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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 ...


3

niki, he keeps the assumption that that $\mathfrak{m}$ is $\sigma$-weakly closed. So the third line should read: If, in addition, $\mathfrak{m}$ is a two-sided ideal, then.... Of course, when you deal with norm-closed ideals it is very easy to come up with a counter-example: take the compact operators inside $B(H)$. This is also an illustration to ...


3

The zero representation of an algebra on a Hilbert space $H$ is the map that sends every element of the algebra to the zero operator on $H$. Murphy's book gives the following definitions: If $A$ is a C*-subalgebra of $B(H)$, it is said to be irreducible, or to act irreducibly on $H$, if the only closed vector subspaces of $H$ that are invariant for ...


2

The version Murphy gives is correct too, and incorporates the fact that $0$ is an element of $\sigma_A(b)$ in this situation. If $b$ were invertible in $A$, then the C*-algebra generated by $b$ would automatically contain $1_A$. There was a question and answer about this fact here. (Actually what is shown there is stronger, as it is not assumed that $B$ ...


2

Here is the argument: Since $d$ is selfadjoint, so is $c^{-1/2}dc^{-1/2}$. Then the spectrum of $1+ic^{-1/2}dc^{-1/2}$ is of the form $$ \{1+i\lambda:\ \lambda\in\sigma(c^{-1/2}dc^{-1/2})\}. $$ The $\lambda$ are real, so the spectrum does not contain $0$, and thus $1+ic^{-1/2}dc^{-1/2}$ is invertible. As this equals $c^{-1/2}(c+id)c^{-1/2}$ we conclude ...


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 ...


2

Any $a \in A$ is canonically a (complex) linear combination of self-adjoint elements $\frac{a + a^{\ast}}{2} + i \frac{a - a^{\ast}}{2i}$, so to show that any element can be written as a linear combination of unitary elements it suffices to show that any self-adjoint element can. By rescaling such a thing we can assume WLOG that it has norm at most $1$. What ...


2

I'm writing this as an answer to have a little more space to write. What you want to prove is not true: for a $*$-homomorphism to be necessarily contractive, you need the domain to be a C$^*$-algebra. For instance, let $\mathcal A=C[0,1]$, $\mathcal B=\mathbb C$, $\mathcal D=\{\text{polynomials}\}$, and $\pi(p)=p(2)$. Then $\pi$ is clearly a ...


2

This is not true in general, e.g. let $a_1=\begin{pmatrix}\sqrt{2}/2&-\sqrt{2}/2\\\sqrt{2}/2&\sqrt{2}/2\end{pmatrix}$ and $a_2=a_1^*$ in $M_2(\mathbb{C})$. Since $a_1$ is unitary, then $A=C^*(1,a_1,a_2)=C^*(a_1)$. But $a_1$ has two eigenvalues (so $a_2$ also has two eigenvalues), thus $\Omega(A)=\sigma(a_1)$ has two elements, but ...


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 space $C_0$ is a Banach space. This implies that absolutely convergent sequences are convergent, i.e. if $$\sum_n \Vert \frac{p_n}{3^n} \Vert_\infty$$ is finite, then $\sum_n \frac{p_n}{3^n} \in C_0$. But projections have norm at most one, which means $$\sum_n \Vert \frac{p_n}{3^n} \Vert \leq \sum_n 3^{-n} = \frac{1}{1-1/3} < \infty.$$ This ...


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 ...


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

Take any $B$. Then $B+\mathbb C\,1_A$ is a C$^*$-subalgebra of $A$ that contains $1_A$, so $\sigma_{B+\mathbb C\,1_A}(b)=\sigma_A(b)$. Now by page 44, applied to the inclusion $B\subset B+\mathbb C\,1_A$, you have $\sigma_B(b)=\sigma_{B+\mathbb C\,1_A}(b)$, and you are done.


1

What you need to know is that a positive operator $a$ can be written as $a = s^*s$ for some element $s$ in the $C^*$-algebra. Particularly, it means that every positive element is indeed self-adjoint since $a^* = (s^*s)^* = s^*s = a$ so that $a^*a = aa^*$. This in turn gives you that the $C^*$-algebra of a positive element is commutative since $a$ commutes ...


