Identification of $T_v(T_pM)$ with $T_pM$ In some passages of Do Carmo's Riemannian geometry book he identify $T_v(T_pM)$ with $T_pM$, my question: How one see $T_pM$ as a manifold? who is the atlas? What is the expression of a vector $x \in T_v(T_pM)$ in local coordinates?
 A: Since $T_pM$ is a real vector space, it is isomorphic to $\mathbb{R}^n$ (where $n = \dim M$).  The atlas on $T_pM$ is just the atlas on the standard $\mathbb{R}^n$.  A choice of local coordinates induces an isomorphism $T_pM\leftrightarrow\mathbb{R}^n$, and under this isomorphism you see that a vector $x\in T_v(T_pM)$ is just a vector from multivariable calculus or real analysis with tail at $v$.
The important point here is that $T_pM$ is a finite-dimensional real vector space, which carries a natural manifold structure.
A: The vector space $\mathbb{R}^n$ has a standard structure as a smooth manifold. The tangent space $T_p \mathbb{R}^n$ is also an $n$-dimensional vector space, which can be identified with $\mathbb{R}^n$ by classical multi-variable calculus.
In fact, the entire tangent bundle $T \mathbb{R}^n$ is isomorphic as a smooth manifold to $\mathbb{R}^n \times \mathbb{R}^n$. $T_p \mathbb{R}^n$ is the subspace $\{p\} \times \mathbb{R}^n$.
In an $n$-dimensional differentiable manifold $M$, the space $T_pM$ is isomorphic to $\mathbb{R}^n$. For a system of coordinates, the vectors $\partial/\partial x^i$ give an explicit basis for the vector space $T_pM$, and thus an isomorphism $\mathbb{R}^n \to T_pM$. By the above, there is thus an identification of $T_v(T_p M)$ with $T_p M$, and it doesn't depend on the basis chosen for $T_p M$.
More generally, the tangent bundle $TM$ is sewn together from local copies of $T \mathbb{R}^n$ in exactly the same way that $M$ is sewn together from local copies of $\mathbb{R}^n$.
$T_v TM$, or generally $TTM$ is actually interesting thing to consider, but my differential geometry isn't strong enough to give an explanation.
