# Tag Info

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If we assume that $\varphi(a^2)=\varphi(a)^2$ for all $a$, then for $a,b$, we have, using linearity and the square-preserving property, $\varphi((a+b))^2=(\varphi(a)+\varphi(b))^2=\varphi(a)^2+\varphi(a)\varphi(b)+\varphi(b)\varphi(a)+\varphi(b)^2.$ Now, observe that $\varphi((a+b)^2)=\varphi(a^2+ab+ba+b^2)=\varphi(a^2)+\varphi(ab+ba)+\varphi(b^2)$. By ...

2

I can't bring myself to write $fg$ for the convolution of $f$ and $g$. So I'm going to write $f\mapsto f'$ for the involution, so I can write $f*g$ for the convolution. Are you certain you got the definition of $f'$ straight? What would make much more sense to me would be $$f'(t)=\overline{f(-t).}$$ That seems to me is the "standard" involution on $L^1$. ...

2

If you represent $C$ in $B (H)$ faithfully, then it is well-known that $$M (C)\simeq\{x\in B (H):\ xc\in C,\ cx\in C,\ \forall c\in C\}.$$

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Consider abelian C*-algebra $C^*(a)$ which is infinite dimensional ( because $\sigma(a)$ is infinite). Also $C^*(a) \subset A$ which implies that $A$ is infinite dimensional.

2

The part of theorem 3.31 that is used here is the equality $(3)$; for closed paths $\Gamma_1,\,\Gamma_2$ that are homologous in $\Omega$ (that is, they have the same winding number around all $w \in \mathbb{C}\setminus \Omega$), and any (weakly) holomorphic function $f$ on $\Omega$ we have $$\int_{\Gamma_1}f(\zeta)\,d\zeta = \int_{\Gamma_2} ... 2 You can repeat the proof with \phi/\|\phi\|, which does satisfy the inequality. If the authors didn't mention that their were sloppy, but the argument still works. What they mean by "hereditary" is that if a,b are orthogonal and positive and a'\leq a, b'\leq b, then a' and b' are orthogonal. One way to see this last assertion is by ... 1 It is easier to see without the clutter notation. What you want to show is that if in some C^*-algebra ab=0 with a\geq0, then a^{1/2}b=0; there is no positivity requirement for b. Here are two proofs: From ab=0, you get$$ (a^{1/2}b)^*a^{1/2}b=b^*ab=0,$$so a^{1/2}b=0. From ab=0, you get a^nb=0 for all n\geq 1, so p(a)b=0 for all ... 1 Approximate Unit Regard ball cone:$$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$Order elements:$$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$Then one has:$$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$(That is the hard part!) Quotient Norm Note that it holds:$$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$... 1 Assume k field and A a finite dimensional k algebra. Then the spectrum of any element is finite. Step 1. Every element a of A satisfies a polynomial equation of degree at most n= \dim_{k} A. Indeed, the elements 1, a, \ldots, a^n are linearly dependent. Step 2. Let a satisfying a polynomial equation P(a) = 0, where P \in k[X] , ... 1 Let \lambda_a \colon b \mapsto a\cdot b. Then it's easy to see that the operator norm of the matrix is at most$$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}. For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$. Still, even if we take the operatornorm of the ...

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You should study Morita equivalence of C*-algebras (in the sense of Rieffel): the fact is that $C(X,M_2)$ is Morita equivalent to $C(X)$. And it is a general fact that Morita equivalent C*-algebras have same closed (two-sided) ideals. It follows that the closed ideals of $C(X,M_2)$ are all of the form $C_0(U,M_2)$ (functions vanishing outside $U$) for ...

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