A discussion that has been going recently is that supersymmetry corresponds to grading over the sphere spectrum, coming from an insight due to Kapranov.

To formalise such a statement, one needs a generalised notion of grading allowing the sphere spectrum as a possible candidate. Section 2 of Bunke–Nikolaus's Twisted differential cohomology provides such a definition: given a monoidal $\infty$-category $\mathcal{C}$, we define a $\mathcal{C}$-graded(-commutative) $\mathbb{E}_k$-ring to be a lax (symmetric) $\mathbb{E}_k$-monoidal functor $(\mathcal{C},\otimes_{\mathcal{C}},\mathbf{1}_{\mathcal{C}})\to(\mathsf{Sp},\otimes_{\mathbb{S}},\mathbb{S})$.

For $\mathcal{C}$ an ordinary category, we replace $(\mathsf{Sp},\otimes_{\mathbb{S}},\mathbb{S})$ by $(\mathrm{N}_{\bullet}(\mathsf{Ab}),\otimes_{\mathbb{Z}},\mathbb{Z})$. Now, we have a bijection between the following data:

  • Lax (symmetric) $\mathbb{E}_k$-monoidal functors $(\mathcal{C},\otimes_{\mathcal{C}},\mathbf{1}_{\mathcal{C}})\to((\mathrm{N}_{\bullet}(\mathsf{Ab}),\otimes_{\mathbb{Z}},\mathbb{Z})$;
  • Lax (symmetric) $\mathbb{E}_k$-monoidal functors $(\mathsf{Ho}(\mathcal{C}),\otimes_{\mathcal{C}},\mathbf{1}_{\mathcal{C}})\to(\mathsf{Ab},\otimes_{\mathbb{Z}},\mathbb{Z})$.

so in this case an $\mathbb{S}$-graded(-commutative) ordinary ring corresponds simply to a $\tau_{\leq1}\mathbb{S}$-graded ring. From the description of $\tau_{\leq1}\mathbb{S}$ as a symmetric monoidal category given here, one sees that such an object corresponds to the following definition:

A $\tau_{\leq1}\mathbb{S}$-graded ring is a pair $(R_\bullet,\{\sigma_k\}_{k\in\mathbb{Z}})$ with

  • $R_\bullet$ a $\mathbb{Z}$-graded ring (corresponding to $\pi_0(\mathbb{S})\cong\mathbb{Z}$);
  • $\{\sigma_k\colon R_k\to R_k\}_{k\in\mathbb{Z}}$ a family of order $2$ automorphisms (corresponding to $\pi_1(\mathbb{S})\cong\mathbb{Z}_2$);

such that, for each $k,\ell\in\mathbb{Z}$, each $a\in R_k$, and each $b\in R_\ell$, we have $$\sigma_{k+\ell}(ab)=\sigma_{k}(a)\sigma_{\ell}(b).$$ Moreover, $R$ is $\tau_{\leq1}\mathbb{S}$-graded commutative if 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 $a,b\in R_\bullet$.

Similarly, a $\tau_{\leq0}\mathbb{S}\cong\mathbb{Z}_{\mathsf{disc}}$-graded (commutative) ring (corresponding to "$0$-supercommutativity") is just an ordinary (commutative) ring with a $\mathbb{Z}$-gradation. The same discussion goes through replacing $\mathsf{Ab}$ by $\mathsf{Mod}_R$, defining a notion of $\tau_{\leq1}\mathbb{S}$-graded (commutative) $R$-algebras.

Main Question. How far are $\tau_{\leq1}\mathbb{S}$-graded commutative rings from ($\mathbb{Z}$-graded) superalgebras, realising Kapranov's insight?

For instance, they contain each of the following classes of rings, all as full subcategories:

  • The category of ordinary commutative rings as those $\tau_{\leq1}\mathbb{S}$-graded rings having everything in degree $0$ and such that $\sigma_k=\mathrm{id}_{R_k}$ for all $k\in\mathbb{Z}$;
  • The category of $\mathbb{Z}$-graded supercommutative rings (i.e. satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$) as those $\tau_{\leq1}\mathbb{S}$-graded rings with $\sigma_k(a)=-a$ for each $k\in\mathbb{Z}$ and each $a\in R_k$.

Other questions:

  1. Has this specific notion appeared before in the literature?
  2. What are interesting examples of $\tau_{\leq 1}\mathbb{S}$-graded commutative algebras that genuinely do not come from ordinary algebras or $\mathbb{Z}$-graded-commutative algebras?
  • $\begingroup$ Related: In search of lost graded rings. $\endgroup$
    – Théo
    Sep 10, 2021 at 22:13
  • $\begingroup$ I don't understand your comment about how to embed $\mathbb{Z}/2\mathbb{Z}$ graded supercommutative rings as a full subcategory. What happens when you multiply two elements in grading 1? $\endgroup$ Sep 29, 2021 at 13:44
  • $\begingroup$ @VivekShende Oh right, that certainly doesn't work. Thanks! $\endgroup$
    – Théo
    Sep 29, 2021 at 14:37
  • $\begingroup$ $\newcommand{\a}{\tau_{\leq1}\Sigma^{\infty-1}\mathbb{RP}^\infty}$(Somewhat tangentially related, but I think that if one replaces $\tau_{\leq1}\mathbb{S}$ by $\a$, then $\mathbb{Z}/2$-graded supercommutative rings will embed fully faithfully into the category of "$\a$-graded rings", by a similar reason that $\mathbb{Z}$-graded supercommutative ones embed into that of $\tau_{\leq1}\mathbb{S}$-graded ones; i.e. because $\pi_0(\a)\cong\mathbb{Z}/2$ and $\pi_1(\a)\cong\mathbb{Z}/2$.) $\endgroup$
    – Théo
    Sep 29, 2021 at 14:42
  • $\begingroup$ Is $\tau_{\le1}\Sigma^{\infty-1}\mathbb R{\mathrm P}^\infty$ a unit of some monoidal structure? $\endgroup$ Sep 29, 2021 at 16:55


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