In a textbook, I have found the following relation. $$ d \stackrel{2}{\omega_V} = \frac{1}{3!} (\mathrm{div} V) \epsilon_{ijk} dx^i \wedge dx^j \wedge dx^k . $$ It is cool, but I don't know how to show it in an elegant way, please show me the way.
Here are definitions:
It is a relation in a Euclidean space $\mathbb{R}^3$. There is a vector field in that space $V \in \Gamma (T\mathbb{R}^3)$. With the use of components of that vector, one can define a two-form $$ \stackrel{2}{\omega_V} := \frac{1}{2} V^i \epsilon_{ijk} dx^j \wedge dx^k, $$ where $\epsilon$ is completely antisymmetirc and $\epsilon_{123} =1$. Divergence of a vector is $$ \mathrm{div}V = V^m_{\hphantom{mk},m} $$ where comma indicates partial derivative.
My attempt:
By taking the derivative of $\stackrel{2}{\omega_V}$ I get $$ d\stackrel{2}{\omega_V} = \frac{1}{2} V^i_{\hphantom{m},l} \epsilon_{ijk} dx^l \wedge dx^j \wedge dx^k $$ clearly, only terms where $i=l$ are non-zero. Consequently, by substituing $V^i_{\hphantom{m},l}$ with $V^i_{\hphantom{m},i}$ and dividing by number of extra repetition which is $3$, I get what I want, but its v hand-wavy. Can someone can show me how to do in a decent way?
You can make a remark that $$ dx^l \wedge dx^j \wedge dx^k = \epsilon^{ljk} {\rm vol} $$ where ${\rm vol} = \frac{1}{3!}\epsilon_{ijk}dx^i\wedge dx^j\wedge dx^k$ is the volume form. Then use the fact that $$ \epsilon_{ijk}\epsilon^{ljk} = 2\delta^l_i$$ That may make the final transformations a bit smoother.