# Calculating Pull-Back of a $1$-form.

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

• $df$ is the dual of $f^*$, so if you take the matrix calculated for $f^*$ you need to transpose it in order to obtain $df$ – user84976 Mar 2 at 23:42
• @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 $f^{*}\omega$ – Lucas Corrêa Mar 3 at 0:13
• 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. – Ted Shifrin Mar 3 at 0:23
• @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? – Lucas Corrêa Mar 3 at 3:22

## 2 Answers

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.

• 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 $\omega(f(p))$? – Lucas Corrêa Mar 3 at 15:06
• 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. – BenCWBrown Mar 3 at 18:06
• I see. Thank you! It's very helpful! – Lucas Corrêa Mar 4 at 0:10

(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

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