How is $\mathrm{d}(\alpha \cdot \beta)$ defined when $\alpha$ and $\beta$ are vector valued forms? Let $\alpha$ and $\beta^p$ be respectively $0$ and $p$ vector valued forms. Then $\alpha\cdot \beta^p$ is a scalar valued $p$ form. How is
$$
\begin{equation}
\mathrm{d}\left(\alpha\cdot \beta^p \right)
\end{equation}
$$
defined in terms of $\mathrm{d}\alpha$ and $\mathrm{d}\beta^p$ ? Here $\cdot$ is the standard scalar product (assuming that these are vectors in $\mathbb{R}^3$ but maybe it's possible to generalize). Moreover, how do we calculate
$$
\begin{equation}
\left(\alpha\cdot \beta^p \right)\wedge\left(\gamma\cdot \delta^q \right)
\end{equation}
$$
where $\gamma$ and $\delta^q$ are respectively $0$ and $q$ vector valued forms ?
I've tried simply expanding in vector components in a general basis but it's no so clean notationally and in the end it cannot be written a single dot product.
 A: As far as I understand, $\boldsymbol{\alpha}$ is a vector valued $0$-form, in other words, a vector valued function, or, a vector field:
$$
\boldsymbol{\alpha}=\alpha^\rho\boldsymbol{e}_\rho
$$
where the summation convention is used. ​Further, $\boldsymbol{\beta}$ is a vector valued $p$-form:
$$
\boldsymbol{\beta}={\beta^\rho}_{|\mu_1...\mu_p|}\boldsymbol{d}x^{\mu_1}\,\wedge...\wedge \,\boldsymbol{d}x^{\mu_p}
$$
where the summation convention is used and $|\mu_1...\mu_p|$ stands for index combinations with $\mu_1<...<\mu_p\,.$
When you say that $\boldsymbol{\alpha}\cdot\boldsymbol{\beta}$ is a scalar valued $p$-form it seems you mean
$$
\boldsymbol{\alpha}\cdot\boldsymbol{\beta}=\underbrace{\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|}}_{\text{scalar}}\,\boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,.
$$
The exterior derivative of this should then be (as always)
$$
\boldsymbol{d}(\boldsymbol{\alpha}\cdot\boldsymbol{\beta})=\partial_\sigma(\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|})\,\boldsymbol{d}x^\sigma\wedge \boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,.
$$
You are asking how this can be expressed in terms of $\boldsymbol{d}\boldsymbol{\alpha}$ and $\boldsymbol{d}\boldsymbol{\beta}\,.$ Clearly,
$\boldsymbol{d}\boldsymbol{\alpha}$ is a vector valued $1$-form.
In chapter 14 of [1] we find the expressions
\begin{align}
\boldsymbol{d}\boldsymbol{\alpha}&=\boldsymbol{e}_\rho\boldsymbol{d}\alpha^\rho +\alpha^\rho \boldsymbol{de}_\rho\,,\\
\boldsymbol{de}_\rho&=\boldsymbol{e}_\nu{\boldsymbol{\omega}^\nu}_\rho
\end{align}
where the coefficients ${\boldsymbol{\omega}^\nu}_\rho$ by which
the vector valued $1$-form $\boldsymbol{de}_\rho$ is expanded
are closely related to curvature, i.e., to the difference of the partial
derivative from the covariant derivative.
As far as I can tell we will get a nice result when we assume that all $\boldsymbol{\omega}$ vanish. Then
$$
\boldsymbol{d}\boldsymbol{\alpha}=\boldsymbol{e}_\rho\boldsymbol{d}\alpha^\rho=\boldsymbol{e}_\rho(\partial_\sigma\alpha^\rho)\boldsymbol{d}x^\sigma\,.
$$
since each $\alpha^\rho$ is a scalar.
Handling the vector valued $(p+1)$-form $\boldsymbol{d\beta}$ is simpler:
$$
\boldsymbol{d}\boldsymbol{\beta}=\partial_\sigma({\beta^\rho}_{|\mu_1...\mu_p|})\,\boldsymbol{d}x^\sigma\wedge \boldsymbol{d}x^{\mu_1}\wedge...\wedge \boldsymbol{d}x^{\mu_p}\,.
$$
It is also clear that
$$
\partial_\sigma(\alpha_\rho\,{\beta^\rho}_{|\mu_1...\mu_p|})=(\partial_\sigma\alpha_\rho){\beta^\rho}_{|\mu_1...\mu_p|}+\alpha_\rho(\partial_\sigma {\beta^\rho}_{|\mu_1...\mu_p|})\,.
$$
Thus, if there is no curvature we should get the not so surprising result
$$
\boldsymbol{d}(\boldsymbol{\alpha}\cdot\boldsymbol{\beta})=\boldsymbol{\alpha}\cdot\boldsymbol{d\beta}+\boldsymbol{d\alpha}\cdot\boldsymbol{\beta}\,.
$$
[1] Misner, Thorne & Wheeler, Gravitation.
