On a finite lattice, I define an equivalence relation based on automorphisms, and want to determine if the quotient set is a lattice.

Question

Let $(L,\land, \lor)$ be a finite lattice. I consider the relation $\approx$ on L defined by $x \approx y$ iff there exists an automorphism $h$ on $L$ such that $h(x) = y$. $\approx$ is an equivalence relation. I note $|a|$ the equivalence class of an element $a$ of L. Is it true that for a,b,c,d in $L^4$, if $a \approx b$ and $c \approx d$, then $a \land b \approx c \land d$? If this is the case, can I conclude that the quotient set $L/\approx$ is a lattice, by defining $|a| \land |b| = |a \land b|$? Thanks!

Answer 1

No, this doesn't work. Consider the six-element lattice $\{0,x,y,u,w,1\}$ with $0\le x\le y\le 1$ and $0\le u\le w\le 1$ (and no other inequalities other than those required by the foregoing). Then in this lattice we have $x\approx x$ (trivially) and $y\approx w$ but $$x\vee y=y\not\approx 1=x\vee w.$$ Re: your notation, I'm here taking $a=b=x$, $c=y$, $d=w$. (Incidentally, I think there's a typo in the initial question you ask: you don't want to compare $a\vee b$ and $c\vee d$ but $a\vee c$ and $b\vee d$.)

In general, the orbit relation on a structure is rarely a congruence, since even if $a_i\approx b_i$ for $i\in I$ we may not have a single automorphism $f$ with $f(a_i)=b_i$ simultaneously for all $i\in I$. (This old answer of mine treats another case of this.)