Let $M$ be a compact manifold and consider a differential form $\alpha\in\Omega^1(M)$, which we can think of as a map $\alpha:M\to T^*M$. Since $T^*M$ is a smooth manifold we can compute the differential of this map $$D\alpha:TM\to T(T^*M)$$ How does this differential differ from the exterior derivative on 1-forms? I have a feeling that $D\alpha(X) = d(\alpha(X))$, but I'm not sure how to show this.

  • $\begingroup$ What is $X$?${}{}$ $\endgroup$ – Camilo Arosemena-Serrato Jun 2 at 19:13
  • $\begingroup$ Presumably a smooth vector field on M. $\endgroup$ – yousuf soliman Jun 3 at 2:15
  • $\begingroup$ Your feeling can't be right: These objects live in different places. Moreover, $D\alpha(X)$ depends pointwise just on the value of $X$ at the point, whereas $d(\alpha(X))$ involves differentiating the vector field $X$. $\endgroup$ – Ted Shifrin Jun 3 at 17:56

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