What is the weakest idempotent Maltsev condition that implies that abelian tolerances generate abelian congruences?

Abstract

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We answer the question in the title. In the process, we correct an error in our AMS Memoir The Shape of Congruence Lattices.

1 Introduction

Our 2013 AMS Memoir [3] contains the following claim.

Theorem 3.24 of [3]. Assume that $\mathcal V$ satisfies a nontrivial idempotent Maltsev condition. If $\textbf{A}\in \mathcal {V}$ has a tolerance T and a congruence $\delta$ such that $\textbf{C}(T,T;\delta )$ holds, then $\textbf{C}({\textrm{Cg}}^{\textbf{A}} (T),{\textrm{Cg}}^{\textbf{A}} (T);\delta )$ holds.

In the case where $\delta =0$, the conclusion of this theorem asserts that if T is an abelian tolerance, then it generates an abelian congruence.

The recent paper [2] of the first author contains the following claim.

Theorem 4.5 of [2]. The following are equivalent for a variety $\mathcal V$.

• (1) $\mathcal V$ has a weak difference term.

• (2) Whenever $\textbf{A}\in \mathcal V$ and $\alpha \in {{\,\textrm{Con}\,}}(\textbf{A})$ is abelian, the interval $I[0,\alpha ]$ consists of permuting equivalence relations.

• (3) Whenever $\textbf{A}\in \mathcal V$ and $\alpha \in {{\,\textrm{Con}\,}}(\textbf{A})$ is abelian, the interval $I[0,\alpha ]$ is modular.

• (4) Whenever $\textbf{A}\in \mathcal V$ and $\alpha \in {{\,\textrm{Con}\,}}(\textbf{A})$, there is no pentagon labeled as in Figure 1 with $[\alpha ,\alpha ]=0$. 1 with $[\alpha ,\alpha ]=0$. (No “spanning pentagon” in $I[0,\alpha ]$.)

• (5) Whenever $\textbf{A}\in \mathcal V$ and $\alpha \in {{\,\textrm{Con}\,}}(\textbf{A})$ is abelian, there is no pentagon labeled as in Figure 1 where $[\alpha ,\alpha ]=0$ and $\textbf{C}(\theta ,\alpha ;\delta )$.

Fig. 1

Forbidden sublattice if both $[\alpha ,\alpha ]=0$ and $\textbf{C}(\theta ,\alpha ;\delta )$ hold

During the final editing of [2], we realized that Theorem 3.24 of [3] is not consistent with Theorem 4.5 of [2]. A close examination revealed an error in the proof of Theorem 3.24 of [3]. In this paper, we identify the error and correct Theorem 3.24 and correct the results of [3] which depend on Theorem 3.24. We include a short summary of the affected results at the end of Section 4.

2 What is the error?

As we have written above, Theorem 3.24 of [3] is false as stated. In this section we identify the error in the proof of Theorem 3.24. In later sections of this article we will correct Theorem 3.24 and correct the results of [3] that depend on Theorem 3.24.

Throughout this article we use the terminology and notation of the Memoir, [3]. The arguments there and here rely on the basic properties of the centralizer relation, which are enumerated in the following theorem.

Theorem 2.19 of [3]. Let $\textbf{A}$ be an algebra with tolerances S, $S'$, T, $T'$ and congruences $\alpha$, $\alpha _i$, $\beta$, $\delta$, $\delta '$, $\delta _j$. The following are true.

• (1) (Monotonicity in the first two variables) If $\textbf{C}(S,T;\delta )$ holds and $S'\subseteq S$, $T'\subseteq T$, then $\textbf{C}(S',T';\delta )$ holds.

• (2) $\textbf{C}(S,T;\delta )$ holds if and only if $\textbf{C}({\textrm{Cg}}^{\textbf{A}}(S),T;\delta )$ holds.

• (3) $\textbf{C}(S, T; \delta )$ holds if and only if $\textbf{C}(S, \delta \circ T\circ \delta ; \delta )$ holds.

• (4) If $T\cap \delta = T\cap \delta '$, then $\textbf{C}(S,T;\delta )\;\Longleftrightarrow \; \textbf{C}(S,T;\delta ')$.

