5
$\begingroup$

All (pseudo)varieties considered here are (pseudo)varieties of monoids.

It is known that any (finite or infinite) monoid that satisfies the identities \begin{equation} xhxyty = xhyxty, \quad xhytxy=xhytyx \tag{$*$} \end{equation} is finitely based. Therefore every subvariety of the variety $\mathbf{V}$ defined by $(\ast)$ is finitely based.$^\dagger$ How about the pseudovariety $\mathbb{V}$ defined by $(\ast)$? Is every subpseudovariety of $\mathbb{V}$ finitely based?

$^\dagger$Each subvariety of $\mathbf{V}$ can be defined by $(\ast)$ together with finitely many of the following identities:

(1)\begin{equation} x^{e_0} \prod_{i=1}^r (h_ix^{e_i}) = x^{f_0}\prod_{i=1}^r (h_ix^{f_i}), \end{equation} where $e_0,f_0,\ldots,e_r,f_r \geq 0$ and $r \geq 0;$

(2)\begin{equation} x^{e_0} y^{f_0} \prod_{i=1}^r (h_ix^{e_i}y^{f_i}) = y^{f_0} x^{e_0} \prod_{i=1}^r (h_ix^{e_i}y^{f_i}), \end{equation} where $\ \ e_0,f_0 \geq 1,\ $ $e_1,f_1,\ldots,e_r,f_r \geq 0,\ $ $\sum_{i=0}^r e_i \geq 2,\ $ $\sum_{i=0}^r f_i \geq 2,\ $ and $\ \ r \geq 0.$

$\endgroup$
2
  • $\begingroup$ Is there a uniform bound on the number of variables needed to define any subvariety? $\endgroup$ Oct 9, 2014 at 20:44
  • $\begingroup$ @Ben: No, there is no uniform bound. But thanks for the question; I can provide a description of identities defining subvarieties of $\mathbf{V}$. $\endgroup$
    – E W H Lee
    Oct 9, 2014 at 21:50

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.