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Let $\mathcal{C}$ be the $E_k$-monoidal $\infty$-category of left modules over a fixed connective $E_{k+1}$-ring spectrum $A$. Suppose that $M$ is an object of $\mathcal{C}$ which is an $E_0$-algebra, in other words $M$ admits a unit map $A\to M$. There is an adjunction $U\colon Alg_{E_k}(\mathcal{C})\leftrightarrows Alg_{E_0}(\mathcal{C})\colon F$ by which we can construct the free $E_k$-algebra on $M$. Note that the $E_k$ algebra generated by the $E_0$-algebra $M$, i.e. with fixed unit, will not generally be equivalent to the $E_k$-algebra generated by $M$ thought of as object of $\mathcal{C}$ (an algebra for the trivial operad).

The unit of the adjunction gives a map of $E_0$-algebras $u\colon M\to F(M)$. I do not have a good description of $F(M)$ in general, but I am interested in showing that if the cofiber of the unit map $A\to M$ is $d$-connected (so $\pi_\ast(M/A)=0$ for $\ast<d$ ) then $cof(u)$ is $2d$-connected.

My rough (and I believe incorrect) attempt is as follows: we can construct the $E_k$-algebra $F(M)$ as an operadic left Kan extension along the inclusion $E_0^\otimes\hookrightarrow E_k^\otimes$, which tells us that it is (I think!), just as an object of $\mathcal{C}$, i.e. ignoring multiplicative structure, equivalent to a colimit over the diagram of injections $\langle m\rangle\to \langle n\rangle$ in $Fin_\ast$ that live over $\langle 1\rangle$. This is a bit terse, but the idea is that you're supposed to take a colimit over the "active" morphisms in $E_0^\otimes$ that live over $\langle 1\rangle$. Recall that $E^\otimes_0=N(Fin^{inj}_\ast)$, i.e. it's the nerve of the category of finite pointed sets with morphisms $f$ such that $f^{-1}(i)$ has at most one element for each $i$ not the base point in the codomain. In other words, the base point goes to the base point, and everything else either goes to something which isn't hit by anything else, or goes to the base point. Also recall that the active morphisms therein are precisely those which only take the base point to the base point, so we've ruled out all the maps that take non-base-point elements to the base point. Thus an active morphism of $E_0^\otimes$ is exactly the data of an injection $\{1,\ldots,m\}\to\{1,\ldots,n\}$ (in particular necessitating that $m<n$).

This means then that $F(M)$ as an object of $\mathcal{C}$ is the colimit in $\mathcal{C}$ of a diagram whose objects are $M^{\otimes_A n}$ for $n\geq 0$, with $M^{\otimes_A 0}=A$, and with maps in between them coming from injections on the indexing sets using the unit of $M$. Let's call the colimit of this diagram $N$. It seems like $N$ is probably quite complicated, but it obviously admits a map from $M$ since $M$ is in the diagram constructing $N$. Then, using that colimits commute, we can take the cofiber of the inclusion of $M$ into $M^{\otimes_A n}$ for all $n>0$, and produce $N/M$ as $colim(M^{\otimes_An}/M)$ for $n>0$. And here I'm being really shady by just writing that as some kind of quotient, because there are a lot of maps from $M$ to $M^{\otimes_An}$ and we're equalizing all of them. Anyway, while $N/M$ is probably quite complicated, it seems like we can say that, at the very least, it is the colimit of a diagram of $2d$-connected spaces, since $M$ is $d$-connected and every object of that diagram is of the form $M^{\otimes_A n}$ for $n>1$, with possibly a bunch of cells attached when we mod out by all the maps from $M$. Thus at least up to degree $2d-1$, $M$ and $F(M)$ are equivalent.

