Timeline for For which $k$-types are $E_{n,m}$-algebras automatically $E_{n+1}$ algebras?
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| Oct 12, 2019 at 15:24 | comment | added | KotelKanim | You understand the definitions correctly. The way the proof goes is to detect $d$-equivalences by mapping to $(d+1)$-operads (i.e. operation spaces are $d$-types). It might be that the statement that you suggest also holds (changing the connectivity assumption and conclusion "by 1/2"). I will have to think about it a little. | |
| Oct 12, 2019 at 13:33 | comment | added | Noah Snyder | I was surprised that your relative result requires the map to be a d-equivalence when I was expecting just d-connected (ie only subjective at the top level). Have I misunderstood your definitions? Might there be a slightly stronger version of the relative statement? (I will also check that the theorem I claimed is actually correct.) | |
| Oct 12, 2019 at 9:15 | comment | added | KotelKanim | 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. | |
| Oct 12, 2019 at 9:00 | comment | added | KotelKanim | 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. | |
| Oct 11, 2019 at 15:58 | comment | added | Noah Snyder | 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? | |
| Oct 11, 2019 at 14:37 | comment | added | KotelKanim | Actually, all the $A_m$-s are exactly and only $(-1)$-connected. The operation spaces are not connected since the multiplication is not commutative. | |
| Oct 11, 2019 at 14:02 | comment | added | KotelKanim | 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). | |
| Oct 10, 2019 at 19:33 | history | asked | Noah Snyder | CC BY-SA 4.0 |