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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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up vote 7 down vote accepted

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.

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Is there any finite counterexamples? Or an infinite complete lattice? –  user47958 Aug 4 at 13:34
    
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. –  Emil Jeřábek Aug 4 at 14:16
1  
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. –  Emil Jeřábek Aug 4 at 14:51
    
Can the group be abelian? –  user47958 Aug 4 at 14:58
1  
The lattice is not modular. –  Emil Jeřábek Aug 4 at 14:59

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