The Barratt-Priddy-Quillen(-Segal) theorem says that the following spaces are homotopy equivalent in an (essentially) canonical way:

  1. $\Omega^\infty S^\infty:=\varinjlim~ \Omega^nS^n$

  2. $\mathbb{Z}\times ({B\Sigma_\infty})_+$, where $\Sigma_\infty$ is the group of automorphisms of a countable set which have finite support, and $+$ is the Quillen plus-construction.

  3. The group completion of $B\left(\bigsqcup_n \Sigma_n\right)$, where $\Sigma_n$ is the symmetric group on $n$ letters, and $B(\sqcup_n \Sigma_n)$ is given the structure of a topological monoid via the block addition map $\Sigma_n\times \Sigma_m\to \Sigma_{n+m}$.

  4. $\Omega|S^\bullet.\operatorname{FinSet}|$, where $S^\bullet$ is the Waldhausen $S$-construction, and $\operatorname{FinSet}$ is the category of pointed finite sets, given the structure of a Waldhausen category by declaring cofibrations to be injections and weak equivalences to be isomorphisms. I don't want to define this since it's complicated (for a reference, see Chapter IV of Weibel's K Book), but it should be thought as a homotopical version of the Grothendieck ring of finite sets, where addition is given by disjoint union and multiplication is given by the cartesian product. Clark Barwick's answer here makes this more precise.

Now, the homotopy groups of the first space are manifestly the stable homotopy groups of spheres; on the other hand, the last two spaces clearly encode some information about the combinatorics of finite sets. So my question is:

Is there a concrete combinatorial interpretation of the higher stable homotopy groups of spheres in terms of the combinatorics of finite sets or symmetric groups?

For example it is easy to see via (3) or (4) that $\pi_{0+k}(S^k)=\mathbb{Z}$ corresponds to the Grothendieck ring of finite sets. Similarly, (2), or with some theory (4), make it clear that $\pi_{1+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ corresponds to the abelianization of $\Sigma_n$ (via the sign homomorphism). I am interested in concrete interpretations of the higher stable homotopy groups in this style.

A good answer would be, for example, a direct combinatorial interpretation of $\pi_{2+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ and $\pi_{3+k}(S^k)=\mathbb{Z}/24\mathbb{Z}$; a not-so-good answer would be a statement like "the sphere spectrum is a categorification of the integers," which is not the sort of concrete thing I'm looking for.

EDIT: So with the exception of Jacob Lurie's comment on $\pi_{2+k}(S^k)$ below (interpreting it as the Schur multiplier $H_2(\Sigma_\infty, \mathbb{Z})$ of $\Sigma_\infty$), it seems like it might be too much to hope for any reasonably complete combinatorial interpretation of the stable homotopy groups. So I'd settle for something like the following: namely, a sequence of groups $G_n$ defined in some combinatorial way, and maps $f_n: G_n\to \pi_{n+k}(S^k)$ or $g_n: \pi_{n+k}(S^k)\to G_n$ such that

  1. $f_n$ or $g_n$ are nontrivial for infinitely many $n$,

  2. The maps are related in some way to the constructions 2-4 above, and

  3. The $G_n$ are combinatorially interesting.

One such example is $G_n:=H_n(\Sigma_\infty, \mathbb{Z})$ with $g_n$ the Hurewicz map (whence the interpretation of $\pi_{2+k}(S^k)$). But even in this case, the combinatorial meaning is sort of mysterious (to me at least) for large $n$.

