Hi all, is there a way of getting added to the Slack homotopy chat?

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an $E_n$-algebra is, by definition and in particular, $A_\infty$. Is there a good reason for this? I mean, one could consider something which is quite coherently homotopy commutative but not fully-coherently homotopy associative, like Barwick does for example in page 41 here arxiv.org/pdf/1302.5756.pdf .

but somehow, $E_n$-algebras are the popular objects. one possible explanation could be "the interesting examples of quite coherently homotopy commutative objects arising in practice are also $A_\infty$". is there an explanation for this fact?

@BrunoStonek I think you need at least $E_1$ to have a (reasonably well-behaved) category of modules. That alone justifies the focus on them, in my mind.

I see... what breaks down if you want to consider modules over, say, $A_m$-algebras?

@BrunoStonek How do you define them?

A module (over an $E_n$-algebra $R$) is a spectrum $M$ with an $E_1$-map $R→End(M)$

I guess you could consider $A_m$-algebra maps, but it wouldn't reduce to the same thing for $A_m$-algebras which are $A_∞$-algebras, so it doesn't sound as appealing.

you mean an $A_m$-module over an $A_\infty$-algebra wouldn't be an $A_\infty$-module? why does that reduce its appeal?

I'm not even quite sure it deserves the name "module'. Also, I don't know how to prove that the category of $A_m$-modules is, e.g., stable. That said, I'm sure there are people studying nonassociative rings somewhere

It's just not too surprising that the focus is pretty much on associative rings

but they are associative, only not fully coherently :)

@BrunoStonek Ok, "associative" for me means "at least as much structure as the endomorphism space of something" :)

another issue is that R is probably not an "A_m-module" over itself for various reasonable guesses of the definition for A_m-module. Also, if I understand correctly, the proposed definition of A_m-module is just that of a left module over Ind_{A_m}^{A_{infty}}(R), and hence covered by the classical story

The idea of an $A_n$ structure is that it supposed to be exactly what you need to produce a "bar complex up to dimension $n$". So maybe there's a notion of a structure which is exactly what you need to produce a "$\Gamma$-object up to dimension $n$".

in the operadic setup you can just restrict to the symmetric sequence on \le n things and then freely extend the rest (one of two possible definitions of 'arity filtration'). this should agree, in Barwick's operator category setup, with the truncation of operator categories by cardinality (ex 1.5)... presumably in the case of Gamma-objects it'll be functors on Fin_*^{\le n} that satisfy the Segal condition whenever it makes sense

where n might be n +1 or something

schwede's paper on rigidity in stable homotopy theory studies coherent actions of Moore spaces, which are some kind of A_m-modules stucture, in section 2: math.uni-bonn.de/people/schwede/rigid.pdf

i'd suggest that one of the reasons for the focus on E_n-algebras is the classical one: namely, the E_n-operads go with n-fold loop spaces via the recognition principle

one does encounter some things in the literature that are partially commutative and associative, e.g. homotopy commutative and associative H-spaces, or homotopy-commutative A_infty rings

& I know that some people have tried to consider alternative "filtrations" of the E_infty operad with the hope that they would make it easier to construct E_infty structures on certain stubborn things

but by and large i think that the lack of work is due to lack of examples & applications of partial structures, not because E_n is "fundamentally right"

that kind of thinking gets you into trouble

@Dylan Wilson If you restrict to the $n$th stage of the free-arity filtration of an operad $O$, getting a suboperad $O_n\subseteq O$, then you get a corresponding filtration of the identity functor on $O$-algebras, whose $n$th stage sends $A\mapsto | B(O,O_n,A) |:=F_n(A)$. But this is not the same in general as the filtration defined by $A\mapsto |B(O,O,A)|_{\leq n}:=S_n(A)$.

For instance, if $O=$ $E_\infty$-operad, then $F_n(A)/F_{n-1}(A)$ is (I think) supposed to look like the free $O$-algebra on $P_n\otimes_{\Sigma_n}A^{\otimes n}$, where $P_n$ is some version of the partition complex. (Not sure if this is actually a theorem someone has proved.)

