# Example of differentiable manifold of class C^1

1. Nov 3, 2013

### mnb96

Hello,

I read from several sources the statement that the set of points M$\in$ℝ2 given by $(t, \, |t|^2)$ is an example of differentiable manifold of class C1 but not C2.

Is that true?
To be honest, that statement does not convince me completely, because in order for M to be a manifold, we should be able to find an atlas of charts $x_i:U_i \rightarrow M$ such that all the points of M are covered by the atlas.
So how do we cover the point (0,0) ?

Do we need to use 3 charts as follows?

$$x_1(t) = (t,\, t^2) \quad \, t\in(0,+\infty)$$
$$x_2(t) = (t,\, -t^2) \quad \, t\in(-\infty, 0)$$
$$x_3(t) = (t,t^2) \quad \, t\in(-1,1)$$

Last edited: Nov 3, 2013
2. Nov 3, 2013

### Mandelbroth

I don't understand what you're doing. For real numbers $t$, $|t|^2=t^2$, right?

3. Nov 3, 2013

### mnb96

Sorry, I reported the wrong equation for the curve . What I meant was that the points of M are given by $(t, \, \mathrm{sign}(t)t^2)$

4. Nov 3, 2013

### WWGD

Why don't you use a $C^1$ function f that is not $C^2$ and try to find a manifold whose transition functions ae given by f ?

Only one I can think of now is $$f(x)= \frac {-x^2}{2}; x\leq 0$$ and

$$f(x) =\frac{x^2}{2} ; x>0$$

We have f'(x)=|x| , which is not differentiable. I'm not sure this will work; just an idea. Seems like an interesting question: given a (finite) collection of functions, can I find a manifold for which the transition functions are precisely those functions? I'm thinking this is the way one can construct bundles by choosing cocycles satisfying certain properties; can we do something similar for manifolds?

5. Nov 4, 2013

### mnb96

Hi WWGD,

I think the function you suggested is the same as the one I am considering, up to a scalar factor of 1/2.
In my first post I gave three coordinate charts x1, x2, x3 that I regarded as possibly correct candidates to form an atlas.

What I missed were the transition maps.
In this case, it seems that the transition maps are always the identity function on some domain: $$\tau_{1,3}(t)=\tau_{3,1}(t)=t \quad ; \, t\in(0,1)$$
$$\tau_{2,3}(t)=\tau_{3,2}(t)=t \quad ; \, t\in(-1,0)$$

If this last step is correct, it should prove that the curve is indeed a $C^1$-differentiable manifold (but not $C^2$).

Last edited: Nov 4, 2013