I am wondering if there is an easy answer to the following question:

Let us consider a finite partially ordered set $P$. It is clear that there exists a $k$ such that there is an order embedding $P\rightarrow\mathbb{N}^k$. Furthermore, one can even deduce a strategy for finding a minimal $k$. Actually one has to determine the order dimension of $P$ (for instance, defined by realizers).

In the literature, there are many more approaches. However, I am wondering if there is an easy or at least existing answer to the following question:

Consider partial orders on antisymmetric (i.e. no inverse elements) cancellative monoids given by multiplication, say, from the left. So, $ab\leq b$ (or, dually, $b\leq ab$) for all $a,b\in H$. Now, what is the minimal number $l$ such that there exists an antisymmetric cancellative monoid $H$ with $l$ generators and an order embedding $P\rightarrow H$?

Thank you for any hints!

  • $\begingroup$ Do you want G to be an ordered group? Do you want the image to be first-order definable inside G so that you can use interpretability results? Is it enough to inject P? Gerhard "Embedding Can Be Many Things" Paseman, 2018.07.11. $\endgroup$ – Gerhard Paseman Jul 11 '18 at 16:44
  • $\begingroup$ Thank you! I forgot to mention that I mean a special order on $G$. It is updated now. $\endgroup$ – Rene Jul 11 '18 at 16:47
  • $\begingroup$ But that's not a partial order. If $\forall a,b\in G: ab\leq b$ holds and $\leq$ is truly a partial order (in particular if it is antisymmetric), then $G$ is the trivial group. $\endgroup$ – Johannes Hahn Jul 11 '18 at 17:01
  • $\begingroup$ Yes, of course you are right. I changed the notion to antisymmetric monoids and added cancellativity to it (otherwise it is a mess, I guess). $\endgroup$ – Rene Jul 11 '18 at 17:08
  • 1
    $\begingroup$ Maybe you want the lattice of order ideals. I don't know about number of generators though. Gerhard "Recollection Is Far From Ideal" Paseman, 2018.07.11. $\endgroup$ – Gerhard Paseman Jul 11 '18 at 17:38

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.