Relationship between exterior and directional derivative

Question

For a smooth $k$-form $\omega$ I know the formula $$ \begin{split} \mathrm{d} \omega (X_0,\dots,X_k) &= \sum_{i=0}^k (-1)^i X_i(\omega(X_0,\dots,\hat{X_i},\dots,X_k)) \\ &+ \sum_{i<j} (-1)^{i+j} \omega([X_i,X_j],X_0,\dots,\hat{X_i},\dots,\hat{X_j},\dots,X_k), \end{split}$$ which relates the directional derivatives of the functions $p \mapsto \omega(X_0,\dots,\hat{X_i},\dots,X_k)\big{\rvert}_p$ to $\mathrm{d}$. I guess it should be true that if $p\mapsto \mathrm{d} \omega (X_0,\dots,X_k)\big{\rvert}_p$ is a bounded function, then $p \mapsto X_i(\omega(X_0,\dots,\hat{X_i},\dots,X_k))\big{\rvert}_p$ is bounded as well. However, I can't manage to proof this. I checked these two functions in local coordinates and tried to connect them, but it seems that I end up at the above formula. Is this even true and if yes, how to prove it?

Answer 1

There is no reason why this should be true. If you take $\omega$ to be the exterior derivative of some $(k-1)$-form, then $d\omega=0$ but this certainly does not imply that $X_i(\omega(X_0,\dots,\widehat{X_i},\dots,X_k))$ is bounded for arbitrary vector fields $X_j$.