# For which $k$-types are $E_{n,m}$-algebras automatically $E_{n+1}$ algebras?

Recall that an $$E_{n,m}$$ algebra is an $$A_m$$ algebra in $$E_n$$ algebras. Here I index my $$A_m$$ algebras so that an $$A_1$$ algebra is pointed, an $$A_2$$ algebra has a unital multiplication, $$A_3$$ is homotopy associative, etc. In particular, an $$E_n$$ algebra is on the one hand an $$E_{n,1}$$ algebra and on the other hand an $$E_{n-1,\infty}$$ algebra.

Question: Suppose $$X$$ is a homotopy $$k$$-type and an $$E_{n,m}$$-algebra, when does it have a canonical $$E_{n+1}$$ structure? In other words, how connected is the map of operads from the $$E_{n+1}$$-operad to the $$E_{n,m}$$-operad? What if $$m=2$$?

Let me also explain briefly where this question comes from and why I'm particularly interested in the $$m=2$$ case.

A well-known theorem about monoidal categories says:

Theorem: A monoidal category $$\mathcal{C}$$ is braided iff there's a monoidal splitting of the canonical forgetful map from the Drinfeld center $$Z(\mathcal{C}) \rightarrow \mathcal{C}$$.

I was wondering what the appropriate generalization of this statement is to general $$E_n$$ algebras. That is for which $$k$$-types do $$E_{n+1}$$ structures on $$E_n$$ algebras correspond to homotopy splittings of the forgetful map from the $$E_{n+1}$$ center $$Z(A) \rightarrow A$$. If you think through the $$E_0$$ case, it's not difficult to see that a homotopy retract of an $$E_1$$ algebra is only an $$A_2$$ algebra. Applying this observation to $$E_0$$ algebras in $$E_n$$ algebras, we should only expect $$A$$ as above to be an $$E_{n,2}$$ algebra. So the theorem above corresponds to the statement that a homotopy $$1$$-type that is an $$E_{1,2}$$ algebra is automatically an $$E_2$$ algebra. Very roughly this is because of the Eckman-Hilton argument, which says that the two multiplications agree and so the second multiplication is $$A_\infty$$ and not just $$A_2$$.

• I will try to write a fuller answer when I have some more time, but I suggest you look at arxiv.org/pdf/1808.06006.pdf In there we show (together with Tomer Schlank) that the tensor product of a reduced $d_1$-connected operad with a reduced $d_2$-connected operad is $(d_1+d_2+2)$-connected. $E_n$ is $(n-2)$-connected. I can't say from the top of my head what is the connectivity of all operation spaces in $A_m$, I guess it is known or can be worked out (but It of course at least $(-1)$ since they are not empty, so you always get something). – KotelKanim Oct 11 at 14:02
• Actually, all the $A_m$-s are exactly and only $(-1)$-connected. The operation spaces are not connected since the multiplication is not commutative. – KotelKanim Oct 11 at 14:37
• Interesting paper! I'm a bit confused because going from the $E_{0,2}$ question to the $E_{1,2}$ question I get a jump of two dimensions (i.e. an $E_{0,2}$ algebra must be a $(-1)$-type for it to extend to an $E_1$-algebra, while a $1$-type that's an $E_{1,2}$ algebra is an $E_2$-algebra). But your formula predicts a jump by one. Presumably this is something about the map between the operads being one dimension more connected than the operads themselves? – Noah Snyder Oct 11 at 15:58
• So, the more relevant theorem from the paper is not what I said (1.0.1) but indeed the relative version (1.0.2) saying that if $P \to Q$ is a map of reduced operads which is a $d$-equivalence (i.e. equivalence on $d$-truncations of all operation spaces) and $R$ is a $k$-connected reduced operad then the map $P\otimes R \to Q\otimes R$ is a $(d+k+2)$-equivalence. – KotelKanim Oct 12 at 9:00
• Now, if I am not mistaken, the map $A_m \to A_{m+1}$ is an $(m-3)$-equivalence and as I mentioned before, $E_n$ is $(n-2)$-connected. Hence if $k=n+m-3$, then a $k$-type which is $E_{n,m}$ promotes uniquely to an $E_{n,m+1}$ and so on up to $E_{n+1}$. E.g., $A_2 \to A_3$ is not iso. on $\pi_0$ (the multiplication of $A_2$ is not homotopy associative), but it becomes so after tensoring with $E_1$. This means that a $0$-type which is $E_{1,2}$ is uniquely $E_{1,3}$ and so on up to $E_2$. I can't see why this woud be the case for $1$-types though. – KotelKanim Oct 12 at 9:15