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How to characterize equational theories $T$ which have the following property: for any two terms $t(x_1,...,x_n)$ and $t'(x_1,...,x_n)$ in the signature of $T$, if for any closed terms (i. e. terms without variables) $c_1$, ..., $c_n$ the identity $t(c_1,...,c_n)=t'(c_1,...,c_n)$ follows from (the identities of) $T$ then so does $t(x_1,...,x_n)=t'(x_1,...,x_n)$.

In algebraic terms this means to characterize those varieties of algebras which are generated by their initial algebra (the free algebra on the empty set).

The only widely known example of this that I was able to come up with is the theory/variety of Boolean algebras. But in fact any algebra $A$ in any signature generates its own variety of this kind, by adding to the signature a bunch of constants in the well known way and then generating the subvariety by $A$ itself.

So I am interested in any "intrinsic" (say, category-theoretic) characterization of theories/varieties with the above property, as well as in any other familiar examples of such.

May one hope to actually classify such things up to, say, categorical equivalence?

Much later

After so much time I hardly made any progress on this, but here is one feature which might be useful.

I believe I have a (simple) proof that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

მამუკა ჯიბლაძე
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3 Answers3

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I don't know the answer to this question, but I have a correction and a remark.

The question asks "When is a variety $\mathcal V$ generated by its free algebra over the empty set?" It has been asserted that this question is equivalent to the question of when ${\mathbf F}_{\mathcal V}(\emptyset)$ is existentially closed in $\mathcal V$.

The correction. The assertion is not true. There is an implication, but not an equivalence.

Let ${\mathbf F}_0:={\mathbf F}_{\mathcal V}(\emptyset)$, and assume that $\mathbf F_0$ is existentially closed (e.c.) in $\mathcal V$. $\mathbf F_0$ is embeddable is $\mathbf F_m:=\mathbf F_{\mathcal V}(m)$ for every $m$. If the q.f. formula $s(\bar{x})\neq t(\bar{x})$ is satisfiable in $\mathbf F_m$, then it is satisfiable in $\mathbf F_0$ by the e.c. property. This shows that $\mathbf F_0$ fails every identity that fails in $\mathcal V$, while the fact that $\mathbf F_0\in{\mathcal V}$ shows that it satisfies every identity that holds in $\mathcal V$. Altogether this shows that $\mathbf F_0$ e.c. implies $\mathcal V$ is generated by $\mathbf F_0$.

But the converse is false. The variety of commutative rings is generated by ${\mathbf F}_0 = \mathbb Z$, but this ring is not e.c. in the variety of commutative rings. Moreover, there are many nontrivial varieties generated by their initial algebras, $\mathbf F_0$, where these initial algebras happen to be finite. (E.g., the variety of Boolean algebras, or bounded distributive lattices, or the variety generated by the ring of integers modulo $n$, or any variety generated by the constant expansion of a finite algebra.) But a nontrivial finite algebra cannot be e.c.

The remark. The original poster asks if there is a category-theoretic characterization of these varieties. Well, there is one, since you can express Birkhoff's HSP Theorem category-theoretically. To express $\mathcal V = \mathbf{HSP}(\mathbf F_0)$ you just need to say that every object in $\mathcal V$ is the image of an extremal epimorphism from some object that has a monomorphism into some power of the initial algebra. But, I doubt that there is a nontrivial characterization of these varieties. As the original poster noted, any variety generated by the constant expansion of an algebra has the desired property, so this class of varieties is as varied as the class of constant expansions of algebras.

Keith Kearnes
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Aren't these the same as clones containing the constant functions ?

From a clone, you just take the equational theory consisting of all the functions in it, with equations all the relations satisfied by those functions.

From a theory/variety, just take all the operations restricted to the free algebra on the empty set in that varietey.

There are many nontrivial clones on a two-element set, but if I read the diagram right there are only finitely many containing the constant symbols. Among these are:

The clone of monotone functions gives you the variety of bounded lattices.

