Integration along the unit interval for a differential form

Let $\omega \in {\mathrm{\Omega }}^k(M\times [0,1])$ be smooth up to the endpoints. For $k\ge 1$, its interval integral is the $(k-1)$-form $K\omega ={\int }_0^1{\beta }_{t}dt$, where $\omega ={\alpha }_t+dt\wedge {\beta }_t$ and both families are tangential to $M$. Set $K=0$ on degree zero and on zero terms.

Use the product structure of Products of smooth manifolds have a canonical product smooth structure, restricted from $M\times \mathbb{R}$. The families are intrinsically ${\alpha }_t=i_t^{\ast }\omega$ and ${\beta }_t=i_t^{\ast }({\iota }_{{\mathrm{\partial }}_t}\omega )$, using Interior product of a form by a vector field; evaluation on tangential tuples and on $({\mathrm{\partial }}_t,v_1,\dots ,v_{k-1})$ proves existence and uniqueness of the decomposition. The integral is in the fixed finite-dimensional fibre ${\bigwedge }^{k-1}{T}_x^{\ast }M$. Coefficients have smooth local extensions across endpoints. Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral supplies parameter differentiation; coordinate independence and full smoothness are proved in The interval homotopy operator is coordinate independent ↗.

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.