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!
Best Answer
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 $$\iota^*(\alpha)=(x\circ\iota)\iota^{*}dy-(y\circ\iota)\iota^*(dx)=d\theta$$