Let $H$ be a separable infinite dimensional Hilbert space, $M \subset B(H)$ a von Neumann algebra and $A \subset M$ a separable $C^*$-algebra such that $A''=M$.

Let $p,q \in M_{\infty}(A)$ be (Murray–von Neumann) equivalent projections.
Let $U_{p,q} = \{u \in M \otimes B(H) \text{ partial isometry } \vert uu^*=p$, $u^*u=q \}$

Question: Is is true that $U_{p,q} \cap M_{\infty}(A) \neq \emptyset$ ?

Application: let $\Gamma$ a countable torsion-free ICC group, $H=l^2(H)$, $A=C^*_r(\Gamma)$, $M=L\Gamma = A''$.
$M \otimes B(H)$ is a ${\rm II}_\infty$-factor with a trace $Tr$. Two projections $p,q$ are equivalent iff $Tr(p)=Tr(q)$.
If the question above admits a positive answer for $A$: could we deduce that $K_0(A)$ is a subgroup of $\mathbb{R}$?
The problem is that $K_0$ is defined (here p8) for idempotents (more general that projections).

  • $\begingroup$ For $\Gamma$ torsion-free, the Kaplansky-Kadison conjecture (proved e.g. for a-T-menable groups) says that $A=C^*_r\Gamma$ has no projection except 0 and 1. So the answer to your question, in the application, is trivially "yes". $\endgroup$ Dec 20, 2014 at 22:04
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    $\begingroup$ Do you mean "$K_0(A)$ is isomorphic to a subgroup of $\mathbb{R}$"? This seems to be unrelated to the trace. Indeed, for $\Gamma$ a surface group, $K_0(A)=\mathbb{Z}^2$, but $tr_*(K_0(A))=\mathbb{Z}$, as proved by Kasparov in 1983. $\endgroup$ Dec 20, 2014 at 22:08
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    $\begingroup$ For $C^*$-algebras, you may use either idempotents or projections, you get the same group $K_0$. $\endgroup$ Dec 20, 2014 at 22:13
  • $\begingroup$ @AlainValette: Thank you! I've corrected the question. Because $K_0(A) = \mathbb{Z}^2$ for $\Gamma$ a surface group (and so countable torsion-free ICC), this should mean that $\exists p,q \in M_{\infty}(A)$ such that $U_{p,q} \cap M_{\infty}(A) = \emptyset$, because otherwise, we should have $K_0(A)$ a subgroup of $\mathbb{R}$. $\endgroup$ Dec 21, 2014 at 4:21
  • $\begingroup$ Sorry you still didn't clarify what you mean by "being a subgroup of $\mathbb{R}$". Remember that, as a consequence of Baum-Connes and the Chern homomorphism, $K_0(A)\otimes\mathbb{Q}$ is $\oplus_{n\geq 0} H_{2n}(\Gamma,\mathbb{Q})$ - and the trace sees only the 0-dimensional part of $K_0$. Although I know of no example, it might be that $K_0(A)$ has non-trivial torsion, which is of course killed by the Chern character, and that would prevent $K_0(A)$ to embed into $\mathbb{R}$. $\endgroup$ Dec 22, 2014 at 10:42

1 Answer 1


In general the answer is no; here is one example. Let $A=\{\lambda I +K:\lambda\in \mathbb C, \ K \text{ compact }\}$ (the unitization of the compacts in $B(H)$). Then $M=B(H)$. Let $q$ be the identity and $p$ a projection of codimension one. Then every member of $U_{p,q}$ is an isometry which is unitarily equivalent to the unilateral shift, and hence has Fredholm index $-1$. But every Fredholm operator in $A$ has index $0$.

  • $\begingroup$ Thank you! Do you think the answer is also no for the example in application? $\endgroup$ Dec 20, 2014 at 18:09
  • $\begingroup$ It is not quite clear to me what conclusion you are trying to draw in the application. In any case, for any (torsion-free, countable, discrete) group $\Gamma$ for which the Baum-Connes conjecture holds, the reduced $C^*$ algebra $C^*_r \Gamma$ contains no non-trivial projections, so the answer is "yes" in these cases but for a trivial reason. $\endgroup$
    – Mike Jury
    Dec 20, 2014 at 21:41
  • $\begingroup$ (The post has been improved) As Alain wrote in comment above, for $\Gamma$ a surface group (which is countable ICC torsion-free) and $A=C^*_r(\Gamma)$, $K_0(A) = \mathbb{Z}^2 ⊄ℝ$; we can deduce directly that $∃p,q$ as above such that $U_{p,q}∩M_∞(A)=∅$, isn't it? $\endgroup$ Feb 15, 2016 at 10:05

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