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We know that if $A$ is a separable $C^{*}$-algebra then $K_1(A)$ is countable.

Can anybody give an example of a C*-algebra for which $K_1(A)$ is uncountable?

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2 Answers 2

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There must be tons of ways to do this, but a simple one is to start with an uncountable set $X$, equipped with the discrete topology, and consider $c_0(X)$. There are uncountably many pairwise inequivalent minimal projections in this algebra, so its $K_0$ group is uncountable. Now use $K_0(c_0(X)) \cong K_1(Sc_0(X))$ where $SA$ is the suspension of $A$.

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Nik's answer nails it but if you prefer something representable on a separable Hilbert space then you may consider the suspension $SM$ of any ${\rm II}_1$-factor $M$. Indeed, as $M$ is tracial, $K_0(M) \cong \mathbb R$ and the suspension simply reverses the $K$-groups.

Actually you may produce further commutative examples that are representable on separable Hilbert spaces: for example $S\ell_\infty$, since $K_0(\ell_\infty)$ comprises $\mathbb{Z}$-valued continuous functions on $\beta \mathbb N$.

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    $\begingroup$ Very nice answer! $\endgroup$
    – Nik Weaver
    Jul 10, 2020 at 19:51
  • $\begingroup$ Excuse me Sir but in the first paragraph, we $K_1(M) = 0$. Am I wrong? @NikWeaver $\endgroup$ Jul 11, 2020 at 8:33
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    $\begingroup$ @PegLegScott, yes, but $K_1(SM) = K_0(M) = \mathbb R$. $\endgroup$ Jul 11, 2020 at 9:39
  • $\begingroup$ oh, I see now. You mean $K_1(SM)$ as an example. $\endgroup$ Jul 11, 2020 at 9:41

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