# Tag Info

5

An idempotent $e$ is always equivalent to $e^*$. On the unitization of $A$, let $z=1+(e-e^*)^*(e-e^*)$. This is positive and invertible. By noticing that $z=1-e-e^*+ee^*+e^*e$, it is clear that $z$ commutes with $e$ and $e^*$, and a fortiori so does $z^{-1}$. Let $x=ee^*\in A$, $y=e^*z^{-1}e\in A$ (recall that $A$ is an ideal in its unitization). Note ...

4

Suppose $G$ and $H$ are abelian. Then their group C*-algebras are isomorphic iff their Pontryagin duals are homeomorphic. So take, for example, $G$ to be the Prufer 2-group and $H$ to be the Prufer 3-group. Their Pontryagin duals are the 2-adic and 3-adic integers, which are homeomorphic but not isomorphic even as abstract groups. I'm not sure off the top of ...

3

I recently saw it asserted that the second dual of $C[0,1]$ was $\mathcal L^\infty[0,1]$, the space of bounded Borel functions. I was surprised at this, tried to prove it, couldn't quite make the details work. Finally saw a simple proof it was not so. (See the Amusing Note below regarding what $\mathcal L^\infty[0,1]$ actually is.) Started to post something ...

3

Let $A$ be a simple GCR C*-algebra and $\pi:A\to B(H)$ be an irreducible C*-algebra. Since $A$ is simple then $\pi$ is one to one and so it is an isometry. Since $A$ is GCR then $K(H)\subseteq \pi(A)$. Clearly $K(H)$ is a closed ideal in $\pi(A)$ , so if it were proper in $\pi(A)$ then $\pi^{-1}(K(H))$ forms a non-trivial closed ideal in $A$ (since $\pi$ ...

3

This is far from being true, indeed, every symmetric operator has only real eigenvalues: If $\psi\in\ker(T-\lambda),\,\psi\neq 0$, then $$\lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2,$$ hence $\lambda=\bar\lambda$. Now every symmetric operator that is not self-adjoint yields a counterexample to your ...

3

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$\phi(T)+\|T||\in\mathbb R,$$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get ...

2

In my experience, the topic of subalgebras of $M_n(\mathbb C)$ is not part of the usual linear algebra curriculum. What you need to understand first is the form that finite-dimensional C$^*$-algebras have. A finite-dimensional C$^*$-algebra $A$ is always a finite direct sum $$\bigoplus_{k=1}^m M_{n(k)}(\mathbb C).$$ The "blocks" can be identified via the ...

2

Identifying $A$ as a $C^*$-subalgebra of $B(\mathcal H)$ for some Hilbert space $\mathcal H$, and hence identifying $M_n(A)$ as a $*$-subalgebra of $M_n(B(\mathcal H))$, observe that $\sum_{i,j}y_i^*x_{ij}y_j$ can be identified as an operator in $B(\mathcal H)$. So the question boils down to prove that if $\sum_{i,j}y_i^*x_{ij}y_j\in B(\mathcal H)$ is ...

2

First, we establish a simple result. Suppose we have two operators $A,B$ with $[A,B]=1$. Then one can easily prove via induction that more generally $[A,B^k]=k B^{k-1}$ for any natural $k$. Therefore the map $d_B$ defined as $d_B(\cdot)=[A,\cdot]$ acts as a derivative on all such terms $B^k$, and since this map is linear we furthermore have that this extends ...

1

For every normed space $X$, the dual $X^*$ is $1$-complemented in $X^{***}$. Indeed, let $i:X\to X^{**}$ be the canonical embedding; then its adjoint $i^*$ is a projection of norm $1$ of $X^{***}$ to $X^*$. Simply put, it takes a functional $\phi:X^{**}\to \mathbb{C}$ and composes it with $i$. In particular, the above applies to $\mathbb{B}(\mathbb{H})$, ...

1

Assume for simplicity that $A$ is unital. If you had that $K=H^{(n)}$, then you know that the projections onto each copy of the direct sum would be $E_{jj}$, where $E_{kj}$ denote the canonical matrix units in $M_n(A)$ (this where "unital" is useful, but it can be dispensed with). Also, your last formula suggests what you need to do: you can define ...

1

Because $x$ commutes with $e$, then $e$ commutes with $E$ as well, including with $f=E_x(\mathbb{R}\setminus\{0\})$. Because $fx=xf=x$ and $fe=ef$, then $ef$ is also a projection such that $(ef)x=x(ef)=x$. Therefore $ef=fe=e$ follows from the minimality of $e$. Furthermore $x(f-e)=0$, which puts the range of $f-e$ in $\mathcal{N}(x)=\mathcal{R}(E_x\{0\})$; ...

1

I can't blame you, the proof is written horribly (let's hope Vaughan doesn't read me on this...) First you have, from 4.1.3, that $e_{n+1}s_ne_{n+1}=E_{A_{n-1}}(s_n)e_{n+1}$. From 4.2.1 you have $ws_n=w$ for all $w\in A_{n-1}$, and so you deduce that $E_{A_{n-1}}(s_n)\in Z(A_{n-1})$. Also from 4.2.1 you know that $1-s_{n-1}$ is a minimal projection in ...

1

Yes, it is. Condition $u^*u\ge uu^*$ is called hyponormality. Proposition: Let $M$ a subspace of $H$ and $T\in B(H)$. If $M$ is $T$-invariant, then $(T|_M)^*=PT^*|_M$, where $P$ is the orthogonal projection onto $M$. Now, if $S$ is subnormal in $H$, there is a superspace $K\ge H$ and a normal operator $N\in B(K)$ such that $N|_H=S$ and $H$ is ...

1

As Murphy says, the unit ball in $B(H)$ is separable and metrizable in the strong topology (see Remark 4.4.2 on pages 133-4 for details). Since a subspace of a separable metric space is separable, the unit ball in $A$ is separable in the strong topology. Now just take a countable strongly dense subset of the unit ball of $A$ and let $B$ be the C*-algebra ...

1

I detail the way WishBeLeibniz proposes. Since $U$ is normal, there is a unitary $R$ s.t. $R^*UR=diag(\lambda_j)$. Since $U$ is unitary, $\lambda_j=e^{i\theta_j}$ where $\theta_j\in\mathbb{R}$; then $R^*UR=\exp(idiag(\theta_j))$ and $U=\exp(iRdiag(\theta_j)R^*)$; finally, $H=Rdiag(\theta_j)R^*$ is hermitian and satisfies $U=e^{iH}$. Of course, $H$ is not ...

1

An irreducible representation $\pi:M\to B(K)$ has range dense in $B(K)$; but, if $\pi$ is normal, the range is all of $B(K)$. As $\ker\pi$ is a weakly closed ideal of $M$, it is of the form $pM$ for some central projection $p\in M$, and we get $(1-p)M\simeq B(K)$. In other words, $M$ has sufficiently many normal irreducible representations if and only if ...

1

What you do is to define $\phi$ by linearity on $$\mathcal M_\phi=\text {span}\,\{a\in A^+:\ \phi (a)<\infty\}.$$ Then the inequality from your comment shows that $b^*a\in \mathcal M_\phi$ whenever $a,b\in\mathcal N_\phi$.

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