Cauchy surface for sphere

Question

If we have a Lorentzian manifold $\mathbb{R} \times S^n$ with metric $g= -dt^2 + ds^n$ where $ds^n$ is just the standard round metric for spheres. Does this manifold have a Cauchy surface, i.e. is it globally hyperbolic? And if so what is the Cauchy surface?
Basically that would mean that the Cauchy surface is homeomorphic to $S^n$. I think that means that there exists no Cauchy surface but I am not sure.

Answer 1

Every $\{t\}\times \mathbb S^n$ with $t\in \mathbb R$ fixed is a Cauchy surface : let $\gamma: \mathbb R\to \mathbb R\times \mathbb S^n$ be a smooth time-like curve. Write $\gamma(\lambda)= (t(\lambda),x(\lambda))$.

Then : $$ \left[\text dt (\dot \gamma)\right]^2 > [\text ds(\dot\gamma)]^2 \tag{1}$$ so in particular $\text dt(\dot \gamma) = \frac{\text dt}{\text d\lambda} \neq 0$. Therefore, $t(\lambda):\mathbb R\to t(\mathbb R) = (a,b)$ is a diffeomorphism and we can consider the function $x(t): (a,b)\to \mathbb S^n$. Equation $(1)$ gives us : $$\left\|\frac{\text dx}{\text dt}\right\|<1$$

and we can extend $x(t)$ to a map $\mathbb R \to\mathbb S^n$, which meets every $\{t\}\times \mathbb S^n$.

Remark : This works for every space-time of the form $\mathbb R\times S$ metric $-\text dt^2 + \text ds^2$ where $(S,\text ds^2)$ is a complete Riemaniann manifold.