• (5) (Semidistributivity in the first variable) If $\textbf{C}(\alpha _i,T;\delta )$ holds for all $i\in I$, then $\textbf{C}(\bigvee _{i\in I}\alpha _i,T;\delta )$ holds.

• (6) If $\textbf{C}(S,T;\delta _j)$ holds for all $j\in J$, then $\textbf{C}(S,T;\bigwedge _{j\in J}\delta _j)$ holds.

• (7) If $T\cap \big (S\circ (T\cap \delta )\circ S\big ) \subseteq \delta$, then $\textbf{C}(S,T;\delta )$ holds.

• (8) If $\beta \wedge \big (\alpha \vee (\beta \wedge \delta )\big ) \le \delta$, then $\textbf{C}(\alpha ,\beta ;\delta )$ holds.

• (9) Let $\textbf{B}$ be a subalgebra of $\textbf{A}$. If $\textbf{C}(S, T; \delta )$ holds in $\textbf{A}$, then $\textbf{C}(S|_{\textbf{B}}, T|_{\textbf{B}}; \delta |_{\textbf{B}})$ holds in $\textbf{B}$.

• (10) If $\delta '\le \delta$, then the relation $\textbf{C}(S, T; \delta )$ holds in $\textbf{A}$ if and only if $\textbf{C}(S/\delta ', T/\delta '; \delta /\delta ')$ holds in $\textbf{A}/\delta '$.

The proof of [3, Theorem 3.24] is a little longer than a page and a half. Just before the half-page mark we read

Claim 3.25. $\Delta \cap R_i=0$ is the equality relation for both $i = 1$ and 2.

This claim is proved correctly. After the statement of Claim 3.26 of [3], the proof proceeds with

From Claim 3.25 and Theorem 2.19 (8) we get that $\textbf{C}(R_i,\Delta ;0)$ holds for $i = 1$ and 2.

The relations $\alpha ,\beta ,\delta$ that appear in the statement of Theorem 2.19 (8) are congruences. Suppose for a moment that $\Delta$ and $R_i$ from Claim 3.25 are congruences. Then we could label $\alpha :=R_i$, $\beta :=\Delta$, and $\delta :=0$ in Claim 3.25 and derive that

$$\begin{aligned} \beta \wedge (\alpha \vee (\beta \wedge \delta ))&=\Delta \wedge (R_i\vee (\Delta \wedge 0))\\&=\Delta \wedge R_i\\&=0\\&\le \delta , \end{aligned}$$

and we could indeed conclude from Theorem 2.19 (8) that $\textbf{C}(R_i,\Delta ;0)$ holds, and the proof could be concluded as we did in [3]. Unfortunately, in the proof of Theorem 3.24 of [3], $R_i$ is only a tolerance and not necessarily a congruence (it is a reflexive, symmetric, compatible, binary relation, but it need not be a transitive relation). This means that Theorem 2.19 (8) does not apply to the situation. Theorem 2.19 (7) is more appropriate for tolerances. Using it, one may correctly derive from Claim 3.25 (i.e., from the claim that $\Delta \cap R_i=0$ holds) that $\textbf{C}(\Delta ,R_i;0)$ holds. But this is different than $\textbf{C}(R_i,\Delta ;0)$, which is what we needed for the rest of the proof of Theorem 3.24. If we had proved a slightly stronger statement in Claim 3.25, namely that $\Delta \cap (R_i\circ R_i)=0$, then we could correctly derive from Theorem 2.19 (7) that $\textbf{C}(R_i,\Delta ;0)$ holds. But, as it happens, there is no way to correct the proof of Theorem 3.24, since the statement of the theorem is false, as we will see. It will be corrected and strengthened in Theorem 3.1 of the next section.

