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Given a ring spectrum $R$ and an $R$-module $E$, we have the spectral symmetric algebra $\mathrm{Sym}_R(E)$ of $E$ over $R$, defined by $$ \begin{align*} \mathrm{Sym}_R(E) &\overset{\mathrm{def}}{=} \mathrm{colim}_{\mathbb{F}}(\Delta_{E})\\ &\cong \bigoplus_{n\in\mathbb{N}}E^{\otimes_\mathbb{S}n}_{\mathsf{h}\Sigma_{n}}, \end{align*} $$ where $\mathbb{F}\overset{\mathrm{def}}{=}\mathsf{FinSets}^\simeq$ is the groupoid of finite sets and permutations. As A Rock and a Hard Place showed here, the $\mathbb{E}_\infty$-ring $\mathrm{Sym}_R(E)$ comes with a natural grading by the sphere spectrum, inducing a $\mathbb{Z}$-grading on $\pi_0(\mathrm{Sym}_R(E))\cong\mathrm{Sym}_{\pi_0(R)}(\pi_0(E))$. So e.g. picking $R=E=\mathbb{S}$, gives $$ \begin{align*} \pi_0(\mathrm{Sym}_{\mathbb{S}}(\mathbb{S})) &\overset{\mathrm{def}}{=} \pi_0(\mathbb{S}\{t\})\\ &\cong \mathbb{Z}[t], \end{align*} $$ which carries the natural $\mathbb{Z}$-grading.

However, the $\pi_0$ of an $\mathbb{S}$-graded ring can be more complicated than just a commutative $\mathbb{Z}$-grading, and for instance allows for the multiplication on the $\pi_0$ to be supercommutative, satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$. This led me to the following pair of questions:

  1. Is there an "spectral exterior algebra" construction $\bigwedge_RE$, whose $\pi_0$ is the $\mathbb{Z}$-graded supercommutative exterior algebra $\bigwedge_{\pi_0(R)}\pi_0(E)$? If so, does it come with an $\mathbb{S}$-grading?
  2. One of the more homotopy-theoretic points of view on symmetric and exterior algebras is that the passage from the former to the latter corresponds to considering a larger portion of the sphere spectrum. More generally, do we have an $\mathbb{N}$-indexed sequence of "higher exterior algebra" constructions $\mathrm{Sym}_R(E)$, $\bigwedge_R(E)$, $\bigwedge^{\mathbf{2}}_R(E)$, $\ldots$?
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Interesting question! I can't give a real answer, but here are some idle musings:

Note that one way to encode exterior powers is as $\Lambda^i_R(E) = \Sigma^{-i}(\mathrm{Sym}^i_R(\Sigma(E)))$ (since placing the generators in degree one switches, by virtue of the Koszul sign rule from the usual $\Sigma_n$-action to the alternating one). So the symmetric algebra in SAG or DAG on a $1$-shifted module looks like a sheared version of the exterior algebra. That is, if $R$ is an ordinary commutative ring and $M$ an $R$-module, then $$ \mathrm{Sym}^*_R(\Sigma (M)) \simeq \Sigma^{*}(\Lambda_R^*(M)) $$ (though here both symmetric and exterior algebras are understood in the spectral or derived sense, and may not agree with their usual algebraic counterparts unless $M$ is flat). In the DAG context, we similarly have $$ \mathrm{Sym}^*_R(\Sigma^2(M)) \simeq \Sigma^{2*}(\Gamma_R^*(M)), $$ where $\Gamma^*_R(M)$ is the free divided power $R$-algebra generated by $M$. See Section 25.2 of the SAG book for the proofs.