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  • $\begingroup$ I think the proof has the right idea, except that you need to worry that killing the maps $M \to M^{\otimes n}$ doesn't accidentally add in some $d+1$-cells. Probably a way to get a handle on this is to use a May-Milgram/arity type filtration and identify the bottom piece as the inclusion of M and identify the rest of the associated graded as something manageable, like an evident hocolim of at least 2d-connective bits. $\endgroup$ – Dylan Wilson Feb 7 '18 at 12:23
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    $\begingroup$ Actually I'm a little confused: do you really mean $M$ is d-connective and not the cofiber of the map $A \to M$? That seems a little strange. $\endgroup$ – Dylan Wilson Feb 7 '18 at 14:57
  • $\begingroup$ @dylanwilson the cofiber, definitely. $\endgroup$ – Fernando Muro Feb 7 '18 at 17:06
  • $\begingroup$ @dylanwilson yeah that's right, that's a mistake. Or in other words, A and M are the same up through some degree. $\endgroup$ – Jonathan Beardsley Feb 7 '18 at 18:42
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I don't have a general answer but I can confirm your intuition in characteristic $0$. Then we can work with chain complexes and $E_k$-algebras are Poisson $k$-algebras for $k\geq 2$, i.e. commutative DGAs equipped with a compatible Lie bracket of degree $k-1$.

For the sake of simplicity let us consider an $M$ where the unit map splits, $M=A\oplus N$. We can assume that the complex $N$ is trivial in degrees $<d$ (you can even suppose that $N$ has trivial differential if this helps). Then $F(M)$ can be described as follows,

\[ \begin{array}{rcl} F(M)&=&A\oplus S(\Sigma^{1-k}Lie(\Sigma^{k-1}N))\cr &=&A\oplus \bigoplus_{m\geq 1}(\bigoplus_{n\geq 1}\Sigma^{1-k}Lie(n)\otimes_{\Sigma_n} (\Sigma^{k-1}N)^{\otimes^n})^{\otimes^m}_{\Sigma_m} \end{array} \]

Here $S(-)$ and $Lie(-)$ in the first line are the free commutative and Lie algebra functors, and $Lie(n)$ in the second line denotes the Lie operad, which is concentrated in degree $0$.

The map $M\rightarrow F(M)$ is the inclusion of the factor $A$ and the factor obtained for $m=1$ and $n=1$, hence the cofiber is simply obtained by removing these two direct summands.

The complex $\Sigma^{k-1}N$ is $d+k-1$ connected, hence its $n^{\text{th}}$ tensor power is $n(d+k-1)$ connected. Tensoring with $Lie(n)$ doesn't affect connectivity, and if we desuspend $k-1$ times it is $n(d+k-1)-(k-1)$ connected, which is $\geq 2d$ for $n\geq 2$. For $n=1$ the tensor power on $m$ is at least $m\geq 2$, so we get again connectivity degree $\geq 2d$. Note that coinvariants do not reduce connectivity.

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Let me try to turn my comment into an answer. Again, the intuition is to filter $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ by arity, and the result (at least in slightly less generality) is certainly classical, but I'll give a proof in the language you use above. I will also make the following assumptions (which hopefully you are making too):

  • The $\mathbb{E}_{k+1}$-ring $A$ is connective.
  • Let $\overline{M}$ denote the cofiber of the unit $A \to M$. Then $\overline{M}$ is $d$-connective (i.e. $(d-1)$-connected).

Proof: I claim that there is a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ of the induced algebra such that

  • The associated graded is $\bigoplus_{n\ge 0} \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$
  • $F_1=M$ and the map $F_1=M \to \mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the standard one.

I'll build that in a minute but first let's check that it proves the result. It suffices to check that $\mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ is at least $2d$-connective as soon as $n\ge 2$. But, by definition, this object is a colimit over a diagram whose objects all look like $\overline{M}^{\otimes_An}$. Since the $t$-structure is compatible with the tensor product (since $A$ is connective), these are $nd$-connective, and the subcategory of $nd$-connective objects in closed under colimits (since connectivity is checked on underlying spectra and it's true there). So the result is proved assuming we have the filtration. Now let's build it.