3 The main correction

Let $\mathcal V$ be a variety. Let $\mathscr {A}$ be the property “for every $\textbf{A}\in \mathcal {V}$ and every tolerance T on $\textbf{A}$, if T is abelian, then the congruence on $\textbf{A}$ generated by T is abelian.” Let ${\mathscr {T}}$ denote the Maltsev definable property “there exists a Taylor term”. Let ${\mathscr {W}}$ be the Maltsev definable property “there exists a weak difference term”. The incorrect result [3, Theorem 3.24] asserts that for any variety $\mathcal V$, ${\mathscr {T}}\Rightarrow {\mathscr {A}}$. If true, this would imply that the conjunction of the properties ${\mathscr {T}}$ and ${\mathscr {A}}$ is equivalent to the property ${\mathscr {T}}$. We shall learn in Theorem 3.1 that the conjunction of the properties ${\mathscr {T}}$ and ${\mathscr {A}}$ is equivalent to the property ${\mathscr {W}}$. This is enough to disprove [3, Theorem 3.24], since it is known that ${\mathscr {W}}$ is slightly (but properly) stronger that ${\mathscr {T}}$.

In more detail, a Taylor term for a variety $\mathcal V$ is a $\mathcal V$-term $T(x_1,\ldots ,x_{n})$ such that $\mathcal V$ satisfies the identity $T(x,\ldots ,x)\approx x$ and, for each i between 1 and n, $\mathcal V$ satisfies some identity of the form $T({\textbf{w}})\approx T({\textbf{z}})$ with $w_i \ne z_i$. A weak difference term for $\mathcal V$ is a $\mathcal V$-term w(x, y, z) such that for any $\textbf{B}\in \mathcal {V}$, $w^{\textbf{B}}(a,a,b) = b=w^{\textbf{B}}(b,a,a)$ holds whenever the pair (a, b) is contained in an abelian congruence. It is known that the class of varieties with a weak difference term is definable by a nontrivial idempotent Maltsev condition (see [4, Theorem 4.8]). It is known that the class of varieties with a Taylor term is definable by the weakest nontrivial idempotent Maltsev condition. (This is derivable from Corollaries 5.2 and 5.3 of [6].) In particular, this proves that ${\mathscr {W}}\Rightarrow {\mathscr {T}}$. It is known that ${\mathscr {T}}\not \Rightarrow {\mathscr {W}}$ for arbitrary varieties (see [2, Example 4.4]). It is also known that ${\mathscr {W}}\Leftrightarrow {\mathscr {T}}$ for locally finite varieties (see [1, Theorem 9.6]). These facts, (i) ${\mathscr {W}}\Rightarrow {\mathscr {T}}$, (ii) ${\mathscr {T}}\not \Rightarrow {\mathscr {W}}$, (iii) ${\mathscr {W}}\Leftrightarrow {\mathscr {T}}$ for locally finite varieties, are the data supporting our statement that “$\mathscr {W}$ is slightly (but properly) stronger than $\mathscr {T}$”.

The next theorem proves that ${\mathscr {T}}\Rightarrow ({\mathscr {W}}\Leftrightarrow {\mathscr {A}})$. Since ${\mathscr {W}}\Rightarrow {\mathscr {T}}$, it is possible to derive ${\mathscr {W}}\Leftrightarrow ({\mathscr {T}}\; \& \; {\mathscr {A}})$ from this. In the terminology of [2], this means that the property ${\mathscr {A}}$ is Maltsev definable relative to ${\mathscr {T}}$ by the Maltsev condition that defines $\mathscr {W}$.

Theorem 3.1

Let $\mathcal V$ be a variety that has a Taylor term (i.e., $\mathcal V$ satisfies ${\mathscr {T}}$). The following are equivalent properties for $\mathcal V$:

• (1) $\mathcal V$ has a weak difference term. ($\mathcal V$ satisfies ${\mathscr {W}}$.)

• (2) $\mathcal V$ has term w(x, y, z) such that, whenever $\textbf{A}\in \mathcal {V}$, T is an abelian tolerance on $\textbf{A}$, and $(a,b)\in T$, then $w^{\textbf{A}}(a,a,b) = b = w^{\textbf{A}}(b,a,a)$. (w is a “weak difference term for tolerances”.)

• (3) In any algebra of $\mathcal V$, any abelian tolerance is a congruence.

• (4) In any algebra of $\mathcal V$, the congruence generated by an abelian tolerance is abelian. ($\mathcal V$ satisfies ${\mathscr {A}}$.)

