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This is a purely idle question, but one I'm increasingly interested the more thought I put into it:

For $\mathcal{A}$ a universal algebra (that is, nonempty set together with some named functions), a congruence of $\mathcal{A}$ is an equivalence relation on $\mathcal{A}$ which respects the named functions: $$\overline{x}\approx\overline{y}\quad \implies \quad f(\overline{x})\approx f(\overline y).$$ It is immediate that the set of congruences ordered by $\subseteq$ forms a lattice.

An important theme of universal algebra (cf. http://www.math.hawaii.edu/~ralph/Classes/619/willard-ua.pdf) is that interesting structural properties of varieties often translate to interesting equations satisfied by their congruence lattices. For example, the congruence lattice of any group is always modular, and the congruence lattice of any Boolean algebra is always distributive.

For $\mathbb{V}$ any variety, let $\mathbb{V}'$ be the variety generated by the congruence lattices of elements of $\mathbb{V}$. Note that in general, $\mathbb{V}'$ will be a variety of algebras in a signature different from that of $\mathbb{V}$, so we cannot compare $\mathbb{V}$ and $\mathbb{V}'$ directly. However, if $\mathbb{V}$ is a variety of lattices, then $\mathbb{V}$ and $\mathbb{V}'$ are comparable. My question is, when does passing to the variety generated by congruence lattices result in a nicer ( = satisfying more equations) variety?

There are two specific versions of this question I'm interested in.

Generally,

(1) Are there natural conditions on a variety $\mathbb{L}$ of lattices which ensure that $\mathbb{L}\supseteq\mathbb{L}'$?

More specifically,

(2) Is there an example of a reasonably natural variety of lattices $\mathbb{L}$ such that $\mathbb{L}\supsetneq\mathbb{L}'\supsetneq\mathbb{L}''\supsetneq . . . $?

(Note, of course, that $\mathbb{V}$ and $\mathbb{V}'$ are comparable as long as the signature of $\mathbb{V}$ is $\{\wedge, \vee\}$, not only when $\mathbb{V}$ is a variety of lattices; but I'm specifically interested in varieties of lattices.)

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  • $\begingroup$ 1942 result: lattices are a congruence distributive variety; thus V'' stabilizes under '. $\endgroup$ – The Masked Avenger Aug 24 '14 at 23:29
  • $\begingroup$ I didn't know about that result, that's nice! But I don't see why that implies that ' stabilizes. $\endgroup$ – Noah Schweber Aug 24 '14 at 23:31
  • $\begingroup$ If you change ' so that it is the class of congruence lattices and not the variety generated by congruence lattices, you may bump into an open problem. It may be known whether any distr. lattice is the congruence lattice of a distributive lattice but I don't know that. $\endgroup$ – The Masked Avenger Aug 24 '14 at 23:35
  • $\begingroup$ Because any nontrivial lattice will have its congruence lattice be nontrivial. You will have V' be some lattice variety, then V'' will either be trivial or generated by the two element lattice. More ''s does not change that, if ' means variety generated by congruence lattices. $\endgroup$ – The Masked Avenger Aug 24 '14 at 23:39
  • $\begingroup$ I'm sorry, I'm still not following; I think I'm just being slow: given that $V'$ is a lattice of distributive varieties, why can we conclude that $V''$ is either trivial or generated by 2? (Or is it known that these are the only two varieties of distributive lattices? That seems odd.) Also, regardless of this, question 1 still seems potentially interesting. $\endgroup$ – Noah Schweber Aug 24 '14 at 23:57
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A congruence lattice of a universal algebra is a complete (therefore bounded) lattice, and is also algebraic (every element is a join of compact elements). As such, the collection C of congruence lattices of algebras of some class is a variety only when C is the collection of one element lattices.

The ' operation above considers the lattice variety L generated by C. This variety L is either trivial, or it contains the variety generated by the two element lattice. This last variety is the variety D of distributive lattices. If the beginning variety of algebras has a Mal'cev term for distributivity, then C and thus L consists of distributive lattices. Lattices have such a term: take the join of the three terms (x_i meet x_{i+1}) where the subscripts are taken modulo 3. Thus for any lattice variety L, L' is either D or trivial, as D is a minimal variety in the collection (lattice!) of lattice varieties.

Finally, for any starting variety V of algebras, we have V'=L for some lattice variety L. If L is trivial, call this T, then T'=T. Otherwise L'=D=D'. So the ' operation reaches one of two fixedpoints D or T within two steps or one. This shows that L contains L' for all lattice varieties L, with strict containment precisely when L has a nondistributive lattice. This answers Question 1. Question 2 is answered negatively: there are many two element non-ascending chains, all but one ending in D.

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