bio  website  www2.unine.ch/alain.valette 

location  Neuchâtel, Switzerland  
age  56  
visits  member for  3 years, 8 months 
seen  3 hours ago  
stats  profile views  4,565 
19h

answered  What is a “Ramanujan Graph”? 
Dec 22 
comment 
Murray–von Neumann equivalence on C$^*$algebra and von Neumann algebra
Sorry you still didn't clarify what you mean by "being a subgroup of $\mathbb{R}$". Remember that, as a consequence of BaumConnes 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 0dimensional part of $K_0$. Although I know of no example, it might be that $K_0(A)$ has nontrivial torsion, which is of course killed by the Chern character, and that would prevent $K_0(A)$ to embed into $\mathbb{R}$. 
Dec 20 
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Murray–von Neumann equivalence on C$^*$algebra and von Neumann algebra
For $C^*$algebras, you may use either idempotents or projections, you get the same group $K_0$. 
Dec 20 
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Murray–von Neumann equivalence on C$^*$algebra and von Neumann algebra
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. 
Dec 20 
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Murray–von Neumann equivalence on C$^*$algebra and von Neumann algebra
For $\Gamma$ torsionfree, the KaplanskyKadison conjecture (proved e.g. for aTmenable 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". 
Nov 15 
awarded  Custodian 
Nov 15 
reviewed  Approve vectorbundles tag wiki 
Nov 14 
awarded  Necromancer 
Nov 10 
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Counterexample for closed graph theorem in unmetrizable case
In Proposition 4.13 and Example 4.14 of that paper: arxiv.org/pdf/math/0612398.pdf you have a very explicit counterexample to the open mapping theorem, where $X$ is a (nonseparable) Hilbert space, $Y$ is locally convex, and $A:X\rightarrow Y$ is linear continuous, bijective, with noncontinuous inverse. 
Nov 10 
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Counterexample for closed graph theorem in unmetrizable case
Robert, you shot first! ;) 
Nov 2 
awarded  Nice Question 
Nov 2 
accepted  CantorBernstein for quasiisometric embeddings? 
Nov 1 
asked  CantorBernstein for quasiisometric embeddings? 
Nov 1 
answered  About the roots of the matching polynomial 
Oct 11 
revised 
MilnorWolf result on growth of solvable groups
Corrected 2 typos 
Sep 30 
awarded  Explainer 
Sep 26 
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Morita Equivalence of Full Corners in $C^*$algebras
The answer to (2) is yes, as $X=pA$ is an imprimitivity bimodule between $B$ and $A$. So if $E$ is a (left) projective finite type module over $A$, then $X\otimes_A E$ is projective finite type over $B$. 
Sep 26 
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Basics on lattice in classical groups
The determinant maps $GL_n(\mathbb{Z})$ to $\{\pm 1\}$, so it is not a lattice in $GL_n(\mathbb{R})$. About references: as a beginner you may enjoy the book by Dave WitteMorris: people.uleth.ca/%7Edave.morris/books/IntroArithGroups.html 
Sep 20 
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Separability of the C*algebra in the definition of Khomology
I completely agree with Paul. One difficulty of working with dual algebras is illustrated by my old paper: projecteuclid.org/download/pdf_1/euclid.pjm/1102720214 in which I tried to get a new proof of the PimsnerVoiculescu 6terms exact sequence... but I could only get a 5terms sequence! 
Sep 7 
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KadisonSinger problem
If you read french, please consider having a look at: bourbaki.ens.fr/TEXTES/1088.pdf 