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    $\begingroup$ I'd love to know an answer to this! +1 $\endgroup$ Sep 27, 2011 at 20:03
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    $\begingroup$ The homology equivalence of the identity component of QS^0 with the classifying space of the infinite symmetric group gives a homology equivalence of its universal cover with the classifying space of the infinite alternating group G. Hence $\pi_2 QS^0 = H_2(G)$ is the Schur multiplier of G: that is, the kernel of the universal central extension of G (which is also the universal central extension of A_n as soon as n is reasonably large; I think it starts at n=8.) $\endgroup$ Sep 27, 2011 at 23:13
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    $\begingroup$ @Dylan: It's exactly reminiscent, via any of 2-4 above, which should be viewed as exactly analogous to the $+$ or $S^\bullet$ constructions of algebraic $K$-theory. @Jacob Lurie: Another way to see this is to use that Hurewicz is an isomorphism from $\pi_1\to H_1$, and (I think) is a ring homomorphism for ring spectra, and then to use that the generators of $\pi_1$ and $H_1$ square to give the generators of $\pi_2$ and $H_2$. $\endgroup$ Sep 28, 2011 at 0:34
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    $\begingroup$ Daniel - you might be interested in the ``discrete models map'' for the J-homomorphism appearing in papers of Snaith and the book of May-Quinn-Ray-Tornehave. It comes from the forgetful functor from finite-dimensional vector spaces over F_p to finite sets, and gives maps K_n(F_p) --> \pi_n^S[1/p] that capture the image of J away from p. $\endgroup$ Sep 28, 2011 at 5:12
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    $\begingroup$ @Daniel: Have you checked out the work of Jie Wu and collaborators, "Configurations, braids, and homotopy groups" in the JAMS? They give a combinatorial description of the homotopy groups of $S^2$ in terms of Brunnian braids. This might not be quite what you were asking about, but its certainly related. $\endgroup$
    – Mark Grant
    Sep 28, 2011 at 6:57

4 Answers 4


The following construction is due to Jones and Westbury, in a paper titled "Algebraic $K$-Theory, homology spheres and the $\eta$-invariant" (it is a very nice paper).

Let $M$ be a homology $3$-sphere and $\rho:\pi_1 (M) \to GL_N(\mathbb{C})$ a representation. Then the Quillen plus construction gives a map $S^3 = M^+ \to (BGL_N(\mathbb{C}))^+$, in other words, an element $[M,\rho] \in K_3 (\mathbb{C})$. On this group, there is the $e$-invariant $e:K_3 (\mathbb{C})\to \mathbb{C}/\mathbb{Z}$. Jones and Westbury give a formula for $e([M,\rho])$ in terms of the eigenvalues of $\rho$. If $M$ is the Poincare sphere, you get a complicated, but manageable formula.

Now replace $\rho$ by an action of $\pi_1 (M)$ on a finite set $X$. This then gives, by Barratt–Priddy–Quillen, an element of $\pi_{3}^{st}$. The Jones–Westbury formula tells you how to compute the e-invariant of the image of this element in $K_3 (\mathbb{C})$ under the map induced by $\Sigma_N \to GL_N(\mathbb{C})$. As the homomorphism $\pi_{3}^{st} \to K_3 (\mathbb{C})$ is injective, you do not lose information.

As a grad student, I played with these formulae in the similar situation of mapping class groups (instead of general linear or symmetric group). This gives elements in the homotopy of the plus-construction of the classifying space of the mapping class group and I was able to find a generator of $\pi_3 (B\Gamma^{+})=\mathbb{Z}/24$ in this way (by an acion of the tgroup of the Poincare sphere on a surface of rather small genus, see http://wwwmath.uni-muenster.de/mjm/vol3.html. I guess it is possible to get a generator of $\pi_{3}^{st}=\mathbb{Z}/24$ by an action of the fundamental group of the Poincaré sphere on a finite set.

  • $\begingroup$ This is pretty cool! $\endgroup$ Feb 28, 2012 at 17:58

Write $S = \bigsqcup_n BS_n$ for the symmetric monoidal category of finite sets and bijections under disjoint union, and write $\mathbb{S}$ for the sphere spectrum, thought of as a symmetric monoidal $\infty$-groupoid. There is a natural map

$$S \to \mathbb{S}$$

of symmetric monoidal $\infty$-categories. I'll interpret "combinatorics of finite sets or symmetric groups" to mean $S$, and so I'll interpret the question to mean "what does this map tell us about $S$?" The guiding principle for everything that follows is the following pair of universal properties, which in particular tell us where the above map comes from:

$S$ is the free symmetric monoidal $\infty$-category on a point, while $\mathbb{S}$ is the free symmetric monoidal $\infty$-groupoid with inverses on a point.

The punchline is the following:

$S$ naturally acts on all symmetric monoidal $n$-categories, and on symmetric monoidal $n$-groupoids with inverses, this action naturally factors through the $n^{th}$ truncation $\mathbb{S}_{\le n}$ of the sphere spectrum.

