bio | website | www2.unine.ch/alain.valette |
---|---|---|
location | Neuchâtel, Switzerland | |
age | 55 | |
visits | member for | 3 years, 6 months |
seen | 8 hours ago | |
stats | profile views | 4,447 |
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 |
Aug 20 |
awarded | Nice Answer |
Jul 2 |
awarded | Curious |
May 29 |
comment |
quantum states and observables for the non-commutative torus
$A_\theta$ can be obtained as the $C^*$-algebra generated by two unitaries on $\ell^2(\mathbb{Z}^2)$ associated with the formulation of the problem of the Bloch electron (describe the motion of a free electron on the square lattice submitted to a uniform magnetic field orthogonal to the lattice). See a famous paper by D. Hofstadter: zimp.zju.edu.cn/~xinwan/qm2/note/… |
Apr 28 |
revised |
Spectral gap of unitary representation
Precision added |
Apr 28 |
answered | Spectral gap of unitary representation |
Apr 18 |
awarded | Yearling |
Feb 20 |
awarded | Enlightened |
Feb 20 |
awarded | Nice Answer |
Jan 25 |
answered | Learning about Lie groups |
Jan 23 |
comment |
Why aren't fields called “bodies” instead?
A remark on local versions of french: the Belgians use "champ" (= field) for "corps commutatif", while the Swiss say as the French... |
Jan 19 |
comment |
A question on lie groups( Lie algebras)
Take Sasha's example, and embed it into the simple Lie group $SU(3)$ (where $\mathbb{T}^2$ embeds as a maximal torus). |
Jan 12 |
answered | Modern Mathematical Achievements Accessible to Undergraduates |
Jan 8 |
comment |
Reflexive (hyperbolic) graphs
I see. Thanks for clarifying. |
Jan 8 |
comment |
Reflexive (hyperbolic) graphs
Why are you using "hyperbolic" in the title? (for me, hyperbolic graphs are graphs which are hyperbolic when viewed as metric spaces, which seems irrelevant to your question...) |