Maximal Atlases on a Set
 Table of Contents

# Maximal Atlases on a Set

Recall from the Atlases on a Set page that if $M$ is a set then an $m$-dimensional atlas on $M$ is a collection of $m$-dimensional charts $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha}) : \alpha \in \Gamma \}$ (where $\Gamma$ is some indexing set) such that:

• a) $M$ is covered by the charts in $\mathcal A$, i.e., $\displaystyle{M = \bigcup_{\alpha \in \Gamma} U_{\alpha}}$.
• b) For all $\alpha, \beta \in \Gamma$, $(U_{\alpha}, \varphi_{\alpha})$ and $(U_{\beta}, \varphi_{\beta})$ are compatible.

We also said that if $(U, \varphi)$ is a chart on $M$ then $(U, \varphi)$ is said to be compatible with $\mathcal A$ if $(U, \varphi)$ is compatible with every chart in $\mathcal A$. Furthermore, if $\mathcal A$ and $\mathcal A'$ are both atlases on $M$ then $\mathcal A$ is said to be equivalent to $\mathcal A'$ if every chart in $\mathcal A$ is compatible with every chart in $\mathcal A'$.

 Lemma 1: Let $M$ be a set and let $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha} : \alpha \in \Gamma \}$ be an altas on $M$. If $(U, \varphi)$ and $(V, \psi)$ are charts on $M$ such that $(U, \varphi)$ and $(V, \psi)$ are compatible with $\mathcal A$ then $(U, \varphi)$ is compatible with $(V, \psi)$.
• Proof: Let $(U, \varphi)$ and $(V, \psi)$ be charts on $M$ that are compatible with $\mathcal A$. Since $\mathcal A$ is an atlas on $M$ we have that $\displaystyle{M = \bigcup_{\alpha \in \Gamma} U_{\alpha}}$ and so:
(1)
\begin{align} \quad \quad \quad \varphi (U \cap V) = \varphi (U \cap V) \cap \varphi(M) = \varphi (U \cap V) \cap \bigcup_{\alpha \in \Gamma} \varphi(U_{\alpha}) = \bigcup_{\alpha \in \Gamma} \varphi (U \cap V) \cap \varphi (U_{\alpha}) = \bigcup_{\alpha \in \Gamma} \varphi (U \cap V \cap U_{\alpha}) = \bigcup_{\alpha \in \Gamma} \varphi (U \cap U_{\alpha}) \cap \varphi (V \cap U_{\alpha}) \end{align}
• Since $(U, \varphi)$ is compatible with $\mathcal A$ we have that for all $\alpha \in \Gamma$ that $(U, \varphi)$ is compatible with $(U_{\alpha}, \varphi_{\alpha})$, so $\varphi (U \cap U_{\alpha})$ and $\varphi (V \cap U_{\alpha})$ are open sets. From above, we have expressed $\varphi (U \cap V)$ as an arbitrary union of open sets which is open.
• A similar argument can be made to show that $\psi (U \cap V)$ is also open.
• Now take any $\alpha \in \Gamma$. Then $\varphi_{\alpha} \circ \varphi^{-1} : \varphi (U \cap V) \to \varphi_{\alpha} (U \cap V)$ is a diffeomorphism and $\psi \circ \varphi_{\alpha}^{-1} : \varphi_{\alpha} (U \cap V) \to \psi (U \cap V)$ is a diffeomorphism. The composition of two diffeomorphisms is a diffeomorphism and so the following function is a diffeomorphism:
(2)
\begin{align} \quad \psi \circ \varphi^{-1} = ( \psi \circ \varphi_{\alpha}^{-1} ) \circ ( \varphi_{\alpha} \circ \varphi^{-1} ) : \varphi(U \cap V) \to \psi (U \cap V) \end{align}
• So $(U, \varphi)$ and $(V, \psi)$ are compatible. $\blacksquare$
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