Bi-character for finite, commutative monoids?

Question

If I have a finite commutative monoid $M$ (which is not a group), is it possible to get a bi-character on this? By bi-character, I mean a map $\beta:M\times M\rightarrow \mathbb{C}^*$ such that, $\forall m\in M$, both $\beta(m, \,)$ and $\beta(\, ,m)$ are homomorphic map to the multiplicative group of $\mathbb{C}^*$. I undrstand that the linear characters of monoids, though linearly independent, are not orthogonal in general. That is why it doesn't seem possible to get a bi-character. But I may be wrong. Thanks

Answer 1

There is always the trivial bi-character $\beta(m,n)=1$.

There exist finite commutative monoids for which this is the only bi-character.

Take for example the monoid with two elements $\{1,0\}$. It must be the case that $\beta(1,1)=1$. Suppose that $\beta(1,0)=x$ and $\beta(0,0)=y$. Then $xy=\beta(1,0)\beta(0,0)=\beta(0,0)=y$ and so $x=1$. Similarly, $\beta(0,1)=1$. Additionally, $y^2=\beta(0,0)\beta(0,0)=\beta(0,0)=y$ so $y=1$.