3
$\begingroup$

I'm looking for a classification (or just non-trivial examples) of partial orders on monoid $\mathbb{N}^{m}$ that are compatible with addition. That is, partial orders $\leq$ satisfying two additional axioms: $$\forall A \in \mathbb{N}^{m} \;\; (0, ..., 0) \leq A,$$ $$\forall A, B, C, D (A \leq B \text{ and } C \leq D) \rightarrow (A + C \leq B + D).$$

Apart from lexicographical order there exist many examples defined in terms of scalar products i.e. fix collection of vectors $v_1, ..., v_k \in \mathbb{R}_{\geq 0}^{m}$ and declare $A \leq B$ iff $(B - A, v_j) \geq 0 \; \forall j$.

Improve this question
$\endgroup$
4
  • $\begingroup$ en.wikipedia.org/wiki/Monomial_order is very close to what you want at least; that might be a good place to start looking. $\endgroup$
    – Sam Hopkins
    May 11, 2021 at 19:34
  • $\begingroup$ @SamHopkins Thanks. Indeed, that seems very close, but i'm not sure about the relation between the two. $\endgroup$
    – Rybin Dmitry
    May 11, 2021 at 19:44
  • $\begingroup$ You can also just use the pointwise ordering and many many more $\endgroup$
    – Benjamin Steinberg
    May 12, 2021 at 0:21
  • 2
    $\begingroup$ @Sam, it seems equivalent to monomial order (with the $+$ operation). If the order is compatible with addition, and $A \le B$, then taking $D=C$ gives $A+C \le B+C$, so it is monomial. If the order is monomial, and $A\le B$ and $C \le D$, then $A+C \le B+C \le B+D$, where the first inequality is by adding $C$ to $A \le B$, and the second is by adding $B$ to $C \le D$. $\endgroup$
    – Jukka Kohonen
    Jan 26 at 16:54

0

Reset to default

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.