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1

First of all, see this question. If $Tx=xT=0$ for every $x\in I$ then for every $h\in H$ we have $Txh=0$. Now the set $\{xh,\ x\in I, h\in H\}$ is dense in $H$, since the representation is non-degenerate by the question above. Since $\ker(T)$ is closed in $H$ and contains $\{xh,\ x\in I, h\in H\}$, which is dense in $H$ then $\ker(T)=H$ and $T=0$.


5

By using the polar decomposition, we can write $T=V|T|$. So $|T|=V^*T\in J$, and then $J$ contains a positive non-compact operator. On a side note, this argument also shows that $J$ contains all adjoints of its operators, since now $T^*=|T|V^*\in J$. So from now on we assume $T\geq0$, non-compact, $T\in J$. This means that there is $\lambda>0$ with ...


0

Answered in a comment by 1015: Yes, closed ideals are hereditary. So your argument works.


3

The normal functionals of $B(H)$ can be identified with the elements of the predual of $B(H)$, which are the trace-class operators $T(H)$, via the duality $$ T(H)\ni X\longmapsto \text{Tr}(X\ \cdot). $$ So the question is why $T(H)$ is weakly dense in $B(H)$. There are several ways of proving this. The easiest is to notice that $T(H)$ contains all ...


1

It is true for the min-norm, because you can construct the minimal tensor product explicitly by using two faithful representations. Namely, if you fix two faithful representations $\pi:A\to B(H)$ and $\mu:B\to B(K)$, then you have $$ A\otimes_\min B=\overline{\pi(A)\otimes\mu(B)}\subset B(H\otimes K). $$ Now you can use the restrictions $\pi|_{A_1}$ and ...


1

The second statement in Martin's answer is not quite correct - the Cantor space is totally disconnected/zero dimensional but NOT extremally disconnected/stonean. To see this, let us first be clear about our representation of the Cantor space $C$ as, say, real numbers in the interval $[0,1]$ that have ternary expansions without $1$'s. Take a point $x\in C$ ...


5

You can do this in any II$_1$-factor. Note that $Q_j$ is a II$_1$-factor. In any II$_1$-factor $M$ you can always get a sequence of pairwise orthogonal projections that add to the identity (those could be the $q_j$ in your setup). So now you want to embed $M_n(\mathbb C)\hookrightarrow q_jMq_j$. Since you are in a II$_1$-factor, you can divide $q_j$ as a ...


2

Yes. The tensor product of separable algebras is separable. You can construct a dense subset of the tensor product by taking the algebraic tensor of two countable dense subsets of each algebra.


2

The situation is the following: you have $U=\lambda I+T$, with $T$ compact. And $\{P_M\}$ is a sequence of finite-rank projections such that $P_M\nearrow I$. So, given $\varepsilon>0$, by the compactness of $T$ you can write $T=P_MTP_M+T_0$, with $\|T_0\|<\varepsilon$. Then $$ \|P_MU-UP_M\|=\|P_MT_0-T_0P_M\|<2\varepsilon. $$ For your second ...


1

Reposting my comment: in order for the map you describe to be defined it seems like you at least need $D(A)$ to contain $A$ (I don't see the point of the fraktur here; most of the time it just makes things harder to read). I don't see why this needs to hold if $A$ is noncommutative.


2

No. It wouldn't make sense if it were independent of the algebra. The properties you mention depend on the relation of the identity with the rest of the projections. For example, $M_2(\mathbb C)$ is finite because $I$ is not equivalent to any proper projection (because equivalence of projections is given by rank). On the other hand, on ...


0

As with many concepts, I think it's hard to provide a precise starting point because there are many different more-or-less closely related formulations at different times. Basically, the theory of C^* dynamical systems comes from Quantum Mechanics. The Hilbert space formulation of this was developed by von Neumann in the 1930's. But when you want to describe ...


1

I don't think you can get too far with your approach, because you want to deal with the set of all measures on $X$, and there is nothing explicit about it. So you want to use fewer states. 1) Following on what Phoenix87 said, here is an example of a faithful representation (denomination way more common in the literature than "injective"). It is based on ...


1

If I do understand your question correctly, you want a representation $\pi$ of $C_0(X)$ on some Hilbert space, and you want $\pi$ to be injective. As injectivity in this sense is equivalent to "isometric", I'll give you three constructions of isometric representations of $C^*$-algebras that I know of. Construction 1 This is a universal construction, known ...


0

If $\varphi(a)\geq0$ for all states $\varphi$, you can do the following: Note that a state is selfadjoint, i.e. it maps selfadjoints to real numbers. This, because any selfadjoint is a difference of two positives. Write $a=b+ic$ with $b,c$ selfadjoint. Then $\varphi(b)$ and $\varphi(c)$ are real. So $\varphi(a)=\varphi(b)+i\varphi(c)$ is positive, which ...


0

This is not true. For example you could take $A$ unital, $a=\frac12\,1$. Then $$ (1-u_n)^2=\left(1-\frac 1{\frac12+\frac1n}\right)^2\to 1. $$ So $\varphi(b(1-u_n)^2b^*)\to\varphi(bb^*)$ for any $b$.


2

It is not true that a linear functional $f\in A^*$ is positive if it is bounded and $f(I)\geq0$. For instance, let $A=M_2(\mathbb C)$ and define $$ f\left(\begin{bmatrix}x&y\\ z&w\end{bmatrix}\right)=x-w/2. $$ The $f(I)=1/2\geq0$, but $f(E_{22})=-1/2$. What is true is that a linear functional is positive if and only if $f(I)=\|f\|$. To extend ...


0

Asyou remarked, one direction is trivial from definitions. For the other direction you can use the following well known facts: 1) An element is positive if and only if its spectrum is contained in $[0,\infty )$ 2) If $\lambda$ belongs to spectrum of $a$ then there exists a state $f$ such that $f(a)=\lambda$.


3

The basic idea is that $\varphi(f)$ behaves like evaluation of $f$ at $a$ for any continuous function. The immediate application is that we can "evaluate" well-known functions with elements of a $C^{*}$-algebra, e.g. $f(x) = e^x$, logarithms, square roots, etc. They have analogous uses that these functions have when dealing with numbers. For example, if we ...


0

The spectrum is invariant under unitary transformations, thus it is rotational invariant: $$\sigma(W(f\neq0))=\sigma(W(g)W(f\neq0)W(g)^*)=e^{-i\sigma(g,f\neq0)}\sigma(W(f\neq0))$$ Note that this excludes the special case: $$\sigma(W(0))=\sigma(1)=\{1\}$$


0

If $\phi(a)=0$ for some state $\phi$, then $\phi(u_n)=0$ for all $n$, because $0\leq\phi(u_n)\leq2^n \phi(a)$. Now, for any $b\in A_+$, $$ \phi(b)=\lim_n \phi(u_nbu_n)=\lim_n\phi(u_n^{1/2}[u_n^{1/2}bu_n^{1/2}]u_n^{1/2})\leq\limsup_n\|u_n^{1/2}bu_n^{1/2}\|\,\phi(u_n) \\ \leq\|b\|\limsup_n\phi(u_n)=0. $$ So $\phi(b)=0$ for all $b\geq0$, and thus $\phi=0$ as ...



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