Causal structure, time orientability and equivalence classes

Question

Quoting from this Wikipedia article, if $(M,g)$ is a Lorentzian manifold then the tangent vectors at each point in the manifold can be classified into three different types. Using a $(+,-,-,-)$ metric signature, a tangent vector $X$ is

• timelike if $g(X,X)>0$
• null if $g(X,X)=0$
• spacelike if $g(X,X)<0$

The article then states that if $X$ and $Y$ are two timelike tangent vectors at a point $P$ of $M$, then we say that $X$ and $Y$ are equivalent (written $X\sim Y$) if $g(X,Y)>0$. It turns out (this is related to my question) that for each point there are two equivalence classes, which between them contain all timelike tangent vectors at that point. We then call one of these equivalence classes "future-directed" and the other "past-directed".

Question: How do we know that there are exactly two equivalence classes at each point? This could be stated mathematically as

For any three timelike tangent vectors $X$, $Y$ and $Z$ to a point $P$ of a Lorentzian manifold $(M,g)$, if we have $X\nsim Y$ and $X\nsim Z$, is it necessarily true that $Y\sim Z$? (using the equivalence relation defined earlier)

Disclaimer: I suspect this is a rather trivial result of the topology and geometry of Lorentzian manifolds, but I know very little of the mathematics of these two fields, so if your answer uses any terms other than those defined here then I'd greatly appreciate it if you could provide a definition of those terms in your answer.

Answer 1

Judging from the linked wikipedia article I suppose the purpose of this question is to understand the notion of time orientability. My answer will focus on the geometric picture behind it. I will closely follow these excellent lecture notes (note the different sign convention).

At each point $p \in M$ the tangent space $T_p M$ is clearly isometric to regular Minkowski spacetime. We can thus choose an orthonormal basis $\{E_{\alpha}\}$ such that ${g(E_\alpha, E_\beta)} = \eta_{\alpha \beta}$. For a null vector $X$ we now find: $$ g(X,X) = 0 \iff (X^0)^2 = (X^1)^2 +(X^2)^2 +(X^3)^2, $$ thus the set of all null vectors at point $p$ $$ \mathcal{N}_p = \{X \in T_p M : g(X,X) =0 \} $$

geometrically corresponds to a double cone. The set of timelike vectors at $p$

$$ \mathcal{I}_p = \{X \in T_p M : g(X,X) > 0 \} $$

is the interior of the solid double cone enclosed by $\mathcal{N}_p$. We can thus conclude that $\mathcal{I}_p$ is an open disconnected set with two components, which we denote by $\mathcal{I}_p^+$ (future directed timelike vectors) and $\mathcal{I}_p^-$ (past directed timelike vectors). These two open sets correspond to the two equivalence classes.

The time orientation of $M$ can now be defined as a continuous assignment of $\mathcal{I}_p^+$ for all points $p$.

Answer 2

Well, by a process of elimination, if you have your time orientation (a vector field $\tau^\mu$), and you have two equivalence classes, for $g(X,\tau) > 0$ and $g(Y, \tau) <0$, the only remaining possibility for a third class is that $g(Z,\tau) = 0$. But any vector orthogonal to a timelike vector will be spacelike.

It can be proven thusly : if you have a timelike vector $u$ and a vector $v$ such that $g(u,v)$, it is possible to rewrite the basis of the tangent space $(x,y,z,t)$ as $(u,v,x,y)$. The new metric will be $\text{diag}(g(u,u), g(v,v), g(x,x), g(y,y))$. $x$ and $y$ are spacelike, so their norm will be negative. $g(u,u)$ is positive. By Sylvester's law, a change of basis cannot change the signature of the metric, so $v$ has to be a spacelike vector as well.