The Word Problem for Finitely Presented Quandles is Undecidable

Abstract

This work presents an algorithmic reduction of the word problem for recursively presented groups to the word problem for recursively presented quandles. The faithfulness of the reduction follows from the conjugation quandle construction on groups. It follows that the word problem for recursively presented quandles is not effectively computable, in general. This article also demonstrates that a recursively presented quandle can be encoded as a recursively presented rack. Hence the word problem for recursively presented racks is also not effectively computable.

Appendices

A Identities

The right cancellation rules have a very nice symmetry with respect to the operators $*$ and  / . That is, exchanging the roles of $*$ and  /  in one rule yields the other. This also holds for idempotence in the theory of quandles and the right self-distributivity rule in quandles and racks. A complete set of such rules are presented below.

1.1 A.1 Idempotence

Assuming the quandle identities, one may reason as follows:

$$x/x = (x*x)/x = x.$$

Hence, the  /  operator is also idempotent according to the theory of quandles.

1.2 A.2 Right Distributivity

This section presents the remaining right distributive quandle and rack identities.

• 1. $(x*y)/z = (x/z)*(y/z)$: First note that by the second right cancellation rule and right self-distributivity,

  $$\begin{aligned} (x*y)&= ((x/z)*z)*((y/z)*z)\\&= ((x/z)*(y/z))*z. \end{aligned}$$

  Therefore by the first right cancellation rule,

  $$\begin{aligned} (x*y)/z&= (((x/z)*(y/z))*z)/z\\&= (x/z)*(y/z). \end{aligned}$$

• 2. $(x/y)*z = (x*z)/(y*z)$: By right distributivity and the second right cancellation rule,

  $$\begin{aligned} ((x/y)*z)*(y*z)&= ((x/y)*y)*z\\&= x*z. \end{aligned}$$

  This means that

  $$\begin{aligned} (x/y)*z&= (((x/y)*z)*(y*z))/(y*z)\\&= (x*z)/(y*z). \end{aligned}$$

• 3. $(x/y)/z = (x/z)/(y/z)$: By employing second right cancellation rule and the first right distributivity law of this section, one reasons

  $$\begin{aligned} x/y&= ((x/z)*z)/((y/z)*z)\\&= ((x/z)/(y/z))*z. \end{aligned}$$

  Then the first cancellation rule ensures

  $$\begin{aligned} (x/y)/z&= (((x/z)/(y/z))*z)/z\\&= (x/z)/(y/z). \end{aligned}$$

B $\mathbf {R}_{\mathbf {Q}}$ Is the Same Algebra as $\mathbf {Q}$

Let $\mathbf {Q}= \langle {A}|E\rangle$ be a finite quandle presentation and

$$\mathbf {R}_{\mathbf {Q}}= \langle {A}|E\cup \{a*a = a, a/a = a| a \in A\}\rangle $$

be a finite rack presentation.

The additional conditions on $\mathbf {R}_{\mathbf {Q}}$ are both sufficient and necessary to the condition that $q*q = q$ and $q/q = q$ for all $q \in \mathbf {R}_{\mathbf {Q}}$. The proof of such follows by structural induction on the quandle/rack expression q. The base cases in which $q = a \in A$ are direct consequences of the new part of the rack presentation. Next suppose that $q = q_1*q_2$ with induction hypotheses $q_1*q_1 = q_1$, $q_1/q_1 = q_1$, $q_2*q_2 = q_2$, and $q_2/q_2 = q_2$. Then

$$\begin{aligned} q*q&= (q_1*q_2)*(q_1*q_2)\\&= (((q_1*q_2)/q_2)*q_2)*(q_1*q_2)\\&= (((q_1*q_2)/q_2)*q_1)*q_2\\&= (q_1*q_1)*q_2\\&= q_1*q_2\\&= q, \end{aligned}$$

which employs the second right cancellation rule, right self-distributivity, the first right cancellation rule, and the induction hypothesis on $q_1$, in that order. The case $q = q_1/q_2$ proceeds in a similar fashion. It follows that $\mathbf {R}_{\mathbf {Q}}$ is idempotent and so a quandle.

Since $\mathbf {R}_{\mathbf {Q}}$ is a quandle and satisfies the equations in E, there exists a unique quandle homomorphism $\alpha :\mathbf {Q}\rightarrow \mathbf {R}_{\mathbf {Q}}$ that fixes the generators in A. Certainly $\mathbf {Q}$ is a rack and satisfies the equations in $E\cup \{a*a = a, a/a = a| a \in A\}$, so there exists a unique rack homomorphism $\beta :\mathbf {R}_{\mathbf {Q}}\rightarrow \mathbf {Q}$ fixing the elements of A. Of course, a quandle homomorphism is also a rack homomorphism and a rack homomorphism between quandles is a quandle homomorphism. Hence, $\alpha \circ \beta :\mathbf {R}_{\mathbf {Q}}\rightarrow \mathbf {R}_{\mathbf {Q}}$ is a rack homomorphism that fixes the elements of A and $\beta \circ \alpha :\mathbf {Q}\rightarrow \mathbf {Q}$ is a quandle homomorphism that also fixes the generators of A. The universal mapping property on presentations implies that $\alpha \circ \beta = id_{\mathbf {R}_{\mathbf {Q}}}$ and $\beta \circ \alpha = id_{\mathbf {Q}}$, so $\mathbf {R}_{\mathbf {Q}}$ is, for all intents and purposes, the same algebra as $\mathbf {Q}$.