Identities of the Kauffman Monoid $\mathcal{K}_4$ and of the Jones monoid $\mathcal{J}_4$

Abstract

Kauffman monoids Kn and Jones monoids Jn, n = 2, 3, . . . , are two families of monoids relevant in knot theory. We prove a somewhat counterintuitive result that the Kauffman monoids K3 and K4 satisfy exactly the same identities. This leads to a polynomial time algorithm to check whether a given identity holds in K4. As a byproduct, we also find a polynomial time algorithm for checking identities in the Jones monoid J4. ⋆ Supported by the Ministry of Science and Higher Education of the Russian Federation, project no. 1.580.2016, and the Competitiveness Enhancement Program of Ural Federal University.length of w and is denoted |w|. We say that a letter x ∈ X occurs in a word w ∈ X + or, alternatively, w involves x whenever x ∈ alph(w).An identity is an expression of the form w ≏ w ′ with w, w ′ ∈ X + . If S is a semigroup, we say that the identity w ≏ w ′ holds in S or, alternatively, S satisfies w ≏ w ′ if wϕ = w ′ ϕ for every homomorphism ϕ : X + → S. If w ≏ w ′ does not holds in S, we say that it fails in S.The following observations are immediate: if a semigroup S satisfies an identity w ≏ w ′ , so do each subsemigroup and each quotient of S; if semigroups S 1 and S 2 satisfy w ≏ w ′ , so does their direct product S 1 × S 2 .It is well known and easy to see that the free semigroup X + possesses the following universal property: for every semigroup S, every mapping X → S uniquely extends to a homomorphism X + → S. Thus, the homomorphisms X + → S are in a 1-1 correspondence with the mappings X → S, which we call substitutions. Therefore we can restate the fact of w ≏ w ′ holding in S also in the following terms: every substitution of elements in S for letters in X yields equal values to w and w ′ .Given a semigroup S, its identity checking problem 1 Check-Id(S) is the following decision problem. The instance of Check-Id(S) is an arbitrary identity w ≏ w ′ . The answer to the instance w ≏ w ′ is 'YES' whenever the identity w ≏ w ′ holds in S; otherwise, the answer is 'NO'.We stress that here S is fixed and it is the identity w ≏ w ′ that serves as the input so that the time/space complexity of Check-Id(S) should be measured in terms of the size of the identity, that is, in |ww ′ |.Studying computational complexity of identity checking in semigroups (and other 'classical' algebras such as groups and rings) was proposed by Sapir in the influential survey [Kharlampovich and Sapir, 1995], see Problem 2.4 therein. For a finite semigroup S, the problem Check-Id(S) is always decidable. Indeed, given an identity w ≏ w ′ , there are only finitely many substitutions of elements in S for letters in alph(ww ′ ), and one can check whether or not each of these substitutions yields equal values to w and w ′ . Moreover, Check-Id(S) with S being finite belongs to the complexity class coNP: if for some words w, w ′ that involve m letters in total, the identity w ≏ w ′ fails in the semigroup S, then a nondeterministic algorithm can guess an m-tuple of elements in S witnessing the failure and then verify the guess by computing the va...