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?

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    $\begingroup$ @Test123 So the answer would be yes and the charts are global i.e. $(\mathbb{R}^n, Id)$? $\endgroup$ – JDoe Feb 1 '17 at 12:53
  • $\begingroup$ The tangent space at a point topologically is homeomorphic to $\mathbb{R}^n$ let's say via $\phi$. There will be one chart $(\mathbb{R}^n,\phi)$. $\endgroup$ – Kal S. Feb 1 '17 at 12:57

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 $V$.

  • $\begingroup$ I see. Where can I find the result that a vector space can be give n a manifold structure? $\endgroup$ – JDoe Feb 1 '17 at 12:56
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    $\begingroup$ Lee's book, example 1.24. I'll summarize the argument above in a moment $\endgroup$ – Alex Mathers Feb 1 '17 at 12:58
  • $\begingroup$ Alex Mathers, do we need norm? I know only a little functional analysis, but I know inner product implies norm implies metric space. From elementary topology, metric is enough to imply a topology. Is norm used in manifold aspects besides the topology of the manifold such as the smooth/differentiable atlas? I skimmed in Example 1.24 of Lee and did not see any particular reason to use norm. $\endgroup$ – Selene Auckland Apr 17 at 11:04
  • $\begingroup$ Also, are finite real vector space isomorphisms diffeomorphisms just as they are homeomorphisms? $\endgroup$ – Selene Auckland Apr 17 at 11:04

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