In this section, we study some basic properties ofθ-bordered andθ-unbordered words whereθ is a (anti)morphism with the property thatθn = IonΣ∗forn≥2 or any literal (anti)morphism.
In the case whereθn= Iandθis an antimorphism, it is clear thatnhas to be an even number. The following result was proved in [11], and can be easily generalized to the case of mor- phic involutions.
Lemma 4.3.1 [11] Let u∈Σ+\D(1). Then there exists v∈Σ∗
with|v| ≤ |u|
2 such that v<d u.
Lemma 4.3.2 Letθ be a morphic or an antimorphic involution and let u ∈ Σ+\Dθ(1). Then
there exists v∈Σ∗with|v| ≤ |u2| such that v <θd u.
The next two results, Propositions 4.3.3 and 4.3.4, establish some relations between the set ofθ-borders of a wordu, namelyLθd(u), and the set ofθ-borders ofθ(u), namelyLθd(θ(u)).
Proposition 4.3.3 Let u ∈ Σ+. Then for a morphism θ on Σ∗ such that θn = I for n > 2,
4.3. Properties of pseudo-(un)bordered words 71
Proof Letv ∈Lθd(θ(u)) which impliesθ(u)= vx = yθ(v) for some x,y ∈Σ+which further im- pliesθ2(u)= θ(v)θ(x)=θ(y)θ2(v). Continuing in this way, we will getθn(u)= θn−1(v)θn−1(x)=
θn−1(y)θn
(v) and thusu= θn−1(v)θn−1(x)= θn−1(y)θn(v) which impliesθn−1(v)∈Lθd(u) and hence
v∈θ(Lθd(u)). Thus,Lθd(θ(u))⊆ θ(Lθd(u)).
Conversely, let v ∈ Lθd(u) which implies u = vx = yθ(v) for x,y ∈ Σ+ and hence θ(u) =
θ(v)θ(x) = θ(y)θ2(v) which further implies θ(v) ∈ Lθ
d(θ(u)). Also, since v ∈ L θ
d(u), θ(v) ∈
θ(Lθd(u)). Thus, Lθd(θ(u))=θ(Lθd(u)).
However, if θ is literal (anti)morphism that is not bijective, Proposition 4.3.3 does not necessarily hold, as demonstrated by Example 4.1.
Example 4.1 LetΣ ={a,b}andθbe (anti)morphism such that,θ(a)=a, θ(b)=a, u= ababaa. Thenθ(u)= aaaaaa, Lθd(u) = {λ,a,ab},θ(Lθd(u))= {λ,a,aa}, Lθd(θ(u))= {λ,a,aa,· · · ,aaaaa}. Clearly, Lθd(θ(u)), θ(Lθd(u)).
Note that the inclusion θ(Lθd(u)) ⊆ Lθd(θ(u)) holds in case of Example 4.1. Moreover, the inclusion holds in general for any literal morphismθ.
Proposition 4.3.4 Let u ∈Σ+. Then for any literal morphismθonΣ∗,θ(Lθd(u))⊆ Lθd(θ(u)).
Proof Letv ∈ Lθd(u) which impliesu = vx = yθ(v) for x,y ∈Σ+and henceθ(u) = θ(v)θ(x) =
θ(y)θ2(v) which further impliesθ(v) ∈ Lθd(θ(u)). Also, sincev ∈ Lθd(u), θ(v) ∈ θ(Lθd(u)). Thus,
θ(Lθd(u))⊆ Lθd(θ(u)).
It is known, [16], that, for an antimorphic involutionθ, the relation<θdis transitive.
Lemma 4.3.5 [16] Let u∈Σ∗and v,w∈Σ+such that u<θd w and w<θd v. Then for a morphic involutionθ, we have u<d v and for an antimorphic involutionθ, we have u<θd v.
The statement of Lemma 4.3.5 does not necessarily hold in the case whenθis a morphism which is literal and not bijective, as demonstrated by Example 4.2.
Example 4.2 Let Σ = {a,b} and θ be a morphism such that θ(a) = a, θ(b) = a, u = ab, w=abaa, v=abaabbaaaa. Then u<θd w and w<θd v but u ≮d v.
The following proposition demonstrates the transitivity of relation <θd for literal antimor- phismsθ.
Proposition 4.3.6 Ifθis any literal antimorphism onΣ∗, then the relation<θdis transitive, i.e. for u∈Σ∗
and v,w∈Σ+such that u<dθ w and w<θd v, we have u<θd v.
