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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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1 Answer 1

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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.

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  • $\begingroup$ Is there any finite counterexamples? Or an infinite complete lattice? $\endgroup$ Aug 4, 2014 at 13:34
  • $\begingroup$ 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. $\endgroup$ Aug 4, 2014 at 14:16
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    $\begingroup$ 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. $\endgroup$ Aug 4, 2014 at 14:51
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    $\begingroup$ The lattice is not modular. $\endgroup$ Aug 4, 2014 at 14:59
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    $\begingroup$ Ask Alex Pogel. $\endgroup$
    – Tri
    Apr 18, 2015 at 23:24

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