Proof

Assume that Item (1) holds. The class of varieties with a weak difference term is definable by an idempotent Maltsev condition, say $\Sigma$. Let $\mathcal {V}^{\circ }$ be the idempotent reduct of $\mathcal V$. $\mathcal {V}^{\circ }$ also satisfies $\Sigma$, hence must have a weak difference term, say w(x, y, z). Now choose $\textbf{A}\in \mathcal {V}$, let T be an abelian tolerance of $\textbf{A}$, and choose $(a,b)\in T$. Observe that $\{a,b\}^2 = \{(a,a),(a,b),(b,a),(b,b)\}\subseteq T$, since T is reflexive and symmetric. By Zorn’s Lemma, we may extend the subset $\{a,b\}\subseteq A$ to a subset $B\subseteq A$ maximal for (i) $\{a,b\}\subseteq B$ and (ii) $B^2\subseteq T$ (B is a T-block). The maximality of B and the fact that T is a tolerance jointly guarantee that B is closed under all idempotent term operations of $\textbf{A}$. Thus, if $\textbf{A}^{\circ }$ denotes the idempotent reduct of $\textbf{A}$, then B is a subuniverse of $\textbf{A}^{\circ }$. The fact that T is an abelian tolerance of $\textbf{A}$ implies that T is an abelian tolerance of the reduct $\textbf{A}^{\circ }$, hence $T|_{B}$ is an abelian tolerance of $\textbf{B}\;(\le \textbf{A}^{\circ })$. Since $B^2\subseteq T$, it follows that $T|_B$ is the universal binary relation on B, hence $\textbf{B}$ is an abelian algebra. We have that w(x, y, z) is a weak difference term for $\mathcal {V}^{\circ }$, $\textbf{B}\in \mathcal {V}^{\circ }$, and $\{a,b\}\subseteq B$, so $w^{\textbf{B}}(a,a,b)=b=w^{\textbf{B}}(b,a,a)$. Since w(x, y, z) is a term of $\textbf{A}$, we have $w^{\textbf{A}}(a,a,b)=b=w^{\textbf{A}}(b,a,a)$ whenever $(a,b)\in T$. This establishes that w(x, y, z) is a weak difference term for tolerances for the variety $\mathcal V$, proving that Item (2) holds.

Now assume that Item (2) holds. Choose any $\textbf{A}\in \mathcal {V}$ and any abelian tolerance T of $\textbf{A}$. Our goal is to prove that T is transitive, so choose elements satisfying $a\;T\;b\;T\;c$. Since $(a,b),(b,c)\in T$ and T is a reflexive subalgebra of $\textbf{A}^2$ we have $w^{T}((a,b),(b,b),(b,c))\in T$. Since w(x, y, z) is a weak difference term for tolerances and T is an abelian tolerance containing (a, b) and (b, c), we have $w^{T}((a,b),(b,b),(b,c))=(w^{\textbf{A}}(a,b,b),w^{\textbf{A}}(b,b,c))=(a,c)$, proving that $(a,c)\in T$. This shows that Item (3) holds.

The implication (3)$\Rightarrow$(4) holds since the congruence generated by a congruence is itself.

To prove the final implication (4)$\Rightarrow$(1) we explain instead why $\lnot (1) \Rightarrow \lnot (4)$ holds. Assume that Item (1) fails. According to Theorem 4.5 of [2], which is stated in the Introduction of this article, some $\textbf{A}\in \mathcal V$ must have a pentagon in its congruence lattice labeled as in Figure 1, and for this pentagon we may assume that $[\alpha ,\alpha ]=0$ and $\textbf{C}(\theta ,\alpha ;\delta )$. Let $T = (\delta \circ \beta \circ \delta )/\delta$. We will complete the proof that (4) fails by arguing that T is an abelian tolerance on $\textbf{A}/\delta$ which does not generate an abelian congruence on $\textbf{A}/\delta$.

