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I am confused as to how $\mathscr A=\{(\mathbb R^n,id)\}$ forms a smooth structure on $\mathbb R^n$.

Here's why.

Let $U$ be the open ball of unit radius centered at origin of $\mathbb R^n$ and define $\varphi:U\to U$ as $\varphi(x)=x$ for all $x\in U$.

Then $(U,\varphi)$ is a chart on $\mathbb R^n$.

Further, $(U,\varphi)$ is smoothly compatible with $(\mathbb R^n,id)$.

But then $(U,\varphi)$ should lie in $\mathscr A$, which is doesn't.

What am I doing wrong?

caffeinemachine
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2 Answers2

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$\mathscr A$ is an atlas but no a maximal atlas.

Martín-Blas Pérez Pinilla
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  • In John M. Lee's 'Introduction to Smooth Manifolds', version 3.0, on pg 11, it says that ${(\mathbb R^n,id)}$ is a smooth structure on $\mathbb R^n$. In his book, a smooth structure is defined as a 'maximal smooth atlas'. – caffeinemachine Mar 31 '14 at 11:56
  • Maximal in the sense of "there is no $\mathscr B$ with $\mathscr A\subset\mathscr B$..." – Martín-Blas Pérez Pinilla Mar 31 '14 at 12:02
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    Exact quote: "... with the smooth structure determined by the atlas consisting of the single chart..." – Martín-Blas Pérez Pinilla Mar 31 '14 at 12:05
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    I just want to point out that the "version 3.0" of my book that you're looking at is a prepublication draft which is full of errors, and which somebody posted on the internet illegally and without my permission. It seems to have gone viral, which is really unfortunate, because it's full of errors. If you're trying to learn differential geometry from that version, I expect you're going to run into lots of problems. – Jack Lee Apr 23 '14 at 20:56
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$A$ is an atlas of one chart. The smooth structure induced by $A$ is $A$'s equivalence class. We call two atlaces equvalent iff their union is an atlas.

Edit: In terms of your texbook, they mean the atlas you get when you include all compatible charts.

superAnnoyingUser
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