Is any globally-hyperbolic manifold conformally equivalent to one with complete slices?

Question

Let $(M,g)$ be a globally-hyperbolic Lorentzian manifold. By the seminal work of Geroch and Bernal-Sánchez, we know that $$M=\mathbb{R}\times\Sigma,\,\,\,\quad g=-\beta^{2}dt^{2}+h_{t}$$ where $\Sigma$ is a (smooth, spacelike) Cauchy hypersurface and $h_{t}$ a one-paremeter family of Riemannian metrics depending smoothly on $t$. Now, it is well-known that a conformal transformation $g\mapsto\Omega g$ for $\Omega\in C^{\infty}(M,[0,\infty))$ preserves the causal structure and hence also global hyperbolicity.

Is any globally-hyperbolic manifold $(M,g)$ conformally invariant to a globally-hyperbolic manifold $(M,\overline{g})$ with the property that $(\Sigma,\overline{h}_{t})$ is a (geodesically) complete Riemannian manifold for all $t\in\mathbb{R}$?

By the work of Nomizu-Ozeki, it is known that any Riemannian manifold is conformally equivalent to a complete one, so in particular we can find for any $t\in\mathbb{R}$ a conformal transformation $\omega_{t}\in C^{\infty}(\Sigma,[0,\infty))$ such that $\omega_{t}h_{t}$ is a complete metric. So, the question basically boils down to the question whether these functions $\omega_{t}\in C^{\infty}(\Sigma,[0,\infty))$ can be chosen such that $\Omega(t,\cdot):=\omega_{t}(\cdot)$ depends smoothly on $t$.

Answer 1

This is the original Nomizu-Ozeki article:

Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.

Applying their proof to a family of metrics $h_t$, the conformal factors $\omega_t(x)$ need only satisfy the property that $\omega_t(x) > 1/r_t(x)$, where $r_t(x)$ is the sup of all radii of precompact metric balls at $x$. The existence of a smooth $\omega_t(x)$ only relies on the continuity of $r_t(x)$, which they prove (and second countability of the underlying manifold $\Sigma$, which goes hand in hand with Riemannian metrizability).

A priori, we do not know that $\omega_t(x)$ can be chosen to be smooth in $(t,x)$, nor that $r_t(x)$ itself is continuous in $(t,x)$. But the logic of the proof still applies if we replace $\Sigma$ by $M = \mathbb{R}\times \Sigma$ and instead choose a smooth $\Omega(t,x)$ so that $\omega_t(x) = \Omega(t,x) > 1/R(t,x) \ge 1/r_t(x)$ for some $R(t,x)$ that is continuous on $M$. It's simple to define by analogy the function $R(t,x)$ to be the sup of all radii of precompact balls at $(t,x)$ in the metric space $(M,\Delta)$ where $\Delta((t,x), (t',x'))$ is the Riemannian metric distance on $(M, dt^2 + h_t)$. Exercise: repeat the arguments from Nomizu-Ozeki to show that one can still reduce to the case where $R(x,t) < \infty$ everywhere and that $R(x,t)$ is continuous.

Answer 2

The answer by Igor is complete and very clear. Although I'd like to give a much shorter one: Yes!

Miguel Sánchez actually answers this himself in [1, Section 3.2]

[1] Sánchez https://arxiv.org/abs/2110.13672