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19

The first result that you stated is commonly known as the Gelfand-Naimark-Segal Theorem. It is true for arbitrary C*-algebras, and its proof employs a technique known as the GNS-construction. This technique basically allows one to construct a Hilbert space $\mathcal{H}$ from a given C*-algebra $\mathcal{A}$ such that $\mathcal{A}$ can be isometrically ...

15

Take $x \in \ell^1(\mathbb{Z})$ to be $x(0) = 1$, $x(1) = x(2) = -1$, and $x(n) = 0$, otherwise. Then compute $(x^\ast \ast x)(n) = \begin{cases} 3, & n = 0, \\ -1, & n = \pm 2, \\ 0, & \text{otherwise.}\end{cases}$ This gives $\|x^\ast \ast x\|_1 = 5$ while $\|x\|_{1}^2 = 9$. Later: The above example shows that the $C^\ast$-identity isn't ...

11

Edit: I forgot to answer the question in the title: yes the center of a $C^{\ast}$-algebra is indeed a $C^{\ast}$-algebra. As with the other steps, also your argument for step 4. is fine. If $c_n \in Z(A)$ and $c_n \to c \in A$ then $c \in Z(A)$ because for every $a \in A$ we have $\|ac -ca\| = \|a(c-c_n) - (c-c_n)a\|$ which is equal to the left hand side ...

11

No. Every infinite dimensional C*-algebra is a strict subset of its second dual. Edited to add: Perhaps an easier way to see it is to use the fact that the extreme points of the unit ball of a C*-algebra are precisely the partial isometries. It is not too hard to see that any C*-algebra of dimension greater than 1 contains elements of norm one that are not ...

9

(Below by "algebra" I mean "complex algebra.") You can't even always find a norm that turns an algebra into a Banach algebra. For example, the Weyl algebra $\mathbb{C} \langle x, y \rangle/(xy - yx - 1)$ can't embed into a Banach algebra (see this math.SE question). Another constraint on being a Banach algebra is that the spectral radius (a purely ...

8

Let $e \in C_0(X)$ be the unit. For each $x \in X%$ there is a $f \in C_0(X)$ with $f(x) \ne 0$. As $e(x)f(x) = f(x)$ we must have $e(x) = 1$. So $e$ is the constant function $e=1$. But $1$ "vanishes at infinity" only if $X$ as compact: By definition there is a compact $K$ such that $1 = |1(x)| \le \frac 12$ outside $K$, i. e. on $X \setminus K$, so we must ...

8

Roughly, the answer will be that closed C*-subalgebras will correspond to quotient spaces (via pull-back of functions). In your example, the quotient map is one which identifies the two points into a single point. I haven't thought through, though, whether this is a completely correct statement as it stands, or whether one has to add additional caveats. ...

8

Your notation indicates that you seem to assume that $L^{\infty}(m) = C(I)$. This can't be true, of course. Consider Lebesgue measure on $I = [0,1]$. Then $C(I)$ is a separable algebra while $L^{\infty}[0,1]$ is non-separable, so the two can't be isomorphic. Note also that $L^{\infty}(m)$ contains many functions that aren't equal to a continuous function ...

7

Consider an orthonormal basis $\{x_j\}$ of $H$ with $x_1=x$, and consider $\{E_{kj}\}$ the corresponding matrix units (we don't really need matrix units, just the projection onto the span of $x$, but it might help understand). Assume that you can write $F_x=\alpha f + (1-\alpha)g$, for some $\alpha\in[0,1]$ and states $f,g$. Then, since $0\leq E_{11}\leq I$, ...

7

For the proof of first part see $C^*$-algebras and their automorphisms group by G. K. Pedersen pages 11-12, lemmas 1.4.4-1.4.5. For the proof of the second part use previous result with $\alpha=1/3$. Then you get $u\in A$ such that $x=u(x^*x)^{1/3}$ and $u^*u=(x^*x)^{1/3}$. Hence $x=uu^*u$. Proof of uniqueness I leave to you since this is a homework.

7

Let $\phi:A\longrightarrow B$ be a positive linear map between two $C^*$-algebras. We will show that $\phi$ is automatically bounded. We can assume that $B$ is unital without loss of generality, but this does not change anything. The problem is with $A$. Here is a proof I learned in Blackadar's book which works whether $A$ is unital or not, for positive ...

7

Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$. Lemma Let $a\in A$. The following assertions are equivalent. 1 - $p=a^*a$ is idempotent, i.e. $p^2=p$. 2 - $aa^*a=a$ 3 - $a^*aa^*=a^*$ 4 - $q=aa^*$ is idempotent. Proof ...

6

W. Arveson: An Invitation to $C^*$-Algebras, Springer (The presentation is as simple and concrete.) G.J. Murphy: $C^*$- Algebras and operator theory. Academic Press. (Very accessible and readable.) K. Davidson: $C^*$-algebras by example. AMS. (Useful example-based approach.)

