Is there an example of a map $h$ between two lattices such that $h: L_1 \rightarrow L_2$,that is order embedding from $L_1$ to $L_2$ but not an embedding when they are taken as algebras?
I would appreciate a graphic example, I have been trying to come up with one, I know that $L_1$ most likely has a quadrilateral form, while $L_2$ has the quadrilateral form with another point in the middle, but I am unsure as how to define $h$. I would understand this better if I could see it. Thank you for your help
I think you mean an injective function $h$ with the property $u \leq v$ implies $h(u) \leq h(v)$ for all $u, v$, and there exist $u, v$ with $h(u\wedge v) \neq h(u) \wedge h(v)$. Is this what you mean?
If so, here's a suggestion:
Let $L_1$ be 2x2, that is, $L_1$ has elements $a, b, c, d$ with order relations: $a\leq b, a\leq c, b\leq d, c\leq d$ (so b and c are not comparable and $b\wedge c = a$ and $b \vee c = d$).
Let $L_2$ be the 4-element chain $w \leq x \leq y \leq z$.
Now, can you find a one-to-one map $h:L_1 \rightarrow L_2$ that satisfies the required conditions?
Hint: construct $h$ so that
$$w = h(a) = h(b\wedge c) \neq h(b) \wedge h(c)$$