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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 = ...


1

Observe that $Tr(\rho\ \cdot\ )$ defines a state, say $\omega_\rho$, hence you have the estimate $$|\omega_\rho(U^*[U,O])|\leq\Vert U^*[U,O]\Vert\leq\Vert[U,O]\Vert.$$


2

$$\mathrm{Tr}[\rho U^{\ast }[U,O]]=<\rho ,U^{\ast }[U,O]> $$ defines a bounded linear functional on the trace class on some separable Hilbert space. Since the dual of the trace class (norm $||..||_{1}$) is the set of bounded operators (norm $||..||$), we have the inequality $$ |<\rho ,U^{\ast }[U,O]>|\leqslant ||\rho ||_{1}||U^{\ast ...


0

There's something off here, because the left side is positively homogeneous in $\rho$ but the right side doesn't depend on $\rho$. Are you missing an assumption on the trace norm of $\rho$?


0

The argument goes as follows. You take $b\in B\otimes\mathbb C1$, with $0\leq b\leq 1$. For any $a\geq0$ in $B\otimes M_k$, $ab=a^{1/2}ba^{1/2}$; this shows that $0\leq ab\leq a$. Then $$\tag{1} 0\leq \rho(ab)\leq\rho(a),\ \ \ a\geq0,\ \ a\in B\otimes M_k. $$ Now define states $f_b$ and $g_b$ on $B\otimes M_k$ by $$ f_b(x)=\frac1{\rho(b)}\,\rho(bx),\ \ \ ...


2

Yes, a morphism in this category is epic iff it is surjective. Obviously surjective morphisms are epic, so it suffices to show that if $f: X\to Y$ is not surjective, there is a locally compact Hausdorff space $Z$ and distinct $g_1, g_2 \in \hom(Y,Z)$ such that $g_1 \circ f = g_2 \circ f$. I will use the notion of a perfect map as defined in Engelking's ...


1

Let $\mathcal{A}$ be unital and $\omega$ be a state. Then $$ 0 \leq \omega(|a - \omega(a){\bf 1}_\mathcal{A}|^2) = \omega(a^*a - a\omega(a^*) - a^* \omega(a) + |\omega(a)|^2{\bf 1}_\mathcal{A}) = \omega(a^*a) - |\omega(a)|^2. $$


3

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 $, ...


0

This is false. Let $$ x_1=x_2=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ y_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}\ \ y_2=\begin{bmatrix}0&0\\0&1\end{bmatrix}. $$ Then $\|x_1\|=\|x_2\|=\|y_1\|=\|y_2\|=1$, but $$ \left\|\begin{bmatrix}x_1& x_2\\0&0\end{bmatrix}\right\| = ...


1

The following categories are contravariantly equivalent: locally compact Hausdorff spaces with proper continuous maps commutative C$^*$-algebras with non-degenerate $*$-homomorphisms Here, a $*$-homomorphism $f : A \to B$ is non-degenerate if the following equivalent conditions are satisfied: The ideal generated by the set-theoretic image of $f$ is ...


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 ...


0

Not sure what you are asking. If $A=C[0,1]$, $I=\{f\in A:\ f(0)=0\}$, and $\tau(f)=\int f$, then the range projection each of each $f$ is not in $A$, nevermind $I$.


0

Use the spectral theorem for normal operators. $H = -i \log(U)$ (for any branch of the logarithm).


3

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.


0

(please write a complete question so this answer makes sense to those reading it) Since $\pi$ is an irreducible representation $A\to M_2(\mathbb C)$, and $A$ is generated by $p$ and $q$, necessarily $\pi(p)$ and $\pi(q)$ generate $M_2(\mathbb C)$. As such, both need to be rank-one projections, because otherwise one of them would be either $0$ or $1$ and ...


3

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 ...


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 ...


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 ...


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 ...


0

Just to make a link with other things: If A were a von Neumann algebra and $\pi$ a normal representation, then its kernel would be an weakly closed two sided ideal. Then there exists a unique central projection $c$ such that $ \mathrm{Ker}(\pi) = c A = A c$ and thus $\pi(A) \cong (1-c) A$. (cf. for example "C*- algebras and W*-algebras" S. Sakai, 1.10.1 ...


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

The answer is no. By Takesaki I, Theorem V.1.41, the von Neumann algebra generated by P and Q is the direct sum of an abelian part and a type $I_2$ part, in which the projections P and Q have the form $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} , \quad Q = \begin{pmatrix} c^2 & cs \\ cs & s^2 \\ \end{pmatrix}, $$ where $c$ and ...


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 $ ...


6

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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