1
$\begingroup$

Let $\mathbb{S}^1$ have the manifold structure induced by it being a regular submanifold of $\mathbb{R}^2$. Let $\{$ $U_i$,$\phi_i$ $\}_{i=1}^4$ be the differentiable structure on $\mathbb{S}^1$ where $U_i$ are the open semicircles and each $\phi_i$ projects $U_i$ to an axis.

I would like to show that these smooth structures determine the same smooth structure. To do so, I must show that their union is also a smooth atlas.

Let $\{$ $(V_{\alpha}\cap \mathbb{S}^1,\psi_{\alpha}) \}_{\alpha}$ be the smooth structure given by the adapted charts. On $V_{\alpha}\cap \mathbb{S}^1 \cap U_i$ , we have

$\phi_{i}(\psi_{\alpha}^{-1}(\psi(p))=\phi_{i}(\psi_{\alpha}(x^1(p))=\phi_{i}(p)=p=\psi_{\alpha}^{-1}(\psi_{\alpha}(p))$

Moreover, $\psi_{\alpha}({\phi^{-1}(\phi(p))}=\psi_{\alpha}(p)$

Similarly for other charts.

Is this correct? How do I fix this?

$\endgroup$

1 Answer 1

3
$\begingroup$

Let us see which differentiable structure is induced on $S^1$ as a submanifold of $\mathbb R^2$. Define $$\phi : (-1,1) \times \mathbb R \to (-1,1) \times \mathbb R, \phi(x,y) = (x, y - \sqrt{1-x^2}).$$ This map is a diffeomorphism (with inverse $\phi^{-1}(x,y) = (x, y + \sqrt{1-x^2})$) such that

  1. $\phi((-1,1) \times (0,\infty)) = (-1,1) \times [0,\infty) \cup$ open lower half disk of $D^2$ = $W_1$.

  2. $\phi(U_1) = (-1,1) \times \{0\}$, where $U_1$ is the open upper half circle of $S^1$.

Thus $\phi : (-1,1) \times (0,\infty) \to W$ is a chart on $\mathbb R^2$ such that $\phi(S^1 \cap (-1,1) \times (0,\infty)) = W \cap (\mathbb R \times \{0\})$.

This gives us the chart $$\phi_1 : U_1 \stackrel{\phi}{\to} W \cap (\mathbb R \times \{0\}) \stackrel{proj}{\to} (-1,1) $$ on $S^1$ as a submanifold of $\mathbb R^2$. You see that this is nothing else than the map in your question. For the other $U_i$ you can apply the same construction.

$\endgroup$

You must log in to answer this question.