$n$ is, of course, allowed to take the value $\infty$. The $n$-truncated map $S \to \mathbb{S}_{\le n}$ can be thought of as a "higher sign character" following e.g. Ganter-Kapranov.

I apologize in advance for the extreme length of this answer, but I'm making some very formal statements and they won't count for much if I can't make them at least a little concrete.

Let's start with the case $n = 1$. (The case $n = 0$ is an exercise.) Any symmetric monoidal category $(C, \otimes)$ admits a natural family of endofunctors

$$k : V \mapsto V^{\otimes k}$$

which themselves admit a natural family of natural automorphisms given by the symmetric groups

$$S_k \ni \sigma : V^{\otimes k} \to V^{\otimes k}.$$

A concise way of summarizing the compatibilities these things satisfy is that any symmetric monoidal category $C$ is naturally a right module category over the free symmetric monoidal category $S$ on a point (with a second monoidal structure). This is because, by the universal property, there is a natural identification

$$[S, C] \cong C$$

(where here $[S, C]$ denotes the category of symmetric monoidal functors $S \to C$), and so $[S, S]$ naturally acts on $C$ by precomposition. But by the universal property we have $[S, S] \cong S$, with composition giving us a second monoidal structure on $S$ naturally making it a monoid object in symmetric monoidal categories (with respect to a suitable tensor product).

If $C$ is a symmetric monoidal category, let $C^{\times}$ denote the subcategory of invertible objects in $C$ and isomorphisms between these, and let's restrict our attention to this subcategory. For invertible $V$, the endofunctors $k : V \mapsto V^{\otimes k}$ have natural pointwise inverses

$$-k : V \mapsto (V^{-1})^{\otimes k}$$

and so the natural action of $\pi_0(S) \cong \mathbb{Z}_{\ge 0}$ naturally extends to an action of $\mathbb{Z}$. This $\mathbb{Z}$ is, of course, $\pi_0(\mathbb{S})$.

By invertibility, the natural map $\text{Aut}(1) \to \text{Aut}(V^{\otimes k})$ is an isomorphism. By the Eckmann-Hilton argument, $\text{Aut}(1)$ is an abelian group, and so the natural action of $S_k$ on $V^{\otimes k}$ must factor through its abelianization, and hence through the sign character $S_k \to \mathbb{Z}_2$. Moreover, these sign actions on $V^{\otimes k}$ determine and are determined by a sign action on $1$, and naturally give rise to sign actions on $(V^{-1})^{\otimes k}$ by dualizing.

This $\mathbb{Z}_2$ is, of course, $\pi_1(\mathbb{S})$, and a concise summary of the above pair of observations, together with all of the compatibilities relating them, is that the action of $S$ on symmetric monoidal groupoids with inverses factors through the natural map to the $1$-truncation $\mathbb{S}_{\le 1}$ of the sphere spectrum. This says a bit more than just that the natural actions of $S_k$ factor through their abelianizations: it also says that the sign character is compatible with the natural maps $S_k \times S_{k'} \to S_{k + k'}$, as well as that the data of the sign actions on $V^{\otimes k}$ can be packaged into the data of an operation

$$\pi_0(C^{\times}) \ni V \mapsto \sigma \in \text{Aut}(V^{\otimes 2}) \cong \text{Aut}(1) \cong \pi_1(C^{\times})$$

where $\sigma$ is the nontrivial element of $S_2$ acting on $\text{Aut}(V^{\otimes 2})$. This operation can be thought of as assigning every invertible object in $C$ its dimension, and the fact that $\pi_1(\mathbb{S}) \cong \mathbb{Z}_2$ implies in particular that the dimension of an invertible object is always $2$-torsion.

Example. Let $C$ be the symmetric monoidal category of super vector spaces over a field $k$ equipped with the tensor product. This operation assigns to the even invertible super vector space $1 \in \text{Aut}(1) \cong k^{\times}$ and assigns to the odd invertible super vector space $-1$.

The story for general $n$ (including $n = \infty$) is formally identical but takes more work to unpack. Again any symmetric monoidal $n$-category admits a natural action of $S$, and for the same reason. If $(C, \otimes)$ is a symmetric monoidal $n$-category, let $C^{\times}$ denote the subcategory of invertible objects in $C$, isomorphisms between these, isomorphisms between isomorphisms, etc.

