3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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". $\endgroup$ – Alain Valette Dec 20 '14 at 22:04
  • 1
    $\begingroup$ 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. $\endgroup$ – Alain Valette Dec 20 '14 at 22:08
  • 1
    $\begingroup$ For $C^*$-algebras, you may use either idempotents or projections, you get the same group $K_0$. $\endgroup$ – Alain Valette Dec 20 '14 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 $\mathbb{R}$. $\endgroup$ – Sebastien Palcoux Dec 21 '14 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 $\mathbb{R}$. $\endgroup$ – Alain Valette Dec 22 '14 at 10:42
4
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ Thank you! Do you think the answer is also no for the example in application? $\endgroup$ – Sebastien Palcoux Dec 20 '14 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 '14 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? $\endgroup$ – Sebastien Palcoux Feb 15 '16 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.