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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 $A_2$.

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    $\begingroup$ 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). $\endgroup$
    – KotelKanim
    Commented Oct 11, 2019 at 14:02
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    $\begingroup$ Actually, all the $A_m$-s are exactly and only $(-1)$-connected. The operation spaces are not connected since the multiplication is not commutative. $\endgroup$
    – KotelKanim
    Commented Oct 11, 2019 at 14:37
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    $\begingroup$ 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? $\endgroup$
    – Noah Snyder
    Commented Oct 11, 2019 at 15:58
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    $\begingroup$ 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. $\endgroup$
    – KotelKanim
    Commented Oct 12, 2019 at 9:00
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    $\begingroup$ 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. $\endgroup$
    – KotelKanim
    Commented Oct 12, 2019 at 9:15

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