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The tangent space of a circle is a line.

The tangent space of a sphere (in every point) can be thought of as a plane.

Is this a general thing? I mean, having an $n$ dimensional Riemannian manifold, can the tangent space in every point be thought as $\mathbb{R}^n$?

If the answer is yes, does this happen as well with the Lorentzian manifolds of GR? Can the tangent space of any space-time always be regarded as a Minkowski space?

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  • $\begingroup$ The tangent space is always a vector space whose dimension is the dimension of the manifold, by definition. $\endgroup$ – Siminore Mar 9 '14 at 18:09
  • $\begingroup$ The tangent bundle has dimension two times that of the manifold. $\endgroup$ – user122283 Mar 9 '14 at 18:18
  • $\begingroup$ @SanathDevalapurkar: That is true and consistent with Siminore's comment. $\endgroup$ – bradhd Mar 9 '14 at 18:19
  • $\begingroup$ Note that Minkowski space is not isomorphic to $\mathbb{R}^n$ - rather, a Wick rotation takes Minkowski space to $\mathbb{R}^n$. $\endgroup$ – user122283 Mar 9 '14 at 18:19
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The answer is yes, the tangent space at any point of an $n$-dimensional real manifold is isomorphic to $\mathbb{R}^n$. What distinguishes the Riemannian from the Lorentzian case is that in the Riemannian case this vector space comes equipped with a positive definite metric, which can be rescaled so to give the use Euclidean metric, while in the Lorentzian case the tangent space comes equipped with a Lorentzian metric, which can b rescaled to give the Minkowskian metric in $n$ dimensions: diagonal with all pluses and one minus (or the opposite depending on your conventions).

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One can think of a smooth $n$-manifold $M$ embedded in some $\Bbb R^N$ for some $N$, thus, given a point $p \in M$ parametrized by $\psi:U\subset \Bbb R^n \to M$, one of the definitions of the tangent space is the image of $\Bbb R^n$ by $D\psi_{\psi^{-1}(p)}$. Being $D\psi_{\psi^{-1}(p)}$ non-singular ($\psi$ is a diffeomorphism onto its image), this is a $n$-plane in $\Bbb R^N$

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If $M$ is a smooth $n$-dimensional manifold, then for each $p\in M$ the tangent space $T_p M$ is an $n$-dimensional real vector space. This tangent space is therefore isomorphic to $\mathbb R^n$ as a real vector space, though not in a "natural" way, in the sense that $T_pM$ does not have a distinguished basis corresponding to the usual basis of $\mathbb R^n$.

This is true for any smooth manifold, regardless of whether it is equipped with a Riemannian metric or a Lorentzian metric.

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