Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A note I'm studying says its primitive ideals space is $\lbrace \lbrace 0 \rbrace, \mathcal{B}_0(\mathcal{H}) \rbrace $. I think it might be just $\lbrace \lbrace 0 \rbrace \rbrace $ so I'm somewhat confused now. Anyone please help clarify?

share|improve this question
    
Adjoining a unit correspond to one point compactifiction of the primitive ideal space. –  Marc Palm Jul 22 '11 at 7:48
1  
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"?? –  Matthew Daws Jul 22 '11 at 16:38
    
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. –  Chao K. Jul 22 '11 at 22:46
    
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". –  Matthew Daws Jul 25 '11 at 13:12
    
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? –  Chao K. Jul 25 '11 at 20:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.