I could not prove that curl of gradient is zero. How can I do this by using indiciant notation?

Question

I could not find a way to equal this to zero.

$\vec{\nabla}\times(\vec{\nabla}\phi)= \epsilon_{ijk}\partial_{j}(\partial\phi)_{k}= \epsilon_{ijk}(\phi(\partial_{j}\partial_{k}) + \partial_{k}(\partial_{j}\phi)) $

Answer 1

This is virtually answered in the comments. Definitions: ${\let\del\partial}$ $$(\nabla \times F)_i = \epsilon_{ijk}\del_jF_k \\ (\nabla \phi)_k = \del_k\phi $$ Where Einstein summation is used. See also definition of $\epsilon_{ijk}$. then as long as $\phi\in C^2$ (Clairaut's theorem), $$ (\nabla\times\nabla \phi)_i = \epsilon_{ijk} \del_j \del_k \phi \overset{\text{Clairaut}}{=}\epsilon_{ijk} \del_k \del_j \phi = -\epsilon_{ikj}\del_k \del_j \phi = -(\nabla\times\nabla\phi)_i$$ which implies that $\nabla\times\nabla \phi = 0$.

This has answers but they are not accepted - Proving the curl of a gradient is zero

This is closely related, and one answer is just this proof (but phrased more tersely) - why the curl of the gradient of a scalar field is zero? geometric interpretation