Now let's try to run the above arguments again for the action of $S$ on $C^{\times}$. It is again true that for invertible $V$, the natural map $\text{Aut}(1) \to \text{Aut}(V^{\otimes k})$ (now a map of $n$-groups) is an isomorphism. The analogue of Eckmann-Hilton here asserts that the group operation on $\text{Aut}(1)$ is symmetric monoidal, so the natural action of $S_k$ on $V^{\otimes k}$ factors through a map

$$S_k \to \text{Aut}(V^{\otimes k}) \cong \text{Aut}(1)$$

to a symmetric monoidal $n$-group. The "symmetric monoidal $n$-abelianization" of $S_k$ is determined by taking the $n$-truncation of the answer for $n = \infty$, which is the loop space of the free connective spectrum on $BS_k$, or $\Omega \Sigma^{\infty} BS_k$.

I don't know much about that space, but fortunately we don't have to deal with it because we haven't used all of the available structure: we're talking about the individual symmetric groups instead of talking about $S$, so let's talk about $S$. The identification $[S, C] \cong C$ says that symmetric monoidal functors $S \to C$ can be identified with objects of $C$. If we restrict our attention to $C^{\times}$, then symmetric monoidal functors $S \to C^{\times}$ naturally factor through the "group completion" of $S$ (by which I mean the left adjoint to the inclusion of symmetric monoidal $\infty$-groupoids with inverses into symmetric monoidal $\infty$-categories), which Barratt-Priddy-Quillen-Segal tells us is the sphere spectrum $\mathbb{S}$. If $C$ is an $n$-groupoid then we can further factor through the $n$-truncation $\mathbb{S}_{\le n}$. Note that if you believe that $S$ and $\mathbb{S}$ have the universal properties that I said they did, then BPQS is a formal consequence of those universal properties. On the other hand, while it's not hard to show that $S$ is the free symmetric monoidal category on a point, it's less clear why $S$ is the free symmetric monoidal $\infty$-category on a point.

Okay, but that was a lot of formal junk. What am I actually saying for $n = 2$? Here $C$ is a symmetric monoidal $2$-category and we're considering the action of the symmetric groups $S_k$ on the tensor powers $V^{\otimes k}$ of some invertible object in $C^{\times}$. Write

$$\varphi : S_k \to \text{Aut}(V^{\otimes k})$$

for the corresponding morphism of $2$-groups. The starting observation is that part of what it means to be a morphism of $2$-groups is that $\varphi$ does not in any sense just "strictly preserve" the group operation on both sides, in the sense that we don't just have

$$\varphi(\sigma_1 \sigma_2) = \varphi(\sigma_1) \varphi(\sigma_2).$$

What we have instead is a family of $2$-morphisms

$$\eta_{\sigma_1, \sigma_2} : \varphi(\sigma_1 \sigma_2) \to \varphi(\sigma_1) \varphi(\sigma_2)$$

satisfying the coherence condition that the two ways of going from $\varphi(\sigma_1 \sigma_2 \sigma_3)$ to $\varphi(\sigma_1) \varphi(\sigma_2) \varphi(\sigma_3)$ coincide. As previously explained here, in the special case that the only element of $\text{Aut}(V^{\otimes k})$ is $\text{id}_{V^{\otimes k}}$ (this is an evil condition but it's easier to say things this way) this family of $2$-morphisms is precisely a $2$-cocycle on $S_k$ with values in $\text{Aut}(\text{id}_{V^{\otimes k}})$, and the coherence condition is precisely the $2$-cocycle condition.

$\pi_2(\mathbb{S})$ describes how nontrivial these cocycles can be subject to the condition that the natural action of $S_k$ on $\text{Aut}(V^{\otimes k}) \cong \text{Aut}(1)$ is coherently compatible with the symmetric monoidal structure $S_k \times S_{k'} \to S_{k + k'}$. These compatibilities imply, in particular, that we can package all of the morphisms $S_k \to \text{Aut}(V^{\otimes k})$ into a single morphism

$$S_{\infty} \to \text{Aut}(1)$$

of $2$-groups. We can trivialize this map on $\pi_0$ by passing from $S_{\infty}$ to the infinite alternating subgroup $A_{\infty}$, which has the effect of letting us ignore the non-identity elements of $\text{Aut}(1)$. We then get a natural map

