# Open Submanifolds

Let $$M$$ be a smooth $$n$$-manifold and let $$U\subseteq M$$ be any open subset. Define an atlas on $$U$$ $$\mathcal{A}_{U}=\big\{\text{smooth charts}\;(V,\varphi)\;\text{for}\; M\;\text{such that}\;V\subseteq U\big\}.$$

I must prove that $$\mathcal{A}_{U}$$ is a smooth atlas for $$U$$, that is

(1) $$U=\bigcup{V}$$, where $$V$$ is the domain of charts such that $$V\subseteq U.$$

This point is ok.

and

(2) It remains to prove that $$\mathcal{A}_U$$ is a smooth atlas for $$U$$.

My attempt. Let $$\big(V_1,\varphi_1\big)$$, $$\big(V_2,\varphi_2\big)\in\mathcal{A}_U$$, since they are smooth charts for $$M$$, that is are charts of maximal atlas of $$M$$, the maps $$\varphi_2\circ\varphi_1^{-1}\colon\varphi_1\big(V_1\cap V_2\big)\to \varphi_2\big(V_1\cap V_2\big)\quad\text{and}\quad \varphi_1\circ\varphi_2^{-1}\colon \varphi_2\big(V_1\cap V_2\big)\to \varphi_1\big(V_1\cap V_2\big)$$ are $$C^{\infty}$$.

Since $$V_1\subseteq U$$, $$V_1=U\cap V_1$$, then $$V_1$$ is open in $$U$$, similary $$V_2$$ is open in $$U$$, then $$\big(V_1,\varphi_1\big)$$ and $$\big(V_2,\varphi_2\big)$$ are charts of $$U$$, morever $$V_1\cap V_2$$ is open in $$U$$, and then they are $$C^{\infty}$$ compatible.

Question It's correct?

• I just don't see why you need to add that $V_1\cap V_2$ is open at the end, but this proof is correct. May 5 '19 at 9:53

Your proof is correct, but I think it is uncessary to prove (2). If $$\mathcal A$$ denotes the maximal smooth atlas of $$M$$, then any subset $$\mathcal A' \subset \mathcal A$$ is automatically a smooth atlas on $$M' = \bigcup_{(V,\varphi) \in \mathcal A'} V$$ which is an open subset of $$M$$.

It is perhaps worth to mention that $$\mathcal A_U$$ is a maximal smooth atlas of $$U$$. To see this, consider any smooth atlas $$\mathcal B$$ on $$U$$ containing $$\mathcal A_U$$.

Let $$(W,\psi)$$ be any chart in $$\mathcal B$$. It is compatible with all charts in $$\mathcal A_U$$. Now let $$(\varphi,V) \in \mathcal A$$. Its restriction $$(\varphi' = \varphi \mid_{V \cap U}, V' = V \cap U)$$ also belongs to $$\mathcal A$$, and since $$V \cap U \subset U$$, it belongs to $$\mathcal A_U$$. Since $$W \cap V' = W \cap V \cap U = W \cap V$$, the charts $$(W,\psi)$$ and $$(\varphi,V)$$ have the same transition function as the charts $$(W,\psi)$$ and $$(\varphi',V')$$. The latter is smooth, which shows that $$(W,\psi)$$ is compatible with $$(\varphi,V)$$.

Hence $$(\psi,W)$$ is compatible with $$\mathcal A$$ and we conclude $$(\psi,W) \in \mathcal A$$. But this shows $$(\psi,W) \in \mathcal A_U$$ because $$W \subset U$$.

Therefore $$\mathcal B = \mathcal A_U$$.

• @PaulFrostThanks for your answer. But a priori we cannot say that $\mathcal{A}_U$ is a maximal smooth atlas for $U$, but only that it is contained in a unique maximal smooth atlas or not? May 5 '19 at 14:41
• It is a maximal smooth atlas, but this requires a short proof. If you add a chart $(V,\psi)$ with $V \subset U$ which does not belong to $\mathcal A_U$, then certainly $(V,\psi) \notin \mathcal A$ which means that you find a non-smooth transition function with a chart in $\mathcal A$. You have to show that you can also find a chart in $\mathcal A_U$ with the same "defect". May 5 '19 at 14:52
• @PaulFrostI believe I have found a way to show that $\mathcal{A}_U$ is a smooth maximal atlas: let's suppose it is absurd that $\mathcal{A}_U$ is not a smooth maximal atlas, then exists a smooth atlas $\mathcal{B}$ such that $\mathcal{A}_U\subset \mathcal{B}$, therefore exists $(W,\psi)\in\mathcal{B}\setminus \mathcal{A}_U$ with $W\cap U\ne W.$ A chart $(V,\varphi)\in\mathcal{A}_U$ must be compatible with $(W,\psi)$, but $V\subseteq U$, then $V\cap W\subseteq U\cap W\ne W$, absurd. It's correct? May 6 '19 at 14:17
• Not quite correct. $\mathcal B$ is a smooth atlas for $U$, thus for each $(W,\psi) \in \mathcal B$ you have $W \subset U$ and therefore $W \cap U = W$. I shall edit my answer. May 6 '19 at 15:13