# What operator has these algebraic properties?

I am working in a space $V$ of objects that behaves like a vector space with a partial ordering $\preceq$. I have discovered an operator $f:V\times V \rightarrow V$ with the following properties:

For any $a,b,c \in V$ and any scalar $n$:

1. $f(a,a) = a$
2. $f(a,b) = f(b,a)$
3. $f(a+c, b+c) = f(a,b) + c$
4. $f(n\cdot a,n\cdot b) = n\cdot f(a,b)$.
5. $a \preceq f(a,b)$.
6. If $a \preceq b$, then $f(a,b) = b$.
7. If $a\preceq c$ and $b \preceq c$ then $f(a,b) \preceq c$.

I have two questions. First, besides the first two properties (idempotence, symmetry), I'm not sure whether these properties have official names.

Second, I'm wondering whether operators with these properties are familiar in other contexts.

• This operator has the same properties as $\sup$ or $\max$ on $\mathbb{R}$, for example, with the partial ordering being the obvious $\leq$ – b00n heT Jun 14 '16 at 17:36
• Interesting! It also seems a little bit like $\gcd$ under the "is a multiple of" relation, although in that case property (3) doesn't hold. I wonder if there's an analogous property which does, besides (4). – user326210 Jun 14 '16 at 19:37
• Looks a lot like a join to me. If, as you say, your objects are like vectors, maybe you could think of your (commutative, idempotent, binary) operation $f$ as a join on (an analog of) the lattice of subspaces? ...that is, $f(a,b)$ would be the subspace generated by the set $\{a, b\}$. – William DeMeo Jun 14 '16 at 20:25
• @WilliamDeMeo That's a neat idea. In each of the properties, I could replace the so-called vectors $a$, $b$, and $c$ with the subspaces generated by $a$ $b$ and $c$. I don't know much about vector space joins, but as far as I can tell then each of the numbered properties would actually be something like a property held by vector space joins! – user326210 Jun 14 '16 at 21:55