@CharlesRezk I'm confused about whether that actually contradicts what I was saying, but maybe I didn't say it well. Just to be on the same page, here's what I mean by arity filtration: Let SSeq_{\le n} be the category of symmetric sequences up to arity n and give it 'truncated' composition product monoidal structure. Then the forgetful functor SSeq-->SSeq_{\le n} is monoidal and induces a limit preserving forgetful functor on categories of monoids, which in turn has a left adjoint.

so my "O_n" is the composite of the forgetful functor on O with this left adjoint

then it seems like an O_n-algebra is the same as a left module object over the restriction of O to SSeq_{\le n}, where I view that as acting on my category via P\circ A = \bigoplus_{i\le n} P(n) \otimes_{\Sigma_n} A^{\otimes n}

I guess I want to argue that, in the cartesian monoidal setting and when O=E_{infty}, an algebra over O_n is the same as an 'n-truncated Segal object'.

does that not seem right?

er, typo: switch n's with i's in the formula for P\circ A

except in the indexing of the sum

@DylanWilson I think I need to refine or correct what I was saying, which I will think about. My feeling is that for a general operad your filtration won't coincide with what would think of as a "truncation of operator categories", but perhaps I don't understand exactly what that means.

One thing I've wondered before: what do you get when you take Boardman-Vogt tensor products $A_n \otimes A_m$? For example, by Eckmann-Hilton, I'm pretty sure an $A_2 \otimes A_2$-algebra is the same thing as a homotopy-commutative $A_3$-algebra. (recall $A_2$-space = $H$-space, $A_3$ = homotopy associative)

The fact that the associativity and the commutativity both increase in this case suggests to me that maybe $\varinjlim_{n \to \infty} A_2^{\otimes n} \overset ? = E_\infty$. This (or any of the natural variants) would give a somewhat-reasonable (and possibly even interesting) alternative filtration of $E_\infty$ by operads.

For instance, $K(n)$ is homotopy-commutative (when $p \neq 2$, right?). Is $K(n)$ an $A_\infty \otimes A_2$ algebra?

@TimCampion You might be looking for this: arxiv.org/abs/1808.06006

@SaalHardali Hmm... I must admit I was too lazy to look carefully at this paper before! Interestingly, Tomer and Lior's estimate $conn(P \otimes Q) \geq conn(P) + conn(Q) + 2$ doesn't "associate" -- if you take an iterated tensor product, the estimate that pops out depends on the way you associate it. But the $A_2$ operad is $(-1)$-connected and $A_2 \otimes A_2$ is 0-connected (need to prove this by hand with the usual Eckmann-Hilton argument),

but then you get that $A_2^{\otimes n}$ is $(n-2)$-connected. In the limit, you do indeed get that $\varinjlim_n A_2^{\otimes n}$ is $\infty$-connected, and so equals $E_\infty$.

Hmm... maybe I should be more careful with that claim about $A_2 \otimes A_2$, which is key to the argument.

er -- scratch all but the conclusion. The estimate doesn't depend on how you associate things. I'm just bad at addition. We have that $A_2$ is $(-1)$-connected, so $A_2 \otimes A_2$ is $(-1) + (-1) + 2 = 0$-connected, and so forth. Don't need to do anything by hand. We get $\varinjlim_n A_2^{\otimes n} = E_\infty$

In fact, any reduced operad $O$ which has any higher operations at all is $(-1)$-connected, and so we get $\varinjlim_n O^{\otimes n} = E_\infty$.

Even more generally, if $O_1,O_2,\dots$ is any sequence of reduced operads each of which has at least one higher operation, then $\varinjlim O_1 \otimes \dots \otimes O_n = E_\infty$.

So taking tensor products seems like a very easy way to filter $E_\infty$ by operads.

In HA 4.1.8.4, rectification of algebras, can we see that nilpotent algebras correspond? More precisely, if A is an E_1-algebra that is nilpotent, does it correspond to a nilpotent strict algebra?

this thing is awesome: every homotopy commutative ring is a Q_1-algebra (so A_2-algebras are Q_1-algebras), and the mod 2 homology of any Q_1-algebra already has the dyer-lashof operation Q_1 (hence the name)

in fact, the universal Q_1-algebra on a sphere in dimension k is the space Ω^2 S^{k+2}, and so this refines the hopkins-mahowald theorem: the free p-local Q_1-algebra with a nullhomotopy of p is HF_p

and to answer tim's question, the morava K-theory K(n) is a E_1 (x) Q_1-algebra for p odd

btw, the reason for the mod 2 homology of any Q_1-algebra already having the dyer-lashof operation Q_1 is simply that the homology of the second space (Q_1)_n = S^1 is exterior on one generator, and that generator gives Q_1

@skd I like Q_1 also, but where does the claim that the universal Q_1 algebra on a sphere is the universal E_2 algebra on a sphere come from?