What is the group completion of finite sets with respect to cartesian product?

Question

Let $\Sigma_+$ be the groupoid of finite pointed sets. The Barratt-Priddy-Quillen theorem tells us that the group completion of $\Sigma_+$ with respect to the symmetric monoidal structure given by wedge product $\vee$ is the sphere spectrum $\mathbb S$. Since $(\Sigma_+,\vee)$ is symmetric monoidally equivalent to the groupoid $(\Sigma,\amalg)$ of finite sets under disjoint union, $\mathbb S$ is likewise the group completion of $(\Sigma,\amalg)$.

Questions:

1. What is the group completion of $\Sigma_+$ with respect to smash product $\wedge$?

2. What is the group completion of $\Sigma$ with respect to cartesian product $\times$?

3. What is the group completion of $\Sigma_{>0}$ with respect to cartesian product $\times$ (where $\Sigma_{>0}$ is the groupoid of nonempty finite sets)?

Motivation:

$\Sigma$ and $\Sigma_+$ are natural categorifications of $\mathbb N$, and as such, group completion with respect to $\amalg$ / $\vee$ categorifies the passage from $\mathbb N$ to $\mathbb Z$, making $\mathbb S$ a categorification of $\mathbb Z$. I'm interested in categorifying instead the passage from $\mathbb N$ to $\mathbb Q_{\geq 0}$ or perhaps from $\mathbb N_{>0}$ to $\mathbb Q_{>0}$. Naively it seems that one of the above groupoids should be relevant, although I'm not sure precisely which of them is most so. I'd be interested in understanding any of them.

Answer 1

As already addressed in the comments:

• Group completing the groupoid of finite pointed sets under the smash product gives a contractible space.
• The groupoid of finite sets under the cartesian product is isomorphic, and hence group completing it also gives a contractible space.

This leaves us with one remaining case. Here is the result.

• All higher homotopy vanishes when we group complete the groupoid of finite nonempty sets under the cartesian product: it is homotopy equivalent to the discrete space $$K(\Bbb N_{> 0}, \cdot) \cong \bigoplus_{p\text{ prime}} \Bbb Z.$$

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For convenience let me write $X$ for the space you were calling $\Sigma_{> 0}$, with $\pi_0 X \cong \Bbb N_{> 0}$, and $X \to X^\text{gp}$ for the group-completion map. Here are the steps we need in the proof.

1. The group completion $X^\text{gp}$ is a simple space (abelian fundamental group with higher homotopy groups acted on trivially) because it is a grouplike $H$-space.

2. Therefore, by the Hurewicz theorem, the map $X^\text{gp} \to \pi_0(X^\text{gp})$ is a homotopy equivalence if and only if the map $H_* (X^\text{gp}) \to H_*(\pi_0 X^\text{gp})$ is an isomorphism, or equivalently if and only of $H_*(X^\text{gp})$ is trivial for $* > 0$.

3. Quillen's group-completion theorem says that the map $H_* X \to H_* (X^\text{gp})$ is a specific map of graded-commutative rings: it is the localization which inverts all elements in $\pi_0 X \subset H_0(X)$. Therefore, it suffices to show that $(\pi_0 X)^{-1} H_*(X)$ vanishes in positive degrees.

4. By the universal coefficient theorem (and exactness of localization), it is equivalent to show that the mod-$p$ version $(\pi_0 X)^{-1} H_*(X;\Bbb F_p)$ vanishes in positive degrees for all primes $p$. (Normally we'd also have to check rational coefficients, but the homology of the symmetric groups is torsion above degree zero, and so it's still true after localization.)

5. It would suffice to show this after inverting one element. This would follow if we can show that, after inverting $p \in \Bbb N_{> 0}$, the ring $[p]^{-1} H_*(X;\Bbb F_p)$ vanishes in positive degrees.

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Now that we have this outline, we need to know specific calculations. For this I'll discuss what's going on at $p=2$; the story at odd primes is similar.

The mod-$2$ homology of the groupoid $\Sigma$ of finite sets is known explicitly as a ring with a presentation using the Araki–Kudo–DyerLashof operations (you can find this in the book of Cohen–Lada–May). However, we won't need this. We do need that the element $1 \in \pi_0(\Sigma)$ becomes an element $[1] = e \in H_0(\Sigma)$. The ring structure comes from the disjoint union of sets. For example, $[2] = [1+1] = e \cdot e = e^{2}$. (Various references will use different notation for this product.)

The homology of $X$ is the nonunital part: it removes the basis element $[0]$, which is the unit for the ring.

This ring $H_* \Sigma$ is also a Hopf ring: it has the ring structure above, and a second multiplication $\circ$ coming from the Cartesian product of sets. Inverting $[2]$ in $H_*(X)$ is the same as inverting the operator $[2] \circ (-)$ on $H_* X$.

Thus, we are reduced to showing that iteratively applying $[2] \circ (-)$ eventually sends every positive-degree element in $H_* X$ to zero.

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Here are some the properties of this operator $[2] \circ (-)$ that we need.

The "multiplicative" product $\circ$ satisfies a distributivity law over the "additive" product: if the homology coproduct satisfies $\Delta z = \sum z' \otimes z''$, then $$ (x \cdot y) \circ z = \sum (x \circ z') \cdot (y \circ z''). $$ In particular, this implies that $$ [2] \circ x^2 = ([2] \circ x)^2 $$ because $\Delta [a] = [a] \otimes [a]$ for any element $a \in \pi_0 X$.

The element $[1] = e$ is also the unit for $\circ$, and so we have $$ [2] \circ z = (e \cdot e) \circ z = \sum z' \cdot z''. $$ However, the coproduct is symmetric, and so most of these terms go away mod-$2$. To be specific, if we choose a basis $\{f_i\}$ for $H_* X$, then there are coefficients $a^{ij}_k$ in $\Bbb F_2$ such that $$ \Delta f_k = \sum a^{ij}_k f_i \otimes f_j $$ and symmetry of $\Delta$ is equivalent to the identity $a^{ij}_k = a^{ji}_k$. In this basis, $$ [2] \circ (\sum b^k f_k) = \sum b^k a^{ij}_k f_i \cdot f_j = \sum b^k a^{ii}_k f_i^2 $$ because the terms with $i \neq j$ cancel. If we define a linear operator $V$ by $$ V(\sum b^k f_k) = \sum b^k a^{ii}_k f_i, $$ then the formula with a basis says $$ [2] \circ x = (Vx)^2. $$ Moreover, $|Vx| = |x| / 2$, with $Vx = 0$ if $|x|$ is not even. Inductively, we then find $$ [2]^{\circ n} \circ x = (V^n x)^{2^n}. $$ However, if $x$ is in positive degree, then for sufficiently large $n$ the element $V^n x$ lives in degree $|x| / 2^n < 1$. Thus $V^n x = 0$ and $[2]^{\circ n} \circ x = 0$.

(These are features that we tend to see from "graded Dieudonné modules"  —see, e.g., Goerss' monograph.)

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The case at odd primes is similar, except we have to take a $p$-fold coproduct to define the operator $V$ and it divides degree by $p$.