# Tag Info

3

You can write the similarity as $NS=SM$. As $N$ and $M$ are normal, the Fuglede-Putnam theorem guarantees that $N^*S=SM^*$. Taking adjoints, $S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly, $p (S^*S)M=Mp (S^*S )$ for all polynomials; taking limits, $f (S^*S)M=Mf (S^*S)$ for all continuous functions $f$. In particular, if ...

3

For an example of a non-proper $*$-morphism that isn't trivial, consider $A=B=\mathbb C\oplus \mathbb C$, and $\phi(a,b)=(a,0)$. For the existence of $\varphi_*$, the key property is that for any $*$-morphism $\varphi:C_0(X)\to C_0(Y)$, there exists $\varphi_*:Y\to X$, continuous, such that $$\tag{1}\varphi(f)(y)=f(\varphi_*(y))\ \ \ \ \text{ for all }y\in ... 2 Define \phi:(B+I)/I\to B/B\cap I by$$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$Of course we need to check that this is well-defined. If b_1+j_1=b_2+j_2, then$$ b_1-b_2=j_2-j_1\in B\cap I, $$so b_1+B\cap I=b_2+B\cap I. The map is obviously linear, multiplicative, *-preserving, and onto. As for injectivity, if b_1+B\cap I=b_2+B\cap I, ... 2 Let's define for a\in A the left multiplication operator and the right multiplication operator$$ L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$As multiplication is assumed to be continuous we get that L_a and R_a are both continuous for all a\in A. Let B=B_1(0,A) denote the unit ball in A. Then ... 2 I don't know why you say that f(\sigma(x))=F_1. A point in \sigma(x) is either in F_1 or in F_2, and so f(x) is either 0 or 1; and then f(\sigma(x))=\{0,1\}. 1 In a C^*-algebra, the norm is algebraic:$$\tag{1}\|a\|=\text{spr}(a^*a)^{1/2}. When $a$ is normal, we know that $C^*(a)\simeq C(\sigma(a))$ via the Gelfand transform, where $a$ is mapped to the identity function. Also $a^*a$ is mapped to the function $f:t\longmapsto |t|^2$, and it follows that ...

1

This is a corollary of the fact that images of $\ast$-homomorphism of $C^\ast$-algebras are closed. More precisely, denote by $\pi$ the canonical projection $A\to A/I$. Then $\pi(B)\subset A/I$ is closed and so is $B+I=\pi^{-1}(\pi(B))\subset A$ as the preimage of a closed set under a continuous map.

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