Are modular lattices shallow?

Question

Let $A$ be a universal algebra with finitely many finitary operations. Write $F_n$ for the $n$-ary operations.

We define the affine maps on $A$ inductively: $\eta \mapsto \eta$ and $\eta \mapsto c$ where $c \in A$ are affine, and if $f \in F_n$ and $c_i \in A$ are constants, and $g$ is affine, then $\eta \mapsto f(c_1, \ldots, c_k, g(\eta), c_{k+2}, \ldots, c_n)$ is affine.

Affine maps can be thought of as syntactic objects (terms with one variable and constants), and the depth of an affine map is the number of times we use the inductive step in the definition, i.e. "how many layers of parenthesis surround $\eta$". On the other hand, an affine map gives a function $f : A \to A$, and two affine maps $f, g$ are equivalent if they define the same function.

(BTW, I've called these "affine maps" because for some universal algebras these are the affine maps, and I don't know a meaning for that term in universal algebra, but do tell if you know a more standard term for this.)

Now, we say $A$ is $k$-shallow if for every affine map $f$, there exists an affine map $g$ with depth at most $k$, which is equivalent to $f$.

Now let $V$ be a variety of universal algebras (so the class of universal algebras with a fixed set of operations, satisfying a fixed set of laws). We say $V$ is $k$-shallow if every algebra in $V$ is $k$-shallow. We say $V$ is shallow if it is $k$-shallow for some $k$.

Examples:

• Groups are a variety with operations $((x, y) \mapsto x \cdot y) \in F_2$, $(x \mapsto x^{-1}) \in F_1$ and identity $e \in F_0$, and you know the laws. This variety is $3$-shallow: the deepest type of term we need is $\eta \mapsto (x \cdot \eta^{-1}) \cdot y$.

• Lattices have operations $\wedge, \vee \in F_2$ and satisfy laws $x \vee y \equiv y \vee x$, $x \vee (y \vee z) \equiv (x \vee y) \vee z$, $x \vee x \equiv x$, $x \equiv x \vee (x \wedge y$, and the laws obtained by exchanging $\vee \leftrightarrow \wedge$. I know a roundabout way to prove that lattices are not shallow, but I don't have a simple direct proof.

Pre-question. Is there a simple proof that lattices are not shallow?

(According to Wikipedia, in the free lattice on generators $x, y, z$, the elements defined inductively by $p_0 = x$ and $p_{n+1} = (x \vee (y \wedge (z \vee (x \wedge (y \vee (z \wedge p_n))))))$ are distinct; I suppose the affine maps obtained by setting $p_0 = \eta$ are not shallow, but I don't know much about these lattices.)

• Distributive lattices are lattices, with additional laws $x \vee (y \wedge z) \equiv (x \vee y) \wedge (x \vee z)$ and the law obtained by exchanging $\vee \leftrightarrow \wedge$. Distributive lattices are $2$-shallow, with the deepest affine maps equivalent to $\eta \mapsto (\eta \vee x) \wedge y$.

We now get to the modular lattices. These are the lattices where the projection to an interval $[a, b]$ is well-defined in the sense that $a \leq b$ (meaning $a = a \wedge b$) implies $(c \vee a) \wedge b \equiv (c \wedge b) \vee a)$ for all $c$. To make these into a variety, we can rephrase this as the law $(a \wedge b) \vee (x \wedge b) \equiv ((a \wedge b) \vee x) \wedge b$ (this is added on top of the laws of the variety of lattices).

Question. Is the variety of modular lattices shallow?

One can ask about finite variants of these notions, i.e. bounded shallowness of finite algebras. It also interests me whether modular lattices are "finitely shallow", i.e. finite modular lattices are $k$-shallow for some fixed $k$. (Of course, any individual finite algebra is $k$-shallow for some $k$.)

(The motivation for this question is symbolic dynamical, and I don't know how this links to existing research in universal algebra. Feel free to school me on this!)

Answer 1

Terminology.

If $f$ is a fundamental operation of an algebra $A$, then a polynomial of the form $f(c_1,\ldots,c_k,x,c_{k+2},\ldots,c_n)$ is often called a 'basic translation' or a '$1$-translation' of $A$. A composition of $k$ $1$-translations is called a $k$-translation and a 'translation' is a $k$-translation for some $k$. This terminology can be found in Definition 3.1 of

Kirby A. Baker
Finite equational bases for finite algebras in a congruence-distributive equational class
Advances in Mathematics 24, 207-243, (1977)

Baker credits the terminology to

A. I. Mal'cev
On the general theory of algebraic systems
Math. Sb. (N.S.) 35 (77) (1954)

'Translations' are not exactly the same as 'affine maps' since the identity function and the constant functions are not guaranteed to be translations, but if you expand the language trivially to include a new basic operation $t(x,y)$ satisfying $t(x,y) = x$ then you can obtain the identity and every constant as a $1$-translation in the new language.

Math.

