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1. I have a few questions related to the super-algebra page, some of which I’ve also asked at MathOverflow. I think there’s some value on asking them here too (in particular because I’m hoping Urs might see this, who I think might have thought about this before, or, if not, might find this interesting nonetheless).

The “abstract idea” section of the super-algebra page reads:

Superalgebra is universal in the following sense. The crucial super-grading rule (the “Koszul sign rule”, Grassmann 1844, §37, §55)

$a \otimes b = (-1)^{deg(a) deg(b)} b \otimes a$

in the symmetric monoidal category of $\mathbb{Z}$-graded vector spaces is induced from the subcategory which is the abelian 2-group of metric graded lines. This in turn is the free abelian 2-group (groupal symmetric monoidal category) on a single generator. (This point of view is amplified in the first part of (Kapranov 13), whose second part is about super 2-algebra, more details in Kapranov 15). Generally then super-grading and hence super-algebra arises from the 2-truncation (3-coskeleton) of the free abelian ∞-group on a single generator, which is the sphere spectrum $\mathbb{S}$.

and then:

This suggests (as indicated in (Kapranov 13, Kapranov 15)) that in full generality higher supergeometry is to be thought of as $\mathbb{S}$-graded geometry, hence dually as higher algebra with ∞-group of units augmented over the sphere spectrum.

A similar point of view is put forward in the spectral super-scheme page:

Definition. Spectral/$E_\infty$ super-geometry is simply the E-∞ geometry over even periodic ring spectra.

I’ve been exploring these ideas a bit the last few days, and I think I found a possible notion of $\mathbb{E}_{\infty}$-superalgebras and $\mathbb{E}_{\infty}$-supergeometry that differs from the ones suggested by the above quoted pages.

It starts with the following definition, taken from Bunke–Nikolaus, Section 2: given a monoidal category $\mathcal{C}$ and a ring $R$, we define a $\mathcal{C}$-graded $R$-algebra to be a lax monoidal functor $M\colon(\mathcal{C},\otimes_{\mathcal{C}},\mathbf{1}_{\mathcal{C}})\to(\mathsf{Mod}_R,\otimes_R,R)$. If moreover $\mathcal{C}$ is braided, then a $\mathcal{C}$-graded $R$-algebra $M$ is $\mathcal{C}$-graded-commutative if $R$ is a braided lax monoidal functor. More generally, one could allow $\mathcal{C}$ to be a symmetric monoidal $\infty$-category; this is relevant when working with spectra, but for rings only the $1$-truncation is relevant.

These recover monoid-graded algebras as the $A_{\mathsf{disc}}$-graded ones, but there’s also a number of other interesting notions arising from this definition. In particular, following Kapranov’s idea, one can consider rings graded by $k$-truncations of the sphere spectrum. For $k=0$, this recovers $\mathbb{Z}$-graded algebras, but the situation is considerably more interesting for $k=1$:

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 $R_k$;

which is moreover $\tau_{\leq1}\mathbb{S}$-graded commutative if we 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 $a,b\in R_\bullet$.

By choosing $\sigma_k(a)=-a$ for all $a\in R_k$ and all $k\in\mathbb{Z}$, these recover $\mathbb{Z}$-graded-commutative algebras as a special case, and in this definition the Koszul rule comes from $\pi_1(\mathbb{S})\cong\mathbb{Z}_2$, while the $\mathbb{Z}$-grading (vs. $\mathbb{Z}_2$-grading issue that Kapranov mentions) is explained by $\pi_0(\mathbb{S})\cong\mathbb{Z}$!

“Higher supersymmetry” now corresponds to higher and higher $k$-truncations of $\mathbb{S}$ as suggested by Kapranov; though at each $n$-categorical level one can only really see “$n$-supersymmetry” (e.g. for $n=1$ this corresponds to $\infty$-functors $\mathbb{S}\to\mathrm{N}_{\bullet}(\mathsf{Mod}_R)$ being the same as $1$-functors $\mathsf{Ho}(\mathbb{S})\to\mathsf{Mod}_R$, and in this case we have $\mathsf{Ho}(\mathbb{S})\cong\mathsf{Ho}(\tau_{\leq k}\mathbb{S})$ for all $k\geq1$, so only $\tau_{\leq1}\mathbb{S}$ appears here).

This suggests a general definition of “($k$-super)-$\mathbb{E}_{n}$-$R$-algebras” as the $\mathbb{E}_{n}$-monoids in $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{Mod}_R)$, where for $n=\infty$ we recover lax symmetric monoidal functors $\tau_{\leq k}\mathbb{S}\to\mathsf{Mod}_R$. Finally, one possible notion of $\mathbb{E}_{\infty}$-supergeometry would then be the variant of SAG obtained by replacing $\mathbb{E}_{\infty}$-rings as the basic building blocks with these “($k$-super)-$\mathbb{E}_{\infty}$-rings”.

(P.S. There are also other very interesting “universal gradings” leading to variants a bit similar to these. In particular, recalling the characterisation of the sphere spectrum as the free $\mathbb{E}_{\infty}$-group on a point, we may think of considering instead the free $\mathbb{E}_{n}$-group on a point, $\Omega^n S^n$. While $\tau_{\leq k}\mathbb{S}$-gradings give a $\mathbb{Z}$-grading together with actions induced by the first $k$ stable homotopy groups of spheres, $\tau_{\leq k}\Omega^{n}S^{n}$-gradings give a $\mathbb{Z}$-grading again, though this time the induced actions correspond to the unstable homotopy groups of spheres $\pi_{n+1}(S^n)$, $\pi_{n+2}(S^n)$, $\ldots$, $\pi_{n+k}(S^n)$! For ordinary rings, this means that gradings by the free braided $2$-group are given by a $\mathbb{Z}$-grading together with a $\mathbb{Z}$-indexed family of $\mathbb{Z}$-actions, corresponding to $\pi_0(\Omega^2 S^2)\cong\pi_2(S^2)\cong\mathbb{Z}$ for the gradings and $\pi_1(\Omega^2S^2)\cong\pi_3(S^2)\cong\mathbb{Z})$ for the actions!)

