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?


1 Answer 1


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 $f_{c''c'} f_{c'c} \simeq f_{cc''}$.


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