Tangent bundle $TM$ is a submanifold of $\mathbb{R}^{2n}$

Question

Let $M$ be a submanifod of $\mathbb{R}^n$. Show that the tangent bundle $TM$ is a submanifold of $\mathbb{R}^{2n}$.

ps.:I'm trying to use charts, but I'm slowing down, because I've been studying this for a long time.

Answer 1

Let $x\in M$ and $(U,\phi)$ be a chart of $M$ around $x$, then let us define $TU\subset TM$ and $T\varphi\colon TU\rightarrow\mathbb{R}^{2n}$ by: $$TU=\{(x,v):x\in U,v\in T_xM\},T\varphi(x,v)=(\varphi(x),\mathrm{d}_x\varphi(v)),$$ then $\{(TU,T\varphi)\}$ is an atlas for $TM$.

Answer 2

Given a local parametrization $h:U \to V \subseteq M \subseteq \mathbb R^n$, we set $h(u)=x$. Then a vector $v \in \mathbb R^n$ is tangent to $M$ at $x$ if and only if it is a linear combination of vectors $\frac{\partial h}{\partial u_i}(u)$ for $a \leq i \leq n$.

Implicitly, this is a subspace of $TM \subset M \times \mathbb R^n$.

Hence, as a subset of $\mathbb R^n \times \mathbb R^n$. Since $h$ was smooth, its derivative is smooth and gives local co-ordinates to the second co-ordinate by taking the appropriate linear combinations (total derivative.)