posets - Is every lattice the fixed-point set of an order endomorphism of ⋄^n?

For all finite lattices, the answer is Yes.

More generally, for all complete lattices, the answer is Yes, and for all incompleteness lattices, the answer is No.

(Complete = every set has a LUB and GLB.)

Let me first give the positive result. Suppose that L is a complete lattice. Every lattice is naturally a sub-order of the power-set lattice P(L), by associating each point b with it's lower cone b* = { a in L | a <= b }. This map is clearly order-preserving. (Note: it is not a lattice embedding, however, since (b* v c*) is the union of two cones, rather than the cone of (b v c). ) Thus, L is order-isomorphic to the set of lower cones. Define f:P(L) to P(L) by

• f(A) = (sup A)*.

That is, f(A) is the lower cone of (sup A). This is the smallest lower cone containing A. Note that (sup A) is an element of L, since L is complete. The map f is order preserving, since if A is a subset of B, then sup A <= sup B.

Clearly, f(b*) = b* for any b in L, since b is the sup of b*. Conversely, if F(A) = A, then A = b* for b = sup A. That is, A is a lower cone. Thus, the fixed points of f are exactly the lower cones of L, which are order-isomorphic to L, as desired.

Finally, to make the connection with your Diamond lattice, note that P(L) is simply 2^L, a power of the 2 element Boolean algebra. Since Diamond is 2², we can view P(L) as a power of Diamond. (Add a dummy coordinate if L is odd finite size.)

Now, let's consider the negative result. Every power of Diamond is isomorphic as I mentioned earlier to a power set P(J) for some set J. Suppose that f:P(J) to P(J) is an order-preserving map from P(J) to P(J). I claim that the set of fixed points of f must have a smallest element. To see this, let B be the intersection of all A such that f(A) subset A. For any such A it follows that B subset A, so f(B) subset f(A), and so f(B) subset B. So B is the smallest with f(B) subset B. But since applying f gives f(f(B)) subset f(B), it follows that f(B)=B, as desired. By working above any given collection of fixed points, the same argument shows that the collection of fixed points is complete. This establishes:

Theorem. A lattice is complete if and only if it is isomorphic to the set of fixed points of an order-preserving endomorphism of a power set lattice P(J).

Note that many lattices are not complete. For example, the positive integers, as Neel mentioned in his answer. So these lattices are never the set of fixed points of an order-preserving map on a power set lattice, and consequently the same for the powers of Diamond.