A band is a semigroup where every element is idempotent: $a a = a$. A collection of bands is a variety if it is closed under taking subobjects (subsemigroups), quotient objects (images of homomorphisms) and products (including infinite products). As usual in universal algebra, any variety of bands is defined by a collection of identities. For example, there's a variety of semilattices, which are the bands obeying the identity

$$ a b = b a $$

for all $a,b$.

You can define a variety of bands using more than one identity. But amazingly --- to me, at least --- every variety of bands can in fact be defined using just one identity!

(That is, one identity in addition to associativity $(a b) c = a (b c)$ and the idempotence law $a a = a$.)

This was shown here:

  • Charles Fennemore, All varieties of bands, Semigroup Forum 1 (1970), 172-179.

He even showed that there are exactly $8 + 10(n-2)$ varieties of bands that are determined by an identity involving $n$ variables, for $n \ge 2$... and he has a method for listing these identities explicitly.

The set of varieties of bands is partially ordered, based on whether one identity implies another. It's actually a lattice, as usual in universal algebra, and it seems that Fennemore described the lattice operations explicitly in terms of his chosen identities.

A small portion of this lattice is shown on Wikipedia:

lattice of regular bands

I'm wondering if there's a good explanation of 'why' any variety of bands can be defined using just one identity. Fennemore's proof is not easy for me to follow. Are there are other varieties, besides bands, that have this property?

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    $\begingroup$ The variety generated by any finite group is defined by a single identity. This is a deep theorem of Powell - Oates. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 0:09
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    $\begingroup$ Bands are a bit miraculous. I never understood this myself. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 0:10
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    $\begingroup$ If you go to quasivarieties the situation is different. The result of adjoining an identity to a 2-element left zero semigroup has no finite basis of quasi-identities. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 1:01
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    $\begingroup$ You might look at Gerhard, J. A.; Petrich, Mario Varieties of bands revisited. Proc. London Math. Soc. (3) 58 (1989), no. 2, 323–350 which claims to give a more conceptual approach to the lattice of band varieties. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 1:39
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    $\begingroup$ @Steinberg - Thanks, that reference is helpful though itself scary. It starts: "The description of the lattice of all varieties of bands by Birjukov, Fennemore and Gerhard represents a significant achievement in the study of semigroup varieties. Even though their papers have often been quoted, the concrete material in them was rarely referred to in view of the proofs which are either condensed or are too long. The list of representative identities set up by Fennemore was effectively used by Adair and Sukhanov. In brief, their results have been admired from a distance." $\endgroup$ – John Baez Jun 6 '15 at 20:43

Here is a partial answer.

Claim. If $\mathcal V$ is an idempotent variety, then any subvariety of $\mathcal V$ that is finitely axiomatizable relative to $\mathcal V$ is axiomatizable by a single identity and the identities of $\mathcal V$.

Apply the claim to the variety of bands. This doesn't explain why all varieties of bands are finitely axiomatizable, but it does explain why only one identity is needed in the finitely axiomatizable cases.

Here is the idea behind the claim. Suppose that $s=s'$ and $t=t'$ are two identities, where $s, s'$ are terms in, say, two variables, and $t, t'$ are terms in three variables. If $\mathcal V$ is idempotent, then (modulo the identities of $\mathcal V$) the set $\{s=s', t=t'\}$ is equivalent to the single identity
$$ s(t(u,v,w),t(u',v',w'))=s'(t'(u,v,w),t'(u',v',w')). $$

To explain this, let $M$ be the $2\times 3$ matrix $$ M=\left[ \begin{matrix} u&v&w\\ u'&v'&w' \end{matrix} \right] $$ where the entries are distinct variables. Define a 6-ary term $s\diamond t(M)$ by applying $t$ to the rows of $M$ and then $s$ to the row results. I must explain why $\{s=s', t=t'\}$ is equivalent to $\{s\diamond t(M)=s'\diamond t'(M)\}$.

To derive the diamond identity from the original two, argue from $\{s=s', t=t'\}$ that $$ s\diamond t(M) = s\diamond t'(M) = s'\diamond t'(M). $$

To derive $s=s'$ from the diamond identity, just set variables in $M$ equal along rows, say equal to the first entry: from $s\diamond t(M)=s'\diamond t'(M)$ you obtain $s(u,u')=s'(u,u')$. To derive $t=t'$ from the diamind identity, set variables in $M$ equal along columns. (This paragraph is the part that needs idempotence.)

[Note: lattices are defined with two band operations, and some varieties of lattices are not finitely axiomatizable, so there is still some kind of magic involved in the full result about bands.]


There is likely a good technical explanation to be offered; I'm not feeling up to producing one. I do have some handwaving intuition that might help.

Your example is based in a finite language/signature (one binary operation symbol), so I will restrict myself to finite functional languages, even though some things might hold for countable languages. The subvariety lattice for a given finite signature corresponds in a dual fashion to the lattice of closed (first-order) equational theories in that signature, which in turn correspond to the lattice of certain (fully-invariant, I think) congruences in the free term algebra on countably many generators. This lattice can be pretty woolly, but you've found a nice sublattice of it which corresponds to the subvarieties of bands. Further, this nice sublattice has a property of representation that you pointed out: relative to the congruence that defines or describes all bands, any further congruence containing it is principally and singularly defined, needing just one pair of terms to be defined to generate the rest of it.

Why is this? In the case of bands, there is a simple enough set up: reduce all terms using associativity to a nice normal form (semigroup words), then use idempotency to further reduce possibilities. If I recall correctly, idempotency really cuts down on the available distinct words: in particular, bands are a locally finite variety, so for each n there will be only finitely many distinct terms in n variables to choose from.

Why do you get to use exactly one identity? This is less clear to me, but recall that an identity in n variables implies (by substitution) several identities in m variables for m less than n. There may be a way in the case of bands to order the relatively free terms by implicational strength, so that the result is a linear (or not very wide partial) order. Then any collection of identites can be reduced by some process to one.

The last part involved a bit of arm waving as well. I want to emphasize the point that if you set up an initial base of identities in your theory, you can maneuver into a position where there aren't many more steps to x=y, or triviality. If you start high enough (in the lattice of fully invariant congruences of the term algebra), I think you can carve out many theories which represent their completions in such a nice manner.

  • $\begingroup$ You write: "recall that an identity in $n$ variables implies (by substitution) several identities in $m$ variables for $m$ less than $n$." This reminds me of something else that Fennemore states: "There exist exactly $10(n- 1)$ distinct varieties which can be defined by sets of identities which imply an identity in $n$ variables, $n \gt 2$. $\endgroup$ – John Baez Jun 5 '15 at 23:15
  • $\begingroup$ Each proper subvariety of bands is generated by a single semigroup. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 0:55
  • $\begingroup$ @Steinberg - is your remark related to how each variety of bands is defined by a single equation? I don't see how, but I can't help wondering. $\endgroup$ – John Baez Jun 6 '15 at 16:47
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    $\begingroup$ No it is just to help fill in the picture. It is not true that every proper subvariety of a locally finite variety is finitely generated. $\endgroup$ – Benjamin Steinberg Jun 6 '15 at 21:32
  • $\begingroup$ I was wondering about your comment. Do you mean "generated by a single finite semigroup"? Otherwise one could take a certain (infinite) relatively free algebra as the generating algebra. Gerhard "Looking For Some Nontrivial Answers" Paseman, 2015.06.06 $\endgroup$ – Gerhard Paseman Jun 7 '15 at 0:23

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