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
    Commented May 22, 2015 at 3:51
  • $\begingroup$ Yes, and I wrote so in the beginning. $\endgroup$ Commented May 22, 2015 at 4:44
  • $\begingroup$ What is the function of the strict associativity requirement here? $\endgroup$ Commented May 22, 2015 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. $\endgroup$ Commented May 22, 2015 at 15:00

1 Answer 1


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 $\bullet \rightrightarrows\bullet$) $\endgroup$ Commented May 22, 2015 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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