The clone of affine functions gives you the variety of $\mathbb F_2$-vector spaces with a marked point.

The clone of conjunctive or disjunctive functions gives you the variety of bounded partial orders with joins or with meets.

The clone of unary functions gives you the variety of sets with an involution and two marked elements that are switched by the involution.

The clone of all Boolean operations gives you the variety of Boolean algebras.

The clone of constant functions gives you the variety of sets with two marked elements.

Will Sawin
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  • Sorry I feel this answer holds the key but don't quite understand why :D You established a correspondence between what and what? On one side you seemingly have all clones on any sets containing all constant maps from these sets to themselves. On the other side you seemingly have certain algebras in certain varieties. Do you mean that these algebras are in one-to-one correspondence with varieties in question? Specifically I am puzzled by your third line. I would somehow presume that the algebra is itself free on an empty set in the corresponding to be pinned down variety, no? – მამუკა ჯიბლაძე Sep 07 '15 at 05:52
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    @მამუკაჯიბლაძე My first appearance of the word "algebra" in the third line was supposed to be "variety", but I mistyped it. Does that clear everything up? – Will Sawin Sep 07 '15 at 13:45
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    I am on the verge of accepting it :D It took me some time to convince myself and then some more to convince myself that no further details need to be added. There remains just a minor question regarding examples - there might be further identities hidden, no? At least with the first example, as was pointed to me today at the seminar, one seemingly actually gets the subvariety of distributive bounded lattices... – მამუკა ჯიბლაძე Sep 09 '15 at 16:22
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    @მამუკაჯიბლაძე Yes, I was not sure I got all the identities. I guess you can check that you get all of them by checking that every algebra embeds into a proudct of copies of ${0,1}$. Wikipedia just told me that every bounded distributive lattice is a lattice of sets, so that's fine. Clearly every $\mathbb F_2$-vector space lives in a product of copise of $\mathbb F_2$. With join or with meet orders there's a similar embedding into sets (this time respecting only the join or only the meet). The last two are easy. – Will Sawin Sep 09 '15 at 18:34
  • Seems to be fine. Except some might not be embeddable into a product of copies of the initial algebra but rather be a subquotient. But all free ones should be so embeddable - in fact, the collection of all $n$-ary functions from your clone must be the free algebra on $n$ generators ($=$ projections). – მამუკა ჯიბლაძე Sep 10 '15 at 12:06
  • Should I go ahead and just insert the word "distributive"? – მამუკა ჯიბლაძე Sep 10 '15 at 12:08
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This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one, $\mathbf{E}$, is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set. (There is a set containing at least one isomorphic copy of each subdirectly irreducible algebra in $\mathscr{V}$.) We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ cannot be represented by a set. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.

