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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.

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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.

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  • $\begingroup$ ok I think I understand the source of my confusion. So the fact that coefficients in the basis expansion are functions just reflects the fact that the coefficients are changing as we move from point to point? $\endgroup$ – staedtlerr Apr 7 '19 at 6:24
  • $\begingroup$ and that those coefficients should vary smoothly from point to point $\endgroup$ – staedtlerr Apr 7 '19 at 6:42
  • $\begingroup$ @staedtlerr, yes. $\endgroup$ – Camilo Arosemena-Serrato Apr 7 '19 at 19:16

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