Baker–Pixley theorem

Remarks

Every term operation preserves every subalgebra of every power, so only the converse needs work. The idea is interpolation. The operation $f$ agrees with a term operation on each set of at most $d$ points of $A^n$. The reason is that $f$ preserves the subalgebra of $A^d$ generated by the coordinate tuples of those points. The near-unanimity term then glues such local interpolations, by induction on the number $k > d$ of points. Given $k$ points, pick $d + 1$ of them, $p_1, \dots, p_{d + 1}$, and by induction take $d + 1$ terms, the $i$-th agreeing with $f$ on all $k$ points except possibly $p_i$. At each point at most one of these terms disagrees with $f$, so the near-unanimity term applied to them agrees with $f$ on all $k$ points. Since $A^n$ is finite, one reaches a term that agrees with $f$ on all of $A^n$. For a majority term, so $d = 2$, the theorem says that the term operations are the operations that preserve all subalgebras of $A^2$. For example, a finite lattice $L$ has the majority term $(x \wedge y) \vee (y \wedge z) \vee (x \wedge z)$, so an operation on $L$ is a lattice term operation exactly when it preserves all sublattices of $L^2$. For an infinite algebra the same argument only shows that $f$ agrees with some term operation on each finite subset of $A^n$.

Going deeper

For a student

The theorem is a strong form of the statement that the clone of term operations of $A$ equals the set of polymorphisms of the subalgebras of $A^d$. In the language of the Galois connection between operations and relations, the term clone of $A$ is determined by relations of arity at most $d$. It also shows that two finite algebras on the same set with near-unanimity terms have the same term operations once they have the same subalgebras of $A^d$. In an arithmetical variety a Pixley term $p$ gives the majority term $p(x, p(x, y, z), z)$, so the theorem applies there with $d = 2$.