Keith Kearnes
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  • Sorry, could you say a little bit more about your second example? What I figured out is that subdirectly irreducibles are the ones with the center cyclic of prime power order. But why is there a proper class of pairwise nonisomorphic such in $\mathsf{HSP}(G_G)$? – მამუკა ჯიბლაძე Mar 13 '23 at 07:45
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    The paper ''Freese, Ralph; McKenzie, Ralph Residually small varieties with modular congruence lattices. Trans. Amer. Math. Soc. 264 (1981), no. 2, 419-430'' proves that a finite algebra in a congruence modular variety that fails the commutator condition that is now called (C1) generates a residually large variety. Any constant expansion of a finite nilpotent group will fail (C1). If you look at page 422, line 17, set $\mu=1$ and $\nu$= the center, you will see how the condition fails. – Keith Kearnes Mar 13 '23 at 09:30
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    I don't see why the center of $G^I/C_{\iota}$ has two elements. Suppose $g\in G-C$ and ${u}\notin \iota$. Let $f\in G^I$ be the function defined by $f(u)=g$ and $f(x)=1$ if $x\neq u$. Then $f\notin C_{\iota}$ since $f\notin C^I$, but $[f,h]\in C_{\iota}$ for any $h\in G^I$. Therefore $f/C_{\iota}$ is a nontrivial central element in $G^I/C_{\iota}$. This method should produce a lot of central elements. – Keith Kearnes Mar 13 '23 at 18:41
  • Yes, thank you, and sorry. Could you please have a look at another attempt: choose any homomorphism ${\hat\prod}:C^I\to C$ that extends the homomorphism ${\prod}:C^{(I)}\to C$ from finitely supported $C$-valued functions to $C$, with ${\prod}(f):=\prod_{i\in I}f(i)$, and let $K$ be the kernel of ${\hat\prod}$. Then $G^I/K$ seems to work as expected. – მამუკა ჯიბლაძე Mar 13 '23 at 19:08
  • Too bad I erased my previous attempt, I should probably wait for your reaction first. – მამუკა ჯიბლაძე Mar 13 '23 at 19:15
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    @მამუკაჯიბლაძე: I think that your second construction of arbitrarily large SI's in the variety generated by the quaternion group is correct. The existence of $\hat{\prod}$ follows from the fact that $C^I$ is elementary abelian, hence may be viewed as a vector space. $G^I/K$ is large since it has a map onto $(G/C)^I$, which is large if $I$ is large. The center of $G^I/K$ can be seen to be $C^I/K(\cong C)$, which has size $2$. These observations are enough to see that your construction produces arbitrarily large SI groups in the variety. – Keith Kearnes Mar 14 '23 at 03:16
  • Thank you. In fact there seems to be a simpler way - just take $G^{(I)}/\operatorname{Ker}(\prod)$, it has cardinality at least that of $I$ anyway. Maybe it is worth adding this as an explicit example to the final part of this answer? – მამუკა ჯიბლაძე Mar 14 '23 at 05:26
  • Oh no $G^{(I)}$ is not in the variety of $G_G$ unless $I$ is finite! Maybe $G^{(I)}/\operatorname{Ker}(\prod)$ still is?... – მამუკა ჯიბლაძე Mar 14 '23 at 06:48
  • Well more precisely $G^{(I)}\subset G^I$ is not a $G_G$-subalgebra. Maybe it still is in $\mathsf{HSP}(G_G)$, I don't know – მამუკა ჯიბლაძე Mar 14 '23 at 06:54
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    @მამუკაჯიბლაძე: If you are trying to build large subdirectly irreducible algebras in the variety $\mathscr{V}(G_G)$ (instead of in $\mathscr{V}(G)$), then your construction $G^I/K$ works, but you have to be more careful about the choice of $\hat{\prod}$. You can take $\hat{\prod}$ to be any extension of $\prod$ to $C^I$ provided that $\hat{\prod}({z}^I) =z$, where $z\in C-{1}$. – Keith Kearnes Mar 14 '23 at 07:23
  • Absolutely, many thanks for this correction! – მამუკა ჯიბლაძე Mar 14 '23 at 07:24
  • So in fact when $I$ is finite this only works right away when its cardinality is odd – მამუკა ჯიბლაძე Mar 14 '23 at 07:33
  • Freese and McKenzie actually pick an element $0\in I$ and take instead of $K$ the subgroup of those $f$ with $f(0)=\prod_{i\ne0}f(i)$, but I believe in our case this is just the same, right? – მამუკა ჯიბლაძე Mar 14 '23 at 07:42
  • One more question, if I may. Seems like just having a diamond in the lattice of congruences suffices for their construction, while in the event of congruence distributivity C1 trivially holds. This makes me wonder - are there any interesting cases of C1 without congruence distributivity? – მამუკა ჯიბლაძე Mar 14 '23 at 07:49
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    @მამუკაჯიბლაძე: A finite group generates a variety satisfying (C1) iff its Sylow subgroups are abelian. – Keith Kearnes Mar 14 '23 at 08:40