Timeline for Fredholm $C^*$-algebras
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 24, 2021 at 8:57 | vote | accept | Ali Taghavi | ||
| Oct 23, 2021 at 10:25 | comment | added | Ali Taghavi | @AlainValette I just realize that the last part of my previous two comment was not necessary since your argument is giving a complete answer to my question. So your argument also gives a complete answer to the whole question of my previous two comment. Thank you again for your attention to my question and your answer. | |
| Oct 22, 2021 at 7:20 | comment | added | Ali Taghavi | Moreover I think that you use the fact that for every liminal $C^*$ algebra every irreducible representation is necessarilly a finit dimensional representation. yes? | |
| Oct 22, 2021 at 7:16 | comment | added | Ali Taghavi | @AlainValette Thank you very much for your answer. Does every $C^*$ algebra admits an irreducible representation $\pi:A \to B(H)$ such that $A/\ker \pi$ is simple? I think we need to this statement to apply your proof. | |
| Oct 21, 2021 at 18:01 | comment | added | YCor | Ah, this proposition boils down from general case to (non-negative) self-adjoint case, for which (B2 in Appendix B) it send to the book Rickart, General theory of Banach algebras. | |
| Oct 21, 2021 at 17:50 | comment | added | Alain Valette | @YCor My favorite reference is Proposition 1.3.10 in Dixmier's $C^*$-book. | |
| Oct 21, 2021 at 17:24 | comment | added | YCor | Ah OK; actually I wasn't aware of this fact that if an element $M$ is invertible in the big $C^*$-algebra then it's invertible in the smaller one, and could only check it ((2) above) under special assumptions. | |
| Oct 21, 2021 at 17:16 | comment | added | Alain Valette | @YCor: denote by $q$ the quotient map to the Calkin algebra. If $x\in\pi(A)$ is Fredholm, then $q(x)$ is invertible in the Calkin algebra, hence also in $q(\pi(A))$. As $q$ is injective on $\pi(A)$, the element $x$ is also invertible in $\pi(A)$. | |
| Oct 21, 2021 at 17:09 | comment | added | YCor | The proof I can think of: (1) if $A\in B(H)$ is Fredholm and has nonzero index, then either $A^*A$ or $A^*A$ is Fredholm, self-adjoint $\ge 0$ and non-invertible. (2) in a $C^*$-algebra, if $M$ is self-adjoint invertible $\ge 0$ then $M^{-1}\in C^*(M)$. Indeed, approximate $x\mapsto x^{-1}$ by polynomials on the spectrum of $M$. (3) If $N$ is self-adjoint $\ge 0$, Fredholm and non-invertible then $N$ is invertible in restriction to the orthogonal of Ker$(N)$, say with inverse $N'$, extended to $0$ on Ker$(N)$, by (2) $N'\in C^*(N)$. So $I-NN'\in C^*(N)$ is nonzero of finite rank. | |
| Oct 21, 2021 at 16:44 | comment | added | YCor | Where is the assertion located in Nik Weaver's answer (or thread of comments)? (anyway I think it's true) | |
| Oct 21, 2021 at 16:12 | history | answered | Alain Valette | CC BY-SA 4.0 |