Therefore, perhaps the candidates for your degree $n$ higher exterior algebras might be some kind of $n*$-fold desuspensions of the symmetric algebra $\mathrm{Sym}^*_R(\Sigma^n(R))$. For $n=0$, that would recover (flatness issues notwithstanding) the usual symmetric algebra, for $n=1$ the exterior algebra, for $n=2$ the free divided power algebra, and for $n\ge 3$, who knows? :)

Operating on this provisional understanding of what the higher exterior algebras are, let us discuss the $S$-grading. We see from the construction that the "unshearing" of it is the symmetric algebra $\mathrm{Sym}^*_R(\Sigma^n(R))$, which does indeed carry an $S$-grading. The question about whether the higher exterior algebras themselves carry a canonical $S$-grading now becomes a question about how, if at all, some kind of a "shearing" functor $(M_n)\mapsto (\Sigma^n M_n)$ interacts with $S$-gradings. I believe this going to be a very non-trivial question.

Even if we were not talking about the much-more-complicated $S$-gradings, but rather $\mathbf Z$-gradings, the shearing functor $\Sigma^*$ is indeed an automorphism of the $\infty$-category of $\mathbf Z$-graded spectra $\mathrm{Fun}(\mathbf Z,\mathrm{Sp})$, but I don't think it is symmetric monoidal. Indeed, consider rather its inverse squared $\Sigma^{-2*}$. If it were symmetric monoidal, it would send the polynomial $\mathbb E_\infty$-algebra $S[t]\simeq \bigoplus_{n\ge 0}S$, with its usual $\mathbf Z$-grading, into the "$2$-shifted polynomial $\mathbb E_\infty$-algebra" $S[\beta] \simeq \bigoplus_{n\ge 0}\Sigma^{-2n}S$. The latter thing does indeed exist as an $\mathbb E_2$-ring, studied in Section 2.8 of this paper by Lurie. But I have been told before that, due to some obstructions one can calculate, it can be shown that said $\mathbb E_2$-ring does not support an $\mathbb E_\infty$-structure. That would seem to imply that the shearing functor $\Sigma^{-2*}$ (and consequently $\Sigma^*$) can not be symmetric monoidal.

And that was just in the $\mathbf Z$-graded world, let alone all the additional complications that the $S$-graded setting might bring!

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    $\begingroup$ The shearing functor on $\mathbf{Z}$-graded spectra is $\mathbf{E}_2$-monoidal, but provably not $\mathbf{E}_3$-monoidal. Indeed, there is a nontrivial obstruction for $S[\beta]$ to admit an $\mathbf{E}_3$-structure. Consider $\beta^2\in \pi_{-4} S[\beta]$; if $S[\beta]$ did admit an $\mathbf{E}_3$-structure, we'd get a map $\mathrm{Conf}_2(\mathbf{R}^3)_+ \otimes_{\Sigma_2} (S^{-2})^{\otimes 2} \to S[\beta]$ which extends $\beta^2$. The source can be identified with $\Sigma^{-2} \mathbf{R}P^0_{-2}$. $\endgroup$
    – skd
    Sep 6, 2021 at 3:57
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    $\begingroup$ Because the $(-4)$-cell of $S[\beta]$ is unattached, the resulting map $\Sigma^{-2} \mathbf{R}P^0_{-2} \to S[\beta] \to S^{-4}$ of spectra would give a splitting of the bottom cell of $\Sigma^{-2} \mathbf{R}P^0_{-2}$. But, noting that $\mathbf{R}P^0_{-2} \simeq \Sigma^{-1} (\mathbf{R}P^1_{-1})^\vee$, one sees that this is impossible by looking at the cell structure of $\mathbf{R}P^1_{-1}$. (Namely, the top cell is attached by $\eta$ to the $(-1)$-cell.) $\endgroup$
    – skd
    Sep 6, 2021 at 3:57
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    $\begingroup$ One other interesting fact is that the shearing $\mathrm{sh}\ S[\beta]$, whose underlying spectrum additively looks like $\bigoplus_{n\geq 0} S^{-4n}$, admits an $\mathbf{E}_4$-algebra structure in graded spectra. This basically comes from an $\mathbf{E}_4$-coalgebra structure on $\mathbf{H}P^\infty$, viewing it as the $4$-fold bar construction on $\Omega^3 S^3$. But I don't think $\mathrm{sh}^2\ S[\beta] \simeq \bigoplus_{n\geq 0} S^{-8n}$ can be anything more than an $\mathbf{E}_2$-ring (from $\mathrm{sh}$ being $\mathbf{E}_2$-monoidal and $S[\beta]$ being an $\mathbf{E}_2$-ring). $\endgroup$
    – skd
    Sep 6, 2021 at 4:04
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    $\begingroup$ @Emily One way to predict the appearance of divided powers in this story is the following. Let $k$ be a field (say), and let $k[t]$ denote the polynomial ring on a generator in degree $0$. Then the bar construction $k \otimes_{k[t]} k$ (in $\mathbf{E}_\infty$-rings) is $k[S^1]$, whose homotopy can be identified with the polynomial ring on a generator in degree $1$. Note that the homology of $S^1$ is an exterior algebra on a generator in degree $1$; this is the topological instantiation of the fact that $\Sigma^\ast \mathrm{Sym}^\ast_k(k) = \mathrm{Sym}^\ast_k(\Sigma k)$. $\endgroup$
    – skd
    Sep 6, 2021 at 14:40
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    $\begingroup$ Taking the bar construction again (i.e., $k\otimes_{k\otimes_{k[t]} k} k$), you get the $\mathbf{E}_\infty$-$k$-algebra $C_\ast(\mathbf{C}P^\infty; k)$; now observe that the homology of $\mathbf{C}P^\infty$ is a divided power algebra. In general, $\mathrm{Sym}^\ast_k(\Sigma^n k)$ will be related to the homology of $K(\mathbf{Z}, n)$. $\endgroup$
    – skd
    Sep 6, 2021 at 14:40
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This is not an answer, but rather a comment to A Rock and a Hard Place's answer.

