ON SEPARATING SETS OF WORDS IV

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1 ON SEPARATING SETS OF WORDS IV V. FLAŠKA, T. KEPKA AND J. KORTELAINEN Abstract. Further propertes of transtve closures of specal replacement relatons n free monods are studed. 1. Introducton Ths artcle s an mmedate contnuaton of [1], [2] and [3]. References lke I.3.3 (II.3.3, III.3.3, resp.) lead to the correspondng secton and result of [1] ([2], [3], resp.) and all defntons and prelmnares are taken from the same source. 2. Complementary sequences Troughout ths note, let Z A + be a strongly separatng set of words and let ψ : Z A be a mappng wth ψ(z) z for every z Z. Notce that then the correspondng replacement relaton ρ (= ρ Z,ψ ) s rreflexve. Two sequences p 0, p 1,..., p m and q 0, q 1,..., q m, m 1, of words wll be called ((Z, ψ) or ρ-) complementary f, for every 0 < m, ether (p, p +1 ) ρ and q = q +1 or p = p +1 and (q, q +1 ) ρ. Notce that due to the rreflexvty of ρ, just one of the two cases holds. Lemma 2.1. Let p 0, p 1,..., p m and q 0, q 1,..., q m be complementary sequences. Then: () Both the sequences are λ-sequences. () (p 0, p m ) ξ and (q 0, q m ) ξ. () If (p 0, p m ) / τ ((q 0, q m ) / τ, resp.), then p 0 = p 1 = = p m (q 0 = q 1 = = q m, resp.), q 0, q 1,..., q m (p 0, p 1,..., p m, resp.) s a ρ-sequence and (q 0, q m ) τ ((p 0, p m ) τ, resp.). (v) Ether (p 0, p m ) τ or (q 0, q m ) τ. Proof. Easy. Let w 0, w 1,..., w m be a ρ-sequence and let z Z. Furthermore, let = (p, z, q ) Tr(w ) (so that w = p zq ) for all = 0, 1,..., m. We wll say that the ρ-sequence s (z, 0,..., m )-fluent f the sequences p 0, p 1,..., p m and q 0, q 1,..., q m are complementary. The work s a part of the research project MSM fnanced by MŠMT and the second author was supported by the Grant Agency of Czech Republc, No. 201/09/

2 ON SEPARATING SETS OF WORDS IV 2 Lemma 2.2. Let (w 0, w 1 ) ρ and = (p 0, z, q 0 ) Tr(w 0 ). w 0 = p 0 zq 0 and at least one of the followng two cases holds: Then (1) w 0 w 1, (w 0, w 1 ) ρ z and w 1 = p 0 ψ(z)q 0 ; (2) w 1 = p 1 zq 1 and the sequences p 0, p 1 and q 0, q 1 are complementary (and hence the sequence w 0, w 1 s (z,, β)-fluent, β = (p 1, z, q 1 )). Proof. Assume that (1) s not true. Then there s γ = (r, z 1, s) Tr(w 0 ) such that γ and w 1 = rψ(z 1 )s. We have p 0 zq 0 = w 0 = rz 1 s, where p 0 r and q 0 s. Consequently, p 0 r and q 0 s. Frst, assume that p 0 < r. Then r = p 0 r 1, r 1 ε, zq 0 = r 1 z 1 s, q 0 > s, q 0 = s 1 s, s 1 ε and zs 1 = r 1 z 1. From ths, r 1 = zt and s 1 = tz 1 and we get w 0 = p 0 zq 0 = p 0 r 1 z 1 s = p 0 ztz 1 s, q 0 = tz 1 s, r = p 0 zt, w 1 = rψ(z 1 )s = p 0 ztψ(z 1 )s = p 1 zq 1, where p 0 = p 1, q 1 = tψ(z 1 )s and (q 0, q 1 ) ρ. Next assume that r < p 0. Then p 0 = rr 1, r 1 ε, r 1 zq 0 = z 1 s, r 1 z 1, r 1 = z 1 t, s = tzq 0, p 0 = rz 1 t. Now, w 0 = rz 1 tzq 0, w 1 = rψ(z 1 )s = rψ(z 1 )tzq 0 = p 1 zq 1, where q 0 = q 1, p 1 = rψ(z 1 )t and (p 0, p 1 ) ρ. The lemma follows easly from I.6.4 as well. Lemma 2.3. Let w 0, w 1,..., w m be a ρ-sequence and let 0 = (p 0, z, q 0 ) Tr(w 0 ) (so that w 0 = p 0 zq 0 ). Then at least one of the followng two cases holds: (1) w 0 0 w1, (w 0, w 1 ) ρ z and w 1 = p 0 ψ(z)q 0 ; (2) There are 1 n m and = (p, z, q ) Tr(w ) (so that w = p zq ), 0 n, such that the sequence w 0, w 1,..., w n s (z, 0, 1,..., n )-fluent and ether n = m or n < m and w n+1 = p n ψ(z)q n (so that (w n, w n+1 ) ρ z and w n n wn+1 ). Proof. Assume that (1) s not true and proceed by nducton on m. If m = 1 then 2.2 apples. If m 2, we consder the sequence w 1, w 2,..., w m. Remark 2.4. Consder the stuaton from 2.3 (2) and assume that n < m. Put v 0 = p 0 zq 0 = w 0, v 1 = p 0 ψ(z)q 0, v 2 = p 1 ψ(z)q 1,..., v n = p n 1 ψ(z)q n 1 and v n+1 = p n ψ(z)q n = w n+1. Clearly, w 0 = v 0, v 1,..., v n, v n+1 = w n+1 s a ρ-sequence, v 0 = w 0 0 v1, w 1 1 v2,..., w n n vn+1 = w n+1 and w 0 = v 0, v 1,..., v n, v n+1 = w n+1, w n+2,..., w m s a ρ-sequence. In partcular, (p 0 ψ(z)q 0, w m ) = (v 1, w m ) τ. 3. Auxlary results (a) Lemma 3.1. Let z 1, z 2 Z and r A. Then z 1 rψ(z 2 ) = ψ(z 1 )rz 2 ff at least (and then just) one of the followng two cases holds: (1) There are s, t A + such that sr = rt (see I.3.5) and z 1 = ψ(z 1 )s, z 2 = tψ(z 2 );

3 ON SEPARATING SETS OF WORDS IV 3 (2) There are s, t A + such that sr = rt and ψ(z 1 ) = z 1 s, ψ(z 2 ) = tz 2. Proof. Easy. Corollary 3.2. The followng two condtons are equvalent: () z 1 rψ(z 2 ) ψ(z 1 )rz 2 for all z 1, z 2 Z and r A. () sr rt for every r A whenever s, t A + and z 1, z 2 Z are such that ether z 1 = ψ(z 1 )s and z 2 = tψ(z 2 ) or ψ(z 1 ) = z 1 s and ψ(z 2 ) = tz 2 (the latter case does not take place when ψ s strctly length decreasng). Lemma 3.3. Let (w 0, w 1 ) ρ and = (p 0, z, q 0 ) Tr(w 0 ) (see 2.2). If the equvalent condtons of 3.2 are satsfed, then just one of the cases 2.2 (1), (2) holds. Proof. If both 2.2 (1), (2) are true, then p 0 ψ(z)q 0 = w 1 = p 1 zq 1 and ether p 0 = p 1 and (q 0, q 1 ) ρ or (p 0, p 1 ) ρ and q 0 = q 1. Assume the frst case, the other one beng smlar. Then ψ(z)q 0 = zq 1, q 0 = rz 1 s, q 1 = rψ(z 1 )s and ψ(z)rz 1 = zrψ(z 1 ). The rest s clear from 3.2. Remark 3.4. Assume that the equvalent condtons of 3.2 are satsfed and let (w 0, w 1 ) ρ. Then t follows from 3.3 that w 0 0 w1 for a unquely determned nstance Tr(w 0 ). Remark 3.5. Assume that the equvalent condtons of 3.2 are satsfed and consder the stuaton from 2.3. Then just one of the cases 2.3 (1), (2) holds. Furthermore, f 2.3 (2) s true, then the number n and the nstances 0, 1,..., n are determned unquely. 