If $\mathcal{O}$ is a (classical) topological operad with unit $1\in \mathcal{O}(1)$, $\mathcal{O}(0)=\{0\}$ and multiplications $(m;a\_1,\dotsc,a\_r)\mapsto m(a\_1,\dotsc,a\_r)$. Let $X$ be an algebra over $\mathcal{O}$. Then each choice of $m\in \mathcal{O}(2)$ gives us a binary product
$$X\times X\to X, (x,x')\mapsto x\cdot x' := m\cdot (x,x').$$
It is well-known that the topological structure of $\mathcal{O}$ characterises the product:


* If $1,m(0,1),m(1,0)\in \mathcal{O}(1)$ lie in the same component, then $e:=0\cdot ()\in X$ is an $H$-unit.
* If $m(m,1),m(1,m)\in \mathcal{O}(3)$ lie in the same component, then the product is $H$-associative.


I thought about an analogous construction for coloured operads $\mathcal{O}\binom{n}{k\_1,\dotsc,k\_r}$ and algebras $(X\_n)$ over it. Each choice $m^n\_{k\_1,k\_2}\in \mathcal{O}\binom{n}{k\_1,k\_2}$ gives a binary product
$$X\_{k\_1}\times X\_{k\_2}\to X\_n, (x,x')\mapsto x\cdot\_n x' := m^n\_{k\_1,k\_2}\cdot (x,x').$$
A system $(m^n\_{k\_1,k\_2})$ may be called *multiplication system*. We see:


* If all $\mathcal{O}\binom{n}{n}$ are connected, then $e\_k:=0\_k\cdot ()\in X\_k$ is an $H$-unit in the sense that 
$$(-\cdot\_n e\_k)|\_{n}\simeq (e\_k\cdot\_n -)|\_{n}\simeq \mathrm{id}\_n$$
* If all $\mathcal{O}\binom{n}{k\_1,k\_2,k\_3}$ are connected, then the restricted product is $H$-associative.


My hope was that if we fix such a multiplication system, we get an interesting $H$-something. As a coloured operad is the natural “horizontal categorification” of an operad, my first idea was that we obtain an $H$-category with colours as objects, and the above graded product as compositions and the $e\_k$ as identities, but apparently, this is not the kind of structure we have.


(If you prefer algebraic categories, apply singular homology $H\_\*(-;R)$ to the above system and look at the Pontryagin structure. I first thought that the result is an $R$-algebroid, but again, the structure looks different.)


Does someone see what this is? A “monoid” where we can choose which product we want to use dependind on where we want to land?



$H$-space structure on coloured algebras
This is the homotopy version of the following data:


* a collection of spaces $\{X\_c\}\_c$ for all colors $c$;
* units $e\_c \in X\_c$;
* multiplications $- \cdot\_c - : X\_d \times X\_{d'} \to X\_c$;


satisfying the conditions:


* $x \cdot\_c e\_d = x$ for $x \in X\_c$;
* $(x \cdot\_b y) \cdot\_a z = x \cdot\_a (b \cdot\_c z)$ for $x,y,z$ in the relevant components and colors $a,b,c$.


Clearly $(X\_c, \cdot\_c, e\_c)$ defines a monoid for all $c$. Now define $f\_{cd} : X\_c \to X\_d$ by $f\_{cd}(x) = x \cdot\_d e\_d$. Then I claim than:


* Each $f\_{cd}$ is a morphism of monoids. Indeed, for $x,y \in X\_c$, we have:
\begin{align}
f\_{cd}(x) \cdot\_c f\_{cd}(y) & = (x \cdot\_d e\_d) \cdot\_c (y \cdot\_d e\_d) \\
& = x \cdot\_d (e\_d \cdot\_d (y \cdot\_d e\_d)) \\
& = x \cdot\_d (y \cdot\_d e\_d) \\
& = (x \cdot\_c y) \cdot\_d e\_d = f\_{cd}(x \cdot\_c y).
\end{align}
* For each colors $c,c',c''$, we have $f\_{c'c''} f\_{cc'} = f\_{cc''}$. Indeed:
$$f\_{c'c''}(f\_{cc'}(x)) = (x \cdot\_{c'} e\_{c'}) \cdot\_{c''} e\_{c''} = x \cdot\_{c''} (e\_{c'} \cdot\_{c'} e\_{c''}) = x \cdot\_{c''} e\_{c''} = f\_{cc''}(x).$$


In particular $f\_{cc} = \operatorname{id}\_{X\_c}$ so $f\_{cc'}^{-1} = f\_{c'c}$. So in the end all you have is a monoid $X\_c$ for each color $c$ and a compatible system of isomorphisms $X\_c \cong X\_d$. This is uniquely determined (up to a isomorphism) by the choice of one monoid for a given color; for the other colors, take the same monoid. In other words, you get a diagram (in the category of monoids) over the indiscrete groupoid with objects given by your colors.


In your "homotopy" version, what you get (intuitively) is thus just a bunch of H-monoids $X\_c$, H-morphisms between them $f\_{cd} : X\_c \to X\_d$ such that $f\_{cc} \simeq \operatorname{id}\_{X\_c}$ and $f\_{c''c'} f\_{c'c} \simeq f\_{cc''}$.