$$A_{\infty} \to B \text{Aut}(\text{id}_1) \cong B \pi_2(C^{\times})$$

of $2$-groups, and the fact that $\pi_2(\mathbb{S}) \cong \mathbb{Z}_2$ says that the universal map from $A_{\infty}$ to a connected $2$-group (a $2$-group of the form $BA$, for $A$ an abelian group) is a map

$$A_{\infty} \to B \mathbb{Z}_2.$$

By universal coefficients, this is in turn equivalent to Lurie's assertion in the comments that

$$\pi_2(\mathbb{S}) \cong H_2(BA_{\infty}, \mathbb{Z}) \cong \mathbb{Z}_2$$

is the Schur multiplier of $A_{\infty}$. In the same way that passing from $S_{\infty}$ to $A_{\infty}$ universally trivializes the action of $S_{\infty}$ at the level of $\pi_0$, passing from $S_{\infty}$ to $A_{\infty}$ to the universal central extension $\widetilde{A}_{\infty}$ universally trivializes the action of $S_{\infty}$ at the level of $\pi_0$ and $\pi_1$.

The generator of $\pi_2(\mathbb{S})$ can also be thought of as describing an operation

$$\pi_0(C^{\times}) \to \pi_2(C^{\times})$$

which assigns to every invertible object of $C$ its "$2$-dimension." This is, in a very precise sense, the dimension of the dimension: the ring spectrum structure on $\mathbb{S}$ induces in particular a multiplication

$$\pi_1(\mathbb{S}) \times \pi_1(\mathbb{S}) \to \pi_2(\mathbb{S})$$

and it's known that the generator of $\pi_2(\mathbb{S})$ is the square of the generator of $\pi_1(\mathbb{S})$. This means the above operation factors as a composite

$$\pi_0(C^{\times}) \to \pi_1(C^{\times}) \to \pi_2(C^{\times})$$

where in the first step we take the dimension of an invertible object $V$, which is itself an invertible object in $\text{Aut}(1)$, and then take the dimension again.

Example. Let $C$ be the symmetric monoidal $2$-category of $k$-algebras, $k$-bimodules, and $k$-bimodule homomorphisms, where $k$ is a commutative ring. It turns out to be possible to describe the dimension of any object of $C$, not just the invertible ones, and it turns out to be the zeroth Hochschild homology

$$H_0(A) = A / [A, A].$$

$H_0$ respects the tensor product of $k$-algebras in the sense that $H_0(A \otimes_k B) \cong H_0(A) \otimes_k H_0(B)$, so in order for a $k$-algebra to be invertible (which here means Morita invertible) it must in particular have invertible $H_0$. The sequence of abelian groups given by taking $H_0$ and then taking the dimension

$$\pi_0(C^{\times}) \to \pi_1(C^{\times}) \to \pi_2(C^{\times})$$

can in fact be identified with a natural sequence of abelian groups

$$\text{Br}(k) \to \text{Pic}(k) \to k^{\times}.$$

Subexample. To see a case where the composite of the two maps above is nontrivial, let $k = \mathbb{R}$, but now consider super $k$-algebras, super $k$-bimodules, and super $k$-bimodule homomorphisms. Then

$$\pi_0(C^{\times}) \cong \text{SBr}(\mathbb{R}) \cong \mathbb{Z}_8$$ $$\pi_1(C^{\times}) \cong \text{SPic}(\mathbb{R}) \cong \mathbb{Z}_2$$ $$\pi_2(C^{\times}) \cong \mathbb{R}^{\times}.$$

where the $S$ stands for "super." The super Brauer group is generated by a Clifford algebra

$$\text{Cl}(1) \cong \mathbb{R}[e]/(e^2 = -1)$$

where $e$ has degree $1$. The zeroth Hochschild homology of this Clifford algebra is

$$H_0(\text{Cl}(1)) \cong \mathbb{R}[e]/(e^2 = -1, [e, e] = 2 e^2 = 0) \cong \mathbb{R} e;$$

that is, it's a $1$-dimensional odd vector space, and hence its image in both $\text{SPic}(\mathbb{R})$ and $\mathbb{R}^{\times}$ is $-1$ as above.