4. Auxlary results (b) In ths secton, let z 1, z 2 Z, z 1 z 2, r 1, r 2, s 1, s 2 A, t 1 = r 1 z 1 s 1 and t 2 = r 2 z 2 s 2. Lemma 4.1. t 1 t 2 n each of the followng sx cases: (1) r 1 = r 2 ; (2) s 1 = s 2 ; (3) r 1, s 1 are reduced; (4) r 2, s 2 are reduced; (5) r 1, r 2 are reduced; (6) s 1, s 2 are reduced. Proof. Easy to see (use I.6.4). Lemma 4.2. Assume that the mappng ψ s length decreasng. Then (t 1, t 2 ) / τ n each of the followng three cases: (1) r 1 + s 1 r 2 + s 2, z 1 z 2 and at least one of these nequaltes s sharp (equvalently, t 1 < t 2 );

4 ON SEPARATING SETS OF WORDS IV 4 (2) r 1, s 1 are reduced, r 1 + s 1 r 2 + s 2, ψ(z 1 ) z 2 and at least one of these nequaltes s sharp; (3) r 1, s 1 are reduced and ψ(z 1 ) < z 1. Proof. Easy (f r 1, s 1 are reduced and (t 1, t 2 ) τ, then (r 1 ψ(z 1 )s 1, t 2 ) ξ). Corollary 4.3. Assume that the mappng ψ s strctly length decreasng and the words r 1, r 2, s 1, s 2 are reduced. Then (t 1, t 2 ) / ξ and (t 2, t 1 ) / ξ. 5. Auxlary results (c) In ths secton, let z Z, r, s A and t = r z s, = 1, 2, be such that (t 1, t 2 ) / ξ and (t 2, t 1 ) / ξ (see the precedng secton). Put (P (t 1, t 2 ) =) P = {w A (w, t 1 ) ξ, (w, t 2 ) ξ} and denote by (Q(t 1, t 2 ) =) Q the set of w P such that w = w whenever w P and (w, w ) ξ. Lemma 5.1. () If w P, then (w, t 1 ) τ, (w, t 2 ) τ and t 1 w t 2. () If w P, then w Q f and only f (w, w ) / ρ for every w P. Proof. Easy. Remark 5.2. Assume that P. By III.6.4 (), there exsts at least one word t A wth (t 1, t) ξ and (t 2, t) ξ. Then (t 1, t) τ and (t 2, t) τ. Furthermore, f r, s are reduced, then (r 1 ψ(z 1 )s 1, t) ξ and (r 2 ψ(z 2 )s 2, t) ξ. Lemma 5.3. Assume that the relaton ρ s regular (e.g., f ψ s strctly length decreasng see III.7.7). Then for every w P there exsts at least one w Q wth (w, w ) ξ. Proof. Put R = {v P (w, v) ξ} and M = {dst(v, t 1 )+dst(v, t 2 ) v R}. Then M s a non-empty set of postve ntegers and f w R s such that dst(w, t 1 ) + dst(w, t 2 ) s the smallest number n M, then w Q. Notce that f ψ s strctly length decreasng and w R s such that w s the smallest number n R, then w Q. Now, take w Q and let w () 0, w () 1,..., w m (), m 1, = 1, 2, be ρ-sequences such that w () 0 = w and w m () = t. Lemma 5.4. If (w ( 1) j, w ( 2) k ) ξ for { 1, 2 } = {1, 2} and some 0 j m 1, 0 k m 2, then w ( 1) j = w. Proof. We have (w ( 1) j, t 1 ) ξ and (w ( 2) k, t 2 ) ξ. Snce 1 2, t follows that w ( 1) j P. But w Q and (w, w ( 1) j ) ξ. Consequently, w ( 1) j = w.

