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.

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

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: enter image description here

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.


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