Talk:Order isomorphism

Untitled
I'm not sure if this is a mistake or if I misunderstand, but:


 * It should also be remarked that order isomorphisms are necessarily injective. Hence, yet another characterization of order isomorphisms is possible: they are exactly those monotone bijections that have a monotone inverse.

It seems to me that a "monotone bijection" always has a "monotone inverse". This is because:


 * 1) a bijection always has an inverse (being surjective and injective); and
 * 2) that inverse is also a monotone, because it will always map right back to the original ordering (being bijective)

I can't see any counter-examples. The closest is a graph of an increasing function, as in fig 1 in Monotonic function, in which several values of x give the same f(x) (i.e. it's flat). The inverse of this would have a vertical slope, which is not a function - but a bijection doesn't have this problem (since an inverse exists).

I think I must be missing something. The author seems to be getting at something to do with the injectivity... the following would make sense to me (but I really don't know):


 * those monotone surjections that have a monotone inverse.

E.g.
 * A1 → C3      (The index gives the ordering, so A<B, C==D)
 * B2 → D3

The inverse:
 * C3 → A1
 * D3 → B2

EDIT: Although the monotone bijection is reversible, because the orderings are partial, the second set might have extra orderings between elements that (the corresponding elements in) the first set doesn't. In this case, the inverse could not be monotonic, because those orderings could not be preserved (that is, an ordering on the right side wouldn't imply an ordering on the left side) - it's not really a fault of the mapping, but that it's impossible to do for those two sets. This is not a problem for the original monotone, because all that is required is if two elements are ordered, then their corresponding elements are also mapped - when there is no ordering, there is no constraint, and so those extra orderings have no impact.

In other words, the issue isn't whether the relation between two elements is "equal" or "less than", but whether an ordering exists at all. I got a hint from the page on Order type, where it said totality was important:


 * order isomorphic, that is, when there exists a bijection f: X → Y such that both f and its inverse are monotone (order preserving). (In the special case when X is totally ordered, monotonicity of f implies monotonicity of its inverse.)

Have I got it right?

— Preceding unsigned comment added by 124.168.73.107 (talk • contribs) 05:38, 20 November 2011‎

Ordered group
, the reference you added in this edit is unclear. Does the cited source use the phrase "order po-group isomorphism" and definition you are giving? This phrase sounds dubious to me, and also the accompanying description strikes me as slightly misleading, so it would be helpful if you clarified what exactly the citation supports. --JBL (talk) 15:58, 24 July 2016 (UTC)


 * My edit had a small typo (reversed words, should read "order isomorphism of partially ordered groups" or "order isomorphism of po-groups"). The cited source refers to order isomorphisms as "o-isomorphisms" for notational compactness and adds a qualifier to describe the posets being mapped. So an order isomorphism between partially ordered groups is a "o-isomorphism of po-groups", an order isomorphism between partially ordered rings is an "o-isomorphism of po-rings," etc. I just caught a grammatical mistake as well (correcting). How is the accompanying description misleading? Bggoode (talk) 16:20, 24 July 2016 (UTC)


 * Hi, thanks for your response. Let me elaborate, with respect to the current version, quoted here:
 * "When an additional algebraic structure is imposed on the posets $(S,\le_S)$ and $(T,\le_T)$, a function from $(S,\le_S)$ to $(T,\le_T)$ must satisfy additional properties to be regarded as an order isomorphism. For example, given two po-groups, $(G, \le_G)$ and $(H, \le_H)$, an order isomorphism of po-groups from $(G,\leq_G)$ to $(H,\le_H)$ is a group isomorphism that is also an order-embedding, not merely a bijection that is an order-embedding."
 * First, the phrase "order isomorphism of po-groups" sounds redundant: in my experience, the usual thing in mathematical writing would be to write "isomorphism of po-groups," with the understanding that an "isomorphism of [things]" is a map that preserves the defining properties of the [things]. Thus, an "isomorphism of Coxeter groups" is one that preserves the group structure but also preserves the additional structure of a special generating set; an "isomorphism of ordered fields" should preserve both the order and field structures; and by analogy an "isomorphism of po-groups" should preserve both the order and group properties, i.e., it should be both an order isomorphism and a group isomorphism.
 * I also find the first sentence misleading, for the following reason. Given two po-groups, it is possible to have four kinds of maps between them: maps that preserve both order and group properties, maps that preserve order but not group properties, maps that preserve group but not order properties, and maps that preserve neither.  I find it highly doubtful that your source asserts that only the first of these four count as an order-isomorphism (that is, that a map that is an order-isomorphism if you don't know about the group structure suddenly stops being one when there is a compatible group structure).  In fact I do not see either the terminology in question or the misleading claim in the given source, at least not in the first 6 pages on Google Books.
 * Thanks, JBL (talk) 21:37, 25 July 2016 (UTC)


