Let $\mathcal C$ be a permutative category, that is a symmetrical monoidal category with strict associativity. One can then define the $K$-groups of $\mathcal C$, for $n >0$ by $$K_n(\mathcal C) = \pi_n(\Omega B |\mathcal C|),$$ where $|C|$ denotes the realization of the nerve of $\mathcal C$ that inherits a multiplication coming from the monoidal structure.

My question is: If all Hom-sets in $\mathcal C$ are finite, are the groups $K_n(\mathcal C)$ then also all finite?

Here are there three examples I have seen that motivated this question:

(1) If $\mathcal C$ is given by an abelian group $G$, we get $\Omega B |\mathcal C| = BG$.

(2) If $\mathcal C$ is the category of finite sets with inclusions, we have $\Omega B |\mathcal C| = (B\Sigma_{\infty})^+ = Q_0S^0$ by Baratt-Priddy-Quillen, hence we get the stable homotopy groups of spheres.

(3) For $R$ a (commutative) ring, we can take $\mathcal C$ to be free modules of finite rank and get the higher algebraic $K$-groups of $R$. Quillen computed them for $R = \mathbb F_q$ a finite field, they are $K_{2i} = 0$ and $K_{2i-1} = \mathbb Z/(q^i-1)$.

Disclaimer: I have just started to learn about higher K-theory and this question is motivated by my ignorance, so feel free to close it if it's stupid.

  • $\begingroup$ I guess you want to assume $n>0$ in your question. $\endgroup$ – Dan Ramras May 22 '15 at 3:51
  • $\begingroup$ Yes, and I wrote so in the beginning. $\endgroup$ – Jens Reinhold May 22 '15 at 4:44
  • $\begingroup$ What is the function of the strict associativity requirement here? $\endgroup$ – Qiaochu Yuan May 22 '15 at 5:15
  • $\begingroup$ @QiaochuYuan Certain infinite loop space machines use permutative categories rather than symmetric monoidal ones as a technical nicety; it's not an important distinction. $\endgroup$ – Tyler Lawson May 22 '15 at 15:00

First, I do not think that strict associativity makes a difference, so I will ignore it.

Next, let $G$ be a finite group, and let $\mathcal{C}G$ be the category of finite $G$-sets and equivariant bijections (which is symmetric monoidal under disjoint union). Then $$ K(\mathcal{C}G)=\Omega^\infty\Sigma^\infty\left(\coprod_{(H)} BW_GH\right)_+ = \Omega^\infty (S_G)^G. $$ Here $H$ runs over conjugacy classes of subgroups, and $W_GH$ is $(N_GH)/H$, where $N_GH$ is the normaliser of $H$. Also, $S_G$ is the equivariant sphere spectrum, and $(S_G)^G$ is the fixed point spectrum in the sense of Lewis and May. This gives $$ \pi_1 K(\mathcal{C}G) = \mathbb{Z}/2\oplus \bigoplus_{(H)} \pi_1^S(BW_GH) = \mathbb{Z}/2\oplus\bigoplus_{(H)} (W_GH)_{\text{ab}}. $$ (The factor $\mathbb{Z}/2$ is $\pi_1$ of the sphere spectrum, contributed by the disjoint basepoint.)

This is of course finite. However, we can also consider the category $\mathcal{C}\mathbb{Z}$ of finite sets with an action of $\mathbb{Z}$. Such an action must factor through $\mathbb{Z}/n!$ for some $n$, so $\mathcal{C}\mathbb{Z}$ is the colimit of the sequence of categories $\mathcal{C}\mathbb{Z}/n!$, for which we have $$ \pi_1 K(\mathcal{C}\mathbb{Z}/n!) = \mathbb{Z}/2\oplus\bigoplus_{d|n!} \mathbb{Z}/d. $$ I think it works out that the terms in the colimit assemble in the obvious way to give $$ \pi_1 K(\mathcal{C}\mathbb{Z}) = \mathbb{Z}/2\oplus\bigoplus_{d>0} \mathbb{Z}/d, $$ and this is infinite, even though all hom sets in $\mathcal{C}\mathbb{Z}$ are finite.

Of course, the monoid of isomorphism classes in $\mathcal{C}\mathbb{Z}$ is infinitely generated. It would be more subtle (or perhaps even impossible?) to find a counterexample without that property?

  • $\begingroup$ I believe that the (generalized) Barratt-Priddy-Quillen would allow you to get an example with finitely generated set of components with the free symmetric monoidal category on an appropriate category (e.g. the coequalizer diagram $\bullet \rightrightarrows\bullet$) $\endgroup$ – Tyler Lawson May 22 '15 at 14:58

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