Here's another idle musing: what about replacing the relation $ab=(-1)^{\deg(a)\deg(b)}ba$ with coherent homotopies, having an analogue of $\mathbb{E}_{\infty}$ for graded-commutativity?

I'm not sure what's the best way to do this, but I see a possible (probably not totally satisfying) one. The basic idea is to generalise from the characterisation of exterior algebras as free graded commutative algebras.

1. The case of classical exterior algebras.

Recall that:

  1. A $\mathbb{Z}$-graded $R$-algebra $A_\bullet$ is $\mathbb{Z}$-graded commutative if we have $$ab=(-1)^{\deg(a)\deg(b)}ba$$ for each $a,b\in A_\bullet$.
  2. The exterior algebra $\bigwedge_RM$ on an $R$-module $M$ is the free $\mathbb{Z}$-graded commutative algebra on $M$: the assignment $M\mapsto\bigwedge_RM$ defines a functor $$\textstyle\bigwedge_R\colon\mathsf{Alg}_R\to\mathsf{CommGr}_{\mathbb{Z}}\mathsf{Alg}_R$$ that is left adjoint to the forgetful functor $\mathsf{CommGr}_{\mathbb{Z}}\mathsf{Alg}_R\hookrightarrow\mathsf{Alg}_R$.
  3. The category of $\mathbb{Z}$-graded commutative algebras embeds fully faithfully into that of $\tau_{\leq1}\mathbb{S}$-graded commutative algebras, which are lax symmetric monoidal functors $(\tau_{\leq1}\mathbb{S},\otimes,0)\to(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$.

Question: is $\bigwedge_RM$ not only the free $\mathbb{Z}$-graded commutative algebra on $M$, but also the free $\tau_{\leq1}\mathbb{S}$-graded commutative algebra on $M$?

2. The graded commutativity condition.

As I mentioned in my other question, a $\tau_{\leq1}\mathbb{S}$-graded $R$-algebra is a lax monoidal functor from the $1$-truncation of the sphere spectrum to $(\mathsf{Mod}_R,\otimes_{R},R)$, and it is $\tau_{\leq1}\mathbb{S}$-graded commutative when this functor is symmetric lax monoidal.