6

In this context $C^*(a)$ is defined to be the (not-necessarily-unital) C*-algebra generated by $a$. If $a$ is invertible, then $C^*(a)=C^*(1,a)$, $\sigma(a)=\sigma(a)\setminus\{0\}$, and $C_0(\sigma(a))=C(\sigma(a))$. If $a$ is not invertible, then $0\in\sigma(a)$, and $C_0(\sigma(a)\setminus\{0\})$ can be identified with the ideal $M_0=\{f\in ... 6 Take for example any odd, continuous and real-valued$f\in L^1(\mathbb R)\cap L^2(\mathbb R)$which is positive on$(0,\infty)$. (The requirement$f\in L^2$ensures that$f*f^*(x)$is defined everywhere). Then$f^*=-f$, and so $$\vert f*f^*(x)\vert=\left\vert \int_{\mathbb R} f(t)f(x-t)\, dt\right\vert \leq \int_{\mathbb R} \vert f(t)\vert \,\vert ... 6 I am teaching myself representation theory, and it seems to me irreducible representations are nice in two ways. (Since you are talking about representations of C^*-algebras, we might as well restrict our attention to *-representations, that is, the representation respects the involution.) Extrinsically, irreps are nice just as prime numbers are nice ... 6 In order for an ordered vector space to be a Banach lattice, among other things, it is necessary that any two elements of the space have a greatest lower bound and least upper bound in the space. This property usually fails in C^* algebras with their usual C^*-algebraic ordering. Consider for example the C^* algebra A of 2 \times 2 matrices over ... 6 This is more an example than an application, but, among many other non-obvious ordered groups that arise from AF algebras is the group of polynomials with integer coefficients, with positivity meaning having non-zero positive values on the open unit interval (unless the polynomial is the zero polynomial). As shown by Renault, this ordered group arises from ... 6 I will first assume that the C^*-algebra is unital. If it is not one-dimensional, then because every element of the algebra has the form a+ib with a and b self-adjoint, there is a self-adjoint element a that is not a scalar multiple of the identity. Then the spectrum of a contains at least 2 distinct elements. Otherwise, if say \lambda were ... 6 This norm comes from considering M_n(\mathcal A) as acting as operators on the Hilbert C*-module \mathcal A^n, and no Hilbert space representation is required. I will try to give a fairly minimal overview of the situation in this special case, and more details can be found in the first chapter of Lance's Hilbert C*-modules: a toolkit for operator ... 6 Hints: Consider the ideal$$I = \begin{pmatrix} \mathcal{K} & \mathcal{K} \\ \mathcal{K} & \mathcal{K} \end{pmatrix}=\mathcal{K}\otimes\mathbb M_2\mathbb C\cong\mathcal{K} $$in A. What is A/I? Write down the six-term exact sequence in K-theory for the extension$$ 0\to I\to A\to A/I\to 0. $$Plug in what you know about the K-theory of I and ... 5 Every dual Banach space X^{\ast} is complemented in its bidual X^{\ast\ast\ast} with a projection of norm 1: Indeed, the adjoint (\iota_X)^{\ast}:X^{\ast\ast\ast} \to X^{\ast} of the canonical inclusion \iota_{X}: X \to X^{\ast\ast} is left inverse to the canonical inclusion \iota_{X^{\ast}}: X^{\ast} \to X^{\ast\ast\ast}, thus p = ... 5 Take A = L^{\infty}[0,1] with pointwise multiplication and let B = C[0,1] be the closed C^{\ast}-subalgebra of continuous functions. The \ast-subalgebra D \subset A consisting of the simple functions (finite linear combinations of characteristic functions of measurable sets) is dense in L^{\infty}[0,1] but D \cap B = \{\text{constant ... 5 In general, the answer is no. It is maybe easiest to see for your second question: let \Omega be an uncountable set and consider the C^*-algebra B(\Omega) consisting of all bounded complex-valued functions on \Omega (with the supremum norm). If \ell were a faithful state, we would have to have \ell(1) \ge \sum_{\omega \in \Omega} ... 5 There is a minimal way to imbed a nonunital C^*-algebra A into a unital C^*-algebra. As a *-algebra this is A\oplus \mathbb C with componentwise addition, with multiplication (a,s)(b,t)=(ab+ta+sb,st), and with involution (a,s)^*=(a^*,\overline s). But to give this algebra a C^* norm, one method is to identify it with \{L_a:a\in A\}+\mathbb ... 5 No, B needs not be a C^*-algebra in general. Just take A=B with a C^* norm on one side and a non C^*-norm on the other side. But note the following more interesting fact. If A and B are *-isomorphic C^*-algebras, then the *-homomorphism is automatically isometric. This follows from the fact that$$ \|x\|^2=\|x^*x\|=\rho(x^*x) $$where ... 5 Let I_a denote the set of all definite states \omega at a. By what you have proved, which simply follows from Cauchy-Schwarz inequality applied to \omega(x^*y) (see here, question a, for a proof), we get$$ ac-ca\in\bigcap_{\omega\in I_a}\;\mbox{Ker}\;\omega\qquad\forall c\in A. $$Let c in A. If b=ac-ca\neq 0, there exists \omega \in I_a ... 5 If the set of self-adjoint elements was stable under multiplication, we would have AB=(AB)^*=B^*A^*=BA, i.e. any two self-adjoint elements would commute. Writing any element C=\frac{C+C^*}{2}+i\frac{C-C^*}{2i} as a linear combination of two self-adjoint elements, we see that this would imply that \mathcal A be commutative. The converse is clear. Hence ... 5 This result holds in more general case of Banach algebras. By A_+:=A\oplus_1 \mathbb{C} we denote unitalization of Banach algebra A. Multiplication in A_+ is given by the formula$$ (a,\lambda)(b,\mu ) =(ab+\lambda b+\mu a,\lambda\mu)$$The algebra$A_+$is unital with unit$e_{A_+}=(0_A,1_{\mathbb{C}})$. For a given Banach algebra homomorphism ... 5 The fact that GCR C$^*\$-algebras contain the compact operators implies that they have minimal projections; this makes them behave in a way that somehow resembles matrices. Also, their spectrum is Hausdorff. These characteristcs made them easier to understand, and so in the 60s and 70s people studied them. So, it is not that much that they are interesting ...

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