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A lattice $L$ is a partially ordered set such that any two elements $a$ and $b$ have a least upper bound $a\lor b$ and a greatest lower bound $a\land b$. A congruence relation on $L$ is an equivalence relation $\sim$ on $L$ that is compatible with the lattice operation in the sense that $a_1\sim b_1$ and $a_2\sim b_2$ implies $a_1\lor a_2\sim b_1\lor b_2$ and similarly for $\land$.

In any lattice there are always two trivial congruence relations, the congruence relation where each element is its own equivalence class (block), and at the other extreme the congruence relation with a single block.

What's an example of a lattice with exactly three congruence relations?

This is Ex 3.40 in Graetzer (Lattice Theory: Foundation). So it is asking to find a lattice with exactly one non-trivial congruence relation. Any example I can think of always has at least two non-trivial congruence relations.

For example, for the linear poset with three elements $a\prec b\prec c$, one congruence relation has $a,b$ in one block and $c$ alone, and another has $a$ alone and $b,c$ together in one block. Another example, the standard diamond lattice with four elements (consisting of all subsets of a two element set ordered by inclusion) also has two non-trivial congruence relations.

Martin Sleziak
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PatrickR
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    Lattices don't have to have tops or bottoms, though, so it appears we should think about unbounded ones. A screenshot of the text might be handy as well, PatrickR. – Kevin Carlson May 13 '13 at 04:49
  • Please clarify, what are the two non-trivial congruence relations on the diamond lattice? – vadim123 May 13 '13 at 17:09
  • @vadim123 If the diamond lattice has 4 elements $0$, $a$, $b$, $1$ with $0$ as minimum element, $1$ as maximum element, and $a$ and $b$ in between and not comparable, one congruence relation has the two blocks ${0,a}$ and ${b,1}$. The other congruence relation has the blocks ${0,b}$ and ${a,1}$. – PatrickR May 14 '13 at 03:59
  • Why wouldn't ${0}, {a,b}$ and ${1}$ be a congruence? You're gluing the middle pieces together and preserving order. – John Douma May 14 '13 at 05:02
  • @user69810 Because $a\sim b$ implies $a\lor a\sim b\lor a$, that is, $a\sim 1$. – PatrickR May 14 '13 at 05:06
  • In general, each congruence class of a congruence relation on a lattice is a sublattice, i.e., closed under $\lor$ and $\land$. – PatrickR May 14 '13 at 05:09
  • I see. If you tried the lattice with $a$, $b$ and $c$ in between $0$ and $1$ you'd have too many relations. – John Douma May 14 '13 at 05:10
  • You may be on to something. It seems that the lattice $M_3$ with 5 elements $a$, $b$, $c$ and $0$, $1$ that you mention has only the trivial congruences. So maybe a variation of this would have a single non-trivial congruence? – PatrickR May 14 '13 at 05:19

2 Answers2

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Found the following example: lattice with a single non-trivial congruence

The only non-trivial congruence relation has $a$ and $b$ in one block and all the other elements alone.

PatrickR
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A lattice with a single nontrivial congruence.

Proof: The elements of the diamond right under the top must be either separated or in a single class, because diamond has only trivial congruences.

If they are in a single class, then $d\sim b$ implies $a\sim 0$ and $b\sim e$ implies $c\sim 0$. Thus, $a\sim c$ and $0\sim a\sim a\vee c=1$ and this clearly implies that the congruence is trivial.

Similarly, it is easy to see that $a,b,c$ must be separated, otherwise some pair of elements in the diamond is not separated and we have a trivial congruence again.

So the only possible pairs we can relate are $a\sim d$, $b\sim 0$ $e\sim c$. It is easy to see that any of them forces the other two, so we are left with the single equivalence that is a candidate for a congruence, namely $\{\{0,b\},\{a,d\},\{c,e\},\{f\},\{1\}\}$.

The remaining tedious checking that this is indeed a congruence is omitted.

Gejza Jenča
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  • Diamond has two nontrivial congruences. – Ittay Weiss Apr 02 '19 at 17:42
  • @IttayWeiss There is probably some misunderstanding. This is the diamond lattice, it is a standard lattice theoretic term. – Gejza Jenča Apr 04 '19 at 12:59
  • I always thought that this is simply $M_3$ while the diamond lattice refers to just the four element lattice. In any case, certainly $M_3$ has only trivial congruences. – Ittay Weiss Apr 04 '19 at 18:04
  • @IttayWeiss It is true that not everybody always refers to $M_3$ as ``the diamond''. However, Birkoff and Mac Lane did call $M_3$ diamond in their book. Birkhoff then refrained from this term in his monograph "Lattice theory". However, George Grätzer uses "diamond" in his books and he is The Person With Most Authority In Lattice Theory nowadays. – Gejza Jenča Apr 04 '19 at 19:34
  • @IttayWeiss OTOH, I have never heard anybody to refer to $M_2$ as ``the diamond''. – Gejza Jenča Apr 04 '19 at 19:35
  • I'm very happy to accept Grätzer's authority :) – Ittay Weiss Apr 05 '19 at 09:38