Non-diffeomorphic but homeomorphic (under Lorentzian topology) Lorentzian manifolds

Question

$\newcommand{\lorentzian}{\mathrm{lorentzian}}\newcommand{\lorentzian}{\mathrm{lorentzian}}\newcommand{\diff}{\mathrm{diff}}\newcommand{\manifold}{\mathrm{manifold}}$Take a time-oriented Lorentzian manifold $(M, g)$ that is not strongly causal.

The Lorentzian metric $g$, however can define a topology, that is strictly coarser than that of the manifold topology. Say: $\tau_{\lorentzian} \subsetneqq \tau_{\manifold}$.

The topology can be found at this post: Lorentzian Topology

Assume I have two Lorentzian time-oriented manifolds $(M_1,g_1)$, $(M_2,g_2)$ that are not strongly causal, but are homeomorphic under their Lorentzian topologies $(M_1, \tau_{\lorentzian}) \cong (M_2, \tau_{\lorentzian} )$.

My 1st question:

Is it possible that $M_1$ and $M_2$ are not diffeomorphic as differentiable manifolds (with the manifold topology and NOT the Lorentzian topology) $(M_1, \diff) \not\equiv (M_2, \diff) $ while $(M_1, \tau_{\lorentzian}) \cong (M_2, \tau_{\lorentzian} )$ as topological spaces?

My 2nd question:

Is it possible for the $\tau_{\lorentzian}$ not to be antidiscrete?

NOTE: the differential structure in $(M, \diff)$ is exactly the one underlying the Lorentzian manifold.

NOTE: to be more physically sound, one can name the $\tau_{lorenzian}$ topology, also $\tau_{observers}$ topology.

Even an example of such a scenario would be extremely of use!

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Physical motivation:

If such distinct differentiable manifolds that induce the same topology via their Lorentzian metric exist, one can define an entropy for the Lorentzian topology by counting the distinct diffeomorphism classes say $w$ and write $S=k_B \ln(w) $ by Boltzmann's relation. And causality will be emergent rather than fundamental, so to speak.

This can potentially give rise to a cosmological arrow of time or entropic time if not also quantum non-locality in case one interprets the differential and topological degrees of freedom as the Hidden Variables of Quantum Mechanics!

Answer 1

I would like to argue that the situation considered in the comments is "close to generic".

Let $(M,g)$ be a Lorentzian manifold that is not strongly causal; this implies that $(M,g)$ is also not stably causal, and hence a small perturbation of it will admit a closed time-like curve. We may therefore assume generically that our $(M,g)$ admits a closed time-like curve $\gamma$.

Choose $p\in \gamma$, we have therefore $\gamma\subset I^+(p)$ and $\gamma\subset I^-(p)$; hence the set $U_\gamma = I^+(p) \cap I^-(p)$ is non-empty, and must be contained in every "Lorentzian neighborhood" of $p$. In particular, the "Lorentzian topology" must be non-Hausdorff.

As $\gamma$ is compact, it has a tubular neighborhood within $U_\gamma$ that is topologically $\mathbb{S}^1 \times B^n$ where $n$ is the spatial dimension. Let $t\in \mathbb{S}^1$ and $x\in B^n$ provide a set of coordinates for this tubular neighborhood; we can arrange for $\partial_t$ to be time-like on this neighborhood.

Now take an $n$-dimensional manifold $D$ with boundary, whose boundary is $\mathbb{S}^{n-1}$, with non-trivial topology. Equip $D$ with a Riemannian metric. Carve out the tubular neighborhood of $\gamma$ and replace it (smoothly glue in) a copy of $\mathbb{S}^1\times D$, equipped with the product Lorentzian metric.

As this modification is entirely within $U_\gamma$, and one easily sees that for every $q_1, q_2\in \mathbb{S}^1\times D$, there exists a time-like curve connecting the two, we see that the Lorentzian topology of the modified manifold is the same as the original. But as long as $M$ doesn't have too crazy a topology the modified manifold should not be diffeomorphic to the original.