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Consider the following construction (which came up recently in a question about "spectral exterior algebras"):

  1. Pick a ring spectrum $R$ and consider the $\infty$-category $\mathsf{Mod}_R$ of module spectra over $R$;
  2. Endow the $\infty$-category $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{Mod}_R)$ with the Day convolution monoidal structure;
  3. Given an $R$-module $M$, take the constant functor $\Delta_{M}\colon\tau_{\leq k}\mathbb{S}\to\mathsf{Mod}_R$;
  4. Apply the free $\mathbb{E}_{\infty}$-monoid functor to $\Delta_M$, turning it into a symmetric lax monoidal functor $\mathsf{Free}^{k}(M)\colon\tau_{\leq k}\mathbb{S}\to\mathsf{Mod}_R$, i.e. into a $\tau_{\leq k}\mathbb{S}$-graded $\mathbb{E}_{\infty}$-$R$-algebra.

Then it turns out that $\mathsf{Free}^k(R)$ interpolates between

  • For $k=0$, the polynomial $\mathbb{E}_{\infty}$-ring spectrum $R[t]$ on $R$, which in SAG is the flat affine line $\mathbb{A}^{\flat,1}_{R}$;

and

  • For $k=\infty$, the free $\mathbb{E}_{\infty}$-ring spectrum $R\{t\}$ on $R$, which in SAG is the smooth affine line $\mathbb{A}^{1}_{R}$.

I find however this a particularly hard to grasp construction because its classical counterpart works in a really sneaky way: replacing the $\infty$-category $\mathsf{Mod}_R$ of $R$-module spectra by the nerve $\mathrm{N}_\bullet(\mathsf{Mod}_R)$ of the category of $R$-modules over an ordinary ring $R$ and applying the analogous construction to $R$ gives

  • $R[t]\overset{\text{(1-cat. coincidence)}}{\cong}\mathrm{Sym}_R(R)$ for $k=0$, and
  • $\bigwedge^\bullet_R(R)$ for $k=1$ (I think),

at which point it stabilises and returns $\bigwedge^\bullet_R(R)$ also for all $k\geq 1$. More generally, the $n$-categorical version of this process stabilises at $k=n$. But for spectra we only get symmetric algebras for $k=\infty$, as noted above, so perhaps looking at a "finite $n$-categorical level" isn't really a good thing to do to try to grasp $\mathsf{Free}^k(R)$ better...

Question. How should one understand $\mathsf{Free}^k(R)$, besides that it has this interpolation feature? (E.g. can we nicely describe the case $R=\mathrm{H}A$ for $A$ an ordinary ring? Is $\mathsf{Free}^k(R)$ flat/smooth?)

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I believe there are no exterior algebras in sight. To see this, let us think through the $1$-categorical case carefully. We have a commutative ring $R$ and the ordinary category of $R$-modules $\mathrm{Mod}_R^\heartsuit$. We are considering symmetric monoidal functors $\tau_{\le k}(\Omega^\infty(S))\to \mathrm{Mod}_R^\heartsuit.$

For $k=0$, we are looking at symmetric monoidal functors from $\tau_{\le 0}(\Omega^\infty(S))=\pi_0(S) =\mathbf Z$ into $R$-modules, hence we are discussing cmmutative $\mathbf Z$-graded $R$-algebras. The free one, generated by $R$, is certainly the algebra $R[t^{\pm 1}]$. This is not the polynomial $R$-algebra $R[t]$, as claimed in the OP, which would instead be obtained if, rather than $\mathbf Z$-gradings, we were considering $\mathbf Z_{\ge 0}$-gradings.

For $k\ge 1$, since $\mathrm{Mod}_R^\heartsuit$ is a $1$-category, functors from $\tau_{\le k}(\Omega^\infty(S))$ are equivalent to those from $\Omega^\infty(S)$ itself. The latter is the free grouplike $\mathbb E_\infty$-space on a single generator, hence a functor $\Omega^\infty(S)$ is equivalent to specifying an invertible $R$-module. For any such choice of $M\in \mathrm{Pic}(R)$, the corresponding symmetric monoidal functor $F^{\Omega^\infty(S)}_M:\Omega^\infty(S)\to\mathrm{Mod}_R^\heartsuit$ will send object-wise $\mathbf Z\ni n\mapsto M^{\otimes n}$. Any symmetric monoidal functor $F:\Omega^\infty(S)\to\mathrm{Mod}^\heartsuit_R$ will be of this for, i.e. $F=F^{\Omega^\infty(S)}_M$ for some $M\in \mathrm{Pic}(R)$!

Conversely, if we were instead looking at symmetric monoidal functors $F:\mathcal F\mathrm{in}^\simeq\to\mathrm{Mod}_R^\heartsuit,$ any $R$-module $M$ would do, with the functor $F_M^{\mathcal F\mathrm{in}^\simeq}$ now going object-wise $\mathbf Z_{\ge 0}\ni n\mapsto M^{\otimes n}$. In this case, we can even see the full functoriality here: a finite set $I$ is sent to the tensor power $M^{\otimes I}.$ To compute its underlying commutative $R$-algebra, we must take the colimit $\varinjlim_{I\in \mathcal F\mathrm{in}^\simeq}M^{\otimes I}$. Using the presentation $\mathcal F\mathrm{in}^\simeq \simeq \bigoplus_{n\ ge 0}\mathrm B\Sigma_n$, we find that the underlyign commutative ring is $\bigoplus_{n \ge 0}(M^{\otimes n})/\Sigma_n = \bigoplus_{n\ge 0}\mathrm{Sym}^{\heartsuit, n}_R(M) = \mathrm{Sym}^{\heartsuit, *}_R(M)$, the usual symmetric algebra on $M$.

The Barrat-Priddy-Quillen equivalence $\Omega^\infty(S) \simeq (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}$ now implies a relationship between the functors $F^{\Omega^\infty(S)}_M$ and $F^{\mathcal F\mathrm{in}^\simeq}_M$, or more precisely, their underlying commutative $R$-algebras. Indeed, the commutative $R$-algebra $\varinjlim(F^{\Omega^\infty(S)}_M)$ is a localization of the commutative $R$-algebra $\varinjlim(F^{\mathcal F\mathrm{in}^\simeq}_M) = \mathrm{Sym}^{\heartsuit, *}_R(M)$. It is the localizations along the elements in degree one, i.e. along $\mathrm{Sym}^{\heartsuit, 1}_R(M) = M$ - we must invert all these elements to obtain the underlying commutative $R$-algebra of $F^{\Omega^\infty(S)}_M$. That is not too hard to do, but I am not aware of it having a name in general.

In the special case when $M = R$ though, things become easier! We have $\mathrm{Sym}^{*, \heartsuit}_R(R) = R[t]$, the polynomial $R$-algebra, and to invert things in degree $1$, we need just invert $t$. Hence we find that the underlying commutative $R$-algebra of $F^{\Omega^\infty(S)}_R$ is again just $R[t^{\pm 1}]$., same in the $t=\infty$ase as in the $t=0$ one.

And this is also the underlying commutative $R$-algebra of the symmetric monoidal functors $F^{\tau_{\le k}(\Omega^\infty(S))}_R$, for all $k\ge 1$. Indeed, as we discussed before, all these functors are equivalent to one another, and for $k=\infty$, we clearly get the constant functor with vale $R$. We are therefore getting the same constant functor $R$ for all $1\le k \le \infty$. The underlying commutative $R$-algebra is obtained as colimit in $\mathrm{Mod}_R^{\heartsuit}$, but since that is a $1$-category, it is oblivious to all the higher structure, and the colimit will be the same for all $1\le k\le \infty$ as well.

Long story short: for both $k=0$ and $k\ge 1$, we are getting the underlying commutative $R$-algebra be $R[t^{\pm 1}]$, never anything resembling exterior algebas. If you ask me, the "$1$-categorical miracle" we are running into is probably that $R[t]$ is both $\bigoplus_{n\ge 0}R$, i.e. the colimit of the constant $R$-valued functor $\mathbf Z_{\ge 0}\to\mathrm{Mod}_R^\heartsuit$, and $\mathrm{Sym}^{\heartsuit, *}_R(R) = \bigoplus_{n\ge 0} (R^{\otimes n})/\Sigma_n$, i.e. colimit of the constant $R$-valued functor $\mathcal F\mathrm{in}^\simeq\to\mathrm{Mod}_R^\heartsuit.$

That would be false if we were working in $R$-module spectra, where the $\infty$-categorical structure allows us to be more sensitive to the $\mathrm B\Sigma_n$-components of $\mathcal F\mathrm{in}^\simeq$, leading to quotients $(R^{\otimes n})_{h\Sigma_n}\simeq R_{h\Sigma_n}$ not needing to be just $R$. Indeed, the homotopy groups $\pi_i(R_{h\Sigma_n}) = \mathrm H_i(\Sigma_n; R)$ coincide with group homology, which might be non-trivial if $n$ divides the characteristic of $R$.

But even then, we have two choices:

  1. We follow the OP, and consider symmetric monoidal functors $\tau_{\le k}(\Omega^\infty(S))\to\mathrm{Mod}_R$. That will be some kind of thing interpolating between the localizations $R[t^{\pm 1}]$ for $k=0$, and $R\{t^{\pm 1}]$ for $k=\infty$. I don't have much insight about what to say about the intermediate stages, other than to remark that these are intermediate "localized symmetric" algebras, still not "exterior" in any way.
  2. But if instead we wish to compare $R[t]$ and $R\{t\}$, then we will be looking at symmetric monoidal functors $\mathbf Z\to\mathrm{Mod}_R$ and $\mathcal F\mathrm{in}^\simeq\to \mathrm{Mod}_R$ instead. Then the "$k$-truncation" idea will not be as fruitful, ad $\mathcal F\mathrm{in}^\simeq$ is a $1$-category, so we only have two options: $k=0$ for which we get $R[t]$, and $1\le k\le \infty$ for which we get $R\{t\}$.
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    $\begingroup$ Perhaps you are confused about the kind of graded commutativity we are considering here. We are talking about 'naive' graded commutativity, e.g. $ab = ba$ regardless of the degrees of $a$ and $b$, as opposed to the graded-commutativity of the Koszul sign rule $ab = (-1)^{|a||b|}ba$. The notion under discussion here is relevant for instance for doing Proj in algebraic geometry. In this sense, the exterior algebra isn't actually commutative, so it surely can't be a free commutative graded algebra. Instead, that role is played by the symmetric algebra. $\endgroup$ Sep 8 '21 at 1:22
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    $\begingroup$ About the first question though, no, a lax symmetric monoidal functor from $\Omega^\infty(S)$ is not the same thing as an invertible object. You're right, my discussion of symmetric monoidal functors above isn't perfect, as we really should be talking about lax symmetric monoidal ones instead. $\endgroup$ Sep 8 '21 at 1:28
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    $\begingroup$ When I wrote this idea up on my other question, I was thinking about this Koszul kind of graded-commutativity (which agrees with the naive one for $k=0$): $\mathsf{Free}^{k}(R)$ is the free $\mathbb{E}_{\infty}$-monoid on $\Delta_R$ with respect to Day convolution on $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{Mod}_R)$, and is hence not necessarily $\mathbb{E}_{\infty}$ in the naive sense, meaning... $\endgroup$
    – Théo
    Sep 8 '21 at 3:31
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    $\begingroup$ ...that its underlying monoid in $\mathsf{Mod}_R$ (i.e. its left Kan extension along the terminal functor $\tau_{\leq k}\mathbb{S}\to\Delta^0$) is not necessarily an $\mathbb{E}_{\infty}$-ring (and probably isn't one for $k>1$). When you wrote $\mathsf{Free}^\infty(\mathbb{S})\cong\mathbb{S}\{t^\pm\}$, did you mean the usual $\mathbb{E}_{\infty}$-commutativity? $\endgroup$
    – Théo
    Sep 8 '21 at 3:31
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    $\begingroup$ Yes, the correspondence to lax symmetric monoidal functors $\mathbf Z\to\mathrm{Mod}_R$, and commutative graded $R$-algebra wants you to consider the naive commutativity. To get the Koszul sign rule, you can instead use the shearing trick mentioned a few questions back: that really amounts to equipping $\mathrm{Fun}(\mathbf Z, \mathrm{Mod}_R)$ with a different symmetric monoidal structure. The $1$-shearing not being symmetric monoidal is good, because Koszul and naive sign rules are different. $2$-shearing not being symmetric monoidal in char $p$ or spectrally is less intuitive though ... :) $\endgroup$ Sep 8 '21 at 13:10

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