5 Lemma 5.5. w (1) 1 w (2) 1. ON SEPARATING SETS OF WORDS IV 5 Proof. If w (1) 1 = v = w (2) 1, then v P, (w, v) ρ and v = w, snce w Q. Thus (w, w) ρ, a contradcton wth the rreflexvty of ρ. Lemma 5.6. Assume that ether τ s rreflexve (see III.3.3) or that the sum m 1 + m 2 s mnmal (for the word w). Then: () (w ( 1) j, w ( 2) k ) / ξ for all { 1, 2 } = {1, 2}, 0 j m 1, 0 k m 2. () All the words w = w (1) 0 = w (2) 0, w () j, j = 1, 2,..., m, = 1, 2, are par-wse dfferent. Proof. Easy (use 5.4). Lemma 5.7. tr(w) 2 (. e., w s not meagre). Proof. Clearly, w s not reduced. On the other hand, f tr(w) = 1, then w = pzq, where z Z and p, q are reduced. Consequently, w () 1 = pψ(z)q, w (1) 1 = w (2) 1, a contradcton wth 5.5. Lemma 5.8. alph(w) alph(t 1 ) alph(t 2 ). Proof. Let, on the contrary, w = pzq, where z Z and z / alph(t 1 ) alph(t 2 ). Usng 2.3 and 2.4, we get ρ-sequences v () 0, v () 1,..., v m (), = 1, 2, such that v () 0 = w, v () 1 = pψ(z)q and v m () = t. Then v () 1 = v, where (v, t ) ξ, v P and v = w a contradcton wth the rreflexvty of ρ. Lemma 5.9. Assume that z 1 z 2 and z 1 / alph(r 2 ) alph(s 2 ) (. e., z 1 / alph(t 2 )). If w = p 0 z 1 q 0, then the sequence w = w (1) 0, w (1) 1,..., w (1) t 1 s (z 1, 0,..., m1 )-fluent, where 0 = (p 0, z 1, q 0 ), 1 = (p 1, z 1, q 1 ), m1 = (p m1, z 1, q m1 ), p m1 = r 1, q m1 = s 1 (then (p 0, r 1 ) ξ and (q 0, s 1 ) ξ). Proof. Proceedng by contradcton, assume that our result s not true. Accordng to 2.3 and 2.4, there s a ρ-sequence w = v (2) 0, p 0 ψ(z 1 )q 0 = v (2) 1, v (2) 2..., v m (2) 2 = t 2. Thus v (1) 1 = v = v (2) 1, (v, t 1 ) ξ, (v, t 2 ) ξ, v P and v = w, (w, w) ρ, a contradcton wth the rreflexvty of ρ. Lemma Assume that z 1 z 2 and z 1 / alph(r 1 ) alph(r 2 ) alph(s 2 ). Then w y 0 z 1 y 1 z 1 y 2 for all y 0, y 1, y 2 A. Proof. Let, on the contrary, w = y 0 z 1 y 1 z 1 y 2. Then (y 0 z 1 y 1, r 1 ) ξ and (y 2, s 1 ) ξ by 5.9 and 2.1 (). Snce z 1 / alph(r 1 ), we have (y 0 ψ(z 1 )y 1, r 1 ) ξ by 2.4, and therefore (y 0 ψ(z 1 )y 1 z 1 y 2, t 1 ) ξ. On the other hand, z 1 / alph(t 2 ), and so (y 0 ψ(z 1 )y 1 z 1 y 2, t 2 ) ξ as well. Thus y 0 ψ(z 1 )y 1 z 1 y 2 P, a contradcton wth w Q and ψ(z 1 ) z 1. m 1 =

6 ON SEPARATING SETS OF WORDS IV 6 Proposton Assume that z 1 z 2 and r, s are reduced, = 1, 2. Then there exst reduced words x 0, x 1, x 2 A such that just one of the followng two cases takes place: (1) w = x 0 z 1 x 1 z 2 x 2, x 0 = r 1, (x 1 z 2 x 2, s 1 ) τ, (x 0 z 1 x 1, r 2 ) τ and x 2 = s 2 (then w = r 1 z 1 x 1 z 2 s 2, (x 1 ψ(z 2 )s 2, s 1 ) ξ, (r 1 ψ(z 1 )x 1, r 2 ) ξ and r 1, s 2 are reduced); (2) w = x 0 z 2 x 1 z 1 x 2, x 0 = r 2, (x 1 z 1 x 2, s 2 ) τ, (x 0 z 2 x 1, r 1 ) τ and x 2 = s 1 (then w = r 2 z 2 x 1 z 1 s 