I want to prove that there exist a non time-orientable Lorentzian metric on the manifold $M=\mathbb{S}^1\times\mathbb{R}.$ I have an idea on how to do this but since it's the first time I approach this kind of exercises I would like to have a check of my reasoning…
Let $\theta$ be the coordinate on $\mathbb{S}^1$ and $t$ be the coordinate on $\mathbb{R}:$ my intuition suggests to consider the metric given by the matrix
$$\begin{pmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{pmatrix}$$
on the tangent space $T_{(\theta,t)}M$ for each $\theta$ and $t:$ this metric is Lorentzian, since for each choice of the coordinates this matrix has signature $(1,-1).$
I know that a metric is time-orientable if and only if there exists a vector field $X$ on $M$ such that $X_p$ is timelike for each $p\in M:$ to conclude I then have to prove that such a $X$ cannot exist. Working in coordinates such an $X=(X_1,X_2)$ should verify the relation
$$(X_1,X_2)\begin{pmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{pmatrix}\begin{pmatrix}X_1\\X_2\end{pmatrix}<0$$
for each $\theta$ and $t$ (the metric is $t-$indipendent but $X_1,X_2$ are functions of $\theta$ and $t$): a simple computation then gives
$$(X_1^2-X_2^2)\cos\theta+2X_1X_2\sin\theta<0$$
The function on the LHS is continuous: since for $\theta=0$ and $\theta=\pi$ we respectively get
$$X_1^2-X_2^2$$
$$X_2^2-X_1^2$$
we find that this value cannot alway be negative, hence a timelike vector field $X$ cannot exists.
Is this reasoning correct or are there some mistakes? In this case how can I correct them?
Best Answer
If $(M,g)$ is a spacetime and $\Gamma$ is a group of isometries acting freely and properly discontinuously on $M$, the quotient $M/\Gamma$ inherits a Lorentzian metric, which is time-orientable if and only if all isometries in $\Gamma$ preserve the time-orientation.
So, consider the Lorentzian metric $$g = \cos(2\pi x) ({\rm d}x^2-{\rm d}y^2)-2\sin(2\pi x)\,{\rm d}x\,{\rm d}y$$on the plane $\Bbb R^2$. The resulting $(\Bbb R^2,g)$ is time-orientable because the vector field $$X_{(x,y)} = \sin(\pi x)\frac{\partial}{\partial x}\bigg|_{(x,y)} +\cos(\pi x)\frac{\partial}{\partial y}\bigg|_{(x,y)}$$is globally defined and (unit) timelike. Now, consider the isometry $f\colon \Bbb R^2\to \Bbb R^2$ given by $f(x,y) = (x+1,y)$. Given $(x,y)\in \Bbb R^2$, we compute the $g$-product at the image point $f(x,y)$ between
$${\rm d}f_{(x,y)}(X_{(x,y)}) = \sin(\pi x)\frac{\partial}{\partial x}\bigg|_{(x+1,y)} +\cos(\pi x)\frac{\partial}{\partial y}\bigg|_{(x+1,y)} $$
and
$$X_{(x+1,y)} = -\sin(\pi x)\frac{\partial}{\partial x}\bigg|_{(x+1,y)} -\cos(\pi x)\frac{\partial}{\partial y}\bigg|_{(x+1,y)}. $$
Since it equals
$$\cos(2\pi x)(-\sin^2(\pi x)+\cos^2(\pi x)) - \sin(2\pi x)(-2\sin(\pi x)\cos(\pi x))=1>0,$$the metric induced in the quotient $\Bbb R^2/\Bbb Z = \Bbb S^1\times \Bbb R$, for $\Bbb Z \cong \langle f^n\rangle_{n\in \Bbb Z}$, is not time-orientable.
After a moment of thought, I realize I should explicitly say what result I am using here: if you define an isometry between spacetimes to be orthochronous if it takes future-directed timelike vectors to future-directed timelike vectors, then the following criterion holds: if $(M,g)$ is a spacetime, $X$ a timelike field giving its orientation, and $\Phi \in {\rm Iso}(M,g)$, then $\Phi$ is orthochronous if and only if there is a point $z\in M$ such that $g_{\Phi(z)}({\rm d}\Phi_z(X_z), X_{\Phi(z)}) < 0$. In other words, to check whether an isometry is orthochronous, you only need to test it on a single future-directed timelike field, and at a single point.