Okay, but so far I've more or less repackaged things that other people have already said. To really justify having written all of that let's do $n = 3$.

As far as combinatorics goes, we could talk about coherence data for compatible families of maps from the symmetric groups $S_k$ to symmetric monoidal $3$-groups, and we'd eventually learn (I think) that $\pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$ can be identified with the third homology

$$H_3(B \widetilde{A}_{\infty}, \mathbb{Z})$$

of the universal central extension of $A_{\infty}$, but at this point it seems prudent to introduce a better language for extracting something understandable from all of the structure available here. I'll use the language of framed cobordism. The natural map $S \to \mathbb{S}$ naturally factors through a third object

$$S \to \text{Bord} \to \mathbb{S}$$

where $\text{Bord}$ is the symmetric monoidal $\infty$-category of framed cobordisms. That is, it has objects framed points, morphisms framed $1$-cobordisms between points, etc. Like $S$ and $\mathbb{S}$, it has a universal property explaining where the above maps come from, which is a variant of the cobordism hypothesis:

$\text{Bord}$ is the free symmetric monoidal $\infty$-category with duals on a point.

The point is that a symmetric monoidal $\infty$-groupoid with inverses is in particular a symmetric monoidal $\infty$-category with duals.

The natural map $\text{Bord} \to \mathbb{S}$ induces a map

$$\pi_n(\text{Bord}) \to \pi_n(\mathbb{S})$$

which is precisely the Pontryagin-Thom isomorphism from the group of cobordism classes of closed framed $n$-manifolds to the $n^{th}$ stable homotopy group of spheres. A generalization of the claims I made about dimensions and "$2$-dimensions" above is that if $C$ is any symmetric monoidal $\infty$-category, then there is a natural map

$$\pi_n(\mathbb{S}) \times \pi_0(C^{\times}) \to \pi_n(C^{\times})$$

but since the generator of $\pi_1(\mathbb{S})$ does not cube to a generator of $\pi_3(\mathbb{S})$ and its higher powers vanish, it is not possible to interpret these operations as arising from iterating the construction of the dimension, and we need something new. That something new is the observation that the same argument that shows that every symmetric monoidal $\infty$-groupoid with duals is naturally a module over $\mathbb{S}$ shows that every symmetric monoidal $\infty$-category with duals is naturally a module over $\text{Bord}$. Hence if $C$ is any symmetric monoidal $\infty$-category, then the above map refines to a natural map

$$\pi_n(\text{Bord}) \times \pi_0(C^{fd}) \to \pi_n(C^{fd})$$

where $C^{fd}$ denotes the subcategory of fully dualizable objects, isomorphisms between these, etc. In terms of the usual cobordism hypothesis, this map comes from taking a fully dualizable object $V$, making it the value at a point of an $n$-dimensional fully extended topological field theory, and then evaluating the TFT on a framed $n$-manifold $M$ representing an element of $\pi_n(\text{Bord})$. If $C$ is chosen to be a suitable Morita category of $E_n$ algebras then this operation recovers factorization homology. For lack of a better term I'll call it factorization homology in general and use factorization homology notation for it:

$$\pi_n(\text{Bord}) \times \pi_0(C^{fd}) \ni (M, V) \mapsto \int_M V \in \pi_n(C^{fd}).$$

Hence (with the appropriate framings) $\int_{S^1} V$ is the dimension / Hochschild homology and $\int_{S^1 \times S^1} V$ is the $2$-dimension. This is all we get when $n = 1, 2$ because $\pi_1(\mathbb{S}) \cong \mathbb{Z}_2$ is generated by $S^1$ with the Lie group framing and $\pi_2(\mathbb{S}) \cong \mathbb{Z}_2$ is generated by $S^1 \times S^1$. But $\pi_3(\mathbb{S})$ is not generated by $(S^1)^3$: instead it's generated by $S^3 \cong \text{SU}(2)$ with the Lie group framing, and so the operation we get from a generator of $\pi_3(\mathbb{S})$ is taking factorization homology on $S^3$:

$$\pi_0(C^{fd}) \ni V \mapsto \int_{S^3} V \in \pi_3(C^{fd}).$$

Since $\pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$ and the cube of the generator of $\pi_1(\mathbb{S})$ represents the class $12 \in \mathbb{Z}_{24}$, factorization homology of a fully dualizable object on $S^3$ is $24$-torsion and moreover represents a distinguished $12^{th}$ root of the $3$-dimension $\int_{S^1 \times S^1 \times S^1} V$, which is $2$-torsion.

