# varieties of bi-unary algebras

A bi-unary algebra $\mathbf{A} = \langle A, f, g \rangle$ is an algebra whose operations are unary.

## The problem:

For $I \subseteq \mathbb{N}$, define the set of identities in the language of bi-unary algebras: $$\Sigma_I = \{ fgf^ng^2(x) \approx x : n \in I \} \cup \{ fgf^ng^2(x) \approx fgf^ng^2(y) : n \notin I \}.$$ Prove that $Mod(\Sigma_I)$ is not trivial. (This is equivalent to prove that there exists a bi-unary algebra $\mathbf{A}$ such that $|A| \geq 2$ and $\mathbf{A} \vDash \Sigma_I$; also equivalent to proove that not all identities in this language are logical consequences of the identities in $\Sigma_I$.)

Help would be appreciated :)

## A non-working tentative of solution

There is a paper of Stanley Burris in Algebra Universalis, vol.1 (1971) p.386-392 with title "Models in Equational Theories of Unary Algebras" (note that is it not about bi-unary algebras, but unary algebras in which the number of operations can be any).

It seems (in Theorem 1) to exclude the possibility of coming up with a finite algebra satisfying that set of identities. Indeed, in his proof, when there is a finite algebra satisfying some set of identities, there is actually a two-element algebra in which each operation is either the identity or it is constant.
Clearly, that doesn't work in this example.

## Something that might work...

What could perhaps be another tip is that if $Mod(\Sigma_I)$ is non-trivial, then it has to have a non-trivial subdirectly irreducible algebra.

I don't know of any complete characterization of subdirectly irreducible bi-unary algebras (for mono-unary algebras it's easy...), but I know of a necessary condition.
In his MSc Thesis (available here:http://www.collectionscanada.gc.ca/obj/thesescanada/vol2/002/MR87540.PDF), Jesse Mason proves that for bi-unary algebras, the subdirectly irreducible ones are always connected or pseudo-connected.
Here, connected means that the undirected version of the digraph made from $(a,b)$ if and only if $b \in \{ f(a),g(a) \}$ is connected; pseudo-connected means there are two connected components, one of which is a singleton.
Pseudo-connected algebras are of no interest here, since in the singleton class $x \approx y$ is valid.

• Perhaps to start out, you can try to find a nontrivial model for $I = \{1\}$. – John Coleman Aug 3 '16 at 16:05

## 1 Answer

Let $$\Sigma_I' = \{ fgf^{n}g^{2}(x) \approx x : n \in I \} \cup \{ fgf^{n}g^{2}(x) \approx fgf^{m}g^{2}(y) : n, m \notin I \}.$$ It is clear that $W=Mod(\Sigma_I') \subseteq Mod(\Sigma_I)$.

The following is a tentative of giving a sketch of a non-trivial algebra $\mathbf{F} \in W$, showing that $W$ is non-trivial, thence $V$ is non-trivial.
Actually, I think $\mathbf{F} = \mathbf{F}_W(x)$.

Consider the diagram, which corresponds to the case in which $I$ is the set of even numbers: The continuous lines correspond to $f$, and the dashed ones to $g$.
The $0$ element is not part of the signature, but it is the element that corresponds to the images of all $fgf^ng^2(y)$, with $y \in F$ and $n \notin I$ (notice in $Mod(\Sigma_I)$ we might have different ''zeros'' $(z_n)_{n \notin I}$, one for each $n \notin I$, but we're taking a shortcut here...).
So in this diagram, for each $n \in I$, from $f^ng^2(x)$, $g$ turns left to an element whose image by $f$ is $x$;
for $n \notin I$, from $f^ng^2(x)$, $g$ turns right to an element whose image by $f$ is $0$.

The equations, of course, must be satisfied by every element of the algebra, and not only by $x$ (hence the picture is only a sketch).
In order to achieve this, it is enough to add, for each $y \in F$ a whole new bunch of elements making another diagram isomorphic to this one, and which only repeats the elements $y$ and $0$, and proceed for each newly generated element.

This doesn't seem like a complete description of the algebra, but I find it reasonably satisfactory (at least it convinces me). Yet, I would be happy to see a better solution.