I'm studying Differential Forms for the first time. I'm stuck on a problem that seems simple.
My book definition. Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be a differentiable function. Then $f$ induce an aplication $f^{*}$ that map $k$-forms into $k$-forms.
Let $\omega$ a $k$-form in $\mathbb{R}^{m}$. By definition, $f^{\ast}\omega$ is a $k$-form in $\mathbb{R}^{n}$ given by
$$(f^{*}\omega)(p)(v_{1},…,v_{k}) = \omega(f(p))(df_{p}(v_{1}),…,df_{p}(v_{k}))\tag{1}$$
where $p \in \mathbb{R}^{n}$, $v_{1},…,v_{k} \in T_{p}\mathbb{R}^{n}$ and $df_{p}: T_{p}\mathbb{R}^{n} \to T_{f(p)}\mathbb{R}^{n}$ is the differential aplication of $f$.
Here, $T_{p}$ is the tangent plane at $p$.
After that, the book give an example.
Example. Let $\omega$ a $1$-form in $\mathbb{R}^{2}\setminus\{(0,0)\}$ given by
$$\omega = -\frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy.$$
Let $U = \{(r,\theta) \mid r>0,0<\theta<2\pi\}$ and $f:U \to \mathbb{R}^{2}$ given by
$$f(r,\theta) = \begin{cases}
x = r\cos\theta\\
y = r\sin\theta
\end{cases}.$$Let's calculate $f^{*}\omega$.
Since
$$dx = \cos\theta dr – r\sin\theta d\theta,$$
$$dy = \sin\theta dr + r\cos\theta d\theta,$$
we get
$$f^{*}\omega = -\frac{r\sin\theta}{r^{2}}(\cos\theta dr – r\sin\theta d\theta) + \frac{r\cos\theta}{r^{2}}(\sin\theta dr + r\cos\theta d\theta) = d\theta.$$
I think that I don't completely understood the definition.
Using (1),
$$\omega(f(r,\theta)) = -\frac{r\sin\theta}{r^{2}}(\cos\theta dr – r\sin\theta d\theta) + \frac{r\cos\theta}{r^{2}}(\sin\theta dr + r\cos\theta d\theta)$$
But, what about $df_{(r,\theta)}(v)$ with $v \in T_{(r,\theta)}U$?
Best Answer
A $1$-form belongs to the dual space of the tangent space (at a point $p\in U$, say), that is $(T_{p}U)^{\ast}$. Hence its elements (the $1$-forms) are linear maps $\omega_{p}: T_{p}U \rightarrow \mathbf{R}$ which vary smoothly to get a family of $1$-forms $\omega:TU \rightarrow \mathbf{R}$ (i.e. I just drop the $p$ subscript). To be explicit, for some tangent vector $v\in T_{p}U$, we have that $\omega_{p}(v) \in \mathbf{R}$, or again as one varies the point to get a vector field $V\in TU$, $\omega(V)\in \mathbf{R}$.
Now given a smooth map $f:U\rightarrow V$, its differential at a point $p\in U$ is a linear map $d_{p}f:T_{p}U\rightarrow T_{f(p)}V$ which when one varies the point $p$, is usually written as $f_{\ast}$ (called the pushforward of $f$). This in turn induces a dual map $f^{\ast}:(TV)^{\ast} \rightarrow (TU)^{\ast}$ defined as follows: for a $1$-form $\alpha\in (T_{f(p)}V)^{\ast}$ we get a new $1$-form $f^{\ast}\alpha \in (T_{p}U)^{\ast}$ by precomposition, i.e. let $v\in T_{p}U$ then $$ f^{\ast}\alpha(v)|_{p} = (\alpha \circ f_{\ast})(v)|_{p} = (\alpha \circ d_{p}f)(v)|_{p} = \alpha(d_{p}f(v))|_{f(p)} $$ where $(\alpha\circ d_{p}f)$ is ''at $p$'' since $v\in T_{p}U$, yet $\alpha$ is ''at $f(p)$'' because now $d_{p}f(v)$ belongs to $T_{f(p)}V$. This construction can then be extended to $k$-forms.
To get to answering your question, $df_{(r,\theta)}(v)$ for $v\in T_{(r,\theta)}U$ hasn't appeared yet since the $k$-form is not being evaluated on any vectors (otherwise you would just get a real number). The $\omega(f(p))$ part of $\omega(f(p))(d_{p}f(v_{1}),\ldots d_{p}f(v_{k}))$ just means that your $k$-form is at the point $f(p)$, and that no vectors are being eaten up by it. In my notation above it would be $\omega|_{f(p)}$ so distinguish between being an argument of the differential form and referring to the point it is associated to. Apologies if this seems like a rather long-winded answer - a lot of the introductory theory of differential forms is unwinding definitions and remembering what spaces things live in.