# Tangent space of pseudo-Euclidean manifold

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]

• This is true for affine spaces. The signature is irrelevant. – Qmechanic May 20 '19 at 5:11

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

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

and

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

Which are the same value indeed.