# Every differentiable atlas is in an unique maximal atlas

I want to proof the following statement:

Every differentiable atlas is in an unique maximal atlas.

Proof:

Let $\mathcal{A}=\{h_\alpha: U_\alpha\to U'_\alpha\subset\mathbb{R}^n|\alpha\in A\}$ be an arbitrary differentiable atlas of $M$.

Define an atlas $\mathcal{A}':=\mathcal{A}\cup\{\underbrace{h_{\beta\alpha}}_{=:h_\beta\circ h^{-1}_\alpha}:h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)\}$.

I have to show, that $\mathcal{A}'$ is maximal. Therefore every map of $M$ in $\mathcal{A}'$ has to be compatible with every map of $\mathcal{A}'$.

The transition map $h_{\beta\alpha}:h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)$ is a diffeomorphism, since $\mathcal{A}$ is a differentiable atlas. Hence two arbitrary maps are compatible, and $\mathcal{A}'$ is maximal.

By construction $\mathcal{A}'$ is unique.

What do you think? Is my proof correct?

Thanks in advance for any comment.

The set $\mathcal{A}'$ is not an atlas because the maps $h_{\beta \alpha}$ are not charts, they're transition maps (you can see that they are maps from subsets of $\mathbb{R}^n$ to $\mathbb{R}^n$ rather from subsets of $M$ to $\mathbb{R}^n$). Additionally, saying "by construction $\mathcal{A}'$ is unique" only works if you show that somehow you were forced to construct $\mathcal{A}'$ in that way.

What you want to show is that the maximal atlas containing $\mathcal{A}$ is the set of all charts that are compatible with $\mathcal{A}$ (all the transition maps are diffeomorphisms).

• Thank you. Therefore I should define $\mathcal{A}':=\mathcal{A}\cup\{h_\beta|h_\beta\circ h^{-1}_\alpha:(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)\quad\text{is a diffeomorphism for every}\,h_\alpha\in\mathcal{A}\}$ and now show, that this is truly a maximal atlas, right? – Cornman Apr 25 '17 at 23:50
• You can take $\mathcal{A}' = \{ h_\beta \mid h_\beta \circ h_\alpha :h_\alpha(U_\alpha \cap U_\beta) \to h_\beta(U_\alpha \cap U_\beta) \text{ is a diffeomorphism for all } h_\alpha \in \mathcal{A} \}$. You don't need to include $\mathcal{A}$ because everything in $\mathcal{A}$ is compatible with everything else in $\mathcal{A}$ by definition. You also missed an $h_\alpha$ and included an extra 'and' in your definition. One should also specify somewhere in the definition that $h_\beta : U_\beta \to \mathbb{R}^n$ is a chart on $M$. – Trevor Gunn Apr 26 '17 at 0:03
• That makes sense. I am going to write a new proof based on this set later. Thanks! – Cornman Apr 26 '17 at 0:08

Every differentiable atlas is in an unique maximal atlas.

Proof:

Let $\mathcal{A}$ be a differentiable atlas. For every homeomorphism $h_\beta: U_\beta\to U'_\beta\subset\mathbb{R}^n$ define

$\mathcal{A}':=\{h_\beta|\, h_\beta\circ h^{-1}_\alpha\colon h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)\,\,\text{is a diffeomorphism for all}\, h_\alpha\in\mathcal{A}\}$.

Obviously $\mathcal{A}'$ is a differentiable atlas, and every $h_\beta\in\mathcal{A}'$ is compatible with every $h_\alpha$ by definition.

Therefore $\mathcal{A}'$ is a maximial atlas.

It seems like with this definition of $\mathcal{A}'$ everything becomes trivial. Thanks in advance for every comment.

• Being maximal and being unique are easy with this definition. Being an atlas is not. What you need to show is that if $h_\alpha$ is compatable with $\mathcal{A}$ and $h_\gamma$ is compatable with $\mathcal{A}$ then $h_\alpha$ is compatible with $h_\gamma$. To do this, you look at each point in $U_\alpha \cap U_\gamma$ and find a chart $h_\beta \in \mathcal{A}$ in a neighbourhood of that point. Let $V = U_\alpha \cap U_\beta \cap U_\gamma$ and then look at the transition maps $h_\alpha(V) \to h_\beta(V) \to h_\gamma(V)$. – Trevor Gunn Apr 26 '17 at 17:16
• Could you explain what do you mean with the transition maps $h_\alpha(V)\to h_\beta(V)\to h_\gamma(V)$? We defined a transition map $h_{\beta\alpha}:=h_\beta\circ h^{-1}_\alpha: h_\alpha (U_\alpha\cap U_\beta)\to h_\beta (U_\alpha\cap U_\beta)$. – Cornman Apr 26 '17 at 17:40
• These maps: $$h_\alpha(V) \xrightarrow{h_\beta \circ h_\alpha^{-1}} h_\beta(V) \xrightarrow{h_\gamma \circ h_\beta^{-1}} h_\gamma(V).$$ – Trevor Gunn Apr 26 '17 at 17:42
• Thank you. I try to proof this with your hint now, and will edit it above. – Cornman Apr 26 '17 at 17:47