Exercise 2.01

Just because a manifold is topologically nontrivial doesn't necessarily mean it can't be covered with a single chart. In contrast to the circle $S^1$, show that the infinite cylinder $\mathbf{R}\times S^{1}$ can be covered with just one chart, by explicitly constructing the map.

We can construct a chart for the infinite cylinder of radius 1 centered around the $x^{3}$-axis in two successive steps:
• first, we project the point to be at the distance $e^{x^{3}}$ from the $x^{3}$-axis,
• second, we flatten everything onto the $(x^{1},x^{2})$-plane.

$$ \begin{pmatrix} x^{1}\\x^{2}\\x^{3} \end{pmatrix} \to \begin{pmatrix} x^{1}e^{x^{3}}\\x^{2}e^{x^{3}}\\x^{3} \end{pmatrix} \to \begin{pmatrix} x^{1}e^{x^{3}}\\x^{2}e^{x^{3}}\\0 \end{pmatrix} $$ This gives us the following map. $$ \phi(x^{1},x^{2},x^{3})\equiv(y^{1},y^{2})=(x^{1}e^{x^{3}},x^{2}e^{x^{3}}) $$ $\phi(\mathbf{R}\times S^{1})$ is indeed open on $\mathbb{R}^{2}$ by lacking the origin and infinity.

It's then easy to go back to the 3D coordinates of the point on the cylinder. $$ \begin{pmatrix} y^{1}/\sqrt{ (y^{1})^{2}+(y^{2})^{2} }\\ y^{2}/\sqrt{ (y^{1})^{2}+(y^{2})^{2} }\\ \log \sqrt{ (y^{1})^{2}+(y^{2})^{2} } \end{pmatrix} = \begin{pmatrix} x^{1}e^{x^{3}} / \sqrt{ (x^{1}e^{x^{3}})^{2}+(x^{2}e^{x^{3}})^{2} }\\ x^{2}e^{x^{3}} / \sqrt{ (x^{1}e^{x^{3}})^{2}+(x^{2}e^{x^{3}})^{2} }\\ \log \sqrt{ (x^{1}e^{x^{3}})^{2}+(x^{2}e^{x^{3}})^{2} } \end{pmatrix} = \begin{pmatrix} x^{1}\\ x^{2}\\ x^{3} \end{pmatrix} $$