1
$\begingroup$

Let $S$ be a set and $\cdot$ a partial binary operation on $S$. A subset $F\subseteq S$ is $\cdot$-closed if the following condition holds:

  • for all $f,g\in F$, if $(f,g)\in\mathrm{dom}(\cdot)$, then $f\cdot g\in F$.

Fact: the intersection of two closed subsets of $S$ is closed.

A coordinate map for $S$ is a pair $(U,\varphi,G)$ such that $U\subseteq S$ is a closed non-empty subset, $G=(G,\circ)$ is a semicategory, and $\varphi\colon U\rightarrow \mathrm{mor}(G)$ is a bijection such that $(u,v)\in \mathrm{dom}(\cdot)$ iff $(\varphi(u),\varphi(v))\in\circ$ and $(u,v)\in\mathrm{dom}(\cdot)$ implies $\varphi(u\cdot v)=\varphi(u)\circ\varphi(v)$ for all $u,v\in U$.

An atlas for $(S,\cdot)$ is a collection of coordinate maps $(U_\alpha,\varphi_\alpha,G_\alpha)$ such that $S = \cup_\alpha U_\alpha$, $\alpha\in I$ an index set.

Fact: the transition maps $\tau_{\alpha,\beta}\colon \varphi_\alpha(U_\alpha \cap U_\beta)\rightarrow \varphi_\beta(U_\alpha \cap U_\beta)$, $\tau_{\alpha,\beta}=\varphi_\beta\circ\varphi_\alpha^{-1}$ are semigroupoid $\circ$-isomorphisms, i.e. a composability- and composition-preserving correspondence of the morphisms.

The pair $(S,\cdot)$ together with an atlas is an algebraic manifold.

Has such an "algebraic manifold" been defined before? Are there any non-trivial examples?

Improve this question
$\endgroup$
8
  • 5
    $\begingroup$ Perhaps a better term is "locally a semicategory", as "algebraic manifold" already refers to a smooth variety... $\endgroup$
    – Kevin Casto
    Jan 4, 2021 at 0:51
  • 1
    $\begingroup$ Would you first list what examples you have in mind and which ones you consider as "trivial"? $\endgroup$
    – YCor
    Jan 4, 2021 at 1:26
  • $\begingroup$ Hi Kevin, I'll consider "locally a semicategory" for future questions. @YCor I will leave "trivial" up to the reader to define. I mean a non-semicategory. For example the object set and binary relation $(\{a,b,c,d,e\},\{((a,b),c),((d,e),c)\})$ has an atlas with two charts as each of the closed subsets $\{a,b,c\}$ and$\{c,d,e\}$ is isomorphic to the semicategory $(\{a,b,c\},\{((a,b),c)\})$. However it's also a semicategory in its own right, hence trivial. It is possible that with these conditions, $S$ is always a semicategory, i.e. the morphisms of some semicategory. $\endgroup$
    – David Pokorny
    Jan 4, 2021 at 2:14
  • $\begingroup$ cf ncatlab.org/nlab/show/pseudogroup#definition $\endgroup$
    – David Roberts ♦
    Jan 4, 2021 at 3:18
  • 1
    $\begingroup$ You can edit the question rather than adding info in comments. $\endgroup$
    – YCor
    Jan 4, 2021 at 8:36

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.