First recall the construction of $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$. We're meant to consider all the ways to build tensor powers like $M^{\otimes_An}$ (one for each rectilinear embedding $\coprod_n I^k \hookrightarrow I$) and then glue things together via the various maps $M^{\otimes_An} \to M^{\otimes_Am}$ induced by the unit when $n\le m$. Explicitly, let $D$ denote the $\infty$-category whose objects are rectilinear embeddings $\coprod_n I^k \hookrightarrow I$ ($n$ arbitrary) and whose morphisms are embeddings $\coprod_n I^k \hookrightarrow \coprod_m I^k$ compatible with the structure map up to isotopy and which are injective on path components. (A rigorous definition is the fiber product in HA.3.1.3.1 in the case $\mathcal{C} = \mathsf{LMod}_A$, $\mathcal{A}^{\otimes} = \mathbb{E}_0^{\otimes}$, and $\mathcal{B}^{\otimes} = \mathbb{E}_k^{\otimes}$). Then $M$ extends to a diagram $M^{(-)}: D \to \mathsf{LMod}_A$ and the module underlying $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the colimit of this diagram.

Now, $D$ admits an evident filtration by $n$, $D_{\le n} \subset D_{\le n+1}$. Restricting and taking colimits defines a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ and I want to show it has the two properties above. The second should be straightforward from the definition, so I'll just do the first.

Notice that there is a natural map $F_n/F_{n-1} \to \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ coming from the natural maps $M^{\otimes_An} \to \overline{M}^{\otimes_An}$. We want this to be an equivalence. We'll prove this by induction on $n$. When $n=1$ the result is true by inspection: we're looking at the map $F_0=A \to F_1=M$ and taking the cofiber to get $\overline{M}$. I think one can prove the general case by induction on $n$, but I would have to come back and write the details if you want them.

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The following is an expansion of Dylan's suggestion to use an arity filtration. The reader's digest version is: there's a filtration (which I'm going to construct below) whose associated graded is the free $E_k$-algebra on the cofiber $M/A$, this free algebra has a decomposition into homotopy colimits of diagrams of terms $(M/A)^{\otimes_A n}$, and the "new" terms in $F(M)$ are the quadratic and higher terms with $n > 1$ whose connectivity is roughly $n$ times the connectivity of $M/A$.


Let $\widetilde F$ be the simpler adjunction between left $A$-modules and $E_k$-algebras. We have $F(A \oplus X) \simeq \widetilde F(X)$ naturally. That helps a little, because $\widetilde F$ has a simpler decomposition into the "symmetric power" terms: $$ \widetilde F(X) = \bigoplus_{n \geq 0} {\rm Sym}^n_{E_k}(X) $$ Moreover, any map $A \oplus X \to A \oplus Y$ induces a map $\widetilde F(X) \to \widetilde F(Y)$ that only decreases symmetric power filtration, because the terms involving $A$ all become unit terms. Thus, there's a natural filtration objects of the form $F(A \oplus X)$ whose associated graded is $\widetilde F(X)$.

Let's take our general $E_0$-algebra $A \to M$ now and write it as the homotopy colimit of a simplicial object: $$ M \simeq {\rm hocolim} \{ A \oplus M \leftleftarrows A \oplus A \oplus M \dots\} $$ which is basically saying that $M$ is equivalent to the mapping cylinder of $A \to M$. Then we can apply $F$ to this because $F$ commutes with sifted colimits: $$ F(M) \simeq {\rm hocolim} \{ F(A \oplus M) \leftleftarrows F(A \oplus A \oplus M) \dots \} = {\rm hocolim} \{\widetilde F(M) \leftleftarrows \widetilde F(A \oplus M) \dots\} $$ Applying our natural filtration, we find that there's a filtration of $F(M)$ whose associated graded has terms $$ {\rm hocolim} \{{\rm Sym}^n_{E_k}(M) \leftleftarrows {\rm Sym}^n_{E_k}(A \oplus M) \dots \} \simeq {\rm Sym}^n_{E_k} (M/A) $$ because ${\rm Sym}^n_{E_k}$ also commutes with sifted hocolims.

