# $H$-space structure on coloured algebras

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?

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