Proof Letθbe any literal antimorphism such thatu<θd wandw<θd vwhich impliesw= ux= yθ(u) andv = wα = βθ(w) for some x,y, α, β ∈ Σ+, hence v = uxα = βθ(ux) which further impliesv= uxα= βθ(x)θ(u). Henceu<θd v.
Corollary 4.3.7 Let v ∈ Lθd(u) and w ∈ Σ+. Then for any literal antimorphism θ on Σ∗, if w<θd v then w∈Lθd(u).
The converse of the Corollary 4.3.7 does not hold in general. In fact, in the case of an antimorphism, Proposition 4.3.9 holds.
The next results describe relations between theθ-borders of a worduwhenθis a morphism withθn = I,n> 2,(Proposition 4.3.8) or literal (anti)morphisms (Proposition 4.3.9).
Proposition 4.3.8 Let u,v,w ∈Σ+, u , v and u <θd w,v < θ
d w. Ifθis a morphism onΣ
∗
such thatθn= I for n>2, then either v<d u or u<d v.
Proof Letθbe a morphism such thatθn = Iandu<θd w,v<θd wwhich impliesw= ux=yθ(u) and w = vα = βθ(v) for some x,y, α, β ∈ Σ+. If|u| > |v|, then u = vp andθ(u) = qθ(v) for some p,q ∈Σ+ which implyθn(u) = θn−1(q)θn(v) = θn−1(q)v. Thus, we getu = vp = θn−1(q)v
which impliesv<d u. Similarly, if|u|<|v|thenv =up0 andθ(v) =q0θ(u) for some p0,q0 ∈Σ+ which implyθn(v)= θn−1(q0
)θn(u)=θn−1(q0
)u. Thus, we getv= up0 =θn−1(q0
)uwhich implies
4.3. Properties of pseudo-(un)bordered words 73
Proposition 4.3.8 does not necessarily hold ifθis a literal (anti)morphism that is not bijec- tive, as demonstrated by Example 4.3.
Example 4.3 LetΣ = {a,b}, andθbe a morphism or antimorphism such thatθ(a) = a, θ(b) =
a, u=ab, v=abaa, and w =abaabbaaaa. Then u<θd w,v<θd w but neither v<d u nor u<d v.
Proposition 4.3.9 Let u,v,w∈Σ+, u, v and u<θd w,v<θd w. Then for any literal morphismθ onΣ∗, eitherθ(v) <d θ(u)orθ(u) <d θ(v). Ifθis any literal antimorphism, then either v <p u
or u<pv.
Proof Let θ be any literal morphism and u <θd w,v <θd w which imply w = ux = yθ(u) and
w = vα = βθ(v) for some x,y, α, β ∈ Σ+. If|u| > |v|, then u = vpand θ(u) = qθ(v) for some
p,q∈Σ+which implyθ(u)=θ(v)θ(p)=qθ(v). Thus, we getθ(v)<d θ(u). Similarly, if|u|< |v| thenv = up0 andθ(v) = q0θ(u) for some p0,q0 ∈ Σ+ which implyθ(v) = θ(u)θ(p0) = q0θ(u). Thus, we getθ(u)<d θ(v).
Letθbe any literal antimorphism andu<θd w,v<dθ wwhich imply thatw=ux= yθ(u) and
w = vα= βθ(v) for some x,y, α, β∈ Σ+. Hence, we have, ux = vα. If|u| > |v|, v <p uand if
|v|> |u|thenu<pv.
Corollary 4.3.10 Let u,v,w ∈ Σ+, u , v and u <θd w, v < θ
d w. Then for any literal antimor-
phismθonΣ∗, eitherθ(v)<
s θ(u)orθ(u)<sθ(v).
Corollary 4.3.11 Let u ∈Σ+. Then
1. For any morphismθonΣ∗ such thatθn = I for n > 2, Lθ
d(u)is a totally ordered set with
<d, i.e. Lθd(u)={λ <d u1 <d u2 <d · · · <d ui−1}.
2. For any literal morphismθonΣ∗,θ(Lθd(u))is a totally ordered set with<d.
3. For any literal antimorphismθonΣ∗, Lθ
d(u)is a totally ordered set with<p, i.e. L θ d(u) =
Proof Statement 1 follows from Proposition 4.3.8, statement 2 from Proposition 4.3.9 and statement 3 from Proposition 4.3.9 and Corollary 4.3.10, respectively.
The next two propositions (Proposition 4.3.12, 4.3.13) list some properties ofθ-unbordered words for (anti)morphismsθsuch thatθn = I,n>2.
Proposition 4.3.12 Let θ be a morphism on Σ∗ such that θn = I for n > 2. Then for all x,y∈Dθ(1)such that x, y, we have that xy,θn−1(y)x.