For the congruences depicted in Figure 1, we have $[\alpha ,\alpha ]=0$ and $\beta \le \alpha$, so $[\beta ,\beta ]=0$ by monotonicity of the commutator. In terms of the centralizer relation, this means $\textbf{C}(\beta ,\beta ;0)$ holds. Now $\textbf{C}(\beta ,\beta ;0)$ is equivalent to $\textbf{C}(\beta ,\beta ;\delta )$ by Theorem 2.19 (4) above, so $\textbf{C}(\beta ,\beta ;\delta )$ must also hold. We also have $\textbf{C}(\delta ,\beta ;\delta )$ by Theorem 2.19 (8). The statements $\textbf{C}(\beta ,\beta ;\delta )$ and $\textbf{C}(\delta ,\beta ;\delta )$ may be combined to $\textbf{C}(\beta \vee \delta ,\beta ;\delta )$ by Theorem 2.19 (5). By Theorem 2.19 (3) we derive that $\textbf{C}(\beta \vee \delta ,\delta \circ \beta \circ \delta ;\delta )$ holds. If S is the tolerance $\delta \circ \beta \circ \delta$ on $\textbf{A}$, then since $S\subseteq \beta \vee \delta$ we derive from $\textbf{C}(\beta \vee \delta ,\delta \circ \beta \circ \delta ;\delta )$ and Theorem 2.19 (1) that $\textbf{C}(S, S; \delta )$ holds. Now, by Theorem 2.19 (10), we derive that $\textbf{C}(S/\delta ,S/\delta ;0)$ holds. But $T=S/\delta$, so $\textbf{C}(T,T;0)$ holds. We have proved that T is an abelian tolerance of $\textbf{A}/\delta$.

We argue now that the congruence generated by T is not abelian. The congruence generated by $T=(\delta \circ \beta \circ \delta )/\delta$ is its transitive closure, which is $(\beta \vee \delta )/\delta =\alpha /\delta$. To prove that this congruence on $\textbf{A}/\delta$ is not abelian, it suffices to prove that $\textbf{C}(\alpha ,\alpha ;\delta )$ fails in $\textbf{A}$. Since $\beta \le \alpha$ and $\theta \le \alpha$, and the centralizer relation is monotone in its first two variables (Theorem 2.19 (1)), it suffices to prove that $\textbf{C}(\beta ,\theta ;\delta )$ fails in $\textbf{A}$. This is the one place in the proof where we invoke the fact that $\mathcal V$ has a Taylor term. It is proved in the first part of [3, Theorem 4.16 (2)] that if $\textbf{A}$ is an algebra in a variety with a Taylor term which has a pentagon in its congruence lattice labeled as in Figure 1, then $\textbf{C}(\beta ,\theta ;\delta )$ cannot hold. This establishes $\lnot (4)$.

We emphasize the most relevant aspect of the previous theorem in the form of a corollary.

Corollary 3.2

Theorem 3.24 of [3] is correct if the hypothesis that $\mathcal V$ has a Taylor term (or “$\mathcal V$ satisfies a nontrivial idempotent Maltsev condition”) is strengthened to the hypothesis that $\mathcal V$ has a weak difference term.

4 Consequences

Some of the later results in [3] depend on the incorrect Theorem 3.24. We identify them here and indicate whether they are true as stated, and explain how to adjust the statements that are not true as stated.

• 1. ([3, Theorem 3.27]) This theorem examines the following property of a variety $\mathcal V$: given two perspective congruence intervals $\alpha /(\alpha \wedge \beta )$ and $(\alpha \vee \beta )/\beta$ of some $\textbf{A}\in \mathcal V$, one interval is abelian if and only the other is abelian. Call this property $\mathscr {B}$. [3, Theorem 3.27] asserts that ${\mathscr {T}}\Rightarrow {\mathscr {B}}$. This implication is false. Using Corollary 3.2, we can correct Theorem 3.27 by strengthening the hypothesis $\mathscr {T}$ to the hypothesis $\mathscr {W}$. The corrected version then expresses that ${\mathscr {W}}\Rightarrow {\mathscr {B}}$. In fact, it is not hard to see that ${\mathscr {W}}\Leftrightarrow ({\mathscr {T}}\; \& \; {\mathscr {B}})$. (See Theorem 5.1 (2) below.)

• 2. ([3, Theorem 4.16]) This theorem claims that certain finite lattices with some specified centralities cannot occur as sublattices of congruence lattices of algebras in varieties with a Taylor term. The theorem has three parts and each of these parts has more than one claim. Part (1) of the theorem involves three claims.

