My textbook says (without any justification that I can see) that

"since Minkowski spacetime is pseudo-Euclidean, the tangent space $T_P$ at any point $P$ coincides with the manifold itself".

My question is how do we prove/understand this?

What I understand as the tangent space at any point P on a manifold is that it is a Euclidean space with the same dimension. So if we have a 3D Euclidean space, I can see how the tangent space would be the same.

[The textbook is "General Relativity, an Introduction for Physicists" by Hobson pg 115]

  • 1
    $\begingroup$ This is true for affine spaces. The signature is irrelevant. $\endgroup$
    – Qmechanic
    May 20, 2019 at 5:11

1 Answer 1


You can prove it a number of ways so here's a simple-ish one :

Take Minkowski space $M$, with the manifold atlas $(\mathbb{R}^n, \operatorname{Id})$. I think it's no big secret that, for cartesian coordinates, the geodesic equation is

$$\ddot{x}^\mu(\tau) = 0$$

with solution

$$x^\mu(\tau) = x_0^\mu + v_0^\mu \tau$$

If we consider the tangent space at a point $p$ with coordinates $x_0$, any geodesic starting at $p$ and with tangent vector $v^\mu_0$ has this form. What we want is a map from every point of $T_pM$ to $M$ from there. If we take any point $q \in M$ with coordinates $y^\mu$, we have that

$$x^\mu(1) = y^\mu$$

for some $v_0 \in T_pM$. It can be shown fairly easily by considering, component-wise

$$v_0^\mu = y^\mu - x_0^\mu$$

There is a map from some subset of $T_p M$ to $M$. It's not too hard to show the reverse and that overall, this maps $T_p M$ to $M$, giving you some map

$$\phi : T_pM \to M$$

If you also add the Minkowski metric on $M$ as a metric space, we may want to check that $g_p(v,v) = \langle \phi(v), \phi(v) \rangle$, which is just

$$\eta_{\mu\nu} v^\mu v^\nu = (v^0)^2 - \vec{v} \cdot \vec{v}$$


$$\langle \phi(v), \phi(v) \rangle = (x_0^0 - y^0)^2 - (\vec{x}_0 - \vec{y}) \cdot (\vec{x}_0 - \vec{y})$$

Which are the same value indeed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.