Pullback of a 1-form from $\mathbb{R}^2$ onto $S^1$

Question

I am really confused with the following question:

Consider the manifold $\mathbb{R}^2$ with its usual smooth structure and let $\iota : S^{1} \to \mathbb{R}^2$ be the inclusion map. Let $\alpha \in \Omega^1 (\mathbb{R}^2)$ be the one-form: $$\alpha = xdy - ydx $$ In any single chart on $S^1$ of your choice, find the expression for the on-form $\iota^* \alpha$.

My attempt:

Take the chart $U = S^{1} \text \ (-1,0)$ and $\tilde{U} = (\pi, \pi)$ and $f^{-1} : \theta \mapsto (\cos \theta, \sin \theta)$. Then: $$\iota^* (\alpha) = \iota^* (xdy) - \iota^*(ydx)$$

Now I get stuck when trying to describe the inclusion map. I don't understand why the inclusion map can't be described as simply the identity, i.e.:

$$ \iota(\cos \theta, \sin \theta) =(\cos \theta, \sin \theta)$$ Surely if I have a point on $S^1$ its coordinates are $(\cos \theta, \sin \theta)$ so if I want to now 'include' this in $\mathbb{R}^2$ why would it's coordinates change? We would still describe it by $(\cos \theta, \sin \theta)$ because that's where it's located.

I believe I have a misunderstanding with what the inclusion map really does.

I would appreciate any guidance on how to solve this!

Answer 1

Let F be a local parametrization of $\mathbb{S}^1$ s.t $F:(-\pi,\pi)\to \mathbb{S}^1$ and $\theta\mapsto (\cos\theta,\sin\theta)$. Let $G$ be a local parametrization of $\mathbb{R}^2$ such that $G:\mathbb{R}^2\to\mathbb{R}^2$ and $(x,y)\mapsto (x,y)$.

Clearly $\mathbb{S}^1\subset \mathbb{R}^2$, then we can define an inclusion $\iota:\mathbb{S}^1\to \mathbb{R}^2$ s.t $(\cos\theta,\sin\theta)\mapsto (x,y)$. Then we have $x=\cos\theta$ and $y=\sin\theta$. In other words, we find the cartesian coordinate on $\mathbb{S}^1$.

By chain rule, we have $$\iota^*(dx)=d(\iota(x))=d(\cos\theta)=\dfrac{\partial x}{\partial\theta}d\theta=\dfrac{\partial}{\partial \theta}\cos\theta d\theta=-\sin\theta d\theta$$ Similarly $$\iota^*(dy)=\cos\theta d\theta$$

Back to your question $$\alpha=xdy-ydx$$ then $$\iota^*(\alpha)=(x\circ\iota)\iota^{*}dy-(y\circ\iota)\iota^*(dx)=d\theta$$

Answer 2

You have that

$$ i^*(xdy)=(i^*x)(i^*dy)=(x\circ i)(d(i^*y))=(x\circ i)d(y\circ i)=i_1di_2 $$

for $i=(i_1,i_2)$, what means that $i_1(p,q)=p$ and $i_2(p,q)=q$ for every $(p,q)\in S^1$. Then the last expression just means that $i^*(xdy)$ is the form $xdy$ restricted to points of $S^1$.

Now suppose that $\varphi (t)=(\varphi _1(t),\varphi _2(t))=(x,y)$ is a chart in $S^1$, then

$$ \varphi ^*i^*(xdy)=(i\circ \varphi )^*(xdy)=\varphi ^*(xdy)=\varphi _1d\varphi _2=\varphi _1\dot \varphi _2dt $$

as every point on the image of $\varphi $ belongs to $S^1$ so $i\circ \varphi =\varphi $.