  • Each of the three claims of 4.16 (1) are true as stated.

  • The first and third of the three claims of 4.16 (2) are true as stated. The second claim of 4.16 (2) is false.

  • The two claims of 4.16 (3) are derived from the false claim of 4.16 (2), hence we withdraw these two claims. We have neither proof nor counterexample for these claims.

  • All claims of all items in Theorem 4.16 are true when the hypothesis “$\mathcal {V}$ has a Taylor term” is strengthened to “$\mathcal {V}$ has a weak difference term”. (However, the two claims of 4.16 (3) were already known to be true in the presence of a weak difference term, cf. [5, Corollary 5.8].)

  We give more detail about the second bullet point. Using the labeling in Figure 1, it is stated in [3, Theorem 4.16 (2)] that a pentagon of congruences where $\textbf{C}(\beta ,\theta ;\delta )$ holds cannot occur as a sublattice in a variety with a Taylor term. The proof given for this is correct. But then it is stated that a pentagon of congruences where $\textbf{C}(\beta ,\beta ;\beta \wedge \delta )$ holds cannot occur as a sublattice. Call this property ${\mathscr {C}}$. The proof given that ${\mathscr {T}}\Rightarrow {\mathscr {C}}$ depends on [3, Theorem 3.27], which (we have seen) is false without a weak difference term. In fact, it can be shown by arguments like those above that ${\mathscr {W}}\Leftrightarrow ({\mathscr {T}}\; \& \; {\mathscr {C}})$. (See Theorem 5.1 (3) below.) To summarize, the part of [3, Theorem 4.16 (2)] that refers to $\textbf{C}(\beta ,\beta ;\beta \wedge \delta )$ is false as stated, but becomes true when the hypothesis “$\mathcal {V}$ has a Taylor term” is strengthened to “$\mathcal {V}$ has a weak difference term”.

• 3. ([3, Lemma 6.7]) This lemma cites [3, Theorem 3.24] in its proof, but one of the hypotheses of [3, Lemma 6.7] is that the variety has a weak difference term. In this setting, [3, Theorem 3.24] has been shown to hold in Corollary 3.2 above. Consequently, [3, Lemma 6.7] is true as stated.

• 4. ([3, Theorem 6.25]) The proof of [3, Theorem 6.25] refers to Theorem 3.27 immediately before Claim 6.26 to complete part of the argument. An alternative argument is suggested in that proof which uses Lemma 6.10 instead. This alternative argument does not rely on Theorem 3.24. Moreover, the hypothesis of [3, Theorem 6.25] includes that $\mathcal V$ has a weak difference term, and in this setting Corollary 3.2 allows us to leave the proof as it is. Thus, [3, Theorem 6.25] is true as stated.

• 5. ([3, Theorem 8.1]) This theorem refers to [3, Theorem 3.24] in its proof of (4)$\Rightarrow$(5). The reference can be eliminated, as we now explain. Item (4) is the assertion that $\mathcal V$ contains no algebra with a nontrivial abelian congruence. Item (5) is the assertion that $\mathcal V$ contains no algebra with a nontrivial abelian tolerance. What we need to establish, therefore, is that any variety which omits nontrivial abelian congruences also omits nontrivial abelian tolerances. We can use Theorem 3.1 above to do this. Assume that Item (4) of [3, Theorem 8.1] holds. Since $\mathcal V$ contains no algebra with a nontrivial abelian congruence, the ternary projection $w(x,y,z):=x$ is a weak difference term for $\mathcal V$. Any variety with a weak difference term has a Taylor term, so Theorem 3.1 applies to $\mathcal V$. In particular, from the implication (1)$\Rightarrow$(3) of Theorem 3.1 we see that an abelian tolerance on an algebra in $\mathcal V$ is a congruence. Since abelian congruences in $\mathcal V$ are trivial, abelian tolerances in $\mathcal V$ are also trivial, so Item (5) of [3, Theorem 8.1] holds. This shows that [3, Theorem 8.1] is true as stated.

