Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Waldhausen K-Theory takes as input a Waldhausen category C and produces a spectrum K(C). I would like to know what is known about generalized (co-) homology theories that can be realized by this spectra. I guess that stable homotopy groups are known to be representable this way? Can you help me to get an overview?

share|improve this question
add comment

2 Answers

Thomason's paper "Symmetric monoidal categories model all connective spectra" claims to show exactly what the title claims - namely, you can model all generalized homology theories E with En(*) = 0 for n < 0 by taking the spectrum associated to a symmetric monoidal category. So in principle, these are what you might feel like you should get.

However, Waldhausen categories are more restrictive - they require that the symmetric monoidal structure is actually the underlying categorical coproduct. I don't know of any results along this line.

It appears that Thomason's proof is something like the category of weakly contractible spaces over X, but - I will be blunt - I have never managed to sort through Thomason's paper. It seems conceivable that the object he constructs might be equivalent to something coming from a Waldhausen category or its opposite, but this might be optimistic.

share|improve this answer
    
Thanks for the nice reference. I think there might be a good chance to build a Waldhausen category that models Thomason's construction. Does anyone know how to fill the gap? –  user2146 Dec 19 '09 at 12:53
add comment

The Waldhausen category C which is the category of finite pointed sets, with cofibrations the monomorphisms and weak equivalences the isomorphisms, has K(C) = the sphere spectrum—or so I am told.

share|improve this answer
    
Oblig: This is called the Barrat-Priddy-Quillen theorem. –  Tyler Lawson Dec 19 '09 at 3:54
add comment

Your Answer

 
discard

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.