Also, are there other nice classes of $\tau_{\leq1}\mathbb{S}$-graded commutative rings besides ordinary rings and $\mathbb{Z}$-graded-commutative ones?

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 10th 2021
• (edited Sep 10th 2021)

Thanks for writing in.

Regarding the second but last question: My last thinking about this issue is still that recorded at spectral super-scheme, which you have seen. I have not considered the specific construction you describe, which certainly looks like a natural implementation of Kapranov’s suggestion!

But does it not – as you point out above – still suffer from the shortcoming of yielding (higher refinements of) only $\mathbb{Z}$-graded commutativity instead of $\mathbb{Z}/2$-graded commutativity?

It’s only in the latter generality that most of the interesting effects associated with supersymmetry appear (e.g. the super-Poincaré-algebras are only $\mathbb{Z}/2$-graded).

• CommentRowNumber3.
• CommentAuthorDavid_Corfield
• CommentTimeSep 10th 2021

Is it possible to exploit

But ordinary $\mathbb{Z}/2$-graded supercommutative superalgebra is equivalently $\mathbb{Z}$-graded supercommutative superalgebra over the free even periodic $\mathbb{Z}$-graded supercommutative superalgebra (spectral super-scheme)?

What would higher refinements of $\mathbb{Z}$-graded commutativity over the free even periodic $\mathbb{Z}$-graded supercommutative superalgebra look like?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeSep 10th 2021

So exploiting this fact is what led to the proposal at “spectral super-scheme”: There it is argued that these higher refinements of $\mathbb{Z}/2$-graded commutative superalgebras are $E_\infty$-algebras over even periodic ring spectra!

On the one hand, this connects the idea of higher super-algebra to established practice and examples of algebraic topology. (Notably Charles Rezk had already run into this perspective purely for alg-top reasons.) On the other hand, it still lacks more connection to topics in supersymmetry as appearing in mathematical physics.

2. Hi Urs, thanks for the reply!

I think it does have this shortcoming in that $\tau_{\leq1}\mathbb{S}$-graded-commutative rings are not exactly $\mathbb{Z}_2$-graded-commutative rings. However, every $\mathbb{Z}_2$-graded-commutative ring can be naturally regarded as a $\tau_{\leq1}\mathbb{S}$-graded ring by putting everything into degree $0$ and $1$ and by choosing the $\sigma_k$ automorphisms to be given by $a\mapsto-a$.

I have yet to work out how things go on the level of morphisms, though. (I hope to do this tomorrow, and then give an update here; I’m not very hopeful with the Koszul rule for morphisms $f(a b)=(-1)^{\deg(f)\deg(a)}f(a)f(b)$, however :/)

My current intuition on this construction is that it regards $0$-supercommutativity as ordinary commutativity, i.e. $a b=b a$, together with an additional $\mathbb{Z}$-gradation (with no extra conditions). Then $\mathbb{Z}_2$-graded commutativity, i.e. $a b=(-1)^{\deg(a)\deg(b)}b a$, is enforced as $1$-supercommutativity via $\pi_1(\mathbb{S})$. The higher refinements are all torsion―we never see anything like $\mathbb{Z}$ again. So e.g. a $2$-supercommutative ring in $n$-groupoids for $n\geq2$ (we need to consider homotopical versions of rings; the ordinary ones don’t see the higher torsion) might feel like something “having two distinct minus signs” (or more precisely: two order $2$ automorphisms), each one satisfying a Koszul rule. Similarly, $3$-supercommutativity involves additional automorphisms $\sigma_k\colon A_k \to A_k$, one for each $k\in\mathbb{Z}$, satisfying $\sigma^{24}_k=\mathrm{id}_{A_k}$, and so on.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeSep 10th 2021
• (edited Sep 10th 2021)

However, every $\mathbb{Z}_2$-graded-commutative ring can be naturally regarded as a $\tau_{\leq1}\mathbb{S}$-graded ring by putting everything into degree $0$ and $1$ and by choosing the $\sigma_k$ automorphisms to be given by $a\mapsto-a$.

Absolutey, it’s the formalization of this phenomenon, via that Proposition, which yields the definition proposed at spectral super-scheme!

I am not saying this cannot be integrated into the perspective you adopted, just saying that it would need to be integrated to be a satisfactory reply to “What is higher superalgebra?” (The same remark applies to Kapronov’s proposal! and in fact it applies to a sizeable community of “NQ-manifold” theorists out there.) In fact, it seems pretty clear for how to integrate this into you scheme: You can enhance it to speak of higher graded algebras over higher graded base rings, and then you just have to require the higher graded base ring to be 2-periodic. I think.

What I found fascinating about the proposal at spectral super-scheme is that the identification of super-grading with foundations of algebraic topology ranges even deeper than in Kapranov’s proposal: It’s really the notion of $E_\infty$-algebra as such that already captures “higher $\mathbb{Z}$-graded super-algebra” and then the choice of even periodic base ring spectra (which are known to play a pivotal role in chromatic theory etc.) makes it genuinely super-algebraic in the sense of physics.

• CommentRowNumber7.
• CommentAuthorDavid_Corfield
• CommentTimeSep 10th 2021

In case it’s of interest to this discussion, I started a stub at 2-periodic sphere spectrum.

3. I have good news and bad news: the bad news is that the Koszul rule $f(ab)=(-1)^{\deg(f)\deg(a)}f(a)f(b)$ for morphisms of superalgebras $f\colon A\to B$ isn’t enforced by the notion of a “morphism of $\tau_{\leq1}\mathbb{S}$-graded rings”, defined as a monoidal natural transformation $f_\bullet\colon R_\bullet\to S_\bullet$. In detail, the latter is as follows:

The good news is that with this definition, we do have a fully faithful embedding of categories $\mathsf{Gr}_{\mathbb{Z}_2}\mathsf{Rings}\hookrightarrow\mathsf{Gr}_{\tau_{\leq1}\mathbb{S}}\mathsf{Rings}$!

So, how important is it to have the Koszul rule for morphisms $f(a b)=(-1)^{\deg(f)\deg(a)}f(a)f(b)$, instead of just $f(a b)=f(a)f(b)$?

(There’s another interesting point here: I haven’t straightened this out, but I think if we go one level higher, considering

• $2$-rings, modelled as ring categories, which are equivalent to $1$-truncated ring spectra, and
• $k=2$, i.e. $\tau_{\leq2}\mathbb{S}$;

then “$2$-supercommutative $2$-rings” are lax pseudomonoidal pseudofunctors $\tau_{\leq 2}\mathbb{S}\to\mathsf{2Ab}$, and the morphisms between those will contain $2$-cells verifying lax monoidal pseudonaturality, which will be indexed by $\pi_2(\mathbb{S})\cong\mathbb{Z}_2$, needing to satisfy a different kind of Koszul rule.)

Absolutey, it’s the formalization of this phenomenon, via that Proposition, which yields the definition proposed at spectral super-scheme!

This is something I’ve been thinking about too: commutative rings embed fully faithfully into $\mathbb{E}_{\infty}$-rings, and for this reason we view higher algebra as a faithful extension of ring theory. With the same phenomenon happening for $\mathbb{Z}_{2}$-graded rings and $\tau_{\leq1}\mathbb{S}$-graded rings, there again being a fully faithful embedding of categories, are there reasons for not “just” considering them as a generalised form of ordinary supersymmetry (completely forgetting for the moment about any homotopy groups higher than $\pi_1$)?

You can enhance it to speak of higher graded algebras over higher graded base rings, and then you just have to require the higher graded base ring to be 2-periodic. I think.

I’m not sure how to extend the definition I mentioned of a “grading by a symmetric monoidal $\infty$-category” to a “grading by nonconnective spectra” (I’ve been saying “$\tau_{\leq1}\mathbb{S}$-graded rings”, but its really “$\tau_{\leq1}(Q S^0)$-graded rings”), however it seems to me that replacing $\mathbb{S}$ by the $2$-periodified sphere spectrum might lead to a grading even worse than a $\mathbb{Z}$-grading, since $\pi_0(\mathbb{S}[\beta])$ is completely crazy. Is this correct?

What I found fascinating about the proposal at spectral super-scheme is that the identification of super-grading with foundations of algebraic topology ranges even deeper than in Kapranov’s proposal: It’s really the notion of $E_\infty$-algebra as such that already captures “higher $\mathbb{Z}$-graded super-algebra” and then the choice of even periodic base ring spectra (which are known to play a pivotal role in chromatic theory etc.) makes it genuinely super-algebraic in the sense of physics.

There is another point related to this that I’ve been wanting to raise: in spectral super-scheme, you wrote:

But more is true: the $E_\infty$-analog of the integers $\mathbb{Z}$ is the sphere spectrum $\mathbb{S}$, and every E-infinity ring $E$ is canonically $\mathbb{S}$-graded (Sagave-Schlichtkrull 11, theorem 1.7-18).

There’s a number of inequivalent definitions of $\mathbb{S}$-graded, though:

• $\mathbb{S}$-graded as in Sagave–Schlichtkrull;
• $\mathbb{S}$-graded meaning a lax symmetric monoidal functor $QS^0\to\mathsf{Sp}$;
• $\mathbb{S}$-graded in the sense that’s useful for $\mathrm{Proj}$’s in SAG. Rok Gregoric has done (and is doing) work on this, and gives some examples of these here.

One last point (sorry for the really long message!): FWIU, exterior algebras play the role of polynomial algebras for superalgebras. Rok proposed what I think is the correct definition of a “spectral exterior algebra” here, and it agrees with your definition here up to a delooping, coming from $\mathrm{Sym}^\bullet_R(\Sigma (M)) \simeq \Sigma^{\bullet}(\bigwedge_R^\bullet(M))$.

A potentially interesting variant definition (which I believe is not the correct analogue of “spectral (higher) exterior algebras”, but seems interesting nonetheless) would be the free $\tau_{\leq k}\mathbb{S}$-graded commutative $R$-algebra on an $R$-module $M$, defined as the image of $\Delta_{M}\colon\tau_{\leq k}\mathbb{S}\to\mathsf{ModSp}_R$ under the free $\mathbb{E}_{\infty}$-monoid functor for the Day convolution monoidal structure on $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{ModSp}_R)$.

In the $1$-categorical world, we consider instead the free commutative monoid on $\Delta_{M}\colon\tau_{\leq 1}\to\mathsf{Mod}_R$ for the Day convolution monoidal structure on $\mathsf{Fun}(\tau_{\leq 1}\mathbb{S},\mathsf{Mod}_R)$, i.e. the free symmetric lax monoidal functor $\tau_{\leq 1}\mathbb{S}\to\mathsf{Mod}_R$. This looks very interesting: is it the exterior algebra on $M$ again? If not, is it relevant to superalgebra and supersymmetry?

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeSep 12th 2021

Hi, am just back from an offline family weekend.

I should say that I don’t have the leisure right now to follow you more deeply into this discussion (due to other projects active right now) but I am still interested. The topic is potentially of utmost relevance and may be just waiting for a breakthrough insight in order to take off.

On the MO discussion: Thanks for the pointer! Have added the link to the bottom of the entry spectral super-scheme.

Regarding incorporating $\mathbb{Z}/2$-grading into your perspective: Don’t you get a symmetric monoidal $\infty$-category of modules over any given $\tau_n \mathbb{S}$-graded algebra in your sense? If so, just repeat the process and next consider such graded algebras in the infinity category of modules over a 2-periodic one.

Regarding relaxing/generalizing the rules of super-algebra: Part of the interest in the topic here is exactly to understand if such generalizations are natural. So seeing these appear may be a feature instead of a bug. Depends. To sort this out one should try to find good examples and applications.

4. Hi, Urs! I hope you had a great time with your family :)

I should say that I don’t have the leisure right now to follow you more deeply into this discussion (due to other projects active right now) but I am still interested. The topic is potentially of utmost relevance and may be just waiting for a breakthrough insight in order to take off.

No problem! Feel free to drop another message here in the future and restart the discussion when you’re less time-pressed, if you’d like to :)

Regarding incorporating $\mathbb{Z}/2$-grading into your perspective: Don’t you get a symmetric monoidal $\infty$-category of modules over any given $\tau_n \mathbb{S}$-graded algebra in your sense? If so, just repeat the process and next consider such graded algebras in the infinity category of modules over a 2-periodic one.

I think we do, but in this case the unpleasant grading will still be here: in the 1-categorical case already a lax symmetric monoidal functor $\tau_{\leq1}\mathbb{S}\to\mathsf{Mod}_{\mathbb{Z}/2}$ will still carry a $\mathbb{Z}$-grading, again because of $\mathrm{Obj}(\tau_{\leq1}\mathbb{S})=\mathbb{Z}$. The problem really is the $\pi_0$ :/

To get a $\mathbb{Z}/2$-grading under the Bunke–Nikolaus definition, we would need to somehow universally deform $\mathbb{S}$ into a spectrum $\tilde{\mathbb{S}}$ with $\pi_{0}(\tilde{\mathbb{S}})\cong\mathbb{Z}/2$ (in particular 2-periodification doesn’t work in this setting, as we instead get a $\pi_0(\mathbb{S}[\beta])\cong\bigoplus_{k\in\mathbb{Z}}\pi_{2k}(\mathbb{S})$-grading).

Regarding relaxing/generalizing the rules of super-algebra: Part of the interest in the topic here is exactly to understand if such generalizations are natural. So seeing these appear may be a feature instead of a bug. Depends. To sort this out one should try to find good examples and applications.

I agree completely! I’ve been trying to find some interesting examples of these things during the past few days, but have yet to succeed (though admittedly I haven’t thought much about them yet. One thing I’m planning to do soon-ish is to compute what the free $\tau_{\leq1}$-ring on a module looks like, I think that has a good chance of being neither an ordinary ring nor a graded-commutative one)

5. Hi Urs,

A small update unrelated to gradings: I think I found an object $\mathbb{S}/2$ which one may call the “mod 2 sphere spectrum”. It satisfies an analogous universal property to $\mathbb{Z}/2$ and has $\pi_0(\mathbb{S}/2)\cong\mathbb{Z}/2$. See here for the UP, here for Maxime Ranzi’s proof that it exists and also a description (it is $\Omega Q\mathbb{RP}^\infty$), and lastly here for a table with its first homotopy groups, together with a comparison with those of $\mathbb{S}$.

(I have no idea whether this is the right object to consider in Kapranov’s context, but I thought you might like to know about it nonetheless! =)

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeSep 21st 2021
• (edited Sep 21st 2021)

Thanks, looks interesting. BTW, one might denote this $\Sigma^{\infinity-1}_+ B (\mathbb{Z}/2)$.

This construction loses the $\mathbb{Z}/24$ in degree 3, which was a major motivation for Kapranov. That 24 should still be present in the 2-periodic sphere spectrum that David C. highlighted in comment #7 above.

Neither of the two spectra has $E_\infinity$ ring structure, though. I am not sure what either is doing for us in regards to superalgebra, but that’s mostly because I haven’t tried to think about it.

6. For what it’s worth I think $\Sigma^{\infty-1}_+\mathbf{B}\mathbb{Z}/2$ (thanks for the suggestion!) isn’t the right thing to look at. OTOH there’s an argument for using precisely $\mathbb{S}$: we can only put a $(-1)^{\deg(a) \deg(b)}$ in the Koszul rule for $\mathbb{Z}$-graded $R$-modules leading to supercommutativity because abelian groups have inverses, carrying an action $\mathbb{Z}\otimes A\to A$ of the integers given by $a\mapsto k a$. Moreover, the integers are universal with respect to this property: they are the monoid corepresenting the functor $\mathsf{CMon}\to\mathsf{Sets}$ sending a monoid to its set of invertible elements.

Similarly, $Q S^0$ is the corepresenting object for the invertible objects functor $\mathsf{Pic}\colon\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})\to\mathcal{S}$, and indeed this gives spectra an action $\mathbb{S}\otimes E\to E$ of the sphere spectrum. So one possibly interesting thing we might want to look at is what kind of symmetric monoidal structures the $\infty$-category of $\mathbb{Z}$-graded spectra might have.

In the classical case of $\mathbb{Z}$-graded abelian groups (/$\mathbb{Z}$-graded *$\mathbb{Z}$*-modules) there are only two symmetric monoidal structures: the trivial one and the Koszul one, corresponding to $1,-1\in\mathbb{Z}$. But maybe there are more for $\mathbb{Z}$-graded spectra (/$\mathbb{Z}$-graded *$\mathbb{S}$*-modules)!

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeSep 22nd 2021

I certainly think the sphere spectrum $\mathbb{S}$ is the right thing to use. Allow me to recall, as you seem to be revolving around this:

1. Fascinatingly, every $E_\infty$-ring is automatically $\mathbb{S}$-graded and thereby already is $\mathbb{Z}$-graded in a higher homotopical sense;

2. the only aspect missing for this homotopified $\mathbb{Z}$-grading to be a genuine super-grading is that it be 2-periodified;

3. this is provided exactly by restriction to those $E_\infty$-rings which are algebras over 2-periodic $E_\infty$-spectra.

• CommentRowNumber15.
• CommentAuthorDavid_Corfield
• CommentTimeSep 22nd 2021
• (edited Sep 22nd 2021)

Should examples listed at periodic ring spectrum all be $E_{\infty}$? In which case I should remove ’2-periodic sphere spectrum’. Or can an $E_2$-ring spectrum count?

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeSep 22nd 2021

Oh, at “periodic ring spectrum” we don’t need to require $E_\infty$. But the story at spectral super-scheme seems to require $E_\infty$-structure.

• CommentRowNumber17.
• CommentAuthorDavid_Corfield
• CommentTimeSep 22nd 2021
• (edited Sep 22nd 2021)

Vague speculation: I wonder if there’s a multi-initial collection of 2-periodic $E_{\infty}$-rings.

• CommentRowNumber18.
• CommentAuthorDavid_Corfield
• CommentTimeSep 23rd 2021

Re #14, does that tell us something interesting, in terms of point 3, that the 2-periodic spectra are complex oriented (even cohomology theory)?

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeSep 23rd 2021

Yes, this is what I am thinking of when I say (#4) that:

this connects the idea of higher super-algebra to established practice and examples of algebraic topology.

Because the ingredients of spectral super-schemes according to the proposed definition there are not outlandish, but are exactly what one cares about in chromatic homotopy theory.

Conversely, this is the reason why Rezk 09, Sec 2 already ran into this kind of higher super-algebra in investigations of Morava E-theories!

All this makes me think that this AlgTop/super-connection is the right one, and that it now just takes someone to pick this up and run with it.

• CommentRowNumber20.
• CommentAuthorThéo de Oliveira S.
• CommentTimeSep 23rd 2021
• (edited Sep 23rd 2021)

1. Fascinatingly, every $E_\infty$-ring is automatically $\mathbb{S}$-graded and thereby already is $\mathbb{Z}$-graded in a higher homotopical sense;

This is something I’ve been meaning to understand lately. FWIU there are at least three natural definitions of “$\mathbb{S}$-graded” in the literature, one of them being the one given by Sagave–Schlichtkrull, mentioned in spectral super-scheme. Is it correct to say that a “Sagave–Schlischtkrull $\mathbb{S}$-grading” on an $\mathbb{E}_{\infty}$-space $A$ is the same thing as a morphism of $\mathbb{E}_\infty$-spaces $A\to Q S^0$ (where one departs from the classical identification $\{\text{gradings on a monoid }\,A\}\cong\{\text{monoid morphisms }\,A\to(\mathbb{Z},+,0)\}$)?

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeSep 24th 2021

Yes, that’s what their Thm. 1.7 on p.3 is saying (the notation is a little intransparent, and I keep having to sift through their definitions to remind me about what it all means, but luckily they clarify this in the text right beneath).

Now $Q(-)$ is old notation for $\Omega^\infty \Sigma^\infty (-)$ – the underlying space in degree 0 of the suspension spectrum – so that $Q S^0 \;\simeq\; \Omega^\infty \Sigma^\infty S^0 \;\simeq\; \Omega^\infty \mathbb{S}^\infty$ is the underlying space of the sphere spectrum, which we might just as well denote by $\mathbb{S}$ itself, since the sphere spectrum is connective.

In conclusion, the conceptual content of that Thm. 1.7 should be that: just as a $\mathbb{Z}$-grading on a monoid $A$ is equivalently a monoid homomorphism $A \xrightarrow{\;} \mathbb{Z}$ to $\mathbb{Z}$ with its additive abelian monoid structure, so every $E_\infty$-monoid $\mathcal{A}$ already comes equipped with an $E_\infty$-monoid homomorphism $\mathcal{A} \xrightarrow{\;} \mathbb{S}$ to the additive $E_\infty$-monoid underlying the sphere spectrum.

So I think this is direct the $E_\infty$-analog of the classical identification of $\mathbb{Z}$-gradings (plus the fascinating claim that for $E_\infty$-monoids this exists canonically). Why do you say it “departs” from that?

7. Thank you very much, Urs! This cleared up most of my misunderstandings :)

There’s a single small point about which I’m still a bit confused: I see that to every symmetric spectrum $E$ one can associate an $\mathbb{S}$-graded $\mathbb{E}_\infty$-space $\Omega^\mathcal{J}(E)$ (which is indeed very fascinating!). Why are general $E_\infty$-monoids canonically (and non-trivially) $\mathbb{S}$-graded, though?

(From what I understand, Theorem 1.7 says that the model category of commutative $\mathcal{J}$-spaces is Quillen equivalent to that of $\mathbb{E}_\infty$-spaces over $\mathbf{B}\mathcal{J}\simeq\Omega^\infty\mathbb{S}$, but does it also give a way to go from $I$-spaces (/$\mathbb{E}_\infty$-spaces) to $\mathcal{J}$-spaces?)

Why do you say it “departs” from that?

I was thinking about the other notions of gradings for $E_\infty$-spaces: they all “depart” from more or less equivalent characterisations of $\mathbb{Z}$-gradings of monoids (such as lax monoidal functors $\mathbb{Z}_\mathsf{disc}\to A$ or morphisms of monoids $A\to\mathbb{Z}$) which however differ in the $\infty$-setting.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeSep 25th 2021
• (edited Sep 25th 2021)

Why are general $E_\infty$-monoids canonically (and non-trivially) $\mathbb{S}$-graded, though?

I assume that by asking “Why?” you do not mean “How does Sagave-Schlichtkrull’s proof work?”, but you do mean to ask for a conceptual explanation.

I would turn this around: The observation that they are is a peek into the fundamental nature of reality, and is a hint for a deep relation between higher- and super- geometry/algebra, a hint that remains to be fully followed up on.

At the coarse level of homotopy groups, this is an old observation in algebraic topology, embodied in classical phenomena such as the super Lie algebra structure of Whitehead products.

My gut feeling here is that ultimately algebraic topology/homotopy theory is part of the explanation of “Why super-mathematics?” Namely: Super-algebra (in an enhanced form) seems to be the algebra/geometry automatically appearing in homotopy-theoretic foundations.

but does it also give a way to go from $\mathcal{I}$-spaces to $\mathcal{J}$-spaces?

Good question, I am wondering about this each time I pick up their article. I don’t know. (Maybe this is explained somewhere in their writing, but I haven’t spent more time with it, I have to admit.)

The picture I have is:

While each $E_\infty$-spectrum carries a canonical $\mathbb{S}$-grading in that it maps to an $E_\infty$-monoid in $\mathcal{J}$-spaces (via their right adjoint functor $\Omega^{\mathcal{J}}$ in their equation (4.4)), the plain morphisms of $E_\infty$-monoids need not respect this $\mathbb{S}$-grading. Instead, the $E_\infty$-morphisms that respect the canonical $\mathbb{S}$-grading are those in the category of $E_\infty$-monoids in $\mathcal{J}$-spaces.

8. I assume that by asking “Why?” you do not mean “How does Sagave-Schlichtkrull’s proof work?”, but you do mean to ask for a conceptual explanation.

Thanks, Urs! I had something else in mind: Sagave–Schlichtkrull prove that any symmetric spectrum $E$ has an underlying $\mathbb{S}$-graded $\mathbb{E}_\infty$-space $\Omega^\mathcal{J}(E)$, and by the equivalence between connective spectra and grouplike $\mathbb{E}_\infty$-spaces, we see that so does any grouplike $\mathbb{E}_\infty$-space. I’m a bit confused though about whether this applies also to not necessarily grouplike $\mathbb{E}_\infty$-spaces: do we have an $\mathbb{S}$-graded $\mathbb{E}_\infty$-space $\Omega^\mathcal{J}(X)$ for any such not necessarily grouplike $\mathbb{E}_\infty$-space $X$?

(To me this seems a bit unlikely―I’d expect them to be graded over $\mathbb{F}\overset{\mathrm{def}}{=}\coprod_{n=0}^\infty\mathbf{B}\Sigma_{n}$, though not over $\Omega^\infty\mathbb{S}$. OTOH, there’s also $\Omega^\mathcal{J}(X^{\mathrm{grp}})$, but that seems way too lossy…)

Regardless, thank you very much for your explanation, Urs! In particular I didn’t know about the super Lie algebra structure of Whitehead products you mentioned―this seems very nice, and I’ll really enjoy learning about it! :)

• CommentRowNumber25.
• CommentAuthorMarc Hoyois
• CommentTimeSep 25th 2021

The result of Sagave-Schlichtkrull seems slightly misleading to me. A more natural statement would be a grading by the Picard $\infty$-groupoid of invertible spectra. For any presentably symmetric monoidal $\infty$-category $C$, there is by the universal property of Day convolution a symmetric monoidal adjunction between $Fun(Pic(C),Spc)$ and $C$. The right adjoint sends an ($E_\infty$-)object $c$ to a $\Pic(C)$-graded ($E_\infty$-)space, whose unit component is $Map(1,c)$. Now given a particular invertible object $L\in Pic(C)$, there is a unique $E_\infty$-map $\Omega^\infty\mathbb{S}\to Pic(C)$ sending $1$ to $L$ by the universal property of $\Omega^\infty\mathbb{S}$, so any $Pic(C)$-graded space gives rise to an $\Omega^\infty\mathbb{S}$-graded space.

If I understand correctly, the Sagave-Schlichtkrull $\Omega^\infty\mathbb{S}$-graded space is obtained via this construction applied to $L=\Sigma\mathbb{S}$. But going from the $Pic(Sp)$-grading to the $\Omega^\infty\mathbb{S}$-grading seems to forget a lot of information. For example at the level of homotopy groups the $E_\infty$-map $\Omega^\infty\mathbb{S}\to Pic(Sp)$ is a map $\pi_n\mathbb{S}\to \pi_{n-1}\mathbb{S}$ (for $n\geq 2$).

On the other hand the map $\Omega^\infty\mathbb{S}\to Pic(Sp)$ is an equivalence on $1$-truncations, so for graded objects in 1-categories there is no difference.

Sagave–Schlichtkrull prove that any symmetric spectrum $E$ has an underlying $\mathbb{S}$-graded $\mathbb{E}_\infty$-space $\Omega^\mathcal{J}(E)$

$E$ should be an $E_\infty$-ring for this, otherwise $\Omega^\mathcal{J}(E)$ is only an $\Omega^\infty\mathbb{S}$-graded space.

9. Hi Marc,

by the universal property of Day convolution a symmetric monoidal adjunction between $Fun(Pic(C),Spc)$ and $C$.

Could you please elaborate on this a bit? I’m unsure about how to construct this adjunction (which universal property of Day convolution are you using?)

Also, (if I understand it correcty) the construction you describe builds $\Omega^\mathcal{J}(E)$ as a lax symmetric monoidal functor $\Omega^\infty\mathbb{S}\to Spc$, though Sagave–Schlichtkrull seem to view an $\Omega^\infty\mathbb{S}$-grading as instead a morphism of $\mathbb{E}_\infty$-spaces $\Omega^\mathcal{J}(E)\to\Omega^\infty\mathbb{S}$, “going in the opposite direction”. Do you know how these two definitions compare with each other?

$E$ should be an $E_\infty$-ring for this, otherwise $\Omega^\mathcal{J}(E)$ is only an $\Omega^\infty\mathbb{S}$-graded space.

Thanks!

• CommentRowNumber27.
• CommentAuthorMarc Hoyois
• CommentTimeSep 26th 2021

Could you please elaborate on this a bit? I’m unsure about how to construct this adjunction (which universal property of Day convolution are you using?)

The universal property is HA 4.8.1.10 (4) (in the special case of Cor. 4.8.1.12): If $A$ is symmetric monoidal, the Day convolution on $PSh(A)$ is the universal cocomplete symmetric monoidal category under A where the tensor product preserves colimits in each variable.

Also, (if I understand it correcty) the construction you describe builds $\Omega^\mathcal{J}(E)$ as a lax symmetric monoidal functor $\Omega^\infty\mathbb{S}\to Spc$, though Sagave–Schlichtkrull seem to view an $\Omega^\infty\mathbb{S}$-grading as instead a morphism of $\mathbb{E}_\infty$-spaces $\Omega^\mathcal{J}(E)\to\Omega^\infty\mathbb{S}$, “going in the opposite direction”. Do you know how these two definitions compare with each other?

They are equivalent by a symmetric monoidal version of straightening/unstraightening. Namely, if $K$ is an $E_\infty$-space, the equivalence $Spc/K = Fun(K,Spc)$ is symmetric monoidal. I don’t know a direct reference for this but I guess it can be deduced from HA 4.8.1.12 which characterizes the RHS.

10. They are equivalent by a symmetric monoidal version of straightening/unstraightening. Namely, if $K$ is an $E_\infty$-space, the equivalence $Spc/K = Fun(K,Spc)$ is symmetric monoidal. I don’t know a direct reference for this but I guess it can be deduced from HA 4.8.1.12 which characterizes the RHS.

Thanks! This is really nice!

The universal property is HA 4.8.1.10 (4) (in the special case of Cor. 4.8.1.12): If AA is symmetric monoidal, the Day convolution on PSh(A)PSh(A) is the universal cocomplete symmetric monoidal category under A where the tensor product preserves colimits in each variable.

My apologies for asking about this again, but how should we apply the universal property of Day convolution to get the symmetric monoidal adjunction $Fun(Pic(C),Spc)\rightleftarrows C$? (I’ve been trying to figure this out for a while, but I’m still unsure about how to proceed :/… Are we using that $C$ is presentable for there to be a left adjoint to the Yoneda embedding?)

• CommentRowNumber29.
• CommentAuthorMarc Hoyois
• CommentTimeSep 26th 2021

My apologies for asking about this again, but how should we apply the universal property of Day convolution to get the symmetric monoidal adjunction $Fun(Pic(C),Spc)\rightleftarrows C$? (I’ve been trying to figure this out for a while, but I’m still unsure about how to proceed :/… Are we using that $C$ is presentable for there to be a left adjoint to the Yoneda embedding?)

The universal property gives a unique symmetric monoidal colimit-preserving functor $Fun(Pic(C),Spc)\to C$ extending the inclusion $Pic(C)\subset C$. It has a right adjoint, which is the restricted Yoneda embedding. There is no need for C to be presentable, it just needs colimits and a compatible symmetric monoidal structure.

• CommentRowNumber30.
• CommentAuthorThéo de Oliveira S.
• CommentTimeSep 27th 2021
• (edited Sep 27th 2021)

Hi Marc,

I think I finally understood it! Thank you very much! :)

• CommentRowNumber31.
• CommentAuthorDmitri Pavlov
• CommentTimeSep 27th 2021

Re #30: Would you considering transfering your understanding to the nLab article then? It could certainly benefit from having more details.

• CommentRowNumber32.
• CommentAuthorThéo de Oliveira S.
• CommentTimeSep 28th 2021
• (edited Sep 28th 2021)

Hi Dmitri,

Sure! I’ll add this (and some other things) there in a bit.

11. Still on the $\mathbb{Z}$ vs. $\mathbb{Z}/2$ issue in Kapranov’s proposal: what about replacing the sphere spectrum with its ∞-group of units $(\Omega^\infty\mathbb{S})^\times$?

In the classical case we have $(\mathbb{Z},\times,1)^\times=\{-1,1\}\cong\mathbb{Z}/2$, giving

\begin{aligned} \pi_0((\Omega^\infty\mathbb{S})^\times) &\cong \pi_0(\Omega^\infty\mathbb{S})^\times\\ &\cong \pi_0(\mathbb{S})^\times\\ &\cong \mathbb{Z}^\times\\ &\cong \mathbb{Z}/2. \end{aligned}

Meanwhile, $\pi_n((\Omega^\infty\mathbb{S})^\times))\cong\pi_n(\mathbb{S})$ for $n\geq1$, keeping in particular $\pi_1(\mathbb{S})\cong\mathbb{Z}/2$, $\pi_2(\mathbb{S})\cong\mathbb{Z}/2$, and $\pi_3(\mathbb{S})\cong\mathbb{Z}/24$ (differently from $\Sigma^{\infty-1}\mathbb{RP}^\infty$).

• CommentRowNumber34.
• CommentAuthorUrs
• CommentTimeOct 5th 2021

Hi Théo

Nominally that’s true. But since we mean to be grading by the additive group of integers, or by its additive quotient by 2, it is unclear (to me, at least) what is gained by pointing to the multiplicative subgroup $\{ \pm 1 \} \subset \mathbb{Z}$.

12. Ah, right. I completely failed to noticed that the grading would then be “multiplicative”. Thanks, Urs!

• CommentRowNumber36.
• CommentAuthorUrs
• CommentTimeOct 6th 2021
• (edited Oct 6th 2021)

I don’t want to discourage you from exploring, just saying where I am not immediately seeing where it’s headed.

Maybe there is something to this idea of regarding $GL(1,\mathbb{S})$ as the $E_\infty$-space encoding super-algebra, not sure, one would have to play with it.

I still think what is needed next is a better supply of plausible examples of objects that should qualify as spectral super-algebras.

I also still think that $E_\infty$-algebras over even periodic ring spectra is the class to look at, but back when I thought about this a little I failed to identify examples that would more directly relate to supersymmetry in physics.

As you may know, the idea was to see if the tower of universal central extensions emanating from the ordinary superpoint (here) might have an interesting spectral enhancement if instead one starts with some incarnation of a spectral superpoint.

13. The additive vs. multiplicative grading issue is a good point: repeating the story above for “$\tau_{\leq1}(\Omega^\infty\mathbb{S})^\times$-graded algebras”, i think we don’t quite get supercommutative superalgebras, but rather objects $A_\bullet$ consisting of two abelian groups $A_{-1}$ and $A_1$, multiplication maps $A_i\times A_j\to A_{i j}$, a unit $1_A\in A_1$, and a supercommutativity rule of the form

$a b = (-1)^{\begin{pmatrix}deg(a)\\2\end{pmatrix}\begin{pmatrix}deg(b)\\2\end{pmatrix}}b a,$

saying that the multiplication maps

\begin{aligned} A_{1}\times A_{1} &\to A_1,\\ A_{-1}\times A_{1} &\to A_{-1},\\ A_{1}\times A_{-1} &\to A_{-1} \end{aligned}

are commutative, while the map $A_{-1}\times A_{-1}\to A_{1}$ is anticommutative. Have you ever seen a supercommutativity rule of this form appear anywhere?

Speaking of different sign rules, there’s a second such issue I was wondering about: in signs in supergeometry, you discuss the Deligne and Bernstein conventions for super dgas,

\begin{aligned} \alpha_i\alpha_j &= (-1)^{(n_i n_j+\sigma_i\sigma_j)}\alpha_j\alpha_i,\\ \alpha_i\alpha_j &= (-1)^{(n_i+\sigma_i)\cdot(n_j+\sigma_j)}\alpha_j\alpha_i \end{aligned}

for when we have a grading of the form $\mathbb{Z}\times\mathbb{Z}/2$. This kind of grading appears also when we consider the action of the “multiplicative” tensor product $\otimes\colon Ho(\mathbb{S})\times Ho(\mathbb{S})\to Ho(\mathbb{S})$ on the morphisms of $Ho(\mathbb{S})$, coming from $\mathbb{Z}\times\mathbb{Z}/2\cong\pi_0(\mathbb{S})\times\pi_1(\mathbb{S})$, but there instead we have a rule of the form

$((n_i,\sigma_i),(n_j,\sigma_j))\mapsto n_j\sigma_i+n_i\sigma_j$

(in turn coming from the equality $sgn(\sigma\otimes\tau)=m sgn(\sigma)+n sgn(\tau)$ for products of permutations). Have you maybe wondered before about whether one gets anything interesting from using a sign rule of the form

$\alpha_i\alpha_j = (-1)^{(n_j\sigma_i+n_i\sigma_j)}\alpha_j\alpha_i$

instead of the Deligne or Bernstein ones? (Edit: I just noticed I had swapped $i$ and $j$ above; the correct sign rule isn’t $(-1)^{(n_i\sigma_i+n_j\sigma_j)}$, but rather $(-1)^{(n_j\sigma_i+n_i\sigma_j)}$. This happens to be the “Bernstein rule minus the Deligne one”: $(-1)^{(n_j\sigma_i+n_i\sigma_j)}=(-1)^{(n_i+\sigma_i)\cdot(n_j+\sigma_j)-(n_i n_j+\sigma_i\sigma_j)}$.)

As you may know, the idea was to see if the tower of universal central extensions emanating from the ordinary superpoint (here) might have an interesting spectral enhancement if instead one starts with some incarnation of a spectral superpoint.

Thanks for the pointer, Urs! I saw this a long time ago, found it *very* nice, and made a note to learn it better, but ended up forgetting… I really should go back to this! :)