Which $\infty$-groupoids correspond to simplicial abelian groups?

Kan complexes model $$\infty$$-groupoids, so since every simplicial abelian group is a Kan complex, every simplicial abelian group yields an $$\infty$$-groupoid. What sort of $$\infty$$-groupoids do you get?

(A natural guess would be grouplike symmetric monoidal $$\infty$$-groupoids, but I think that's not the correct answer roughly because $$\mathbb{Z}$$ is not the sphere spectrum.)

I realize that this question can be taken to be tautological, but I'd like an answer that I could say specialize $$2$$-groupoids described by the bicategory axioms.

I'd also like to understand what you get if you translate the functor taking simplicial set to the corresponding simplicial abelian group across the homotopy hypothesis.

• The functor taking a simplicial set to the corresponding simplicial abelian group corresponds to taking the free $H \mathbb{Z}$-module spectrum on a homotopy type $X$. This can be thought of as a tensor product (of spectra) of $H \mathbb{Z}$ and the suspension spectrum of $X$ (which is the free spectrum on $X$). – Qiaochu Yuan Oct 16 at 6:52

Basically, you're asking which (weak) homotopy types can be modelled by simplicial sets which carry the structure of a simplicial abelian group.

Every simplicial abelian group splits as a product of Eilenberg MacLane spaces. $$X \simeq \prod_{n \in \mathbb N} K(\pi_i X, n).$$ This is a consequence of the Dold-Kan correspondence, and the fact that every chain complex of abelian groups is quasi-isomorphic to its homology. Conversely, you can use Dold-Kan to realize every product of Eilenberg-MacLane spaces as a simplicial abelian group.

• Thanks! Of course you're right and this does answer the question, but I was hoping for an answer that would help my understand the Dold-Kan correspondence rather than just saying what Dold-Kan yields. – Noah Snyder Oct 16 at 1:44
• @Noah I see, then if you haven't seen it already the answers to this question are probably relevant: mathoverflow.net/questions/118500/…. – Phil Tosteson Oct 16 at 20:13
• It is a bit miraculous from the POV of homotopy theory that "strictly abelian groups" are well defined (as opposed to "$E_\infty$ groups"). This is closely related to the miracle that symmetric powers are homotopy invariant, since this lets you build a monad for the algebraic structure (this monad is closely related to the Lawvere theory skd discusses). A homotopy-invariant perspective on symmetric powers was discussed here: mathoverflow.net/questions/37647/… – Phil Tosteson Oct 16 at 20:14


Here is one such characterization. Let's model grouplike infinite loop spaces $$X$$ as functors $$X:\mathrm{Fin}_\ast\to \mathrm{Top}$$ such that $$\pi_0 \mathrm{Map_{Top}}(Y,X)$$ is an abelian group for all spaces $$Y$$ (i.e., $$X$$ is grouplike) and such that the map $$X([n])\to X([1])^n$$ is an equivalence. Such an object should be in the image of $$\Mod^{\geq 0}_\Z$$ iff it is somehow "strictly commutative". One way to characterize this is as follows. Let $$\Lambda$$ denote the full subcategory of the category of abelian groups spanned by the groups $$\Z^n$$ with $$n\geq 0$$, so there is a functor $$\mathrm{Fin}_\ast\to \Lambda$$. Then an infinite loop space is in the image of $$\Mod^{\geq 0}_\Z$$ if and only if the functor $$\mathrm{Fin}_\ast\to \mathrm{Top}$$ classifying it factors through a finite-product-preserving functor $$\Lambda \to \mathrm{Top}$$. In other words, $$\Mod^{\geq 0}_\Z$$ is equivalent to the full subcategory spanned by the grouplike objects in the category $$\mathrm{Fun^{prod}}(\Lambda, \mathrm{Top})$$. This is a very strong condition to impose on an infinite loop space; for example, $$\mathbf{C}P^\infty$$ admits such a factorization, but $$BU$$ (with either the additive or multiplicative infinite loop space structure) doesn't.

• I probably should've also mentioned the more elementary characterization provided by Phil Tosteson in his answer; sorry! One advantage of the description I gave is that it can be generalized to define "strictly commutative" $\mathbf{E}_\infty$-objects in a category with finite products (where you replace Top with the category under consideration). – skd Oct 16 at 1:10
• Can I understand your answer to say that the answer is supposed to be "grouplike strictly symmetric monoidal $\infty$-groupoids"? – Noah Snyder Oct 16 at 1:48
• Yep, if your definition of "strict symmetric monoidal" means that the functor from Fin* classifying your E_oo-structure factors through Fin* -> Λ. – skd Oct 16 at 2:07
• I wonder if that agrees with more algebraic definitions of strict symmetric monoidal $1$-groupoid or $2$-groupoid. At least that's something that I can try to check. – Noah Snyder Oct 16 at 16:08

The paper HHA.5(1), 2003, pp.49–52 gives a few categories equivalent to chain complexes, including cubical abelian groups with connections, and cubical $$\omega$$-groupoids with connections. There are also the simplicial $$T$$-complexes of N. Ashley, referenced [Ash88] in the bibliography of NAT-book. A simplicial $$T$$-complex is a simplicial complex with in each dimension a spacial kind of simplex called thin satisfying Dakin's axioms: 1) degenerate elements are thin; 2) every horn has a unique thin filler; 3) if every face but one of a thin simplex is thin then so also is the remaining face. (cf Note 222 p. 476 of the NAT book).

These structures of $$T$$-complex do not model all homotopy types, but only the "linear" ones (no quadratic information, such as Whitehead products) but, like linear algebra, it has its uses!