1

No. The "why" depends on your definition of von Neumann algebra. If you define von Neumann algebra the way it is usually used, it is not an intrinsic notion: both the double commutant and the ultraweak operator topology depend on the environment (i.e. $B(H)$). In that case, take a non-type I factor, say a II$_1$, and take an irreducible representation. ...


1

Definetely not. Just take $B$ to be the kernel of $\pi$. For instance, $A=\mathbb C\oplus\mathbb C$, $B=0\oplus \mathbb C$, $\pi(a,b)=a$. Even if you require $B$ to be unital with the same unit as $A$, the answer is no: let $$ A=M_2(\mathbb C),\ \ \ B=\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix}:\ a,b\in\mathbb C\right\} $$ with $\pi$ the ...


1

Your argument is not correct. You say that $\|\phi (a^*a)\|=\|\phi (a)^*\phi (a)\|$ implies that $\phi (a^*a)=\phi (a)^*\phi (a) $, which makes no sense. Also, without the unital condition the statement is trivially false: take $\phi (x)=-x $. Now, here is an argument using all conditions. Note that $ \phi$ maps selfadjoints to selfadjoints. For a ...


1

As an example for better understanding, consider the case where $A=\mathbb C$. Then, for a fixed $z\in\mathbb C$, you write $$z=\sqrt{a^2+b^2}\left(\frac a{\sqrt{a^2+b^2}}+i\frac b{\sqrt{a^2+b^2}}\right),$$ with $a,b\in\mathbb R$. Since $$ -1\leq \frac a{\sqrt{a^2+b^2}}\leq1, $$ there exists $\theta$ such that $\frac a{\sqrt{a^2+b^2}}=\cos\theta$. Similarly, ...


1

Yes, the two closures are equal. Let $x\in \bar A^{wot}$. We have $\bar A^{wot}=\bar A^{sot}$, so there is a net $\{x_j\}\subset A$ such that $x_j\to x$ sot. By Kaplansky, there is a bounded net $\{x_j'\}\subset A$ with $ x_j'\to x$ sot, and so wot. That is, $x$ belongs to the wot closure of a ball in $A$. But on balls, the wot and ultraweak topologies ...


1

I think your argument is fine. I fail to see why Murphy feels the need to use approximate units in this argument.


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.


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 ...


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

No, it is not possible to prove that this is the case. (non-zero) Counterexample in $\Bbb C^{2 \times 2}$: $$ a = \pmatrix{1&0\\0&0}, \quad b = \pmatrix{1&0\\0&-1} $$ Counterexample to the comment: $$ a = \pmatrix{1&0\\0&0}, \quad b = \pmatrix{1&1\\1&1} $$ note that neither $ab$ nor $ba$ is self adjoint. Note that $ab$ ...


1

Simplest counterexample: $a=0$. Somewhat less trivial: in $C([0,1])$ let $a$ be a function that is $0$ on $[0,1/2]$ and nonnegative. Then $ab = ba \ge 0$ if $b \ge 0$ on $[1/2, 1]$, but you could have $b(t) < 0$ somewhere in $[0,1/2)$.


1

Because if $\tau|_C=0$, then $\tau(1)=0$. So for any $x\in A$, $$ \tau(a)=\tau(1a)=\tau(1)\tau(a)=0. $$


1

For the proof of the author's claim, note that $c \geq 1 = c ^{-1} c$ if and only if $1 \geq c^{-1}$. This example doesn't quite work since it doesn't meet the hypotheses set out by the author. Let's consider the C*-algebra $A := C([2,10])$ and the same function $f$, which you defined above. There are various notions of invertibility in the C*-algebra ...


1

There is no reason for you to expect a representation into the same $H$ to be irreducible. The only way that could happen is when your operator has trivial commutant. You need to decide whether you consider the generated subalgebra to contain the ambient unit or not. If you don't require the unit, $C^*(P)=\{\lambda P:\ \lambda\in\mathbb C\}$, and ...


1

It is the operator norm. It is the only one that satisfies $\|a^*a\|=\|a\|^2$ for all $a$. Actually, you obtain the operator norm precisely when you represent $M_n(\mathbb C)$ as the algebra of linear operators on $\mathbb C^n$.



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