# Show that $\partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$

Let us start from $$\textbf{B}=\nabla \times \textbf{A}$$ and write its components $$B_k=\epsilon_{ijk}\partial_i A_j$$.

I want to show that $$\partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$$. I can sense that it works, but I want to see it directly. How should I start?

• Contract with $\epsilon_{ijk}$ and figure out what the contraction of two $\epsilon$-symbols is. – Nephente Sep 2 at 7:58

Using the identity $$\epsilon_{kij}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$,$$\epsilon_{ijk}B_k=\epsilon_{ijk}\epsilon_{klm}\partial_lA_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_lA_m=\partial_iA_j-\partial_jA_i=F_{ij}.$$