Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $A$ is a C*-algebra whose unitary group is contractible (e.g. $B(H)$ or more generally the stable multiplier algebra of any C*-algebra). It is clear from the definition that $K_1(A) = 0$, but I think it is also true that $K_0(A)$ is necessarily $0$. Is this correct? Is there an elementary proof?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Here is an idea: To show that $K_1(A)=0$ you actually only use that $U(A)$ is path connected. Now it suffices to show that $U(\Sigma A)$ (unitary group of suspension) is path connected if $U(A)$ is contractible.

share|improve this answer
    
Yeah, that looks like it does the job. I was secretly hoping for something that doesn't involve Bott periodicity, but that might just be too much to hope for. –  Paul Siegel Apr 26 '12 at 22:56

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.