Is there known structures/ or has there been studies on $Hom(L_1,L_2)$ of distributive lattices? Could it be made into a lattice naturally? Is there any structure on the set of ring valued functions $R[L]$ for a commutative ring $R$ on a finite lattice $L$, like the theory of group rings?

  • $\begingroup$ More generally, there have been some studies of poset exponentiation for finite posets (and possibly larger), as well as poset arithmetic. I believe Garrett Birkhoff and Ralph McKenzie are two researchers who have done some work in this area (maybe J. D. Farley also?) . I do not know if there are similar studies restricted to distributive lattices. Gerhard "More Power To Your Searches" Paseman, 2018.01.19. $\endgroup$ Jan 19, 2018 at 14:43
  • $\begingroup$ Found two papers by Birkhoff as cited in link.springer.com/article/10.1023/A:1006449213916 . Thank you. $\endgroup$
    – mukhujje
    Jan 19, 2018 at 20:14
  • $\begingroup$ There's a paper by Davey and Priestley, "Lattices of Homomorphisms," J. Austral. Math. Soc. Ser. A 40 (1986), no. 3, 364–406. At least for bounded distributive lattices, I think $Hom(L_1,L_2)$ will have the same ordering as $Hom(P_2,P_1)$ where P_i is the Priestley dual of L_i (assuming your lattice morphisms preserve the bounds). $\endgroup$
    – Tri
    Apr 9, 2021 at 23:39

1 Answer 1


Something more general holds: if $P$ is a poset and $L$ is a lattice, then $\text{Hom}(P,L)$ is also a lattice. If $f,g \in \text{Hom}(P,L)$ we define $f\vee g: P\to L$ by $$(f\vee g)(x) = f(x) \vee g(x) \text { for all } x\in P.$$ It's easy to see that this is the least upper bound; the infimum is constructed dually. Also, a routine verification shows that if $L$ is distributive, so is $\text{Hom}(P,L)$.

  • 5
    $\begingroup$ Yes, but I think OP is looking not just at poset maps $L_1 \to L_2$, but at maps that preserve finite meets and joins. It's not true that your pointwise definition for $f \vee g$ gives a distributive lattice map, because $(f \vee g)(x \wedge y) = (f(x) \wedge f(y)) \vee (g(x) \wedge g(y))$ does not match $(f(x) \vee g(x)) \wedge (f(y) \vee g(y))$. $\endgroup$
    – Todd Trimble
    Feb 9, 2018 at 14:39
  • 1
    $\begingroup$ @ToddTrimble For your first equality, you suppose $f,g$ to preserve meets, am I right ? Otherwise, Dominic's lattice property is all right (it is a sublattice of the lattice product $L^P$). $\endgroup$ Aug 8, 2018 at 20:15
  • 2
    $\begingroup$ @DuchampGérardH.E. Yes, I assume $f, g$ preserve all distributive lattice structure: meets and joins (and also empty meets and joins if those are considered as part of the definition of lattice). The point is that the inequality $(a \wedge b) \vee (c \wedge d) \leq (a \vee c) \wedge (b \vee d)$ typically isn't an equality, as one can easily see by putting the right side in disjunctive normal form, or using Venn diagrams, etc. $\endgroup$
    – Todd Trimble
    Aug 8, 2018 at 21:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .