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I am looking for examples of varieties of Universal Algebras that admit a ternary majority term $p$. For example, boolean-algebras have such a term:

  • $p(x,y,z) = (x \wedge y)\vee (x \wedge z) \vee (y \wedge z)$

After looking through standard references, I have only found a few examples of such varieties, most of them are arithmetical varieties. If you have any examples, they would be much appreciated.

Mike
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1 Answers1

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As Pedro S. Terraf points out in his comment, the term you present is valid for lattices, not just for Boolean algebras.
Likewise, it works for every variety with a lattice reduct.
Some of these varieties are not arithmetical. Example: the variety of ortho-lattices (but the one of ortho-modular lattices is arithmetical). One example of a not congruence-permutable ortho-lattice is the eight-element of length $4$.

If you want some which are not lattice-based, consider the following example, which provides in the most trivial way a ternary majority term:
$\mathbf{A}$ is an algebra with a single ternary operation $g$ which satisfies $g(x,x,y) \approx x$; if $x \neq y$, let $g(x,y,z) \approx z$ (just to give a complete description of the operation). Then clearly $\mathbf{A}$ admits a ternary majority term — the fundamental operation $g$ — and so the same happens with the variety that $\mathbf{A}$ generates. I don't know whether or not this variety is arithmetical.

For another "exotic" example consider the Example 1.4 here: http://spot.colorado.edu/~kearnes/Papers/idem.pdf
There, the author claims that the variety is congruence-distributive (although that doesn't necessarily entail the existence of a majority term, so I don't know if there is one). Again, I also don't know if the variety is arithmetical.

amrsa
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