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I am novice in the algebraic K- theory and don' t know if this is the right place for the following questions. So some people might consider them as basic questions.

Consider an exact monoidal category and its K- groups as introduced by Quillen.

Do $K_n(\mathcal C)$ have a ring structure? Clearly for n=0 the Grothendieck groups have a ring structure induced by the tensor monoidal functor of $\mathcal C$. So bassically the question can be reformulated if the same tensor functor induces also a ring structure in the higher K - groups.

Next natural question would be if a monoidal functor $ F: \mathcal C \rightarrow D$ induces a ring homomorphism at any level of the K- groups, $F^*_n:K_n(\mathcal C)\rightarrow K_n(\mathcal D)$?

I guess these are basic things but unfortunately I don't know any text where they are treated.

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$K_n$ commutes with products. – Martin Brandenburg Apr 22 '13 at 12:58
@ Martin Thank you very much! Does it also follow that any monoidal functor induces a ring morphism? Could you please give me an exact reference for these facts? – Stef Apr 22 '13 at 13:14
Sorry, commuting with the products clearly implies that it induces ring homomorphisms. – Stef Apr 22 '13 at 16:21
up vote 4 down vote accepted

(I'm not sure the term exact monoidal category is a standard one in the literature, so I'll just assume I know what you mean by it.)

Yes, $K_n$ commutes with products, as some have mentioned, but the tensor monoidal functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is not exact, so it doesn't induce a map $ K_n \mathcal C \times K_n \mathcal C \to K_n \mathcal C$. Rather, the functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is bi-exact, in the sense that it is an exact functor in each of its two variables, separately. (That's what makes $K_0 \mathcal C$ into a ring.)

The result is that the collection of abelian groups $ K_n \mathcal C$, indexed by $n$, forms a graded ring, with products $K_m \mathcal C \otimes K_n \mathcal C \to K_{m+n} \mathcal C$. Assuming the tensor product is commutative up to natural isomorphism, the ring is commutative in the graded sense that $ x y = (-1)^{mn} y x$.

The actual construction is intricate, the main idea being to construct a map $K \mathcal C \wedge K \mathcal C \to K \mathcal C$ of spectra. See, for example, Waldhausen's approach in section 9 of Algebraic K-theory of generalized free products. I, II. Ann. of Math. (2) 108 (1978), no. 1, 135–204. It uses the Q-construction of Quillen, but could be simplified by using the S-construction of Segal that Waldhausen developed extensively.

My paper with Gillet, The loop space of the Q-construction, Illinois Journal of Mathematics, 31 (1987) 574-597, gives another approach that avoids both spectra (i.e., delooping) and the phony multiplication trap that Steve mentioned.

PS: My paper Algebraic K-theory via binary complexes, Journal of the American Mathematical Society, 25 (2012) 1149-1167, gives an algebraic description of the elements of the higher K-groups that converts the construction of the product into a simple exercise involving tensor product of chain complexes.

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Thank you very much! That's actually what I needed. – Stef Apr 25 '13 at 5:44

First, see Thomason's paper "Beware the Phony Multiplication on Quillen's $S^{-1}S$" (Proceedings of the AMS, 1990) for why the most obvious construction is wrong. In the bibliography, you'll find references to papers of Loday, Waldhausen, Segal/Wolfson and May with constructions that are more complicated, but correct.

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Nonetheless, the product commutes with the K - groups. What would be a correct reference for this fact? – Stef Apr 22 '13 at 22:10

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