# What conditions are necessary (and sufficient) for the order-dual of a Scott-Ershov domain to also be a domain?

That is, considering the underlying poset of a domain, when does the order-dual poset also comprise a domain?
Below's a little, not strictly necessary, elaboration of that question.

Usual straightforward def of domain (as per, e.g., def 3.1.8, pg 57, "Mathematical Theory of Domains", Stoltenberg-Hansen, et al, CUP 1994, ISBN 0-521-38344-7) is that
(i) the poset is a dcpo (lubs of all directed subsets exist), and is algebraic (every element is the lub of the set of compact elements weaker than itself).
(ii) every consistent set (a set with an upper bound) has a lub, i.e., if an upper bound exists then a least upper bound exists.

So, if a poset satisfies (i) and (ii) above, when does its order-dual also satisfy them?
Also, since there are various slight variations of the definition of a domain, choose any other you prefer. Whatever definition, the same question remains: when is the order-dual poset also a domain?
(This isn't really research, so I started posting it to cs.stackexchange, but only cstheory has a domain-theory tag)

• No, usually not. Typical domains don’t even have a largest element (a domain with a largest element is in fact an algebraic complete lattice), hence there is no way for their dual to be a domain. Furthermore, the dual of an algebraic complete lattice is not in general algebraic either. Jun 3, 2021 at 8:29
• Oh, I missed the question is when, not whether. Well, just follow the definition: a poset and its dual are both domains iff it is an algebraic and dually algebraic complete lattice. Either way, this is not a research-level question. Jun 3, 2021 at 8:40
• @EmilJeřábek Thanks (yeah, I know it's not research-level, which is why I wrote that explanatory parenthetical sentence at the very end). Nevertheless, re the question itself, what I was thinking is that maybe a domain whose underlying poset is orthocomplemented might remain a domain with respect to the order-dual poset. Clearly, $\perp\leftrightarrow\top$ isn't a problem, and I think there's a waybelow<-->wayabove (so to speak) correspondence that preserves the algebraic property, but haven't been able to prove it or to locate a proof. Any ideas along those (or any other) lines? Thanks again. Jun 3, 2021 at 10:21
• This is a research level question. A while ago people organized workshops in which they studied bi-continuous lattices, the OP is asking about an algebraic version of that. Jun 3, 2021 at 10:29
• @AndrejBauer Clearly, you are reading into the question something that I do not see there. Care to explain what? How is the question not answered by “algebraic and dually algebraic complete lattice”, as I wrote above? Jun 3, 2021 at 10:32