Let $I$ be a monoid and $G$ be an $I$-graded monoid, with multiplication

$$ ( - \cdot - ) : G_i \times G_j \to G_{i+j}. $$

I'm interested in the following property of $G$:

P: for any two indices $i, j \in I$, and any $g \in G_{i+j}$, there are unique elements $g_i \in G_i$, $g_j \in G_j$ such that $g = g_i \cdot g_j$.
In other words, multiplication is bijective when restricted to specific degrees.

Examples include:

  • Any monoid indexed by itself.
  • The free monoid on an alphabet $A$, indexed by degree in $\mathbb{N}$.

Has this property been studied anywhere? It's a strong condition; can it be reformulated in more familiar terms, perhaps as a certain freeness property?

Note: I am also interested in answers to this question in other monoidal categories.

  • 2
    $\begingroup$ I don't know the answer(s) to your question, but if $I$ is a free monoid on generators $S$, what you'll get is pretty simple : sets $G_s, s\in S$, and $G$ is a free monoid on $\coprod_{s\in S}G_s$, with grading induced by $I$ $\endgroup$ – Max Jan 28 at 20:38
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    $\begingroup$ Something else : when $I$ is a group, each $G_x\times G_{x^{-1}}$ is in bijection with $G_1$. In particular, if $G_1$ has only one element (I don't know if you have that as a convention in your definition), then so does every $G_x$, and so we're in the first situation you described : it's $I$ indexed by $I$. Then, note that multiplication is a bijection $G_1\times G_1\to G_1$, but $1\in G_1$ so $G_1$ can have only one elemnt since the restriction to $G_1\times \{1\}$ is surjective : so if $I$ is a group you're necessarily in your first situation $\endgroup$ – Max Jan 28 at 20:52
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    $\begingroup$ The condition you wrote is that multiplication is bijective when restricted to specific degrees, not just injective. Is that what you really want? $\endgroup$ – Eric Wofsey Jan 28 at 21:06
  • $\begingroup$ @EricWofsey Indeed, I should have written bijective (corrected it now). I know it's a strong condition, I was hoping for some characterisation along the lines of Max's comments that it would end up being some kind of free object, but subject to the grading. $\endgroup$ – Will Jan 28 at 23:26
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    $\begingroup$ Here is another interesting example: Let $I=[0,\infty)$ and let $A$ be a set, and $G_i=A^{(0,i]}$, with the operation $G_i\times G_j\to G_{i+j}$ being the obvious "concatenation" operation. (More generally, $I$ could be any totally ordered cancellative monoid with $0$ as its least element--taking $I=\mathbb{N}$ recovers your second example.) This would not seem to be "free" in any obvious sense, given the infinite products involved. $\endgroup$ – Eric Wofsey Jan 28 at 23:42

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