4.1 Summary

• 1. ([3, Theorem 3.24]) is false as stated. Theorem 3.24 is true when the hypothesis “$\mathcal {V}$ has a Taylor term” is strengthened to “$\mathcal {V}$ has a weak difference term”.

• 2. ([3, Theorem 3.27]) is false as stated. Theorem 3.27 is true when the hypothesis “$\mathcal {V}$ has a Taylor term” is strengthened to “$\mathcal {V}$ has a weak difference term”.

• 3. ([3, Theorem 4.16])

  • Each of the three claims of 4.16 (1) are true as stated.

  • The first and third of the three claims of 4.16 (2) are true as stated. The second claim of 4.16 (2) is false.

  • We do not know whether the two claims of 4.16 (3) are true.

  • All claims of all items in Theorem 4.16 are true when the hypothesis “$\mathcal {V}$ has a Taylor term” is strengthened to “$\mathcal {V}$ has a weak difference term”.

• 4. ([3, Lemma 6.7]) is true as stated.

• 5. ([3, Theorem 6.25]) is true as stated.

• 6. ([3, Theorem 8.1]) is true as stated.

5 Afterword

We close this note by recording the results of the form ${\mathscr {W}}\Leftrightarrow ({\mathscr {T}}\; \& \; {\mathscr {X}})$, where ${\mathscr {X}} = {\mathscr {A}}, {\mathscr {B}}$ or ${\mathscr {C}}$, which we have considered in this paper, in the form of a theorem statement. Results of the form ${\mathscr {W}}\Leftrightarrow ({\mathscr {T}}\; \& \; {\mathscr {X}})$, assert that Property $\mathscr {X}$ is Maltsev definable relative to $\mathscr {T}$ (= existence of a Taylor term) by the Maltsev condition that defines $\mathscr {W}$ (= existence of a weak difference term).

Observe that ${\mathscr {W}}~\Leftrightarrow ~({\mathscr {T}}\; \& \; {\mathscr {X}})$ is equivalent to the conjunction of (i) ${\mathscr {W}}~\Rightarrow ~{\mathscr {T}}$, (ii) ${\mathscr {W}}~\Rightarrow ~{\mathscr {X}}$, and (iii) $({\mathscr {T}}\; \& \; {\mathscr {X}})~\Rightarrow ~{\mathscr {W}}$. Implication (i) is known to be true: the class of varieties with a weak difference term is definable by a nontrivial idempotent Maltsev condition (cf. [4, Theorem 4.8]) and the class of varieties with a Taylor term is definable by the weakest nontrivial idempotent Maltsev condition (cf. Corollaries 5.2 and 5.3 of [6].) Implication (iii) may be rewritten in the form (iii)’ $({\mathscr {T}}\; \& \;\lnot {\mathscr {W}})\Rightarrow \lnot {\mathscr {X}}$. Altogether, we want to record why

• (ii) ${\mathscr {W}}~\Rightarrow ~{\mathscr {X}}$, and

• (iii)’ $({\mathscr {T}}\; \& \; \lnot {\mathscr {W}})\Rightarrow \lnot {\mathscr {X}}$

hold when ${\mathscr {X}} = {\mathscr {A}}, {\mathscr {B}}$ or ${\mathscr {C}}$.

Theorem 5.1

If $\mathcal V$ has a Taylor term, then any one of the following properties will hold for $\mathcal {V}$ if and only if $\mathcal {V}$ has a weak difference term.

• (1) ($\mathcal V$ satisfies ${\mathscr {A}}$.) In any algebra of $\mathcal V$, the congruence generated by an abelian tolerance is abelian.

• (2) ($\mathcal V$ satisfies ${\mathscr {B}}$.) Given two perspective congruence intervals $\alpha /(\alpha \wedge \beta )$ and $(\alpha \vee \beta )/\beta$ of some $\textbf{A}\in \mathcal V$, one interval is abelian if and only the other is abelian.

• (3) ($\mathcal V$ satisfies ${\mathscr {C}}$.) A pentagon of congruences, labeled as in Figure 2, cannot occur as a sublattice of ${{\,\textrm{Con}\,}}(\textbf{A})$ if $\textbf{A}\in \mathcal {V}$ and $\textbf{C}(\beta ,\beta ;\beta \wedge \delta )$ holds.

