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I'm trying to analyze the causal character of the surface $x^2 + y^2 - z^2 = -1$ in Lorentz-Minkowski space $\mathbb{L}^3$, with the convention $\mathrm{diag[1,1,-1]}$, that is $$\langle \left(x_1, y_1, z_1 \right), \left(x_2,y_2,z_2 \right) \rangle = x_1x_2 + y_1y_2 - z_1z_2$$

First I tried to see the general situation of a parametrized surface $\varphi : D \subset \mathbb{R}^2 \stackrel{C^1}{\rightarrow} \mathbb{L^3}$. We have $\mathrm{d}\varphi_p: T_pD \rightarrow T_{\varphi(p)}\mathbb{L}^3$ for a given $p \in D$, and I thought about the pull-back metric $$\varphi^*\langle,\rangle_p: T_pD \times T_pD \rightarrow \mathbb{R} \qquad \left(\varphi^*\langle,\rangle_p\right)(\zeta, \zeta) = \langle \mathrm{d}\varphi_p (\zeta), \mathrm{d}\varphi_p(\zeta)\rangle$$

Where $\zeta = \left(\zeta_1, \zeta_2\right) \in T_pD$. So far, so good. Expliciting $\varphi(u,v) = \left(x(u,v), y(u,v), z(u,v)\right)$, we would get: $$\mathrm{d}\varphi_p(\zeta) = \left(\begin{array}{c c} x_u(p) & x_v(p) \\ y_u(p) & y_v(p) \\ z_u(p) & z_v(p) \end{array}\right) \cdot \left(\begin{array}{c} \zeta_1 \\ \zeta_2 \end{array}\right) = \left(\begin{array}{c} \zeta_1 x_u(p) + \zeta_2 x_v(p) \\ \zeta_1 y_u(p) + \zeta_2 y_v(p) \\ \zeta_1z_u(p) + \zeta_2 z_v(p) \end{array} \right)$$ Proceeding, we would compute $\langle \mathrm{d}\varphi_p (\zeta), \mathrm{d}\varphi_p(\zeta)\rangle$. Now, I don't think I missed anything here in this tedious computation. If someone asks me to, I'll add here the whole computation. Factoring in matrix form, I got: $$\langle \mathrm{d}\varphi_p (\zeta), \mathrm{d}\varphi_p(\zeta)\rangle = \left(\begin{array}{c c} \zeta_1 & \zeta_2 \end{array} \right) \cdot \left(\begin{array}{c c} \langle \varphi_u, \varphi_u \rangle & \langle \varphi_u, \varphi_v \rangle \\ \langle \varphi_u, \varphi_v \rangle& \langle \varphi_v, \varphi_v \rangle \end{array} \right)_p \cdot \left(\begin{array}{c} \zeta_1 \\ \zeta_2 \end{array} \right)$$

Then, the causal character of $\varphi$ would depend on the positive-definiteness, etc, of the middle matrix above. Now, to the specific problem of the surface $x^2 + y^2 - z^2 = -1$. Geometrically it's easy to see that this hyperboloid is spacelike, but I would like to conclude that from my calculations above. One parametrization is $\varphi(u,v) = \left(\sinh u \cos v, \sinh u \sin v, \cosh u\right)$. Then, $\varphi_u(u,v) = \left(\cosh u \cos v, \cosh u \sin v, \sinh u \right)$ and $\varphi_v(u,v) = \left(- \sinh u \sin v, \sinh u \cos v, 0 \right)$, and we have: $$\begin{array}{c}\langle \varphi_u, \varphi_u \rangle = 1 \\ \langle \varphi_u, \varphi_v \rangle = 0 \\ \langle \varphi_v, \varphi_v \rangle = \sinh^2 u\end{array}$$

and so we get the matrix $\left(\begin{array}{c c} 1 & 0 \\ 0 & \sinh^2 u \end{array}\right)$. With this, I would conclude that if $u \neq 0$, then $\sinh^2 u > 0$ and $\varphi$ is spacelike. But, if $u = 0$, then $\sinh^2 u = 0$ and $\varphi$ would be lightlike. Was that supposed to be possible? I don't know if my approach of the theory was wrong, or my geometric vision, or my calculation, or anything at all. If someone can see this for me, I am very grateful. I hope my notation was clear enough.

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No parametrization of the surface is needed for this problem.

A 1-dimensional linear subspace in $\mathbb L^3$ is, by definition, timelike, if the bilinear form $\langle \cdot, \cdot\rangle$ restricted to the subspace is negative definite. A 2-dimensional subspace is spacelike if the bilinear form restricted to it is positive definite. A surface in $\mathbb L^3$ is spacelike if all its tangent spaces are spacelike.

An easy linear algebra exercise is that the orthogonal complement (with respect to the bilinear form) of a timelike 1-dimensional subspace is a 2-dimensional spacelike subspace.

Now the tangent space at a point $p_0=(x_0,y_0,z_0)$ of your surface is the kernel of the differential at $p_0$ of the defining function of your surface, $f(x,y,z)=x^2+y^2-z^2$ (the surface is the $-1$ level set of $f$). The differential of $f$ at $p_0$ is $df(p_0)=2x_0dx+ 2y_0dy-2z_0dz$, hence the equation for the tangent plane at $p_0$ is $x_0x+ y_0y-z_0z=0$. In other words, the tangent space to your surface at $p_0$ is the orthogonal complement of $p_0$. Since $\langle p_0,p_0\rangle=-1<0$, $p_0$ is timelike, hence by the linear algebra exercise above the tangent space is spacelike.

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To answer your specific question: your parametrisation is bad. When $u = 0$ your vector $\varphi_v$ is actually the zero vector. That tells you that your parametrisation is not regular, and the induced metric, expressed relative to this parametrisation, can have singularities that are related to the poor choice of parametrisation rather than to actual causal character of the surface.


To give an example: you can parametrise the unit sphere in $\mathbb{R}^3$ by

$$ \varphi(u,v) = (\sin u \cos v, \sin u \sin v, \cos u) $$

with $u\in [0,\pi]$ and $v\in [0,2\pi)$. Clearly the surface is Riemannian. But if you compute naively the induced metric in the $u,v$ coordinates you find that

$$ g = \begin{pmatrix} 1 & 0 \\ 0 & (\sin u)^2 \end{pmatrix} $$

which is the exact analogue of your problem above. And here your geometric vision should be clear: when $u = 0,\pi$ you are at the north/south pole and there the longitudinal lines all intersect, so $\varphi_v = 0$.

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