7,044 reputation
2040
bio website www2.unine.ch/alain.valette
location Neuchâtel, Switzerland
age 56
visits member for 3 years, 8 months
seen 3 hours ago

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 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 $\mathbb{R}$.
Dec
20
comment 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
comment 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
comment Murray–von Neumann equivalence on C$^*$-algebra and von Neumann algebra
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".
Nov
15
awarded  Custodian
Nov
15
reviewed Approve vector-bundles tag wiki
Nov
14
awarded  Necromancer
Nov
10
comment 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 counter-example to the open mapping theorem, where $X$ is a (non-separable) Hilbert space, $Y$ is locally convex, and $A:X\rightarrow Y$ is linear continuous, bijective, with non-continuous inverse.
Nov
10
comment Counterexample for closed graph theorem in unmetrizable case
Robert, you shot first! ;-)
Nov
2
awarded  Nice Question
Nov
2
accepted Cantor-Bernstein for quasi-isometric embeddings?
Nov
1
asked Cantor-Bernstein for quasi-isometric embeddings?
Nov
1
answered About the roots of the matching polynomial
Oct
11
revised Milnor-Wolf result on growth of solvable groups
Corrected 2 typos
Sep
30
awarded  Explainer
Sep
26
comment 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
comment 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 Witte-Morris: people.uleth.ca/%7Edave.morris/books/IntroArithGroups.html
Sep
20
comment Separability of the C*-algebra in the definition of K-homology
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 Pimsner-Voiculescu 6-terms exact sequence... but I could only get a 5-terms sequence!
Sep
7
comment Kadison-Singer problem
If you read french, please consider having a look at: bourbaki.ens.fr/TEXTES/1088.pdf