Expansion of p-form in terms of a basis of $\Lambda^p(M)$

Question

In Lee's Introduction to Smooth manifolds, as well as the wiki page for exterior algebra, they give the decomposition of a $p$-form in terms of the basis of $\Lambda^p(M)$ as $$ \omega = \omega_I dx^I $$ where $dx^I$. I figured the decomposition would look like this, but I did not anticipate that the $\omega_I$ were not real numbers, but real valued functions. I think this just comes down to a question of "What is $\Lambda^p(M)$ a vector space over?". I assumed it was the reals, and I'm sure it is. But I believe this statement is saying it is also a VS over $C^0(M)$. If so, how is scalar multiplication defined exactly?
The book gives examples like $$ ydx \wedge dz + x dy \wedge dz \ \ \text{on } \mathbb{R}^3 $$ and $$ \text{sin}(xy) dy \wedge dz $$ Can someone give an explanation of how these work? In the first case, it is an element of $\Lambda^3(\mathbb{R})$, so it is supposed to take in $p$ covectors and return a real number. The only way I could think of this happening is by pointwise multiplication, where we evaluate the function part and the wedge parts individually and multiply the real number outputs. I was just very surprised to see continuous maps come in, and I'm kind of not sure why they're allowed to be there.

Answer 1

You're confused with the definitions as $\mathbb R^n$ is treated both as a vector space and as a manifold.

First let us clarify what the symbol $dx_i$ means as a differential form of the manifold $\mathbb R^n$.

$dx_i$ is the functión with domain $\mathbb R^n$ such that $p\mapsto\left(\frac{\partial}{\partial x_i}|_p\right)^\ast$, where $\left(\frac{\partial}{\partial x_i}|_p\right)^\ast$ is the dual element of $\frac{\partial}{\partial x_i}|_p$ in $(T_p\mathbb R^n)^\ast$.

Then as $\frac{\partial}{\partial x_1}|_p,\ldots,\frac{\partial}{\partial x_n}|_p$ is a basis of $T_p\mathbb R^n$ for each $p$, we get that $dx_1(p),\ldots, dx_n(p)$ is a basis of $(T_p\mathbb R^n)^*$ for all $p$.

A $1$-form $\omega$ of the manifold $\mathbb R^n$ is a function which assings to each $p$ an element of $(T_p\mathbb R^n)^*$. Thus for each we have $$\omega(p)=f_1(p)dx_1(p)+\cdots+f_n(p)dx_n(p),$$ for some $f_1(p),\ldots,f_n(p)\in\mathbb R$, as $dx_1(p),\ldots, dx_n(p)$ is a basis of $(T_p\mathbb R^n)^*$ for all $p$. Hence each $1$-form of the manifold $\mathbb R^n$ comes along with $n$ functions $\mathbb R^n\rightarrow\mathbb R$. Such a form $\omega$ is a differential $1$-form if these functions are smooth. Conversely $n$ smooth functions $\mathbb R^n\rightarrow\mathbb R$ determine a differential $1$-form.

If $f:\mathbb R^n\rightarrow \mathbb R$ is a smooth function and $\omega$ is a $1$-differential form of $\mathbb R^n$ we can get another differential $1$-form by setting for each $p$ the element of $(T_p\mathbb R^n)^\ast$, $f(p)\omega(p)$, given by $v\mapsto f(p)\cdot\omega(p)(v)$ for all $v\in T_p\mathbb R^n$. Thus, the set of all differential $1$-forms is a $C^\infty(\mathbb R^n)$-module, and in particular, it is a vector space over $\mathbb R$; where $C^\infty(\mathbb R^n)$ is the ring of all smooth functions $\mathbb R^n\rightarrow\mathbb R$.

Notice, however, that the differential forms $dx_1,\ldots,dx_n$ are a basis of this set over the ring $C^\infty(\mathbb R^n)$, but they clearly cannot be a basis over $\mathbb R$. Similarly, the set of all differential $k$-forms of $\mathbb R^n$ is also an $C^\infty(M)$-module with basis $\{dx_{i_1}\wedge\cdots\wedge dx_{i_k}:1\leq i_1<\cdots<i_k\leq n\}$.

Similarly, for any smooth manifold $M$ the set of all differential $k$-forms of $M$ is a $C^\infty(M)$ module, where $C^\infty(M)$ is the ring of all smooth functions $M\rightarrow\mathbb R$, and a basis of size $n\choose k$ using an atlas of this manifold and a partition of unity subordinated to this atlas.