Are there finite lattices which are not order isomorphic to a sublattice of $\mathbb Z^n$?

Question

Consider the lattice $\mathbb Z^n$ for some finite $n$ ordered by $x\leq y \leftrightarrow \forall i: x_i \leq y_i$. I'm having difficulty thinking of all lattice which is not a sublattice of this.

For example, the diamond lattice is isomorphic to:

        (2,2,2)
      /    |     \
(1,0,0)  (0,1,0)  (0,0,1)
     \     |     /
        (0, 0, 0)

Are there counterexamples? If not, is there a name for the theorem that they are all isomorphic?

Answer 1

While not every finite lattice is a sublattice of $\mathbb Z^n$, as amrsa pointed out, it is true that every finite poset is a subposet of $\mathbb Z^n$.

Given a finite poset $P$ consider the map $L\colon P\to\mathbb Z^P$ given by $$ L(p)_q = \begin{cases} 1 & \text{if $q\le p$}, \\ 0 & \text{otherwise}. \end{cases} $$ Then $L$ is an order-preserving and order-reflecting injection and hence $P\cong L(P)$ as posets.

Note that this map factors through the boolean lattice on $P$ by sending $p$ to its downset $p_\downarrow=\{\ q\ |\ q\le p\ \}$ which then gets sent to the characteristic function $L(p) = \chi_{p_\downarrow}$.

Answer 2

Every non-distributive lattice is not isomorphic to any sub-lattice of $\mathbb Z^n$.

In particular, the diamond can be represented as you did, but that representation is not a sub-lattice of $\mathbb Z^n$. Indeed, in $\mathbb Z^3$, we have, for example, that $(1,0,0) \vee (0,1,0) = (1,1,0)$, not $(2,2,2)$ as you put it.

The reason for my first statement is that every sub-lattice of a distributive lattice (such as $\mathbb Z^n$) is again distributive because the joins and meets in the sub-lattice must coincide with those in the original lattice.