1
$\begingroup$

If $(L, \land, \lor, 0, 1)$ is a lattice, and there exists a unary operation $'$ on $L$ such that

  1. $(x \lor x')=1$, and

  2. $(x \land x')=0$

both hold, is the unary operation $'$ an isomorphism between the semilattices $(L, \land)$ and $(L, \lor)$? If we add the condition that $'$ is an involution (i. e. $x''=x$), is $'$ an isomorphism?

1 Answers 1

2

No.

Consider this lattice:

                                    1                                    / \                                   /   \                                  /     \                                 d       w                                / \     / \                               b   c   y   z                                \ /     \ /                                 a       x                                  \     /                                   \   /                                    \ /                                     0 

and make $'$ exchange $0\leftrightarrow 1$, $d\leftrightarrow w$, $b\leftrightarrow y$, $c\leftrightarrow z$, $a\leftrightarrow x$. It is easy to verify that $r\land r' = 0$ and $r\lor r' = 1$ for all $r$; but $'$ does not define an isomorphism from $(L,\land)$ to $(L,\lor)$, since for example $b'\lor c' = y\lor z = w$, but $(b\land c)' = a' = x\neq w$. Nor does it define an isomorphism going the other way, since the map is self-invertible.

Added. A smaller example:

                                     1                                     / \                                    /   \                                   b     y                                   |     |                                   a     x                                    \   /                                     \ /                                      0 

But $a'\lor b' = x\lor y = y$, $(a\land b)' = a' = x$.

  • 0
    Did you use any particular method to find this?2012-02-16
  • 0
    It suggested itself fairly quickly: take a disjoint union of two copies of the same lattice, stick a $0$ and a $1$ to that, then have $'$ swap the copies. Have enough elements to have a nontrivial join and a nontrivial meet, and make it as small as possible. Had I thought a bit more, I could have made it even smaller: remove $d$ and $w$, or remove $a$ and $x$; you don't need *both* a nontrivial join and a nontrivial meet, just one of them.2012-02-16
  • 0
    Or even smaller: remove $d$, $w$, $c$, and $z$.2012-02-16
  • 0
    I don't know what you mean by "nontrivial join (meet)". If we have an unordered pair {x, y} such that (x $\lor$ y)=z, where z $\neq$ y and z $\neq$ x, would you call their meet nontrivial? Or is there more to it?2012-02-16
  • 0
    I meant "Nontrivial meet"$\neq 0$, "nontrivial join$\neq 1$. The second example shows you don't need $x\lor y\notin\{x,y\}$ for a counterexample to work.2012-02-16