# Example of a differentiable structure on R

1. Jul 14, 2010

### mordechai9

Consider the manifold of the real-line R with a differentiable structure generated by the map $$x^3 : M \rightarrow \mathbb{R}$$. This example is given in a textbook I'm looking at, but I don't really understand how this can work. The inverse map is clearly not smooth at x=0.

Do they mean that at points like x=0 you have to take different charts, centered at other points besides x^3=0 in the image or something?

(Edited to say smooth instead of continuous.)

Last edited: Jul 14, 2010
2. Jul 14, 2010

### quasar987

Why do you think x^{1/3} not continuous at x=0? As x approaches from the right, x^{1/3} goes to 0, and as x approaches from the left, x^{1/3} goes to 0.

But maybe that was a typo and you meant that the derivative of x^{1/3} is not defined at x=0? That is true, and it is not a problem; it only means that the smooth structure x-->x^{1/3} on R is a different one from the smooth structure x-->x.

3. Jul 14, 2010

### mordechai9

Ah yea, sorry, I meant to say *smooth*, not continuous.

Regarding your answer -- I don't understand, or maybe I am not understanding what it means for this map to "generate" the differentiable structure. As I understand it, this map x^3 will act just like a chart; so therefore it must be a smooth homeomorphism -- but then that is clearly not true, like we've discussed. What do you mean that the smooth structure is just different? Thanks--

4. Jul 15, 2010

### quasar987

Ok, say you have a topological manifold M. That is, a space locally homeomorphic to R^n, Hausdorff, 2nd countable and whatnot. Then a smooth atlas on that topological manifold is a choice of charts of M that are smoothly compatible: that is, charts such that the transition functions between those charts are diffeomorphisms (smooth with smooth inverse). Now a smooth structure on M is a maximal smooth atlas: that is, a smooth atlas A such that if you try to add any chart of M that is not already in A, then there will be some chart of A with which is will not be smoothly compatible.

And of course any smooth atlas for M is contained in a unique smooth structure: such add all the charts that are smoothly compatible with each other and with those of A. Thus, if an atlas A is contained in a smooth structure S, then A is said to generate S.

Note that to tell whether or not two smooth atlases A, A' generate the same smooth structure, we just have to check wheter or not all the charts of A are smoothly compatible with all the charts of A'.

So! What we have here is 2 atlases for the manifold M=R: A={f:R-->R: x-->x} and A'={g:R-->R:x-->x³}. In particular, they are both smooth atlases since the smooth compatibility conditions is trivially satisfied since both atlases are only made of one silly chart! To check wheter or not A and A' determine the smooth structure, we then have to verify that the transition functions g o f^{-1}(x) = x³ and f o g^{-1}(x)=x^{1/3} are smooth. The first is, but the second is not, as you noted.

5. Jul 15, 2010

### mordechai9

OK, thank you for that review. For my own edification I will make a few comments or remarks now. I thought that a chart just had to be smooth -- I didn't realize that we actually only require the weaker condition that the chart only needs to be smoothly compatible.

Also, I suspected that the atlas generated by the map x^3 would have more charts than just x^3 in the atlas. For example, take a new map $$x': M \rightarrow \mathbb{R}$$ which is just a translation of the old mapping, defined by $$x' = x^3 + p$$, where $$p \in \mathbb{R}$$. It looks like this chart is smoothly compatible with x^-1, since $$x^{-1} \circ x' = (x^3 + p)^{1/3}$$ is differentiable everywhere. Also the other direction $$x'^{-1} \circ x = (x^3 - p)^{1/3}$$ looks to be differentiable at all points in M.

I thought that we would need to check the smoothness of all these maps generated by x but I guess that's not necessary, you just only allow the charts in the atlas to be exactly the ones that are smoothly compatible with the original chart.

6. Jul 15, 2010

### quasar987

Actually, it does not make sense in a general topological manifold M to speak of the smoothness of a chart U-->R^n. Differentiability is something that makes sense for maps between normed vector spaces.

Call $\phi:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto x^3$ and $\psi:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto x^3+p$. To check that these charts are smoothly compatible, you must check that $\psi\circ\phi^{-1}:x\mapsto x+p$ and $\phi\circ\psi^{-1}:x\mapsto x-p$ are smooth. Note that that you wrote the opposite: you checked that $\psi^{-1}\circ\phi$ and $\phi^{-1}\circ\psi$ are smooth (which they are not by the way; there is a problem at x=±p^{1/3}.)