# Bundle and differential manifold

1. Jan 28, 2015

### Calabi

Hello : let be a differential manifold $$C^{\infty}$$ : $$M$$ of dimension n.
I choose a point p.

In this point I can defined the tangent space. It's a vectoirial space of dimension n, I'll talk about it in a precedent thread, .
This space is in bijection with the derivation space : each derivation associated a real to each fonction $$C^{\infty}$$ define on a neighbourhood of p.
Each derivator is a directionnal derivate.

I defined the tangent bundle as : $$TM = \cup_{p \in M} (\{p\} \times T_{p}M)$$. I wanna demonstrate that this space is a differential $$C^{+\infty}$$ manifold of dimension 2n.

I beginn to define a topology on this space. Let $$(U, \phi)$$ et $$(V, \psi)$$ 2 charts, which are compatible $$C^{\infty}$$ and $$U \cap V \neq \varnothing$$. I defined an open as $$\pi^{-1}(U) = \{ \{p\} \times T_{p}M / p \in U \}$$.

Like all the open define a topology which recover M, all the open I defined with my $$\pi^{-1}$$ on $$TM$$ defined a topology which recover $$TM$$.

Now I defined the same things on $$(V, \psi)$$.

Now lets go back to $$(U, \phi)$$ : I defined :
$$\Phi : \begin{pmatrix} \pi^{-1}(U) \rightarrow \phi(U) \times \mathbb{R}^{n} \subset \mathbb{R}^{2n} \\ (p, X_{p}) \rightarrow (\phi(p), d_{\phi(p)}(X_{p})) \end{pmatrix}$$.

I recall that $$d_{\phi(p)}$$ associated to each vectors of $$T_{p}M$$ a vectors from $$T_{\phi(p)}\mathbb{R}^{n}$$. Like $$\forall x \in \mathbb{R}^{n}, T_{x}\mathbb{R}^{n} \simeq \mathbb{R}^{n}$$, I can identified $$d_{\phi(p)}(X_{p})$$ to an elements of $$\mathbb{R}^{n}$$. With this natural components in the natural base.

I do the same things by defining $$\Psi$$.
And $$(U, \phi)$$ et $$(V, \psi)$$ are arbitrarlly choose.

So 2 question how to demonstrate that $$\Phi$$ is an homeomorphism please?

How to demosntrate that $$\Psi o \Phi^{-1}$$ is a $$C^{\infty}$$ diffeomorphisme please?

Thank you in advance and have a nice afternoon.

2. Jan 28, 2015

### lavinia

On ne doit qu'observer que la dérivée d'une fonction $C^∞$ est $C^∞$ elle-même( par définition de $C^∞$) est que la dérivée d'une fonction de transition est un isomorphism lineare de l'espace tangente a chaque point.

Essayez de montrer que l'espace euclidien est une variété $C^∞$.

Bonne journée.

3. Jan 28, 2015

### Calabi

Salut Lavina. Merci à vous. First by the way : you could replace all the $$+\infty$$ by $$k$$.

I'm gonna wright : $$\Psi o \Phi^{-1}(\phi(p), d_{\phi(p)}(X_{p}) )= (\psi o \phi^{-1}(p) , d_{\psi(p)} o d_{\phi(p)}^{-1}(X_{p}))$$ and I've got : $$\Phi(p, X_{p}) = (\phi(p), d_{\phi(p)}(X_{p})$$.
I know that $$\phi$$ like $$\psi$$ are homeomorphism. And that $$\psi o \phi^{-1}$$ is $$C^{+\infty}$$ because of the $$C^{+\infty}$$ differential manifold structure.

So for the first par of the 2 uplets it's done.

What about $$d_{\psi(p)} o d_{\phi(p)}^{-1}$$ is it an diffeormorphism $$C^{+\infty}$$ please?

And about $$d_{\phi(p)}$$, is it an homeomorphism please?

Thank you in advance and have a nice afternoon.

4. Jan 29, 2015

### lavinia

No. The derivative of a k times differentiable function might only be (k-1) times differentiable. There are examples of continuously differentiable functions whose derivative is nowhere differentiable.

As I said before, the differential of a $C^∞$ diffemomorphism is a $C^∞$ diffeomorphism.

- The differential of a smooth diffeomorphism is a diffeomorphism because it is smooth and invertible.

Last edited: Jan 29, 2015
5. Feb 8, 2015

### Calabi

Hello I wright for the probleme I demonstrate my charts is an homeomorphisme, it's continue and a bijection. The reverse fonction is also a continue bijection. I juste have to demonstrate now that my charts changing $$\Psi o \Phi^{-1}$$ is a $$C^{k-1}$$ diffeomorphism if my manifold is $$C^{k}$$. How could I do please?

Thank you in advance and have a nice afternoon.

6. Feb 8, 2015

### Calabi

How to demonstrate it please? And

Thank you in advance and have a nice afternoon.

7. Feb 8, 2015