¿Does a timelike non-orientable surface exist? Consider the Lorentz-Minkowski space of dimension $3$, $\mathbb{L}^3 = (\mathbb{R}^3,\langle \: \cdot \: \rangle)$
$$
\langle u,v\rangle=u_1v_1 + u_2v_2 - u_3v_3
$$
We say that a surface $S$ is timelike if for every point $p$, its tangent space $T_pS$ contains a timelike vector, ie a vector $w$ such that $\langle w, w\rangle < 0$. I am trying to find an example of such a surface which is not orientable however I think that this is not possible and the clues which lead me to believe this are the following results:


*

*If $S$ is compact then it cannot be timelike and all non-orientable surfaces I know of in $\mathbb{R}^3$ are compact.

*I see no problem with finding a unit normal vector field $N$ by simply parametrizing the surface with $\varphi:\mathbb{R}^2 \rightarrow \mathbb{L^3}$ and defining
$$
N=\frac{\varphi_u \times \varphi_v}{||\varphi_u\times\varphi_v||}
$$
This vector field should be smooth.


If you know of an example of such a surface or help me prove that they don't exist it would help me out a lot. 
 A: Equip the space $\mathbb L^3$ with $x$-, $y$-, and $t$-axes,
where the $t$ axis is parallel to the vector $(0,0,1),$
that is, it is the time axis.
I propose the following embedding of a Möbius strip.
To construct the embedding, first construct an oval figure in the $x,t$ plane consisting of two long flat strips parallel to the $t$ axis, connected by a half-annulus on the "upper" end and connected by another half-annulus on the "lower" end. Now replace one of the strips by a strip with a half-twist, parallel to the $x,t$ plane at either end. 
The strip must be narrow enough that the normal to the surface is always space-like, so the tangent space at that point contains at least one time-like vector.
I think this surface has the property that for any path that makes a full circuit around the surface (thereby exhibiting the fact that it is unorientable), at least part of the path must be in a timelike direction.
I believe that in order to be non-orientable, a surface needs every path that exhibits its lack of orientation to have a timelike portion.

By the way, a simple compact surface that is timelike everywhere (though also orientable) is a flat disk parallel to the $t$ axis.
The first step in my construction (before inserting the half twist)
is cut from such a disk.

If you add the restriction that every closed path on your surface must either be able to shrink to a point or be able to be deformed to a purely spacelike path, then I think it is true that the surface must be orientable.
