Usbrth

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
$$
ydx wedge dz + x dy wedge dz texton mathbbR^3
$$
and
$$
textsin(xy) dy wedge dz
$$
Can someone give an explanation of how these work? In the first case, it is an element of $Lambda^3(mathbbR)$, 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.

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 $pmapstoleft(fracpartialpartial x_i|_pright)^ast$, where $left(fracpartialpartial x_i|_pright)^ast$ is the dual element of $fracpartialpartial x_i|_p$ in $(T_pmathbb R^n)^ast$.

Then as $fracpartialpartial x_1|_p,ldots,fracpartialpartial x_n|_p$ is a basis of $T_pmathbb R^n$ for each $p$, we get that $dx_1(p),ldots, dx_n(p)$ is a basis of $(T_pmathbb R^n)^*$ for all $p$.

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

If $f:mathbb R^nrightarrow mathbb R$ 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 $(T_pmathbb R^n)^ast$, $f(p)omega(p)$, given by $vmapsto f(p)cdotomega(p)(v)$ for all $vin T_pmathbb R^n$. 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 $mathbb R^nrightarrowmathbb R$.

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 $dx_i_1wedgecdotswedge dx_i_k:1leq i_1<cdots<i_kleq n$.

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 $Mrightarrowmathbb R$, and a basis of size $nchoose k$ using an atlas of this manifold and a partition of unity subordinate to this atlas.