Proof Let x,y ∈ Dθ(1). As Dθ(i) ⊆ Σ+ for i ≥ 1, both x and y are non-empty. Suppose
xy=θn−1(y)x, then we have following three cases to consider.
Case 1: |x|= |y|. Thenx=θn−1(y) andy= x, which is a contradiction sincex, y.
Case 2: |x|> |y|. Then there exists p∈ Σ+such thatx = θn−1(y)pand x= pywhich imply thatx= θn−1(y)p= pθn(y), which is a contradiction sincex∈Dθ(1).
Case 3: |y|> |x|. Then there existsq ∈Σ+such thatθn−1(y) = xqandy= qxwhich imply
thaty= qx=θ(x)θ(q), which is a contradiction sincey∈Dθ(1). Since all the three cases leads to a contradictionxy, θn−1(y)x.
Proposition 4.3.13 Let θ be an antimorphism on Σ∗ such that θn = I for n > 2. Then for x∈Dθ(1)and y ∈Σ+such that x,y andθ(x), x, we have that xy, θn−1(y)x.
Proof Letx∈Dθ(1). AsDθ(i)⊆ Σ+fori≥1, xis non-empty. Supposexy= θn−1(y)x, then we have following three cases to consider.
Case 1: |x|= |y|. Thenx=θn−1(y) andy= x, which is a contradiction sincex, y.
Case 2: |x|> |y|. Then there exists p∈ Σ+such thatx = θn−1(y)pand x= pywhich imply thatx= θn−1(y)p= pθn(y), which is a contradiction sincex∈Dθ(1).
Case 3: |y|> |x|. Then there existsq ∈Σ+such thatθn−1(y) = xqandy= qxwhich imply
thaty = qx = θ(q)θ(x), which further impliesθ(q) = q andθ(x) = xwhich is a contradiction sinceθ(x), x.
4.3. Properties of pseudo-(un)bordered words 75
The following lemma provides a necessary and sufficient condition for a word to be θ- bordered, in the case whenθis a literal antimorphism.
Lemma 4.3.14 Let θ be any literal antimorphism onΣ∗. Then x ∈ Σ+ is θ-bordered iff x = ayθ(a)for some a∈Σand y∈Σ∗.
The result below gives several properties ofθ-unbordered words, for literal antimorphisms
θ.
Proposition 4.3.15 Letθbe any literal antimorphism onΣ∗, then 1. For all u,v∈Σ+and w∈Σ∗
, we have uwv∈Dθ(1)iffuv∈Dθ(1).
2. IfΣis an alphabet such that there exist a,b∈Σwithθ(a),b, then Dθ(1)is a dense set.
3. Let a,b∈Σsuch that a ,b. Then for all u∈Σ+, either ua or ub isθ-unbordered.
Proof 1. Suppose uwv ∈ Dθ(1) and uv < Dθ(1) which imply that uv = ayθ(a) for some
a ∈ Σ andy ∈ Σ∗. Ifw = λ, then clearlyuwv
< Dθ(1), a contradiction. Now, ifw , λ,
then we have three possibilities.
Case a: u=a,v= yθ(a), henceuwv=awyθ(a)<Dθ(1).
Case b: u=ay,v= θ(a), henceuwv=aywθ(a)<Dθ(1).
Case c: u = ap,v = qθ(a) wherey = pqfor some p,q ∈ Σ∗
, henceuwv = apwqθ(a) <
Dθ(1).
Since all the three cases leads to a contradiction,uv∈Dθ(1).
Conversely, supposeuwv < Dθ(1) which imply that uwv = ayθ(a) for some a ∈ Σ and
y ∈ Σ∗. Hence, u = au
1 and v = v1θ(a) for some u1,v1 ∈ Σ∗ which further implies,
uv=au1v1θ(a)<Dθ(1), a contradiction. Henceuwv∈Dθ(1).
2. Choosea,b ∈Σsuch thatθ(a) , b. Then for allw∈ Σ∗, there existsa,b∈ Σ∗such that
3. Let us assume that both uaand ubare θ-bordered. Then we have, ua = a1y1θ(a1) and
ub = a2y2θ(a2) for somea1,a2 ∈ Σ andy1,y2 ∈ Σ∗ which impliesu = a1y1 = a2y2 and
a = θ(a1),b = θ(a2). This further implies thata1y1 = a2y2 which implies a1 = a2 and
y1 = y2which further impliesa = θ(a2) = b, a contradiction. Hence, either uaorubis
θ-unbordered.
