Timeline for Primitive Ideals of $\mathcal{B}_0(\mathcal{H})^\sim$ (identity adjoined)
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2011 at 20:54 | comment | added | Chao K. | Off-topic : I just noticed that the usual \{ for a curly brace works in the comment but not in the question. | |
| Jul 25, 2011 at 20:43 | history | edited | Chao K. | CC BY-SA 3.0 |
added 7 characters in body; added 1 characters in body; deleted 8 characters in body
|
| Jul 25, 2011 at 20:37 | comment | added | Chao K. | Oh, yeah! Thank you very much. I somehow only thought of irreps on the same Hilbert space $\mathcal{H}$ and later came to a conclusion that $\mathcal{H}$ must be one-dimensional. So, let's say $x = \{0\}$, $y = A$ and we now have the primitive ideal space $\{x,y\}$. The space is not even $T_1$ since $\clos(\{x\}) = \{x, y\}$. However, the primitive ideal space of $A$ is $\{x\}$. According to pm′s comment, the primitive ideal space of $A^\sim$ is supposed to be the one−point compactification of $\{x\}$ which is $\{x, \infty\}$ with the discrete topology, which is $T_1$. What's wrong here? | |
| Jul 25, 2011 at 13:12 | comment | added | Matthew Daws |
To fix notation, let $A^\sim = \{ (a,t) : a\in A, t\in\mathbb C\}$. Then the irrep you need is $A^\sim\rightarrow\mathbb C; (a,t) \mapsto t$. This is a homomorphism, and has kernel $A$. It's irreducible, as, well, it's non-zero, and $\mathbb C$ is one-dimensional! So your argument is wrong when you claim "This again contradicts the irreducibility".
|
|
| Jul 22, 2011 at 22:46 | comment | added | Chao K. | Say, the algebra is $A$. Then $\pi(I) \in \pi(A)'$ and by irreducibility, $\pi(I) = cI$. Since $\pi(I) = \pi(I)^2$, $c$ must be either 0,1. But $c=0$ contradicts the irreducibility of $\pi$. So, $\pi(I) = I$ and hence any irrep with $\mathcal{B}_0(\mathcal{H})$ as its kernel, would be zero on $\mathcal{B}_0(\mathcal{H})$ and map $I$ to $I$. This again contradicts the irreducibility. Thus, there is no irrep with kernel $\mathcal{B}_0(\mathcal{H})$. I might have misunderstood some big point. | |
| Jul 22, 2011 at 16:38 | comment | added | Matthew Daws | Please read the FAQ on how to ask: why do you think the answer should be $\{\{0\}\}$. Is the confusion over the terms "ideal" or "primitive"?? | |
| Jul 22, 2011 at 7:48 | comment | added | Marc Palm | Adjoining a unit correspond to one point compactifiction of the primitive ideal space. | |
| Jul 22, 2011 at 7:44 | history | edited | Chao K. | CC BY-SA 3.0 |
had trouble typing braces
|
| Jul 22, 2011 at 7:33 | history | edited | Chao K. | CC BY-SA 3.0 |
added 16 characters in body; deleted 16 characters in body
|
| Jul 22, 2011 at 7:20 | history | asked | Chao K. | CC BY-SA 3.0 |