Fig. 2

Forbidden sublattice if $\textbf{C}(\beta ,\beta ;\beta \wedge \delta )$ holds

Proof

Item (1) is Theorem 3.1$(1)\Leftrightarrow (4)$.

It was noted above in Consequence (1) that the erroneous proof in [3] that $\mathscr {T}\Rightarrow \mathscr {B}$ is nevertheless valid as a proof of $\mathscr {W}\Rightarrow \mathscr {B}$. This gives us (ii) ${\mathscr {W}}~\Rightarrow ~{\mathscr {B}}$, as desired. To prove (iii)’, that is, to prove $({\mathscr {T}}\; \& \; \lnot {\mathscr {W}})\Rightarrow \lnot {\mathscr {B}}$, assume that $\mathcal {V}$ has a Taylor term but not a weak difference term. By Theorem 4.5 (5) of [2], there exists $\textbf{A}\in \mathcal {V}$ with a pentagon in its congruence lattice labeled as in Figure 1. Since $[\alpha ,\alpha ]=0$ in this pentagon, we get that $[\beta ,\beta ]=0$ by monotonicity. Equivalently, the congruence quotient $\beta /(\beta \wedge \delta )$ is abelian. For the purpose of obtaining a contradiction assume that $\mathcal {V}$ satisfies $\mathscr {B}$. This allows us to deduce that the perspective interval $(\beta \vee \delta )/\delta = \alpha /\delta$ is also abelian. This is equivalent to the statement that $\textbf{C}(\alpha ,\alpha ;\delta )$ holds, and therefore we derive that $\textbf{C}(\beta ,\theta ;\delta )$ holds from Theorem 2.19 of [3] (i.e., by the monotonicity of the centralizer relation in the first two variables). But according to Theorem 4.2 of [2], no such pentagon can exist in the congruence lattice of any algebra with a Taylor term. This contradiction establishes Item (2).

To prove Item (3), we assert that that the erroneous proof in [3] of Theorem 4.16 (2) (i), which claims that $\mathscr {T}\Rightarrow \mathscr {C}$, is nevertheless valid as a proof of $\mathscr {W}\Rightarrow \mathscr {C}$. Indeed, the proof of $\mathscr {T}\Rightarrow \mathscr {C}$ from [3] is actually a valid proof that $\mathscr {B}\Rightarrow \mathscr {C}$ coupled together with an erroneous reference to $\mathscr {T}\Rightarrow \mathscr {B}$. We have argued above that $\mathscr {W}\Rightarrow \mathscr {B}$, so coupling this with the valid proof that $\mathscr {B}\Rightarrow \mathscr {C}$ yields $\mathscr {W}\Rightarrow \mathscr {C}$.

To complete the proof of this theorem, we must prove (iii)’ $({\mathscr {T}}\; \& \; \lnot {\mathscr {W}})\Rightarrow \lnot {\mathscr {C}}$. Assume that $\mathcal {V}$ fails $\mathscr {W}$, i.e. $\mathcal {V}$ does not have a weak difference term. By Theorem 4.5 (5) of [2] we know that there exists $\textbf{A}\in \mathcal {V}$ with congruences labeled as in Figure 1 such that $[\alpha ,\alpha ]=0=\beta \wedge \delta$ and $\textbf{C}(\theta ,\alpha ;\delta )$. Since $[\alpha ,\alpha ]=0=\beta \wedge \delta$ we have $\textbf{C}(\alpha ,\alpha ;\beta \wedge \delta )$ by the definition of the commutator, hence $\textbf{C}(\beta ,\beta ;\beta \wedge \delta )$ by the monotonicity of the centralizer in its first two variables. Thus, $\textbf{A}$ has a pentagon of congruences that violate $\mathscr {C}$. This establishes that $\lnot \mathscr {W}\Rightarrow \lnot \mathscr {C}$, so the weaker implication $(\mathscr {T}\; \& \;\lnot \mathscr {W})\Rightarrow \lnot \mathscr {C}$ also holds. (We have actually established that $\mathscr {W}\Leftrightarrow \mathscr {C}$.)