Ifθis an antimorphism such thatθn = I,n>2, the following result holds.
Proposition 4.3.16 Letθbe an antimorphism onΣ∗such thatθn = I for n>2. Then u∈Dθ(1)
iffθn−2(u)∈Dθ(1).
Proof Letu∈Dθ(1) and supposeθn−2(u)<Dθ(1) then we haveθn−2(u)= ayθ(a) for somea∈Σ andy ∈ Σ∗which imply thatu = θn(u) = θ2(a)θ2(y)θ3(a) and thusu
< Dθ(1), a contradiction. Henceθn−2(u)∈D
θ(1).
Conversely, supposeθn−2(u) ∈ D
θ(1) andu < Dθ(1). Thenu = ayθ(a) for somea ∈ Σand
y∈Σ∗. Sincenis even andθn= I,n−2 is also even and thusθn−2(u)=θn−2(a)θn−2(y)θn−1(a) <
Dθ(1), a contradiction. Henceu∈Dθ(1).
Lemma 4.3.17 Let θ be a morphic involution on Σ∗ and u ∈ Σ+ such that u ∈ D(1), then
θ(u)∈D(1).
Proof Letu ∈ D(1). Supposeθ(u) < D(1). Thenθ(u) = αβ1 = β2αforα, β1, β2 ∈ Σ+. Thus,
u=θ(α)θ(β1)=θ(β2)θ(α)<D(1), a contradiction. Thus,θ(u)∈D(1).
Along similar lines, we can prove the following result concerningDθ(1) for a morphism of the formθn =I,n≥2.
Lemma 4.3.18 Let θ be a morphism on Σ∗ such thatθn = I, n ≥ 2 and u ∈ Σ+. Then the
following are equivalent: 1. u∈Dθ(1).
4.3. Properties of pseudo-(un)bordered words 77
2. θn−1(u)∈Dθ(1).
3. θ(u)∈Dθ(1).
Proof (1) ⇒ (2): Let u ∈ Dθ(1) and suppose θn−1(u) < Dθ(1). Then θn−1(u) = vx = yθ(v) for somev,x,y ∈ Σ+. This impliesu = θ(v)θ(x) = θ(y)θ2(v), a contradiction sinceu ∈ Dθ(1). Henceθn−1(u)∈Dθ(1).
(2) ⇒ (3): Let θn−1(u) ∈ D
θ(1) and suppose θ(u) < Dθ(1). Then θ(u) = vx = yθ(v) for somev,x,y∈Σ+. This impliesθn−1(u) = θn−2(v)θn−2(x) = θn−2(y)θn−1(v), a contradiction since
θn−1(u)∈D
θ(1). Henceθ(u)∈Dθ(1).
(3) ⇒ (1): Let θ(u) ∈ Dθ(1) and suppose u < Dθ(1). Then u = vx = yθ(v) for some
v,x,y ∈ Σ+. This implies θ(u) = θ(v)θ(x) = θ(y)θ2(v), a contradiction since θ(u) ∈ Dθ(1). Henceu∈Dθ(1).
In fact, the implicationθn−2(u)∈Dθ(1)⇒u∈Dθ(1) of Proposition 4.3.16 and implications (2) ⇒ (3) and (3) ⇒ (1) in Lemma 4.3.18 hold if θ is a literal morphism, not necessarily bijective.
Proposition 4.3.19 Let θbe a morphism on Σ∗ such thatθn = I and u ∈ Σ+. If u ∈ D θ(i)for
some i≥ 2, then for all1≤k< i, Lθd(u)∩D(k), ∅.
Proof By Corollary 4.3.11 we have
Lθd(u)={λ <d u1<d u2 <d · · ·<d ui−1}.
Note thatuk <θd ufor all 1 ≤ k ≤ i−1. Now, sinceuj ∈Ldθ(u) and|uj| < |uk|for all 1 ≤ j< k, by Proposition 4.3.8 we have thatuj <d uk. Hence,
Ld(uk)= {λ,u1,· · ·uk−1}.
Recall that, a u−θ v chain, u = x1 <θd x2 < θ d · · · <
θ
d xn = v is said to beθ-maximal if for
u0 ∈Σ∗,u<θd u0<dθ vimpliesu0= xifor some 1<i< n.
Lemma 4.3.20 [6] Let u ∈ Σ+ be a primitive word. Then u cannot be a factor of u2 in a nontrivial way, i.e., if u2 = xuy, then necessarily either x=λor y =λ.
Proposition 4.3.21 Letθbe an antimorphic involution onΣ∗and f ∈Q. If f ≤θ
d u≤ θ d f 2, then u= f or u= f2, i.e., f ≤θ d f 2 is aθ-maximal chain.
