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Can we efficiently decide whether a finite algebra $\mathbf{A}$ generates a variety with a difference term?
We will outline the proof that the answer is "yes" in the idempotent case.
Definition
A difference term for $\mathcal{V}$ is a term $d$ satisfying, $\forall ; \mathbf A \in \mathcal V$ and $\forall a, b \in A$,
$$d(a,a,b) = b \quad \text{ and } \quad
d(a,b,b) \mathrel{[\theta, \theta]} a$$
where $\theta$ is any congruence containing $(a,b)$ and $[\cdot, \cdot]$ is the commutator.
Problem Statement
Is there a poly-time algorithm that takes a finite
idempotent algebra $\mathbf{A}$ and decides
whether $\mathbb{V}(\mathbf{A})$ has a difference term?
**Theorem** (Kearnes, *J Algebra* 1995)
$\mathbb{V}(\mathbf{A})$ has a difference term
$\; \Longleftrightarrow \;$ $\mathbb{V}(\mathbf{A})$ omits type 1 and type-2 tails
Omitting 1 is poly-time decidable by Valeriote's subtype theorem.
**Reduced Problem**
Is there a poly-time algorithm that takes a finite
idempotent algebra $\mathbf{A}$ and decides whether
$\mathbb{V}(\mathbf{A})$ has nonempty type-2 tails?
Strategy
*Kearnes:*
Having a diff term is characterized by omitting 1's and type-2 tails.
*Valeriote:*
Omitting 1's is poly-time decidable by the subtype theorem.
*To show:*
Type-2 tails, when they occur in $\mathbb{V}(\mathbf{A})$,
are easy to find.
Suppose $\mathbb{V}(\mathbf A)$ omits type 1 and contains a finite algebra $\mathbf{B}$ with a type-2 prime quotient $\alpha \prec \beta$ such that the $\langle \alpha, \beta \rangle$-minimal sets have nonempty tails.
Then there exists a 3-generated subalgebra of $\mathbf A \times \mathbf A$
with this property.
**Conclusion:**
to check for type-2 tails in $\mathbb{V}(\mathbf A)$ it suffices to look in 3-generated subalgebras of $\mathbf A \times \mathbf A$.
Let $S$ be a finite set of finite idempotent algebras that is closed
under taking subalgebras.
Assume $\mathbb{V}(S)$ omits type 1.
Suppose there is a finite $\mathbf{B} \in \mathbb{V}(S)$ with type-2
tails.
WLOG Assume $\mathbf B$ is a subdirect
product of a finite number of members of $S$.
Choose $n$ minimal such that for
some $\mathbf{A}_1$, $\mathbf{A}_2$, $\dots$, $\mathbf{A}_n$ in $S$,
$$\mathbf{B} \leq_s \prod \mathbf{A}_i$$
Under the assumption that $n > 1$ we will prove $n = 2$.
Minimality Assumptions
For this $n$, select the $\mathbf{A}_i$ and $\mathbf{B}$ so that $|B|$ is as small as possible.
Let $\alpha \prec \beta$ be a type-2 prime quotient of $\mathbf B$
whose minimal sets have nonempty tails, and choose $\beta$ minimal with respect to this property.
This implies $\beta$ is join
irreducible and $\alpha$ is its unique subcover (HM Lemma 6.2).
Fourfold Path to a Proof
Lemma 1
Suppose $U$ is an $\langle \alpha, \beta \rangle$-minimal set and $0, 1 \in U$ and $(0,1) \in \beta - \alpha$ and $t \in Tail(U)$
Then $\beta = \operatorname{Cg}(0,1)$ and $\mathbf B$ is generated by ${0, 1, t}$.
**Lemma 2**
For every $s \subset [n]$,
either $\beta \leq \rho_s$ or $\alpha \vee \rho_s = 1_B$.