Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

$A$ 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.

share|improve this answer

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

share|improve this answer
    
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 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 at 12:02
1  
Exact quote: "... with the smooth structure determined by the atlas consisting of the single chart..." –  Martín-Blas Pérez Pinilla Mar 31 at 12:05
1  
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 at 20:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.