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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 $v \in T_{(r,\theta)}U$?

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  • $\begingroup$ $df$ is the dual of $f^*$, so if you take the matrix calculated for $f^*$ you need to transpose it in order to obtain $df$ $\endgroup$ – user84976 Mar 2 at 23:42
  • $\begingroup$ @user84976 I don't know if I understood your comment. For me, seems that $df$ was unnecessary. Just evaluating $\omega$ at $f(r,\theta)$, the author got $f^{*}\omega$ $\endgroup$ – Lucas Corrêa Mar 3 at 0:13
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    $\begingroup$ You should have more useful properties of pullback: e.g., $f^*(\omega\wedge\eta) = f^*\omega\wedge f^*\eta$ (and the pullback of a sum is the sum of the pullbacks). So you need to check with your definition that if $f(r,\theta) = (f_1(r,\theta),f_2(r,\theta)) = (x,y)$, then $f^*dx = df_1$ and $f^*dy = df_2$. Proceed from there. $\endgroup$ – Ted Shifrin Mar 3 at 0:23
  • $\begingroup$ @TedShifrin, thank you for the hints! My book already proved this properties (except $f^{*}dx$, but I will try). I answer the question using this properties. Is correct? $\endgroup$ – Lucas Corrêa Mar 3 at 3:22
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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.

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  • $\begingroup$ Thank you for the answer! If I understood, for get the answer is just necessary to calculate $\omega(f(p))$. Thus, I think a problem with the notation. The $(d_{p}f(v_{k}))$ are used when I'm evaluating in some vector. Otherwise, $f^{*}\omega$ is just $\omega(f(p))$? $\endgroup$ – Lucas Corrêa Mar 3 at 15:06
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    $\begingroup$ Yep exactly! Sadly to keep track of the points $\omega$ and $f^{\ast}\omega$ are at often becomes notationally cumbersome. I've seen some authors use a restriction-like notation $|_{p}$, etc. sometimes to emphasise that the form isn't taking in the point as an argument. $\endgroup$ – BenCWBrown Mar 3 at 18:06
  • $\begingroup$ I see. Thank you! It's very helpful! $\endgroup$ – Lucas Corrêa Mar 4 at 0:10
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(1) $f^{*}(\omega \wedge \varphi) = f^{*}(\omega)\wedge f^{*}(\varphi)$

(2) $f^{*}(\omega + \varphi) = f^{*}(\omega) + f^{*}(\varphi)$

(3) $f^{*}(\omega) = \omega\circ f$ if $\omega$ is a $0$-form

(4) $f^{*}dx = df_{1}$ and $f^{*}dy = df_{2}$

So thinking in $-\frac{x}{x^2+y^2}$ and $\frac{y}{x^2+y^2}$ as $0$-forms, we have

$$\begin{eqnarray*} f^{*}\omega &=& f^{*}\left(-\frac{x}{x^2+y^2}dx + \frac{y}{x^2+y^2}dy\right)\\ & =& -\left(\left(\frac{x}{x^2+y^2}\right)\circ f\right)f^{*}dx + \left(\left(\frac{y}{x^2+y^2}\right)\circ f\right)f^{*}dy\\ & =& -\left(\left(\frac{x}{x^2+y^2}\right)\circ f\right)df_{1} + \left(\left(\frac{y}{x^2+y^2}\right)\circ f\right)df_{2}. \end{eqnarray*}$$

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