Wedge product of $0$-forms?

Question

Given a manifold $M$, and a $0$-form $f$ and a $k$-form $\omega$ on $M$ for $k>0$, it is standard to define $f \wedge \omega$ as simply $f\omega$, where $\bigwedge^k T^*M$ is being considered as a module over $\bigwedge^0 T^*M:=\mathrm{C}^\infty(M)$ in the natural way. My question is very simple: how can one extend this definition to the case where $k=0$ in a consistent way?

Answer 1

The (wedge) product of two zero forms that is of two functions $f,g$ is just $fg$.

Answer 2

An observation that helps to understand the case $k=0$ is the fact that $0$-multilinear forms on a vector space $\mathbb V$ over $\mathbb R$ are scalar functions $$ f\colon\mathbb V^{\otimes^0}\to\mathbb R $$ with no restrictions imposed on them (there is no coordinate where to impose linearity). Since $\mathbb V^{\otimes^0}=\{0_{\mathbb V}\}$, we conclude that $0$-multilinear forms can be identified with constants because: $$ \mathbb R^{\{0_{\mathbb V}\}} \cong \mathbb R. $$ Since constants are also alternating $0$-multiforms, we deduce that $$ \Lambda^0(X)=\coprod_{p\in X}\mathbb R = X\times\mathbb R. $$ Therefore, every section $\omega\colon X\to X\times\mathbb R$ of the projection $\pi\colon X\times\mathbb R\to X$, onto the first coordinate, can be identified with its projection onto the second coordinate, i.e., with a function from $X$ to $\mathbb R$. Thus, the imposition of smoothness on $0$-forms defined on $X$ coincides with the smoothness of scalar functions defined on $X$.