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If $\mathcal{C}$ is a thin category, we call $U \subseteq \mathrm{Ob}(\mathcal{C})$ open if for every object $X \in U$ and any morphism $X \to Y$, we also have $Y \in U$. This declares an Alexandrov topology on $\mathrm{Ob}(\mathcal{C})$.

This defines the Alexandrov topology of a preordered set $P=\mathrm{Ob}(\mathcal{C})$.

My questions:

What is the nerve of this category?

What would be the geometric/topological realization of such simplicial set( in case it has any!)?

The question is motivated by the Lorentzian metric induced Alexandrov topology of a time-oriented not-strongly causal sapcetime.

Such view results in turning spacetime into a thin category that I am curious about its structure.


I do not know if the manifold topology which is finer than that of the Alexandrov topology can be formulated and linked to this category in any easy way at all. Any such functorial relation would be of great importance to me.

My personal superficial/stupid first step would be to first think of the smooth triangulations of the underlying manifold and only later think of any possible relation (bearing in mind that all smooth manifolds can be triangulated).

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  • $\begingroup$ Any (not necessarily triangulable) topological space gives rise to simplicial set, the singular set, see en.wikipedia.org/wiki/Simplicial_set $\endgroup$ Commented Oct 7, 2023 at 14:54
  • $\begingroup$ It was a blunder. Sorry I corrected it. @DonuArapura $\endgroup$ Commented Oct 7, 2023 at 15:35
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    $\begingroup$ Just reading this question, I have no clue what you're talking about. You are forcing the reader to follow links, which is really not a good idea. $\endgroup$ Commented Oct 7, 2023 at 21:02
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    $\begingroup$ In many cases, the category is filtered and the nerve of a filtered category is weakly contractible. For example, this is the case for a Lorentzian manifold with its causality structure. $\endgroup$ Commented Oct 8, 2023 at 1:12
  • $\begingroup$ Just to make sure I follow you. Are you talking about the neighborhood filter of the Lorentzian manifold Alexandrov topology or the a filter over the thin category mentioned in the question above? Because it doesn't seem very immediate to me if the category is a filtered one (in the second scenario). Because the Alexandrov topology defined over the same set $M$ which is a topological manifold does not necessarily match the manifold topology but it's strictly coarser in our case. Two distinct points might turn indistinguishable by a coarser topology and so on. @DmitriPavlov $\endgroup$ Commented Oct 8, 2023 at 15:02

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