Let $X$ be a topological space and $\{(U_{\alpha},\phi_{\alpha})\}$ a smooth atlas. $\{(U_{\alpha},\phi_{\alpha})\}$ is maximal if for all $(U,\phi)$ as above satisfying (6), then $(U,\phi) \in \{(U_{\alpha},\phi_{\alpha})\}$
where (6) is:
Let X be a topological space and $\{(U_{\alpha},\phi_{\alpha})\}$ a smooth atlas. Let $(U,\phi)$ be such that $U \subset X$ is open, $\phi:U \rightarrow V \subset \mathbb{R}^n$ a homeomorphism, such that
$$\phi \ \circ \ \phi_{\alpha}^{-1}, \ \ \ \phi_{\alpha} \ \circ \ \phi^{-1} \text{are $C^{\infty}$ where defined.}$$
Then $\{(U_{\alpha},\phi_{\alpha})\} \cup \{(U,\phi)\}$ is again an atlas.
Why do we care about maximal atlases. Surely this almost trivially always exists? Surely we should care more about minimal atlases (just like in GMT where we care about minimal coverings). Furthermore, why can one easily show the uniqueness of a maximal atlas?
Best Answer
1) It trivially always exists when a smooth atlas exists at all (which is not guaranteed).
2) It is often convenient in practice to say "maximal atlas" so that one may choose whatever chart they like for computations, as opposed to one of a pre-ordained set of charts.
3) Given an atlas $\mathcal A$, and two maximal atlases $\mathcal A_i$ containing $\mathcal A$, the union $\mathcal A_1 \cup \mathcal A_2$ is again an atlas, and hence must coincide with both $\mathcal A_1$ and $\mathcal A_2$ by maximality.
4) We do not really care about these. This is just a matter of defining what a smooth manifold (and smooth maps between them) are. You could do that just as well by saying that a smooth structure is an equivalence class of atlases, or that it's defined by what the smooth functions $U \to \Bbb R$ for open subsets $U \subset M$. Whatever choice you do works more or less the same as the others. In the end, we care about the topology and geometry of smooth manifolds, not whatever particular choice of definition we used.