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One can prove that $p,q$ are Murray -von-Neumann equivalent if they fulfill $$\|aa^*-p\|<\frac 12,\quad \|a^*a-q\|<\frac 12.$$ The proof is standard (for example it is outlined in the comment of @ougao), I was just lucky to find $\|(1/2-p)^{-1}\|=2$. We first assume that the $C^*$-algebra $A$ is unital. First note that $(1-2p)^2=1$, hence ...

3

I had forgotten about this question until Jonas Meyer put a bounty on it. In hindsight, I think I now know how to solve parts of it... I'll write down what I've got and maybe someone can give a complete argument building on it. I'll address the case $1 \in A$ first since it is somewhat simpler. The argument does not seem to be overly difficult, but it is a ...

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Inspired by George Lowther's solution, here is another more elementary example of a (nonunital) C*-algebra $A$ such that $A$ has no nonzero projections, but $M_2(A)$ does have nonzero projections. Let $D$ be the sub algebra of $M_2(\mathbb{C})$ consisting of scalar multiples of the identity. Let $C$ be the sub algebra of $M_2(\mathbb{C})$ consisting of ...

3

The answer to question 2 is negative. It is possible that there are projections in $M_\infty(A)$ which are not equivalent to any projection in $A$, even when $A$ is a $C^*$-algebra (so, representable) and spacious. To do this, I will construct a (non-unitial) $C^*$-algebra which has no non-zero projections, so is trivially spacious. However, $M_2(A)$ will ...

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Yes, $$\mathbb C\,I=\{\lambda\,I:\ \lambda\in\mathbb C\}.$$ It is coherent with the notation $$AB=\{ab:\ a\in A,\ b\in B\},$$ and with $$A+B=\{a+b:\ a\in A,\ b\in B\},$$ etc.

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Oh well, how silly of me not to have thought of the following argument sooner. In what follows, $(e_{i})_{i \in I}$ is a self-adjoint approximate identity of $A$ that is bounded in norm by $1$. It is $C^{*}$-folklore that such an approximate identity exists. Lemma. If $(x_{i})_{i \in I}$ is a convergent net in $A$ whose limit is $x$, ...

2

If $A$ is non-unital, then the spectrum of an element of $A$ is defined via the unitization $A^{+}$ of $A$. Hence, $\lambda \in \sigma(a)$ if and only if $(a,- \lambda) \in A^{+}$ is not invertible. It follows readily that $0 \in \sigma(a)$ for any $a \in A$. If $0 \notin \sigma(a)$, then $(a,0_{\Bbb{C}})$ would be invertible in $A^{+} ... 2 The exponential of an idempotent$R$is very easy to calculate: if$R^2=R$, then $$e^{i\pi R}=\sum_{k=0}^\infty\frac{i^k\,\pi^k\,R^k}{k!} =I+\sum_{k=1}^\infty\frac{i^k\,\pi^k\,R}{k!} =I-R+R\left(\sum_{k=0}^\infty\frac{i^k\pi^k}{k!}\right)\\ =I-R+e^{i\pi}R=I-2R.$$ Then $$U(e^{i\theta})=e^{i\pi E(e^{i\theta})}e^{i\pi F(e^{i\theta})} ... 2 In what follows, we assume that G is a locally compact Hausdorff group that does not have to be abelian. You can certainly define an involution ^{*} on {L^{1}}(G) by$$ \forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}), $$where \Delta denotes the modular ... 2 Edit: Many thanks to Martin for bringing to my attention a flaw in my earlier argument. Here is an entirely new argument. In what follows, n \in \Bbb{N} and [n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n} . Let P_{1},\ldots,P_{n} be projections in B(\mathcal{H}) . Let \lambda_{1},\ldots,\lambda_{n} \in \Bbb{R}_{> 0} satisfy ... 2 I'm not sure if this approach works, but it looks too complicated. The usual way to show what the universal algebra is, is to show that you have the *-homomorphism onto any C^*-algebra with the desired generators and relations. In this case, if x_1,\ldots,x_n\in B(H) are as you postulate, you want to construct a *-homomorphism \rho:C(S^n)\to ... 2 For a\in A,$$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$Which shows that \phi is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if \{u_i\} is approximate unit of C*-algebra A, then \|\phi\|=\lim \phi(u_i). Using this we have$$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ... 2 Here is my own attempt on the problem, which turns out to be essentially San’s argument above (I only realized this when writing my response). I would therefore like to focus on demystifying the continuous functional calculus that takes place in the argument and on making portions of it more transparent. All credit is due to San, of course, for posting the ... 2 Edit: previous answer was wrong. The assertion$\|KT_n\|\to0$does not hold in general. Let$H=\ell^2(\mathbb N)$, and $$T_n(a_1,a_2,\ldots,)=(a_n,a_{n+1},\ldots).$$ Then$T_n\to0$in the strong operator topology. Consider the rank-one operator$P$given by$P(a_1,a_2,\ldots)=(a_1,0,0,\ldots)$. Then $$PT_n(a_1,a_2,\ldots)=(a_n,0,0,\ldots).$$ If ... 1 Yes, your argument is the standard way of proving it. 1 As already mentioned, $$(\spadesuit) \qquad C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} \cong \{ f \in C([0,1] \to {\text{M}_{2}}(\Bbb{C})) \mid \text{ f(0) and f(1) are diagonal} \}.$$ Hence, by the definition of a continuous field of$ C^{*} $-algebras,$ C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} $is a continuous field of$ C^{*} $-algebras over the ... 1 The whole point of the multiplier algebra is that you should be able to multiply. So$\varphi(a)b$has to make sense. If you look at page 39 in Murphy's book, you see that the definition of$M(I)$includes a canonical identification of$I$as an ideal on$M(I)$. As Phoenix mentioned,$\varphi$extends the inclusion$I\to M(I)$. So ... 1 Because you are trying to prove (first) that$\phi$is unbounded on$A^+$. You do this because you want to use that your map is positive, so it makes sense to work on the positive part of$A$. The summands are positive. Because$\|\phi(p)\|$would be an upper bound for the natural numbers. 1 The Axiom Schema of Separation tells us that if$ A $is a set and$ \varphi $is a formula with a free variable, then the collection $$\{ x \mid x \in A ~ \text{and} ~ \varphi[x] \}$$ is a set. Now, fix both$ (G,A,\alpha) $and$ f \in {C_{c}}(G,A) $, and let$ \varphi[r] $denote the sentence There exists a covariant representation$ (\pi,U) $of$ ...