Now, for a functor $(\tau_{\leq1}\mathbb{S},\otimes,0)\to(\mathsf{Mod}_R,\otimes_{R},R)$ to be (symmetric) lax monoidal is the same as for it to be a (commutative) monoid under the Day convolution monoidal structure on $\mathbf{Fun}(\tau_{\leq1}\mathbb{S},\mathsf{Mod}_R)$. So, given an $R$-module $M$, we can consider the constant functor $$\Delta_M\colon\tau_{\leq1}\mathbb{S}\to\mathsf{Mod}_R$$ on $M$ and consider the free commutative $\otimes_{\mathsf{Day}}$-monoid on $\Delta_M$. If the answer to the question above is yes, then this is the exterior algebra of $M$.

3. Spectral exterior algebras.

Now we can repeat the same strategy for module spectra: given a ring spectrum $R$ and an $R$-module $M$, pick the constant functor $$\Delta_M\colon\tau_{\leq k}\mathbb{S}\to\mathsf{ModSp}_R$$ and apply the free $\mathbb{E}_{\infty}$-monoid functor to $\Delta_M$ with respect to the Day convolution monoidal structure on $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{ModSp}_R)$. The result is then a possible candidate for the spectral "$k$th higher" exterior algebra of $M$ over $R$.

4. The catch.

I'm not sure if this really gives a desirable result at all. For instance, does it recover spectral symmetric algebras when $k=0$?

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    $\begingroup$ This is a very interesting idea too, though I believe it's doing something slightly different. Namely, for $k = 0$, we get $\tau_{\le 0}(S) = \mathbf Z$, and so we are looking at the free $\mathbf Z$-graded $\mathbb E_\infty$-$R$-algebra, which is the polynomial $\mathbb E_\infty$-algebra $R[t]$. For $k=\infty$, we obviously get the free $\mathbb E_\infty$-ring $S\{t\}$ back, so what the construction for $0 < k <\infty$ seems to be doing is building in ever more of the higher grading-coherences that are there in an $S$-grading, other than just the underlying $\mathbf Z$-grading.. $\endgroup$
    – A Rock and a Hard Place
    Sep 6, 2021 at 8:41
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    $\begingroup$ So it's not about "higher exteriority", which is to me maybe about changing the commutativity constraint, i.e. symmetric group actions and things like that - as discussed above, that has to do with "free graded algebras" (in whatever sense we're thinking about) on suspensions. Instead, the commutativity constraint is the same, we're just imposing it with varying levels of homotopy coherence. $\endgroup$
    – A Rock and a Hard Place
    Sep 6, 2021 at 8:44
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    $\begingroup$ An interesting thing to ponder is to combine the two constructions: consider the (possibly $n$-times sheared) free $\tau_{\le k}(S)$-graded $\mathbb E_\infty$-$R$-algebra generated by the $n$-fold suspension $\Sigma^n(R)$. No idea what you could say about or do with this $(k, n)$-bigraded objects, but it looks fun! :) $\endgroup$
    – A Rock and a Hard Place
    Sep 6, 2021 at 8:46
  • $\begingroup$ @ARockandaHardPlace Thanks! I'm really shocked at how sneaky a situation this is: if you replace $\mathsf{ModSp}_R$ by $\mathrm{N}_{\bullet}(\mathsf{Mod}_R)$ and apply this construction to $R$, then I think this gives you $R[t]\cong\mathrm{Sym}_R(R)$ for $k=0$ and then $\bigwedge^{\bullet}_RR$ for $k\geq 1$, at which point it stabilises. Similarly, the $n$-categorical version of this stabilises at $k=n$. But when you pass from abelian groups to spectra this gives something totally different, and we get the symmetric algebra only for $k=\infty$! $\endgroup$
    – Emily
    Sep 6, 2021 at 22:52

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