Warning. "Fully dualizable" above is an incredibly strong condition, and in particular it implies being $n$-dualizable for all $n$. Aside from invertible objects, the only examples I can cook up even heuristically are $E_{\infty}$ algebras in a suitable Morita $\infty$-category.

On the other hand, the upshot of the above digression into the cobordism hypothesis is that the operations induced by elements of $\pi_n(\mathbb{S})$, thought of as closed framed $n$-manifolds, make sense on a much more general class of objects than just invertible objects, but in fact on $n$-dualizable objects. So to understand how these operations work we can relax our hypotheses substantially: for example, dimension $\int_{S^1} (-)$ makes sense on $1$-dualizable objects, and these are much easier to find.

Example. Let $C$ be the symmetric monoidal $3$-category of conformal nets in the sense of Bartels-Douglas-Henriques. Then $\pi_3(C^{\times}) \cong U(1)$, and so factorization homology on $S^3$ defines an invariant

$$\pi_0(C^{\times}) \ni V \mapsto \int_{S^3} V \in \pi_3(C^{\times}) \cong U(1).$$

of invertible conformal nets taking values in the $24^{th}$ roots of unity. There is an invertible conformal net called the free fermion $\text{Fer}(1)$, which (conjecturally) is to tmf what the Clifford algebra $\text{Cl}(1)$ is to K-theory, and in Topological modular forms and conformal nets, Douglas and Henriques show that $\int_{S^3} \text{Fer}(1)$ is in fact a primitive $24^{th}$ root of unity.

One last comment on in what sense we can connect the above back to combinatorics, or $S$. Roughly speaking, I want to argue that this question doesn't respect the symmetries of the situation. What I mean is that starting from the action of $S$, we can at best write down some $0$-morphisms, some $1$-morphisms, and some coherence data, and out of this data we can try to say something about $\pi_n(\mathbb{S})$ if we can use this data to put together a closed framed $n$-manifold $M$. But the invariant $\int_M V$ itself doesn't depend on a choice of presentation of $M$, and presenting special cases of it this way blinds us to that: for example, it blinds us to a potentially very rich action of the framed diffeomorphisms of $M$ on $\int_M V$.

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    $\begingroup$ OK, I've voted you up because I like your explanation of $\pi_2$ a lot. But I think much of what you've written is philosophically backwards--the question is not about saying something about computing $\pi_n(\mathbb{S})$, but rather about understanding it. Breaking symmetries (e.g. writing down a basis of a vector space) is a common way to do that...In fact, this is exactly the answer I was trying to avoid by discouraging a discussion of the sphere spectrum as a categorification of the integers. But I do greatly appreciate your attempt at making that philosophy more concrete. $\endgroup$ Feb 10, 2015 at 4:07
  • $\begingroup$ Of course, I do take your point that topological field theories are a good way of understanding stable homotopy groups of spheres; if you gave me a "combinatorial" example of one, where things ended up being somehow computable, I'd probably be happier. $\endgroup$ Feb 10, 2015 at 4:08
  • $\begingroup$ @Daniel: well, roughly speaking there are two ways to try to understand a group: by looking at its structure theory or looking at its representation theory. This answer describes in particular a natural class of representations, in a suitable sense, of the groups $\pi_n(\mathbb{S})$... is the sense in which this is unsatisfactory that I haven't told you how to compute these representations more explicitly? $\endgroup$ Feb 10, 2015 at 4:37
  • $\begingroup$ My complaint is mostly about your comment that "this question doesn't respect the symmetries of the situation." But in the spirit of your comment, I do think that this answer is like saying that "representation theory is a good way to study groups because of the Tannakian formalism." That's true, but once you have the Tannakian formalism in hand, you have to actually do something with it. My question, reinterpreting it in these terms, is: can one do anything very, very concrete? And so far the answer seems to be: not for $\pi_n^{st}$ with $n>2$. $\endgroup$ Feb 10, 2015 at 4:44
  • $\begingroup$ @Daniel: so it seemed like you were happy with $\pi_2(\mathbb{S}) \cong H_2(BA_{\infty}, \mathbb{Z})$. What about $\pi_3(\mathbb{S}) \cong H_3(B \widetilde{A}_{\infty}, \mathbb{Z})$? This can be interpreted as claiming the existence of a "universal $2$-group extension" of the universal central extension $\widetilde{A}_n$ (for $n$ sufficiently large, but I don't know enough about these stabilization results to guess how large) by $B \mathbb{Z}_{24}$. $\endgroup$ Feb 10, 2015 at 4:48

Combinatorial descriptions of homotopy groups of spheres have been studied in the unstable world rather than the stable world.

