5

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?

PhoenixPerson
  • 717
  • The tangent space is always a vector space whose dimension is the dimension of the manifold, by definition. – Siminore Mar 09 '14 at 18:09
  • The tangent bundle has dimension two times that of the manifold. –  Mar 09 '14 at 18:18
  • @SanathDevalapurkar: That is true and consistent with Siminore's comment. – bradhd Mar 09 '14 at 18:19
  • Note that Minkowski space is not isomorphic to $\mathbb{R}^n$ - rather, a Wick rotation takes Minkowski space to $\mathbb{R}^n$. –  Mar 09 '14 at 18:19

3 Answers3

5

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 $\Bbb R^N$

Vinicius M.
  • 1,184
2

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.

bradhd
  • 6,564
2

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).

GFR
  • 5,401