2
$\begingroup$

Let $P$ be a poset on which a partially ordered group $G$ acts by monotone bijections. I call these guys $G$-posets. If I'm not wrong, the semidirect product of two partially ordered groups remains partially ordered. It is then possible to ask:

  1. If $P$ is a $G$-poset and $Q$ a $H$-poset, is there a way to find a third poset that I call for now $P\rtimes_\phi Q$ which has an action of $G\rtimes_\phi H$ (for a given $\phi : H\to Aut(G)$)?

  2. Is there a way to promote this construction to a functor $$ G\text{-Pos}\times H\text{-Pos} \to (G\rtimes_\phi H)\text{-Pos} $$

  3. Given a $(G\rtimes_\phi H)$-poset $P$ is there a way to find when there are $P_G, P_H$ such that $P\cong P_G\rtimes_\phi P_H$?

$\endgroup$
5
  • $\begingroup$ What's a partially order group? A group with a partial order? left-invariant? bi-invariant? What's the order on the semidirect product? $\endgroup$
    – YCor
    May 25, 2017 at 23:10
  • $\begingroup$ $G$ is a group $(G,\cdot)$ and a poset $(G,\le)$, and moreover the group operation is a two-sided congruence: for every $g\le g'$ and $h\in G$ one has $hg\le hg'$ and $gh\le g'h$. $\endgroup$
    – fosco
    May 25, 2017 at 23:15
  • $\begingroup$ I lost the page where I did the computation, but (as soon as you assume that $\phi$ is monotone, which I forgot to mention) the order on $G\rtimes H$ is simply the product-order on $G\times H$ (the semidirect product is obviously the product with a different group operation)... am I wrong? $\endgroup$
    – fosco
    May 25, 2017 at 23:20
  • $\begingroup$ This seems (at least under some further restrictions) obtainable from a combination of two particular cases of the following: given a small category $C$ and an internal category $F$ in $\mathbf{Cat}^C$, there is an equivalence $\left(\mathbf{Cat}^C\right)^F\simeq\mathbf{Cat}^{F\rtimes C}$. $\endgroup$ May 26, 2017 at 5:54
  • $\begingroup$ What is ${\bf Cat}^C$, where is this result stated and what are the further restrictions? $\endgroup$
    – fosco
    May 26, 2017 at 8:03

0

Your Answer

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

Browse other questions tagged or ask your own question.