Is there any self-dual lattice $(X,\le)$ such that there is not any self-duality $f:X\to X$ such that $f\circ f = 1_X$?
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Yes. Let $L$ be the lattice structure on $\mathbb Z$ with the following Hasse diagram:
-6 <----- -2 <---- 2 <---- 6 <---
\ / \ / \ / \
... -5 -3 -1 1 3 5 7 ...
\ / \ / \ / \
---> -4 -----> 0 ----> 4 ----> 8
where all the diagonal arrows go upwards. It is easy to see that the only selfdualities of $L$ are of the form $f(n)=n+c$ for $c\equiv2\pmod4$, and in particular, they are never involutive.
answered Aug 4, 2014 at 11:18
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$\begingroup$ Is there any finite counterexamples? Or an infinite complete lattice? $\endgroup$ Commented Aug 4, 2014 at 13:34
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$\begingroup$ I don’t know. A finite self-dual lattice must have a selfduality whose order is a power of $2$, but I see no particular reason there should be an involution. On the other hand, I don’t see how to construct a counterexample. $\endgroup$ Commented Aug 4, 2014 at 14:16
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1$\begingroup$ In view of your other question, my example can be modified by adding a top element, a bottom element, and a “middle” element separating the two $\mathbb Z$-chains. Then it becomes an algebraic complete lattice, hence (by a result of Tůma) isomorphic to an interval on the subgroup lattice of some group. $\endgroup$ Commented Aug 4, 2014 at 14:51
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