# Generalising supercommutativity as a grading by the $1$-truncated sphere spectrum

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?
• Related: In search of lost graded rings.
– Théo
Sep 10, 2021 at 22:13
• 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? Sep 29, 2021 at 13:44
• @VivekShende Oh right, that certainly doesn't work. Thanks!
– Théo
Sep 29, 2021 at 14:37
• $\newcommand{\a}{\tau_{\leq1}\Sigma^{\infty-1}\mathbb{RP}^\infty}$(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 $\pi_1(\a)\cong\mathbb{Z}/2$.)
– Théo
Sep 29, 2021 at 14:42
• Is $\tau_{\le1}\Sigma^{\infty-1}\mathbb R{\mathrm P}^\infty$ a unit of some monoidal structure? Sep 29, 2021 at 16:55