natural manifold structure on the cotangent bundle If $M$ is a differential manifold with an atlas $\{(U,\varphi)\}$ and $T^*M$ is the cotangent bundle. Then what is the natural atlas on $T^*M$?
I would be inclined to say something like this:
$\{(V,\psi)\}$ where $V:=U\times \mathbb{R}^n$ and $\psi:(u,f)\mapsto (\varphi(u),f(\varphi^{-1}(e_1)),\cdots,f(\varphi^{-1}(e_n))$
 A: *

*Given the basis $\{\partial^i\}_i$ for $T_pM$ corr. to some chart $(U, \varphi)$ then $\{dx^i\}_i$ is a basis for $T^*_pM$. We have a natural bijection $\Phi: T_p^*M \to \varphi(U) \times \mathbb{R}^{\textbf{dim}(M)}$. 

*If we let $\tau$ be the topology on the range space then we can induce a topology on $T_p^*M$

*Given this bijection, we can write any element in $T_p^*M$ as a pair $(p, \xi)$ where $\xi \in T_p^*M$ i.e $\xi = \sum_i \xi_i \ dx^i$. 

*Therefore, we can take as our charts:
$$\Phi: (T_p^*M, \tau') \to (\varphi(U) \times \mathbb{R}^{\textbf{dim}(M)}, \tau)$$
$$ (p, \xi) \mapsto (\varphi(p), \xi(\partial^i_p))$$
A: In some cases, we can write the cotangent bundle explicitly in terms of polynomials (or whatever type of function we choose).  If we have a hypersurface $f(x,y) = x^2 + y^2 = 1$ Then the cotangent bundle looks like this:
$$ \big\{  \big((x,y), (dx,dy)\big): x^2 + y^2 = 1 \text{ and }\; 2x \, dx + 2y \, dy = 0 \big\} $$
In the case of a circle there are two charts:


*

*$(x,y) \mapsto x$ (projection to $y$-axis)

*$(x,y) \mapsto y$ (projection to $x$-axis)


These are not defined everywhere, since these projection functions are not differentiable at certain places.


*

*$(x,y) \mapsto x$ (projection to $x$-axis) $x \in (-1,1)$

*$(x,y) \mapsto y$ (projection to $y$-axis) $y \in (-1,1)$


As long as we're away from the axes, there are transition functions:
$$ x = \sqrt{1 - y^2} \text{ or } y = \sqrt{1 - x^2}$$
These things might lift to transition functions of the cotangent bundle:
\begin{eqnarray*}
x &=& \sqrt{1 - y^2 }\\
2x \, dx &=& 2y \, dy
\end{eqnarray*}
and we can imagine the that the charts are $(-1,1) \times \mathbb{R}$.  This might be unconventional and hopefully it's accurate.
