Relations on free algebra

Question

If we have any class $K$ of algebras (sets with operations and constants) of given type, then there is the well-known construction of free algebra $F_K(X)$ with respect to $K$ generated by given set of variables $X$. If structures of given type in $K$ also have some relations, is there a canonical way, how to define realisations of that relation symbols on $F_L(X)$, where $L$ is the class of reducts of members of $K$ only to such operations and constants?

Answer 1

If we have any class $K$ of algebras (sets with operations and constants) of given type, then there is the well-known construction of free algebra $F_K(X)$ with respect to $K$ generated by given set of variables $X$.

The notion of freeness is a categorical one, and one ought to refer to a category when speaking about free algebras. Let's assume that we are speaking about free algebras in varieties.

If $L$ is a first-order algebraic language, then a variety in this language is a class $K$ of $L$-algebras axiomatizable by identities. $K$ may be considered to be a category if we choose the class of morphisms to be 'all algebra homomorphisms'. Birkhoff used free algebras to show that such classes (=varieties) are precisely the classes of $L$-algebras that are closed under the formation of homomorphic images, subalgebras and products.

Now let's add relations. Assume our language $L$ is an arbitrary first-order language. Let's say that $K$ is a variety of $L$-structures if it is axiomatizable by universally-quantified atomic formulas. (Atomic formula = formula of the form $R(s_1,\ldots,s_k)$ for terms $s_1,\ldots,s_k$.) One can extend Birkhoff's Theorem to show that $K$ is a variety of $L$-structures precisely when it is closed under the formation of homomorphic images, not-necessarily-induced substructures and products. The morphisms in $K$ are the functions that preserve the operations and relations. (We do not require that the morphisms reflect the relations.) The proof uses free algebras constructed as follows.

First, ignore the relations and construct the free algebras $[F]_K(X)$ with respect to the underlying algebra structure. Then interpret the relations on $[F]_K(X)$ as follows. For any $k$-ary relation symbol $R$, say that $R^{F_K(X)}(s_1,\ldots,s_k)$ holds for $s_1,\ldots, s_k\in [F]_K(X)$ if and only if
$R^A(s_1^A(x_1,\ldots,x_n),\ldots,s_k^A(x_1,\ldots,x_n))$ holds universally for every in $A\in K$. Here I am writing $s_i^A(x_1,\ldots,x_n)$ to mean: choose an $n$-ary $L$-term which when applied to $X\subseteq [F]_K(X)$ yields $s_i$. This term has an interpretation $s_i^A(x_1,\ldots,x_n)$ in $A$.

Example. Let $K$ be the class of all lattices. Let $R(x,y)$ be a relation symbol which interprets in every lattice as the order relation. ($R^{A}(x,y)$ holds iff $x\leq y$ for any $A\in K$.) Let $X=\{x_1,x_2,x_3\}$. In $F_K(X)$ we do have $R^{F_K(X)}(x_1\wedge x_2,x_1)$ since $x_1\wedge x_1\leq x_1$ holds universally in every member of $K$, but we do not have $R^{F_K(X)}(x_1\wedge (x_2\vee x_3),(x_1\wedge x_2)\vee(x_1\wedge x_3))$, since
$x_1\wedge (x_2\vee x_3)\leq (x_1\wedge x_2)\vee(x_1\wedge x_3)$ does not hold universally in every member of $K$.

One sentence summary. The relation symbols are interpreted in the free algebras so that they only relate the tuples that they are forced to relate.