks, and I appreciate your time and help!
If you contract the Levi-Civita symbol with a symmetric tensor the result vanishes identically because (using $A_{mji}=A_{mij}$)
$$\varepsilon_{ijk}A_{mji}=\varepsilon_{ijk}A_{mij}=-\varepsilon_{jik}A_{mij}$$
We are allowed to swap (renaming) the dummy indices $j,i$ in the last term on the right which means
$$\varepsilon_{ijk}A_{mji}=-\varepsilon_{ijk}A_{mji}$$
This can only be true if
$$\varepsilon_{ijk}A_{mji}\equiv0$$
Now with $(\nabla \times S)_{km}=\varepsilon_{ijk} S_{mj|i}$ and $S_{mj|i}=a_{m|j|i}$ all you have to investigate is if, and under which circumstances, $a_{m|j|i}$ is symmetric in the indices $i$ and $j$.