TITLE: Is there a "spectral exterior algebra" construction in higher algebra?
QUESTION [5 upvotes]: 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:

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?
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 $\mathrm{Sym}_R(E)$, $\bigwedge_R(E)$, $\bigwedge^{\mathbf{2}}_R(E)$, $\ldots$?

REPLY [2 votes]: 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:

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 $a,b\in A_\bullet$.
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 $\mathsf{CommGr}_{\mathbb{Z}}\mathsf{Alg}_R\hookrightarrow\mathsf{Alg}_R$.
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 $k=0$?