Let $M$ be a smooth manifold, $I:=[0, 1]$ and let $\pi:TM\longrightarrow M$ be the tangent bundle of $M$. We can associate a vector bundle $\pi\times \textrm{id}:TM\times I\longrightarrow M\times I$ whose fiber over $(p, t)\in M\times I$ is $$(TM\times I)_{(p, t)}:=T_pM\times \{t\}.$$ From this, it is clear $TM\times I$ is isomorphic to the pullback bundle $\textrm{pr}_1^* TM$ where $$\textrm{pr}_1:M\times I\longrightarrow M$$ is the projection on the first component. Also, notice the space of time-dependent $k$-forms on $M\times I$, that is, the smooth sections $\Omega^k(M)^I$ of the $k$-th exterior power $\Lambda^k (TM\times I)^*$ satisfies $$\Omega^k(M)^I\simeq C^\infty(M\times I)\otimes_{C^\infty(M)}\Omega^k(M),$$ where $\Omega^k(M)$ stands for the space of smooth sections of $\Lambda^k T^*M$.

How can I use the De Rham differential $d_{DR}:\Omega^k(M)\longrightarrow \Omega^{k+1}(M)$ to induce a differential $$d_{DR}^I:\Omega^k(M)^I\longrightarrow \Omega^{k+1}(M)^I$$

I guess it must be something like:

$$f\otimes \omega\longmapsto f\otimes d_{DR} \omega+ (..)\otimes (...).$$ Thanks.

  • $\begingroup$ As you say, you are really looking at the pullback of $TM$ to $M \times I$. This is actually a subbundle of $T(M \times I)$, so you can just use the de Rham differential there. In product coordinates it should have exactly the form you describe. $\endgroup$ – Charlie Cifarelli Mar 26 '17 at 16:46

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