Jie Wu found a description of homotopy groups of $S^2$ in his thesis and published here. He described $\pi_n(S^2)$ as the center of a combinatorially defined group.

As is well known, there is an analogue of Barratt-Priddy-Quillen theorem for braid groups. By replacing symmetric groups by braid groups, we obtain $\Omega^2 S^2$. And it is natural to expect to have a combinatorial description of $\pi_*(S^2)$ in terms of braid groups. Such a description was obtained by Berrick, Cohen, Wong, and Wu in this paper. They described $\pi_*(S^2)$ as the homotopy groups of a $\Delta$-group (simplicial group without degeneracies) constructed from braid groups.

Recently a papaer by Mikhailov and Wu appeared, in which they extended Wu's description of $\pi_*(S^2)$ in his thesis.


There are papers by Eccles, Freedman, Koschorke, and Koschorke and Sanderson that date circa 1978-1982 which discuss the $n$th-stable stem as the bordism group of oriented codimension 1 immersions of closed $n$-manifolds in $(n+1)$-space. For example the generator of $\pi_1^s$ is represented by an immersed circle in the plane that forms the figure-$8$. The mod-2 number of double points is the invariant. The $2$nd stabe stem is represented by a fully twisted torus that is formed from the figure-$8$ times $[0,1]$, fully twisting one end and attaching to form a torus. This is the Lie framed torus (perturbed from $\mathbb{C}^2$) and projected into space.

The generator of ${\mathbb Z}/(24)$ can be given by the sphere eversion. (Shameless self-promotion) My upcoming book gives this explicitly, not as a generator, but you can compute the self-intersection invariants from the diagrams that I give. In general the Kahn-Priddy maps are the invariants constructed from the self-intersection sets. I have not done it, but it would be a nice exercise to understand your $3$rd point of view from these immersions.

  • $\begingroup$ Maybe I'm being thick, but I don't see how this answers the question. I guess I can see the relation to configuration spaces of points in $\mathbb{R}^n$ as approximations to $B\Sigma_k$ in your mention of bordism groups, which are also approximations to $\Omega^nS^n$ via the electric field map--but that isn't really what I'm asking. Where's the combinatorics? $\endgroup$ Sep 28, 2011 at 0:40
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    $\begingroup$ I agree that I have not fully answered the question! I believe that the combinatorics is found in studying the multiple point manifolds. I should have been more explicit about that, and even then there are gaps in my knowledge. Inside a configuration space, one constructs the multiple point manifolds, and one can also construct a variety of covers thereof. There are symmetric group actions on these. I believe that is where you might find the combinatorial coincidences. $\endgroup$ Sep 28, 2011 at 1:35
  • $\begingroup$ Ah I see--essentially you want to sort of explicitly describe the combinatorics of the "tangle hypothesis" I guess? If you were to expand on this part of the answer a bit it could be really cool. $\endgroup$ Sep 28, 2011 at 1:47
  • $\begingroup$ (+1 by the way): Also, do you have a reference for the fact you mention on the sphere eversion? It's pretty cool. $\endgroup$ Sep 28, 2011 at 4:00
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    $\begingroup$ @Daniel, For the sphere eversion see: southalabama.edu/mathstat/personal_pages/carter/talks.html or in particular, southalabama.edu/mathstat/personal_pages/carter/CU.pdf a draft manuscript is available on the first link (search sphere eversion), but the file is enormous. The book will be coming out in World Science soon. I will write more about the self-intersection sets in a while, but I think looking at the Koschorke/Eccles papers is a good place to start. I am avoiding grading some exams, and the exams are becoming impatient. $\endgroup$ Sep 28, 2011 at 17:02

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