The morphism $\rho$ defined by $\rho(\sa{a})=\sa{ab}$, $\rho(\sa{b})=\sa{ca}$ and $\rho(\sa{c})=\sa{bc}$ complies with the conditions and produces the periodic word $\rho^\infty(\sa{a})=\sa{abcabcabc}\cdots=(\sa{abc})^\infty$. None of the letter is bispecial.
On the contrary, Fibonacci morphism $\phi$, defined by $\phi(\sa{a})=\sa{ab}$ and $\phi(\sa{b})=\sa{a}$, also satisfies the conditions but generates the non-(ultimately) periodic Fibonacci word $\phi^\infty(\sa{a})=\sa{abaababa}\cdots$. In it letter $\sa{a}$ is bispecial since its occurrences are followed either by $\sa{a}$ or by $\sa{b}$, while occurrences of letter $\sa{b}$ are all followed by $\sa{a}$.
Problem 89: Periodicity of morphic words |
The problem shows that it is possible to test whether an infinite word generated by a (finite) morphism is periodic.
An infinite morphic word is obtained by iterating a morphism $\theta$ from $A^+$ to itself, where $A=\{\sa{a}, \sa{b}, \dots\}$ is a finite alphabet. To do so, we assume that $\theta$ is prolongable over the letter $\sa{a}$, that is, $\theta(\sa{a}) = \sa{a}u$ for $u\in A^+$. Then $\Theta=\theta^\infty(\sa{a})$ exists and is $\sa{a}u\theta(u)\theta^2(u)\cdots$. The infinite word $\Theta$ is a fixed point of $\theta$, that is, $\theta(\Theta)=\Theta$.
The infinite word $\Theta$ is periodic if it can be written $z^\infty$ for some (finite) words $z$, $z\neq \varepsilon$.
To avoid unnecessary complications we assume that the morphism $\theta$ is both irreducible, which means that any letter is accessible from any letter (for any $c,d\in A$ the letter $d$ appears in $\theta^k(c)$ for some integer $k$), and is elementary, which means it is not the product $\eta \circ \zeta$ of two morphisms $\zeta:A^+ \longrightarrow B^+$ and $\eta:B^+ \longrightarrow A^+$, where $B$ is an alphabet smaller than $A$. The second condition implies that $\theta$ is injective on $A^*$ and on $A^\infty$.