 * Thanks for the discussion on this, . There's actually two things going on here... first, whether or not the article should specify that order isomorphisms between ordered algebraic structures need to preserve algebraic structure, and second, terminology/wording behind "order isomorphism of ordered [things]."


 * As to the first point (algebraic structure), this was my main motivation for contributing to the article. You actually hit upon my point with your "four types of maps" question. Say we have a pair of po-groups and a group isomorphism between them that (as a function between posets) is an order embedding. Further, say we then forget about the group structure. We both agree we still retain that the map is an order isomorphism between posets. But the converse is not necessarily true. Say we have an order isomorphism between posets such that the posets admit a group structure. The map may not be a group isomorphism once we lift those sets up into the group context... heck, the resulting groups may not even be ordered groups if the order doesn't play nicely with the group operation. That's why I wanted to add the paragraph I added, and that may clear up how you are feeling misled, and put some context into your "four different types of map" question.


 * As to the second point, as far as terminology goes... if I'm understanding you correctly, if we say "isomorphism of po-groups" then the reader should know that we mean "this map preserves group structure and embeds the partial order." I mostly agree, especially from a category theory point of view. However, following that line of argument, if we say "isomorphism of posets" the reader should know that we mean "this map embeds the partial order." However, the phrase "order isomoprhism" was coined specifically to bring to the reader's mind that the map is not just an isomorphism of sets (a bijection) but an isomorphism of ordered sets. I see no good reason why this shouldn't also be the case when we speak outside of set theory, redundancies aside.


 * In other words, if we speak about ordered set theory by modifying the usual language of set theory (by tagging the word "order" onto a definition), then we ought to also speak about ordered [thing] theory by modifying the usual language of [thing] theory. It may be clumsy, but it's consistent.


 * By the way, my reference (Fuchs) defines order isomorphisms between sets on page 2, and order isomorphisms between po-groups on pages 20-21 (I have a physical copy). For what it's worth, his definition does not use the phrase "order embedding" but rather requires that a function and its inverse are both surjective onto the positive cones and order preserving/monotonic/isotonic... his definition is equivalent to the definition of an order-embedding, but requires a small, two or three line proof of equivalency.


 * Thanks again for this discussion, it gives me a lot to think about. Bggoode (talk) 00:14, 26 July 2016 (UTC)


 * Hi, thanks for your remarks. Let me make a concrete wording proposal that deals with my worries.  I suggest to rewrite your paragraph as follows:
 * When an additional algebraic structure is imposed on the posets $$(S,\le_S)$$ and $$(T,\le_T)$$, a function from $$(S,\le_S)$$ to $$(T,\le_T)$$ must satisfy additional properties to be regarded as an isomorphism. For example, given two partially ordered groups (po-groups) $$(G, \le_G)$$ and $$(H, \le_H)$$, an isomorphism of po-groups from $$(G,\leq_G)$$ to $$(H,\le_H)$$ is a an order isomorphism that is also a group isomorphism, not merely a bijection that is an order embedding.


 * What do you think?
 * JBL (talk) 21:38, 1 August 2016 (UTC)
 * Looks good to me, . Bggoode (talk) 15:01, 3 August 2016 (UTC)