Suppose we have the free abelian monoid $S = \{a^m : m \in \mathbb{N}_0\}$ on the set of one element $X = \{a\}$. The binary operation on the monoid is denoted by $\cdot$. If $(T,\ast)$ is another monoid we denote the semigroup homomorphisms from $S$ to $T$ by $Hom(S,T)$.

Question 1: What is the right formulation of a dual monoid $Hom(S,T)$ in the following sense: which $(T,\ast)$ should one take?

I have tried out $T= T_1 = (0,1]$, $T=T_2=[-1,1]$ and $T= T_3 = \Pi = \{z \in \mathbb{C}: |z| = 1\}$ and I always find $Hom(S,T) \simeq T$. So maybe it does not matter?

Question 2: Of course one may iterate the procedure and define the bidual $S^{\ast \ast} = Hom(S^\ast,T)$. Is $S$ reflexive in the sense that $S^{\ast \ast}=S$? Does this depend on the choice of $T$?

Well, and here is a concern I have: for the three cases of $T = T_i$ I considered before the question I got: $S^{\ast \ast} \simeq Hom(T,T)$. But $Hom(T,T)$ depends on $T$. For example $Hom(T_3,T_3) \simeq \mathbb{Z}$, $Hom(T_2,T_2) \simeq \{R \cup sgn\cdot R \cup \{sgn^2\}, \cdot\}$, where $R = \{t \mapsto |t|^\alpha: \alpha \in [0,\infty]\}$ and $sgn = 1_{(0,1]}-1_{[-1,0)}$. Put together: in none of the examples $S$ is reflexive.

Question 3: Is $S = S^{\ast \ast}$ known for topological free monoids, where we consider continuous homomorphisms? In which sense can I make sense of the statement $S^\ast = T^{\# X}$ if $\# X$ is not finite any more, even $S$ not being locally compact?

Are there good references for duality in the semigroup case? I have seen that for groups one may consider the Pontryagin dual, i.e. $T_3=\{z \in \mathbb{C}: |z|=1\}$ but as this is a group, I am not sure if this is the right object for monoids.

  • $\begingroup$ It's a general result that $Hom(S,T) \simeq T$, since each element $t \in T$ corresponds to the hom sending $a$ to $t$. Imitating the definition of eg dual spaces, shouldn't you be considering $T^* := Hom(T,S)$? $\endgroup$ – Morgan Rogers Mar 4 at 12:23

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