1
$\begingroup$

from Theorem 5.6.2 [Murphy-C* algebras and operator theory]:

let (H , φ) be an non zero irreducible representation of C* algebra A, and let I be a closed ideal for A. Denote (H',φ') the restriction of (H , φ) to I. Given A/I and I are postliminal algebras, and I is not contained in Ker(φ):

there are two things i find confusing:

1) Why (H',φ') is a non zero representation?

2) When b∊ I, why is φ(b) compact iff φ'(b) is compact?

Thank you for any help!

$\endgroup$

1 Answer 1

0
$\begingroup$

As far as I can tell,

  1. If $(H,\varphi')$ is zero, then $\varphi'(b)=0$ for all $b\in I$, so $I\subset\ker\varphi$.

  2. the compactness of an operator does not depend on its "complement". And you have $\varphi(b)=\varphi'(b)\oplus 0$ (from $H=H'\oplus(H')^\perp$).

$\endgroup$
2
  • $\begingroup$ Thank you Martin. But i still dont understand something, φ′(b) is defined as the restriction of φ(b) to [φ(I)H]. So why should it provide φ(b)((H′)⊥) = {0} ? thx $\endgroup$ Mar 6, 2019 at 22:47
  • $\begingroup$ Take any $x\in (H')^\perp$, and $y\in H$. Then $$\langle \varphi(b)x,y\rangle=\langle x,\varphi(b^*)y\rangle=0.$$ As this works for all $y\in H$, $\varphi(b)x=0$. $\endgroup$ Mar 6, 2019 at 23:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .