$\mathbb{Z}$-graded rings play an important role in algebra and algebraic geometry, so when moving to derived algebra and spectral algebraic geometry, it's natural to ask about ring spectra graded in the sphere spectrum $\mathbb{S}$.

As discussed in Section 2 of Bunke–Nikolaus's Twisted differential cohomology, such an object may be defined as a symmetric lax monoidal functor $\mathbb{S}\to\mathsf{Sp}$, where $\mathbb{S}$ is considered as a monoidal $\infty$-groupoid.

In particular, given an $\mathbb{S}$-graded ring spectrum $R$, it follows that the ring $\pi_0(R)$ is $\mathsf{Ho}(\mathbb{S})$-graded. By the description here, such a $\tau_{\leq1}\mathbb{S}$-graded ring consists of a pair $(R_\bullet,\{\sigma_k\}_{k\in\mathbb{Z}})$ with

  • $R_\bullet$ a $\mathbb{Z}$-graded ring;
  • $\{\sigma_k\colon R_k\to R_k\}_{k\in\mathbb{Z}}$ a family of order $2$ automorphisms, one for each $R_k$.

Moreover, if $R$ is an $\mathbb{E}_{\infty}$-ring, then $\pi_0(R)$ is "$\mathsf{Ho}(\mathbb{S})$-graded commutative", in that we additionally have $$ ab = \begin{cases} ba &\text{if $\deg(a)\deg(b)$ is even,}\\ \sigma_{\deg(a)+\deg(b)}(ab) &\text{if $\deg(a)\deg(b)$ is odd} \end{cases} $$ for each $a,b\in R_\bullet$.

This includes in particular $\mathbb{Z}$-graded commutative algebras by picking $\sigma_k(a)\overset{\mathrm{def}}{=}-a$ for each $k\in\mathbb{Z}$, as in that case the above condition becomes $$ab=(-1)^{\deg(a)\deg(b)}ba.$$

Question. So, are there any interesting/non-trivial "in nature" examples of $\mathbb{S}$-graded ring spectra?


1 Answer 1


One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads!

1. $S$-grading on $S\{t\}$

For the sake of clarity, allow me to denote the underlying infinite loop space (equivalently: grouplike $\mathbb E_\infty$-space) of the sphere spectrum $S$ by $\Omega^\infty(S)$. Recall that, by the Barrat-Priddy-Quillen Theorem, the ininite loop space $\Omega^\infty(S)$ can be described as the group completion of the groupoid of finite sets and isomorphisms $\mathcal F\mathrm{in}^\simeq$. Consider the constant functor $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$ with value $S$ - since the latter is the monoidal unit, this is a symmetric monoidal functor. Left Kan extension along the canonical map $\mathcal F\mathrm{in}^\simeq \to (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}\simeq \Omega^\infty(S)$ produces a symmetric monoidal functor $\Omega^\infty(S)\to \mathrm{Sp}$, and as such exhibits an $S$-graded $\mathbb E_\infty$-ring.

But which one? To figure out which one, note that the passage to the "underlying $\mathbb E_\infty$-ring" of an $S$-graded $\mathbb E_\infty$-ring is given by passage to the colimit. Thus the underlying $\mathbb E_\infty$-ring, which we have just adorned with an $S$-grading, is $$ \varinjlim_{\Omega^\infty(S)}\mathrm{LKan}^{\Omega^\infty(S)}_{\mathcal{F}\mathrm{in}^\simeq}(S)\simeq \varinjlim_{\mathcal F\mathrm{in}^\simeq}S, $$ where we used that the left Kan extension of a functor does not change its colimit. Now we can use the explicit description of the groupoid of finite sets $\mathcal F\mathrm{in}^\simeq \simeq \coprod_{n\ge 0}\mathrm B\Sigma_n$ to obtain $$ \varinjlim_{\mathcal F\mathrm{in}^\simeq} S\simeq \bigoplus_{n\ge 0}S_{h\Sigma_n}. $$ We may recognize this as the free $\mathbb E_\infty$-ring on a single generator $S\{t\}$, corresponding in terms of spectral algebraic geometry to the smooth affine line $\mathbf A^1$ (as opposed to the flat affine line $\mathbf A^1_\flat = \mathrm{Spec}(S[t])$ for the polynomial $\mathbb E_\infty$-ring $S[t]\simeq \bigoplus_{n\ge 0}S$).

2. $S$-grading on symmetric algebras

The previous example can be easily generalized by observing that $\mathcal F\mathrm{in}^\simeq$ is the free $\mathbb E_\infty$-space ( = symmetric monoidal $\infty$-groupoid) on a single generator. That means that a symmetric monoidal functor $\mathcal F\mathrm{in}^\simeq \to \mathrm{Sp}$ (which always factors through the maximal subgroupoid $\mathrm{Sp}^\simeq\subseteq\mathrm{Sp}$) is equivalent to the data of a spectrum $M\in \mathrm{Sp}$ (the image of the singleton set). The functor is given by sending a finite set $I$ to the smash product $M^{\otimes I}$, and is evidently symmetric monoidal. The same Kan extension game as before now gives rise to an $S$-graded $\mathbb E_\infty$-ring spectrum, this time with underlying $\mathbb E_\infty$-ring $$ \mathrm{Sym}^*(M)\simeq \bigoplus_{n\ge 0}M^{\otimes n}_{h\Sigma_n}, $$ the free $\mathbb E_\infty$-ring generated by $M$. We recover the prior situation by setting $M=S$. For $M = S^{\oplus n}$, we obtain an $S$-grading on $S\{t_1, \ldots, t_n\}$, corresponding to the spectral-algebro-geometric smooth affine $n$-space $\mathbf A^n$.

3. $S$-grading on $S\{t^{\pm 1}\}$

Another way to generalize the example of the $S$-grading on $S\{t\}$ is to take directly the constant functor $\Omega^\infty(S)\to \mathrm{Sp}$ with value $S$, instead of starting with a constant functor on $\mathcal F\mathrm{in}^\simeq$ and Kan-extending it. This produces a perfectly good $S$-graded $\mathbb E_\infty$-ring, which let us denote $S\{t^{\pm 1}\}$.

From the group-completion relationship between $\Omega^\infty(S)$ and $\mathcal F\mathrm{in}^\simeq$, it may be deduced that $S\{t^{\pm 1}\}$ and $S\{t\}$ are related in terms of $\mathbb E_\infty$-ring localization as $S\{t^{\pm 1}\}\simeq S\{t\}[t^{-1}]$, justifying our notation. Here we are localizing $S\{t\}$ at the element $t\in \mathbf Z[t] = \pi_0(S\{t\})$. In terms of spectral algebraic geometry, this is encoding the spectral scheme $\mathrm{GL}_1$, the smooth punctured line.

4. Remark on non-negative grading

In algebraic geometry, we often prefer to think about non-negatively graded commutative rings than graded commutative rings. Just as the latter are equivalent to lax symmetric monoidal functor $\mathbf Z\to\mathrm{Ab}$, so are the former equivalent to lax symmetric monoidal functors $\mathbf Z_{\ge 0}\to \mathrm{Ab}$.

By analogy, the "non-negatively $S$-graded $\mathbb E_\infty$-rings" are lax symmetric monoidal functors $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$. Indeed, just as $\mathbf Z_{\ge 0}$ is the free commutative monoid on one generator, so is $\mathcal F\mathrm{in}^\simeq$ the free $\mathbb E_\infty$-space on one generator. That is the reason why we were encountering such functors above (and Kan extending them along group completion, as we would to view as $\mathbf Z_{\ge 0}$-grading as a special case of a $\mathbf Z$-grading).

5. Some actual "non-tautological" examples though

So far, a not-completely-unreasonable complaint might be that all the examples of $S$-graded $\mathbb E_\infty$-rings were sort of tautological.

For a very non-tautological example, see the main result of this paper of Hadrian Heine. It shows that there exists an $S$-graded $\mathbb E_\infty$-ring spectrum such that its $S$-graded modules are equivalent to the $\infty$-category of cellular motivic spectra. In fact, much more is proved: this situation is very common, and under some not-too-harsh compact-dualizable-generation assumptions, a symmetric monoidal stable $\infty$-category will be equivalent to $S$-graded modules over some $S$-graded $\mathbb E_\infty$-ring. So you may just as well view this result as a wellspring of potentially interesting examples of $S$-gradings "occurring in nature"! :)

  • $\begingroup$ This is perfect! Thank you very much :) $\endgroup$
    – Emily
    Sep 5, 2021 at 19:24
  • $\begingroup$ (In fact, you even managed to read my mind with your point 4! I was curious also about $\mathsf{FinSets}^\simeq$-gradings (defined as lax symmetric monoidal functors $\mathsf{FinSets}^\simeq\to\mathsf{Sp}$), but didn't also ask about them because the induced $\mathsf{Ho}(\mathsf{FinSets}^\simeq)$-grading on $\pi_0$'s is a bit more complicated, with an action of the symmetric group on $R_n$ instead of just an order $2$ automorphism.) $\endgroup$
    – Emily
    Sep 5, 2021 at 19:25
  • $\begingroup$ I'm a bit afraid to ask you more than what I already have (since your answer is beyond amazing!), but do you happen to know an example of an S-graded ring whose $\pi_0$ is not a commutative Z-graded ring, but rather satisfies this funny [commutative if $nm$ is even and else $ba=\sigma_{n+m}(ab)$] condition? $\endgroup$
    – Emily
    Sep 5, 2021 at 19:27
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    $\begingroup$ @Emily: I believe the thing is that the Sagave notion of $S$grading is not the same as the one we are discussing here (though I don't really understand his stuff, so I could be wrong). That said, it is certainly true that any $\mathbb E_\infty$-ring admits a trivial grading, just as any ordinary commutative ring admits a trivial $\mathbf Z$-grading: by placing everything into degree $0$. $\endgroup$ Sep 6, 2021 at 8:29
  • 1
    $\begingroup$ That is to say, an $\mathbb E_\infty$-ring is the equivalent data to a (lax) symmetric monoidal functor $\Delta^0 \to \mathrm{Sp}$ from the single-object category $\Delta^0$, so we may just left Kan-extend along the $\mathbb E_\infty$-map $\Delta^0\to \Omega^\infty(S)$ corresponding to the unit $0\in \mathbf Z =\pi_0(\Omega^\infty(S))$. That gives the "concentrated in degree $0$" grading on the $\mathbb E_\infty$-ring we started off with. $\endgroup$ Sep 6, 2021 at 8:30

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