The variety of modular lattices is not shallow. To see this, consider the sequence $L_1, L_2, L_3, \ldots$ of (simple) modular lattices constructed according to this pattern:

There is an $8$-translation $t(x)$ such that $t(a)=c$ and $t(b)=d$, namely $t(x)=(((((((x\vee u_1)\wedge u_2)\vee u_3)\wedge u_4)\vee u_5)\wedge u_6)\vee u_7)\wedge c$. This translation witnesses the fact that the pair $(c,d)$ belongs to the congruence generated by the pair $(a,b)$.

There is no $(\leq 6)$-translation that maps $(a,b)$ to $(c,d)$ for the following reasons. Assume that $g(x)$ is any $(\leq 6)$-translation that maps $(a,b)$ to $(c,d)$. Each $1$-translation is equivalent to one of the form $x\vee c$ or $x\wedge c$, hence involves only one constant as a parameter. Since $g(x)$ is a $(\leq 6)$-translation, it can involve at most $6$ of the $u$'s as parameters. Since there are $7$ $u$'s in $L_7$, some $u_i$ is not a constant parameter in $g(x)$. Each $u_i$ is doubly irreducible in $L_7$, so $L_7-\{u_i\}$ is a (modular) sublattice of $L_7$. Now, $g(x)$ restricts to a unary polynomial function of $L_7-\{u_i\}$ which is assumed to satisfy $g(a)=c$ and $g(b)=d$. This forces $(c,d)$ to be in the congruence generated by $(a,b)$ in the sublattice $L_7-\{u_i\}$. But it is easy to see that $(c,d)$ is NOT in the congruence generated by $(a,b)$ in the sublattice $L_7-\{u_i\}$.

Now let $7$ go to infinity. The same argument shows that for any odd $n$, $L_n$ will have an $(n+1)$-translation that maps $(a,b)$ to $(c,d)$, but no $(\leq(n-1))$-translation that does this.

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Response to comments.

1. I couldn't find a simple argument that this is modular

To check that a finite lattice is modular, it suffices to check that each of its 'small' intervals is modular. Let me define 'smallness' and sketch a path one could take to prove the criterion of the preceding sentence, and then say how the criterion applies to $L_n$.

Write $x\prec y$ to mean that $x < y$ and there is no $z$ with $x<z<y$. When $x\prec y$, call $x$ a lower cover of $y$ and $y$ an upper cover of $x$ or say that $x$ is covered by $y$ or $y$ covers $x$. Now, when $L$ is a finite lattice, call an interval $I\subseteq L$ 'small' if $I=[x\wedge y, x\vee y]$ where both $x$ and $y$ cover $x\wedge y$ or both are covered by $x\vee y$. I claim that a finite lattice is modular iff all of its small intervals are modular.

To prove that this criterion for modularity is correct for finite lattices, you could follow these steps.

Show that a finite lattice is modular iff it is both upper semimodular [$((x\wedge y)\prec x)\to (y\prec (x\vee y))$] and lower semimodular [$(y\prec (x\vee y))\to ((x\wedge y)\prec x)$].

Show that a finite lattice is upper semimodular iff whenever both $x$ and $y$ cover $x\wedge y$, then both $x$ and $y$ are covered by $x\vee y$. The dual claim holds for lower semimodularity.

Now prove the criterion: a finite lattice $L$ is modular iff whenever $I=[x\wedge y,x\vee y]$ is an interval where both $x$ and $y$ cover $x\wedge y$ or both $x$ and $y$ are covered by $x\vee y$, then $I$ is modular.

To see how the criterion applies to $L_n$, let $u_i^+$ denote the unique upper cover of $u_i$ in $L_n$ and let $u_i^-$ denote the unique lower over of $u_i$ in $L_n$. Convince yourself that the interval $[u_i^-,u_i^+]$ is small and isomorphic to $M_3$, hence is a small modular interval. Then convince yourself that all other small intervals are isomorphic to $M_2$, which is modular. Since all small intervals are modular, $L_n$ is modular.

2. I also don't have a clear argument for that thing that's easy to see

This is referring to my claim that in the sublattice $L_n-\{u_i\}$ the pair $(c,d)$ is not in the congruence generated by the pair $(a,b)$. To prove this, it suffices to exhibit some congruence $\theta$ on $L_n-\{u_i\}$ that contains $(a,b)$ and does not contain $(c,d)$, so let me describe such a $\theta$. I will explain this for the case where $u_i$ has even subscript, but you can modify the argument to handle the case where $u_i$ has odd subscript.

When $i$ is even, the sublattice $L_n-\{u_i\}$ has a unique lattice retraction $\rho$ which has image equal to the interval $[u_i^+\wedge c, u_i^+\vee c]$ and which maps all of the elements in the set $\{a, b, u_1,\ldots,u_{i-1}\}$ to the element $u_i^+$. Let $\theta$ be the kernel of $\rho$. Since $\rho(a)=\rho(b)=u_i^+$, $(a,b)\in\theta$. Since $c$ and $d$ are distinct elements of $\textrm{im}(\rho)$ and $\rho^2=\rho$, we have $(c,d)\notin\theta$.

(To modify this for odd subscripts I would use $u_i^-$ in place of $u_i^+$ in the above argument.)