bio | website | www2.unine.ch/alain.valette |
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location | Neuchâtel, Switzerland | |
age | 56 | |
visits | member for | 3 years, 9 months |
seen | 10 mins ago | |
stats | profile views | 4,670 |
Jan 24 |
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Why is SO(3) not $S^1 \times S^2$? (Where is the mistake?)
An excellent question for Math.SE! |
Jan 19 |
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The kernel of $C^{*}(G)\to C_{r}^{*}(G)$
See "On Isolated Points in the Dual Spaces of Locally Compact Groups", by S. P. Wang, Mathematische Annalen (1975) Volume: 218, page 19-34; available on gdz.sub.uni-goettingen.de/dms/load/img/… |
Jan 17 |
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The kernel of $C^{*}(G)\to C_{r}^{*}(G)$
This is a wide open question. I personally believe that, for $G$ discrete, the question is intractable, because a description of $I(G)$ is very much related to a description of the dual $\hat{G}$. As an example, consider a residually finite group with property (T). It is known that each finite-dimensional rep is isolated in $\hat{G}$, so defines a direct summand of $C^*(G)$ isomorphic to a matrix algebra. So $I(G)$ contains a huge direct sum of matrix algebras, and probably lots of other stuff too. |
Jan 15 |
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Embedding Euclidean buildings into products of trees
You're right, I overlooked the surjectivity issue. I leave my comment however, as a very partial contribution. |
Jan 15 |
awarded | Nice Answer |
Jan 14 |
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Embedding Euclidean buildings into products of trees
I think taking $p=1$ is prevented by the Kleiner-Leeb result on quasi-isometric rigidity of Euclidean buildings: see Theorem 1.1.2 in math.nyu.edu/~bkleiner/symm.pdf |
Jan 7 |
awarded | Enlightened |
Jan 7 |
awarded | Nice Answer |
Jan 7 |
answered | Can the full and reduced group $C^*$-algebras be “noncanonically” isomorphic? |
Jan 5 |
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Are norm-continuous representations smooth?
The 2nd proof is obviously correct, and actually shows that there is a commutative diagram linking $d\rho:\mathfrak{g}\rightarrow A$ and $\rho:G\rightarrow A^{inv}$, where the vertical arrows are given by the exponential maps of $G$ and $A$ respectively. |
Jan 5 |
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MathScinet reviewing: should I check the proofs?
I think we all agree that, as a reviewer for MR, you're not supposed to check the proofs. Trouble may arise if you find a mathematical mistake in the paper, because you are supposed to explain why it is false, or provide a counter-example, and this can result in painfully long and not so informative reviews. |
Jan 5 |
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Is there a manifold with fundamental group $\mathbb{Q}$?
@Igor Belegradek: Is an explicit $n$ known? |
Jan 5 |
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Is this property of the Bell's number evident?
See e.g. that paper for generalizations (and the references within): projecteuclid.org/download/pdf_1/euclid.bbms/1102714984 |
Jan 3 |
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Axiomatic ZFC Set Theory
That's a weird tag... |
Jan 3 |
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Correspondences as generalized morphism between $C^*$-algebras
Ad Q1: Usually homomorphisms are indeed assumed to be *-homomorphisms. Ad Q2: What about taking for $\phi$ the zero homomorphism $A\rightarrow B$? |
Dec 27 |
answered | What is a “Ramanujan Graph”? |
Dec 22 |
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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 |
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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 $\Gamma$ 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". |