Tag Info

Hot answers tagged

3

If $A$ is any unital C*-algebra, the hypotheses are satisfied with $C=A$ and $B=\mathbb C1$.


2

No. For example, $S_3$ contains a characteristic, abelian subgroup not contained in the centre of the group. Both of the automorphisms of this subgroup extend to (inner) automorphisms $G$.


2

1: On a quick look, I don't immediately see how to show that $\psi$ is contractive in general. But note that when $A$ is unital so is $\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case. 2: When you prove that $\psi$ is positive, you don't need to use that $A$ is $A$, just that it is a ...


1

Martini's answer is the canonical one. But in this case one can check things explicitly: We can assume that $A$ is unital, as we can always work on the unitization (we actually need it to define the spectrum). If $p=I$ of $p=0$, it is straightforward to check that the spectra are respectively $\{1\}$ and $\{0\}$. For non-trivial $p$: If $\lambda=0$, $p$ ...


1

We can say that $\sigma(p) \subseteq \{0,1\}$. $p - p^2 = 0$, and hence, by the spectral theorem $$ \{0\} = \sigma(0) = \{\lambda - \lambda^2 \mid \lambda \in \sigma(p) \}$$


1

I think it is simply this fact: take $x\in\ker\pi$. From the relation in the first paragraph of the proof, $$ x=\lim_n\Psi_n(\pi(x))=\lim_n\Psi(0)=0. $$ So $\pi$ is an isometric epimorphism, and so an isomorphism.


1

This looks like an "associativity" property. Both algebras live in $B(H_B\otimes H_A\otimes \ell^2(\Gamma))$, so the topology is the same. At the pre-closure level, you should convince yourself that $C_c(\Gamma,B\odot A)$ and $B\odot C_c(\Gamma,A)$ are equal, and that their closures give the two algebras you want to consider.



Only top voted, non community-wiki answers of a minimum length are eligible