I defined the tangent space $T_xX$ of a manifold $X$ at a point $x \in X$ as the equivalence class of curves $c: (-\varepsilon, \varepsilon) \to X$ such that $c(0) =x$ and $c_1 \sim c_2$ if $(\varphi \circ c_1)'(0) = (\varphi \circ c_2)'(0)$ for a chart $(U, \varphi)$with $x \in U$. I know that $T_xX$ has the structure of a vector space.
Now my question: Is there a way to view $T_xX$ as a manifold itself? If yes, how do the charts look like?
Best Answer
If your manifold has dimension $n$, then $T_xX$ is a vector space of dimension $n$, and vector spaces can always be given a manifold structure (of dimension $n$).
However, what you'll see (and probably find more interesting) if you keep studying is that we can define the "tangent bundle":
$$TX=\coprod_{x\in X}T_x X$$
which can be given the structure of a $2n$-dimensional manifold. You can find this in any book on smooth manifolds; I personally like John M. Lee's book.
Edit: If $V$ is an $n$-dimensional vector space over $\Bbb{R}$, it is a fact that any norm on $V$ determines a topology, which is independent of the norm. Therefore $V$ has a natural topology on it, and any vector space isomorphism $\varphi:V\to\Bbb{R}^n$ actually turns out to be a homeomorphism as well, so $\varphi$ determines a smooth chart for all of $V$.