Example:Smooth map between the real line and the circle (considered as smooth manifolds)

This work is done in the context of the book An Introduction to Manifolds - Loring W. Tu

Example

Let [ilmath]\mathbb{R} [/ilmath] be considered as a smooth manifold (with the differentiable / smooth structure induced by the smooth atlas [ilmath]\{(\mathbb{R},\text{Id}_{\mathbb{R} }:\mathbb{R}\rightarrow\mathbb{R})\} [/ilmath]) and let [ilmath]\mathbb{S}^1[/ilmath] be considered as a smooth manifold with the smooth structure induced by the smooth atlas [ilmath]\{(U^{+},\varphi_+:U^+\rightarrow I),\ (U^-,\varphi_-:U^-\rightarrow I),\ (V^+,\psi_+:V^+\rightarrow I),\ (V^-,\psi_-:V^-\rightarrow I) \} [/ilmath], where:

• [ilmath]I:\eq(-1,1)\subset\mathbb{R} [/ilmath] - the open interval of unit length in each direction
• [ilmath]U^+:\eq\{(x,y)\in\mathbb{S}^1\ \big\vert\ y>0\} [/ilmath] and [ilmath]\varphi_+:U^+\rightarrow I[/ilmath] by [ilmath]\varphi_+:(x,y)\mapsto x[/ilmath]
• [ilmath]U^-:\eq\{(x,y)\in\mathbb{S}^1\ \big\vert\ y<0\} [/ilmath] and [ilmath]\varphi_-:U^-\rightarrow I[/ilmath] by [ilmath]\varphi_-:(x,y)\mapsto x[/ilmath]
• [ilmath]V^+:\eq\{(x,y)\in\mathbb{S}^1\ \big\vert\ x>0\} [/ilmath] and [ilmath]\psi_+:U^+\rightarrow I[/ilmath] by [ilmath]\psi_+:(x,y)\mapsto y[/ilmath]
• [ilmath]V^-:\eq\{(x,y)\in\mathbb{S}^1\ \big\vert\ x<0\} [/ilmath] and [ilmath]\psi_-:U^-\rightarrow I[/ilmath] by [ilmath]\psi_-:(x,y)\mapsto y[/ilmath]

Claim

Under these conditions we claim the map:

• [ilmath]F:\mathbb{R}\rightarrow\mathbb{S}^1[/ilmath] given by [ilmath]F:t\mapsto\big(\cos(t),\sin(t)\big)[/ilmath] is a smooth map between manifolds

Proof

There are two ways we could do this, firstly using the definition where we get to choose a chart such that the domain of the first chart to the first manifold through the map to the second manifold and to that chart's domain is smooth in the usual sense, or we can show this for any charts (covering the necessary point in the domain), we will do it the former way as this is equivalent to the latter anyway.

• Let [ilmath]p\in\mathbb{R} [/ilmath] be given, we will show that [ilmath]F[/ilmath] is [ilmath]C^\infty[/ilmath]/smooth here, since this is arbitrary, smooth everywhere.
  • Choose [ilmath](\mathbb{R},\text{Id}_\mathbb{R})[/ilmath] as the (smooth) chart about [ilmath]p[/ilmath]
    • We now have four cases for where [ilmath](x,y):\eq F(p)[/ilmath] lies, they are [ilmath]y>0[/ilmath], [ilmath]y<0[/ilmath], [ilmath](y\eq 0\text{ and }x\eq-1)[/ilmath] and [ilmath](y\eq 0\text{ and }x\eq 1)[/ilmath]
      1. Suppose [ilmath]y>0[/ilmath], this is covered by the chart [ilmath](U^+,\varphi_+)[/ilmath]
         • We require now that [ilmath]\big(\varphi_+\circ F\circ\text{Id}_\mathbb{R}^{-1}\big):\underbrace{\text{Id}_\mathbb{R}(\mathbb{R}\cap F^{-1}(U^+))}_{\subseteq\mathbb{R} }\rightarrow \underbrace{I:\eq(-1,1)}_{\subseteq\mathbb{R} } [/ilmath] - restricting to common domains as appropriate - be smooth at [ilmath]\text{Id}_\mathbb{R}(p)\eq p[/ilmath]

           This map is given by [ilmath]\big(\varphi_+\circ F\circ\text{Id}_\mathbb{R}^{-1}\big):r\mapsto\cos(r)[/ilmath]

           • We can preform a little bit of analysis (which isn't really needed) to tidy this up as:
             • [ilmath]\big(\varphi_+\circ F\circ\text{Id}_\mathbb{R}^{-1}\big)\eq \varphi_+\circ F:\bigcup_{k\in\mathbb{Z} }\underbrace{\big(2\pi k, 2\pi k+\pi\big)}_{\subseteq\mathbb{R} }\rightarrow \underbrace{I:\eq(-1,1)}_{\subseteq\mathbb{R} } [/ilmath] by [ilmath]\varphi_+\circ F:r\mapsto \cos(r)[/ilmath]
         • As [ilmath]\cos[/ilmath] is known to be smooth (infinitely differentiable) and it is "obvious" that restricted to an open subset of [ilmath]\mathbb{R} [/ilmath] would leave smoothness at a point unhindered.
    • The other cases are similar.

Grade: C

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