Proof Suppose f ≤θd f2is not aθ-maximal chain, i.e.,u, f andu , f2. Since f ≤θd u ≤θd f2, we haveu = f x = yθ(f) and f2 = uα = βθ(u) for x,y, α, β ∈ Σ∗
with|x| = |y|and|α| = |β|. Then,
f2 = f xα=yθ(f)α= βθ(x)θ(f)=βfθ(y). Now, since f2 =βfθ(y), by Lemma 4.3.20 eitherβ=λorθ(y)=λ.
Case 1: Suppose, β = λ. This implies f = θ(y). Since, f xα = f2, we getxα = f = θ(y).
But since,|x|= |y|, x=θ(y)= f and thusu= f x= f2, a contradiction.
Case 2: Suppose,θ(y)= λ. This impliesβ= f. Since, f xα= f2, we get xα= f = β. But since,|α|=|β|,α=β= f which implies f2 =uα=u f and thusu= f, a contradiction.
Since both the cases leads to a contradiction, f ≤θd f2is aθ-maximal chain. Theθ-unbounded annihilatorαub(u) of a worduis defined, [12], as
αub(u)= {v∈Σ+|uv∈Dθ(1)}.
The following results find a relationship between theθ-unbounded annihilator of a wordu
and the set of catenations of suffixes of u, forθ-unbordered words u, and morphisms θ with
θn = I,n≥ 2 (Proposition 4.3.22) or literal antimorphisms (Proposition 4.3.23).
Proposition 4.3.22 Let θ be a morphism onΣ∗ such that θn = I,n ≥ 2. If u ∈ D
θ(1), then (PSuff(u))+ ⊆αub(u).
4.3. Properties of pseudo-(un)bordered words 79
Proof Let u ∈ Dθ(1). Let v = u1u2· · ·um for some ui ∈ PSuff(u) and 1 ≤ i ≤ m. Suppose thatuv< Dθ(1). Then there existsα, α1, β1 ∈ Σ+such thatuv= αα1 = β1θ(α). Then, we have
following two cases:
Case 1: |α| > |v|. Then, we have θ(α) = u00v and u = u0u00 for some u0,u00 ∈ Σ+. This impliesu00 <s u. Fromuv = αα1, we getuv = θn−1(u00)θn−1(v)α1. This impliesθn−1(u00) <p u. This will further imply thatu<Dθ(1), a contradiction.
Case 2:|α| ≤ |v|. Also, we havev=u1u2· · ·umfor someui ∈PSuff(u) for 1 ≤i≤ m. Thus we have following two sub-cases:
Case 2(a): |α|< |um|. Then, we haveθ(α) = um00 andum = um0um00 for someum0,um00 ∈ Σ+.
Since, um ∈ PSuff(u), we have u = u0mum = u0mum0um00 for some u0
m ∈ Σ+. Thus, we have
um00 <s u. Fromuv = αα1, we get uv = θn−1(um00)α1. This implies θn−1(um00) <p u. This will
further imply thatu<Dθ(1), a contradiction.
Case 2(b): |α| ≥ |um|. Then, we haveθ(α) = u00i ui+1· · ·umforui = u0iu00i , u0i ∈ Σ∗, u00i ∈ Σ+ andi = 1,2,· · · ,m−1. Since,ui ∈ PSuff(u), we have u = ui0ui = ui0u0
iu 00 i for someui0 ∈ Σ+. Thus, we haveu00 i <s u. Fromuv = αα1, we getuv= θ n−1(u00 i )θ n−1(u i+1· · ·um)α1. This implies θn−1(u00
i )<pu. This will further imply thatu<Dθ(1), a contradiction. Since all the cases leads to a contradiction, (PSuff(u))+⊆ αub(u).
Proposition 4.3.23 Letθ be any literal antimorphism onΣ∗. If u∈ Dθ(1), then(PSuff(u))+ ⊆
αub(u).
Proof Letv = u1u2· · ·um for someui ∈PSuff(u) and 1 ≤ i ≤ m. Suppose,uv < Dθ(1). Then
uv= ayθ(a) for somea ∈Σandy ∈Σ∗. This further implies,
u= ay1,v= y2θ(a) andy = y1y2
for somey1,y2 ∈Σ∗. Clearly,a<p u. But, since,v= u1u2· · ·um=y2θ(a) whereum∈PSuff(u), we will haveum = um0θ(a) forum0 ∈ Σ∗. Also,u = u0um = u0um0θ(a) and thusθ(a) <s u. This