Product rule (with respect to time) for a smooth family of differential forms

Question

I would like to show that if $\{\omega_t\}$ and $\{\tau_t\}$ are smooth families of $k$-forms and $\ell$-forms respectively on a manifold $M$, then $$\frac{d}{dt}(\omega_t \wedge \tau_t) = \Big(\frac{d}{dt}\omega_t \Big)\wedge \tau_t + \omega_t \wedge \Big(\frac{d}{dt}\tau_t\Big).$$

Here the time derivative is defined as taking the partial derivative with respect to time of the coefficient function when the form is expressed in local coordinates.

In some chart $(U, x^1, \ldots, x^n)$ we have that $$\omega_t = \sum_I a_I(t,p)dx^I\\ \tau_t = \sum_J b_J(t, p) dx^J$$ where $p \in M$ and $I = (i_1, \ldots, i_k$) where $1 \leq i_1 < \cdots < i_k \leq n$ and similarly for $J$. Their wedge product is then $$\omega_t \wedge \tau_t = \sum_{IJ}a_Ib_i(t,p) dx^I \wedge dx^J$$ and so $$\frac{d}{dt}(\omega_t \wedge \tau_t) = \sum_{IJ} \frac{\partial}{\partial t}(a_Ib_J)dx^I \wedge dx^J = \sum_{IJ} \Big(b_J\frac{\partial}{\partial t}(a_I) + a_I\frac{\partial}{\partial t}(b_J)\Big)dx^I \wedge dx^J \\ \color{red}{=} \Big(\frac{d}{dt}\omega_t \Big)\wedge \tau_t + \omega_t \wedge \Big(\frac{d}{dt}\tau_t\Big)$$

I feel like I am abusing notation here and that more has to be shown to justify the last equality above (in red). Is my proof correct?

Answer 1

I think you can do it intrinsically as well; no charts. Write out the definition of the wedge product and then apply product rule with $\frac{d}{dt} \vert_{t=t_0}$.

Let $X_1,\dots,X_{k+l}\in T_pM$ and compute

$$ \frac{d}{dt} \vert_{t=t_0} \left[(\omega_t\wedge \tau_t)(X_1,\dots,X_k,\dots,X_{k+l}) \right] =\\ \frac{d}{dt} \vert_{t=t_0} \left[ \frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sign}(\sigma)\omega_t(X_{\sigma(1)},\dots,X_{\sigma(k)})\tau_t(X_{\sigma(k+1)},\dots,X_{\sigma(k+l)}) \right]=\\ \frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sign}(\sigma)(\frac{d}{dt} \vert_{t=t_0}\omega_t)(X_{\sigma(1)},\dots,X_{\sigma(k)})\tau_t(X_{\sigma(k+1)},\dots,X_{\sigma(k+l)})+\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sign}(\sigma)\omega_t(X_{\sigma(1)},\dots,X_{\sigma(k)})(\frac{d}{dt} \vert_{t=t_0}\tau_t)(X_{\sigma(k+1)},\dots,X_{\sigma(k+l)})=\\ \left((\frac{d}{dt} \vert_{t=t_0} \omega_t) \wedge \tau_t+\omega_t \wedge(\frac{d}{dt} \vert_{t=t_0} \tau_t)\right)(X_1,\dots,X_{k+l}) $$