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A $C^*$-algebra $A$ is isometric to an $L_1$-space (even as Banach space!) iff it is one dimensional. Assume $A$ is isometric to $L_1$ space and $\operatorname{dim}(A)>1$, then $A$ is weakly sequentially complete. By result of Sakai (proposition 2), this is possible only if $A$ is finite dimensional. By classification theorem for $C^*$ algebras we know ...

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It is almost never true that $\overline{aAa}$ is generated by $a$. Note that $a$ is positive, so the algebra it generates is abelian. For the second part of the argument (which is irrelevant to the exercise, but let me address it for your benefit): your write "$\lim\sigma(p_n(a))$"; what do you mean by the limit of a sequence of sets? Also, when you ...

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I don't think you can say anything. For example, let $A_1=K(H)\subset B(H)$ and $A_2\subset B(H)$ be an image of the Cuntz algebra $O_2$ via an irreducible representation; this representation is faithful, because $O_2$ is simple. The intersection $A_1\cap A_2$ is $0$, because $O_2$ has no compact operators. Now form $A=A_1\oplus A_2$, and let $\pi_j:A\to ... 1 If$A$is non-unital, then by definition $$\sigma_A(a) = \sigma_{A_1}(a)$$ where$A_1$is the unitization of$A$, and necessarily one has that$0\in \sigma_A(a)$. One can then define $$I = \{f \in C(\sigma_A(a)) : f(0) = 0\}$$ and obtain a homomorphism$\varphi : I \to A$as you have mentioned. However,$I \neq C_0(\sigma_A(a))$in the sense of ... 1 By continuity of$p, q :X \rightarrow M_n \mathbb(C)$, and (ii), and the total disconnectedess of$X$, find a partition of X into clopen sets$X_1,\cdots, X_k$and complex matrix$v_1,\cdots,v_k$such that$\|v_i^*v_i - p(x)\|<1$and$\|v_iv_i^* - q(x)\|<1$for all$x \in X_i$. Now, define the map$f: X \rightarrow M_n (C)$,$f(x) = v_i$,$x \in ...

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Not exactly. You only need to omit the term with $k=l=0$. As for the unitization yes, if you algebra $B$ is embedded in a unital algebra $A$, then the unitization of $B$ is isomorphic to the subalgebra of $A$ generated by $B$ and $e$. $\sigma_V(v)$ is compact regardless; it is just defined as a the spectrum of $v$ in $V_1$, and this is a compact subset of ...

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In a $C^\ast$ algebra, this is easy. You have $\Vert a \Vert = \Vert a^\ast \Vert$ and (by definition of a $C^\ast$ algebra) you have $$\Vert a^\ast \Vert^2 = \Vert (a^\ast)^\ast a^\ast \Vert = \Vert a \cdot a^\ast\Vert,$$ so that taking $b = \frac{a^\ast}{\Vert a\Vert}$ yields your claim, since the estimate \$\Vert ab \Vert \leq \Vert a \Vert \Vert b ...

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