1
$\begingroup$

In my class notes a differentiable manifold is defined as an ordered pair $(M, \mathcal{A}),$ where $M$ is a set and $\mathcal{A}$ is a maximal atlas of $M.$ There are also the following definitions:

  • We say that two atlases $\mathcal{A}_1$ and $\mathcal{A}_2$ of $M$ are compatible if $\mathcal{A}_1 \cup \mathcal{A}_2$ is an atlas.
  • The maximal atlas $\mathcal{A}^+$ associated with an atlas $\mathcal{A}$ of $M$ is the union of all atlases compatible with $\mathcal{A}.$

My question is: Is it true that $\mathcal{A}_1$ and $\mathcal{A}_2$ are compatible if and only if $\mathcal{A}_1^+ = \mathcal{A}_2^+?$

$\endgroup$
2
$\begingroup$

Yes. If $\mathcal{A}_1$ and $\mathcal{A}_2$ are compatible, then $\mathcal{A}_2 \subset \mathcal{A}_1^+$. Hence $\mathcal{A}_2$ and $\mathcal{A}_1^+$ are compatible and we conclude $\mathcal{A}_1^+ \subset \mathcal{A}_2^+$. Similarly $\mathcal{A}_2^+ \subset \mathcal{A}_1^+$. Conversely let $\mathcal{A}_1^+ = \mathcal{A}_2^+ = \mathcal{A}$. Then $\mathcal{A}_1, \mathcal{A}_2 \subset \mathcal{A}$, hence $\mathcal{A}_1 \cup \mathcal{A}_2 \subset \mathcal{A}$. Since $\mathcal{A}$ is an atlas, also the smaller $\mathcal{A}_1 \cup \mathcal{A}_2$ is one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy