In this this google+ post of Urs Schreiber, he says: "Grading over the sphere spectrum is supersymmetry" and then he redirect us to the abstract idea of superalgebra (in nLab).

Are there some references (other than Kapranov and nLab) about these ideas?

  • 9
    $\begingroup$ As an homotopy theorist, I'm really interested in a good answer to this. One thing to note is that the twist map $\mathbb{S}\otimes \mathbb{S}\to \mathbb{S}\otimes \mathbb{S}$ is -1 (like in supervector spaces), but beyond that I've always heard only vague analogies. In particular I don't know what "graded over the sphere spectrum" means... $\endgroup$ – Denis Nardin Apr 14 '16 at 22:24
  • 3
    $\begingroup$ @DenisNardin Is this an incarnation of the fact that $S^n \wedge S^m \to S^m \wedge S^n$ has degree $(-1)^{nm}$? $\endgroup$ – Saal Hardali Apr 14 '16 at 22:32
  • 2
    $\begingroup$ @Saal Yeah, it's basically the same statement at a stable level (and of course I meant $\Sigma\mathbb{S}\otimes \Sigma\mathbb{S}\to\Sigma\mathbb{S}\otimes \Sigma\mathbb{S}$) $\endgroup$ – Denis Nardin Apr 14 '16 at 22:41
  • 4
    $\begingroup$ I am not sure what the intent is of asking -- regarding an insight genuinely due to Kapranov and well exposed in his very recent article arxiv.org/abs/1512.07042 -- whether there are reference other than Kapranov's? Maybe in 10 years there will be various references reviewing and building on Kapranov's insight, but at the moment where it comes out, I would think that the best reference for Kapranov's insight is Kapranov's article. It's nicely written, too. $\endgroup$ – Urs Schreiber Apr 16 '16 at 6:50
  • $\begingroup$ I would like to add to this discussion Urs' following post. Why Supersymmetry? Because of Deligne's theorem. $\endgroup$ – tttbase Dec 29 '16 at 19:58

Let's agree that whatever "supersymmetry" means it has something to do with working in the symmetric monoidal category of super vector spaces (e.g. we might want to consider Lie algebras or commutative algebras in this category), or something like it. The question is what, if anything, this has to do with the sphere spectrum.

Here is at least the beginning of the story as I understand it. The sphere spectrum has the following universal property: it is the free spectrum, or the free infinite loop space, on a point. Said in a more explicitly higher categorical way,

The sphere spectrum $\mathbb{S}$ is the free symmetric monoidal $\infty$-groupoid with inverses on a point.

Okay, but $\infty$ is a pretty big number, so let's shield ourselves from the full power of this result by $n$-truncating it. Now it says

The $n$-truncation $\Pi_{\le n}(\mathbb{S})$ of the sphere spectrum is the free symmetric monoidal $n$-groupoid with inverses on a point.

Now let's specialize this result to small values of $n$.

$n = 0$ is easy and familiar: the $0$-truncation of the sphere spectrum is (the zeroth stable homotopy group of spheres, which is) $\mathbb{Z}$, and its universal property is that it is the free abelian group on a point.

$n = 1$: the $1$-truncation of the sphere spectrum is a symmetric monoidal groupoid with $\pi_0 \cong \mathbb{Z}$ and $\pi_1 \cong \mathbb{Z}_2$, and its universal property is that it is the free symmetric monoidal groupoid with inverses (sometimes called a "Picard groupoid") on a point. This means that if $C$ is any other symmetric monoidal category (we'll be working with its maximal subgroupoid) and $c \in C$ is an invertible object in $C$, then there's a canonically defined symmetric monoidal functor $\mathbb{S} \to C$ which takes the value $c$ on the generator $1 \in \pi_0(\mathbb{S})$. On objects it sends $n \in \pi_0(\mathbb{S})$ to $c^{\otimes n}$, while on morphisms it sends $-1 \in \pi_1(\mathbb{S})$ to the "sign" of $c$, namely the value of the braiding

$$\beta_{c, c} : c \otimes c \to c \otimes c$$

regarded as an element of $\text{Aut}(c \otimes c) \cong \text{Aut}(1)$, where $1$ is the tensor unit (this identification is canonical given that $c$ is invertible). Notably, this is equal to $-1$ on the odd invertible super vector space and $1$ on the even invertible super vector space. (Several other descriptions of how $-1 \in \pi_1(\mathbb{S})$ acts are possible: for example, it can also be described as $\text{tr}(\text{id}_c)$. This comes from the cobordism hypothesis.)

From here it's possible to give a universal property of a version of super vector spaces: namely,

The symmetric monoidal category of $\mathbb{Z}$-graded vector spaces over a field $k$ equipped with the Koszul sign rule is the free symmetric monoidal cocomplete $k$-linear category on an invertible object of sign $-1$.

One might then hope to, for example, define higher analogues of super vector spaces by increasing the category number from here. Ganter and Kapranov also use the $2$-truncation $\Pi_{\le 2}(\mathbb{S})$ to define higher analogues of the sign character and hence of symmetric and exterior powers here.

However, there are reasons to believe this is not really what's going on. Super vector spaces are arguably not interesting because of how maps out of them behave but because of how maps into them behave: namely, Deligne's theorem about nice symmetric monoidal categories admitting a fiber functor to super vector spaces. This theorem can be intepreted as saying that the symmetric monoidal category of super vector spaces over $\mathbb{C}$ is in some sense "algebraically closed" (compare: every finite-dimensional commutative $k$-algebra admits an algebra homomorphism to the algebraic closure of $k$), or even the "algebraic closure" of, say, the symmetric monoidal category of vector spaces over $\mathbb{R}$.

So a different claim about what "supersymmetry" really is is that it's the study of categorified algebraic closures in this sense. Notably, the analogue of the theorem above with $\mathbb{R}$ replaced by other fields fails in positive characteristic: a conjecture about the correct replacement is given by Ostrik here. I learned this from Theo Johnson-Freyd; see, for example, this paper, where he gives three conjectures for how this theory of categorified algebraic closures of $\mathbb{R}$ generalizes to higher category number, only one of which is related to the sphere spectrum.

| cite | improve this answer | |
  • 2
    $\begingroup$ This is very interesting! I like the interpretation of Deligne's theorem as "supervector spaces are algebraically closed". Do you now if this relates at all to Tannaka duality over a ring spectrum? $\endgroup$ – Denis Nardin Apr 15 '16 at 13:21
  • 1
    $\begingroup$ @Denis: well, one could similarly ask for an analogue of Deligne's theorem about fiber functors for something like symmetric monoidal presentable stable $\infty$-categories satisfying some finiteness condition. I have no idea what to conjecture here though. $\endgroup$ – Qiaochu Yuan Apr 15 '16 at 13:26
  • 3
    $\begingroup$ Kapranov notes a lesser degree of truncation in the physics case: "One thing is worth noticing. Supergeometry, as understood by mathematicians, tackles only the first two columns of Table 1. A similar-sounding concept (supersymmetry) used by physicists, dips into the third column as well: fermions are always wedded to spinors in virtue of the Spin-Statistics Theorem." $\endgroup$ – David Corfield Apr 16 '16 at 6:45

Like Schreiber does in his post, I would advertise the point of view developed by Sagave and Schlichtkrull in their Adv. Math 2012 paper, and used by us to study topological logarithmic geometry. Each symmetric spectrum $X$ has a graded underlying space that is really a $J$-shaped diagram $Y$. Here $J$ is a category with nerve $QS^0$, so $hocolim_J Y$ maps to $QS^0$. If $X$ is a commutative symmetric ring spectrum then $hocolim_J Y$ is an $E_{\infty}$ space over $QS^0$, i.e., it is graded over the sphere spectrum. Sagave started developing this while a postdoc in Oslo. I noted that the nerve of $J$ was not $Z$ but something with $\pi_1 = Z/2$.

| cite | improve this answer | |
  • 2
    $\begingroup$ Thanks for highlighting Sagave-Schlichtkrull arxiv.org/abs/1103.2764 . I had missed that, as well as your comment here until now. So if I understand correctly, this fills in the remaining gap in the story I was trying to highlight in that message which the OP asked about: $\endgroup$ – Urs Schreiber Dec 1 '16 at 15:38
  • 2
    $\begingroup$ With the previous results by Sagave we had only that for $E$ an $E_\infty$-ring, then its $\infty$-group of units is $\mathbb{S}$-graded, via an $E_\infty$-homomorphism $GL_1(E) \to (\mathbb{S}, +)$. The remaining question (at least as far as I had been following) was whether this extends to an $\mathbb{S}$-grading of all of $\Omega^\infty R$. And I suppose that's exactly what Schlichtkrull-Sagave show (discussed in between their theorems 1.7-1.8 and then around formula (4.4) in the main text): there is always $(\Omega^\infty R, \cdot) \to (\mathbb{S}, +)$. That's beautiful. $\endgroup$ – Urs Schreiber Dec 1 '16 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.