MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Eilenberg-Mazur swindle shows that the Grothendieck group of an additive category with countable coproducts is trivial. The strategy is to observe that any "Euler characteristic" $\chi$ on such a category must be zero, because for any object $P$, we have $$P \oplus \bigoplus_{i=1}^\infty P \simeq \bigoplus_{i = 1}^\infty P$$ which implies that $\chi(P) + \chi(Q) = \chi(Q)$ for $Q = \bigoplus_{i=1}^\infty P$.

Is there any analog for the higher $K_i$ (of, say, an exact category in Quillen's sense)? I don't know any simple way of thinking of the higher $K_i$ (e.g. via Euler characteristics), but it would be interesting if, say, the associated K-theory space somehow had to be contractible.

share|cite|improve this question
We'll give a very general answer to this in the class as well. – Clark Barwick Feb 22 '12 at 0:43
up vote 14 down vote accepted

Yes, there is an analogue, see Weibel's book, chapter V, \S 1.9 ( He calls an exact category $A$ "flasque" if there is a functor $\infty: A \to A$ such that $A \coprod \infty(A) \cong \infty (A)$ and proves that the Quillen K-theory space $K(A)$ of a flasque category $A$ is contractible. Your countable coproduct gives an example. Analogous statements hold more generally for Waldhausen categories and Waldhausen K-theory.

Another result that fits into this context is that if $R$ is a ring, the group of ALL automorphisms of $R^{\infty}$ is acylic (Mac Duff, de la Harpe, "Acyclic groups of automorphisms"). This is proven by an Eilenberg-swindle argument as well.

share|cite|improve this answer
Thanks for this answer. – Akhil Mathew Feb 21 '12 at 21:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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