1, (x 1 ψ(z 1 )s 1, s 2 ) ξ, (r 2 ψ(z 2 )x 1, r 1 ) ξ and r 2, s 1 are reduced). Proof. Combnng 5.7, 5.8 and 5.10 (and the dual), we see that tr(w) = 2 and alph(w) = {z 1, z 2 }. Accordng to I.6.4, ether w = x 0 z 1 x 1 z 2 x 2 or w = x 0 z 2 x 1 z 1 x 2, where x 0, x 1 and x 2 are reduced. Assume the former equalty, the latter beng dual. Now, t follows from 5.9 that (x 0, r 1 ) ξ. Snce x 0 s reduced, we get x 0 = r 1. Furthermore, (x 1 z 2 x 2, s 1 ) ξ and, snce z 2 / alph(s 1 ), we have (x 1 z 2 x 2, s 1 ) τ. The rest s smlar. Remark Consder the stuaton from 5.11 (and ts proof) and assume that (1) s true (the other case beng dual). Put u 1 = x 1 ψ(z 2 )s 2 and u 2 = r 1 ψ(z 1 )x 1. We have (u 1, s 1 ) ξ and (u 2, r 2 ) ξ. () If u 1 s reduced, then u 1 = s 1, t 1 = r 1 z 1 x 1 ψ(z 2 )s 2, (w, t 1 ) ρ and (t 1, u 3 ) ρ, where u 3 = r 1 ψ(z 1 )x 1 ψ(z 2 )s 2. () If u 2 s reduced, then u 2 = r 2, t 2 = r 1 ψ(z 1 )x 1 z 2 s 2, (w, t 2 ) ρ and (t 2, u 3 ) ρ, where u 3 = r 1 ψ(z 1 )x 1 ψ(z 2 )s 2. () If the equvalent condtons of II.7.3 are satsfed, then all the words u 1, u 2, s 1, r 2 are meagre. Now, f s 1 s not reduced, then s 1 = y 0 z 3 y 1, z 3 Z, z 1 z 3 z 2, y 0, y 1 are reduced and t 1 = r 1 z 1 y 0 z 3 y 1. If r 2 s not reduced, then r 2 = y 2 z 4 y 3, z 4 Z, z 1 z 4 z 2, y 2, y 3 are reduced and t 2 = y 2 z 4 y 3 z 2 s 2. Remark Assume that P and choose w P such that m 1+m 2 s mnmal, where m 1 and m 2 s the length of a ρ-sequence from w to t 1 and t 2, resp. It s easy to see that 5.5, 5.7, 5.8 and 5.9 reman true. 6. The ultmate consequence Theorem 6.1. Assume that the mappng ψ s strctly length decreasng. Let z 1, z 2 Z and r 1, r 2, s 1, s 2 A be such that z 1 z 2, the words r 1, r 2, s 1, s 2 are reduced and P (t 1, t 2 ), where t 1 = r 1 z 1 s 1 and t 2 = r 2 z 2 s 2. Then Q(t 1, t 2 ) and, f w Q(t 1, t 2 ), then just one of the followng two cases takes place: (1) w = r 1 z 1 xz 2 s 2, (r 1 z 1 x, r 2 ) τ, (xz 2 s 2, s 1 ) τ and x s reduced; (2) w = r 2 z 2 yz 1 s 1, (r 2 z 2 y, r 1 ) τ, (yz 1 s 1, s 2 ) τ and y s reduced. Proof. By 5.3, Q(t 1, t 2 ). The rest follows from 5.11.

7 ON SEPARATING SETS OF WORDS IV 7 References [1] V. Flaška, T. Kepka and J. Kortelanen, On separatng sets of words I, Acta Unv. Carolnae Math. Phys., 49/1(2008), [2] V. Flaška, T. Kepka and J. Kortelanen, On separatng sets of words II, Acta Unv. Carolnae Math. Phys., 50/1(2009), [3] V. Flaška, T. Kepka and J. Kortelanen, On separatng sets of words III, preprnt. Department of Algebra, MFF UK, Sokolovská 83, Praha 8 Department of Informaton Processng Scence, Unversty of Oulu, P. O. BOX 3000 FIN-90014, Oulu E-mal address: flaska@matfyz.cz E-mal address: kepka@karln.mff.cun.cz E-mal address: juha.kortelanen@oulu.f