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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?

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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.

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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

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