In this filtration, the bottom two graded terms are ${\rm Sym^0}(M/A) = A$ and ${\rm Sym^1}(M/A) = M/A$, representing the map $M \to F(M)$ in the associated graded. Thus there is a filtration of the cofiber of $M \to F(M)$ whose associated graded is $$ \bigoplus_{n \geq 2} {\rm Sym}^n_{E_k} (M/A). $$


Great. So how do we get connectivity estimates on these functors ${\rm Sym}^n_{E_k}$?

The functor ${\rm Sym}^n_{E_k}$ has the following description. Let $E_k(n)$ be the $n'th$ space in the $E_k$-operad, and $B_k(n) = {\rm hocolim}_{\Sigma_n} E_k(n)$ be the quotient by the action of the symmetric group. This space is homotopy equivalent to the configuration space of $k$ points in an $n$-disk.

The fact that ${\cal C}$ is $E_k$-monoidal means, in particular, that there's a map $$ B_k(n) \to {\rm Fun}({\cal C}, {\cal C}) $$ from this configuration space to the space of functors ${\cal C} \to {\cal C}$. We get this by starting with a $\Sigma_n$-equivariant map $E_k(n) \to {\rm Fun({\cal C}^n, {\cal C})}$ determined by the monoidal structure, following it with the diagonal ${\cal C} \to {\cal C}^n$, and observing that the target ${\rm Fun}({\cal C}, {\cal C})$ has trivial $\Sigma_n$-action so we can pass to the quotient.

Said another way: the points $\alpha$ of this configuration space parametrize $n$-fold tensor power functors $X \mapsto X^{\otimes_\alpha n}$. All of these are equivalent to $X \mapsto X^{\otimes_A n}$, but there's still a nontrivial space of them. We then have $$ {\rm Sym}^n_{E_k}(X) \simeq {\rm hocolim}_{\alpha \in B_k(n)} (M/A)^{\otimes_\alpha n}. $$ Hoewever, hocolims preserve connectivity and so we only need a lower bound on the connectivity of the inside term, which is equivalent to $(M/A)^{\otimes_A n}$.

As a result (assuming $A$ is connective), we find that if $M/A$ is $d$-connected then $(M/A)^{\otimes_\alpha n}$ is $n(d+1)-1$-connected, and thus the whole sum is at least $(2d+1)$-connected.

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  • $\begingroup$ Whoops! Didn't refresh the page before posting... looks like we did the same thing. Except I thought about reducing to the case $A\oplus \overline{M}$ like that but couldn't make it work because I didn't think $\mathrm{Sym}^n_{\mathbb{E}_k}$ commutes with sifted colimits (which it looks like you're using)... does it? $\endgroup$ – Dylan Wilson Feb 7 '18 at 18:50
  • $\begingroup$ Hey @Dylan, if you believe that the formula for $Sym^n$ is what I described then the tensor power part commutes with sifted colimits, and the hocolim commutes with arbitrary ones. (If $A$ is commutative then this formula is easier because you can write it as $E_k(n)_+ \wedge_{\Sigma_n} (M/A)^{\otimes_A n}$.) Do you have a reason for believing that they don't preserve such colimits in mind? $\endgroup$ – Tyler Lawson Feb 7 '18 at 19:08
  • $\begingroup$ (You're right, I did mean sifted colimits, not filtered ones) $\endgroup$ – Tyler Lawson Feb 7 '18 at 19:11
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    $\begingroup$ @Dylan Right, exactly. Being sifted tells you that $I \to I^n$ is cofinal, so the hocolim over $I^n$ of $M_{i_1} \otimes \dots \otimes M_{i_n}$ is the same as the hocolim over $I$ of $M_i \otimes \dots \otimes M_i$, and so we can safely do either version. $\endgroup$ – Tyler Lawson Feb 7 '18 at 20:24
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    $\begingroup$ So what you can do is take the simplicial object you get after applying F, and take the "degree 0" and then take the "degree $\leq 1$" parts of the filtration levelwise. The first is constant with value $A$, and the second recovers the original simplicial resolution of $M$ that you applied $F$ to. When you take geometric realization these become $A$ and $M$ respectively. $\endgroup$ – Tyler Lawson Feb 14 '18 at 16:11

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