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4

The kernels of nonzero homomorphisms to $\mathbb C$ are modular ideals, terminology that might help you find more references. Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take $\mathbb C$ with the $0$ product, which has maximal ideal $\{0\}$ and no nonzero homomorphisms to $\...


4

No. For any such sequence $\{p_n\}$ with each $p_n$ of finite rank, take $A=B(\mathcal H)$ and $$ B=\overline{ \{b\in B(\mathcal H):\ \exists n,\ b=p_nbp_n\}}.$$ Then $p_nAp_n=p_nBp_n$ for all $n$, but $B$ is separable while $A$ is non-separable.


3

Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense. The right notion to ...


3

There's no simple description of the spectrum of the algebra of bounded holomorphic functions in the disk (known as $H^\infty$). It's an Axiom-of-Choice-ish thing. If $|z|<1$ then $f\mapsto f(z)$ is a complex homomorphism, so the open disk is contained in the spectrum in a natural way. The Corona Theorem says that the disk is dense. This is one of the ...


2

If $\delta\leq1$, then $x^*x$ and $xx^*$ are invertible. In particular, we can do the polar decomposition $x=u(x^*x)^{1/2}$ and we will have $u\in A$. Also, $u$ is a unitary because $$u^*u=(x(x^*x)^{-1/2})^*(x(x^*x)^{1/2} =(x^*x)^{-1/2}x^*x(x^*x)^{-1/2}=1, $$ $$ uu^*=x(x^*x)^{-1/2}(x^*x)^{-1/2}x^*=x(x^*x)^{-1}x^*=1. $$ This last equality is not obvious, but ...


2

Yes. What you do is show that if $A$ is not simple, then there is a non-faithful irrep. If $J\subset A$ is a non-trivial ideal, then consider an irreducible representation of $A/J$ into $B(H_J)$; then $A\to A/J\to B(H_J)$ is an irreducible representation of $A$ with kernel that at least contains $J$, so it is not faithful.


2

You can't find a positive $\gamma$ such that $\gamma \leq \frac{\langle Lu\mid u \rangle}{\langle u\mid u \rangle}$ if $\langle Lu\mid u \rangle=0$. So to find a counterexample, you might look for positive operators such that $Lu$ is sometimes $0$. The simplest example is $L=0$, the operator that sends everything to $0$.


2

Edit: This is an answer to the previous version of this question. This algebra is complete; it is a simple application of Morera's theorem which you may use to show that the uniform limit of such functions is actually holomorphic. This algebra is traditionally denoted by $H^\infty$ and is highly non-separable. For this reason, the maximal ideal space of $H^...


2

Yes, this follows from the monotone convergence theorem for nets (see Theorem IV.15 of Reed and Simon's Functional Analysis, for instance) Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$ and let $(f_{\alpha})$ be an increasing net of continuous functions converging pointwise to $f$. If $\sup \|f_{\alpha}\|_1 < \infty$, then $\...


2

Your questions reveal a typo that Valette was not aware of. The third line of the proof of Proposition 3.3.7 should be:$$\pi_n(\text{GL}_\infty(A)) = \pi_{n - 1}(\Omega \text{GL}_\infty(A)) = \pi_{n - 1}(\text{GL}_\infty(SA)),$$where $\Omega X$ is the loop space of $X$. The first equality is a basic property of loop spaces, see e.g. here. The second follows ...


2

The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$. ...


1

If all you assume about your involution is that it's an involution. that is, $(x+y)^*=x^*+y^*$, $(xy)^*=x^*y^*$, $x^{**}=x$ and $(cx)^*=\overline cx^*$, then most of what you expect doesn't follow. In particular you assume above that $\phi(x^*)=\overline{\phi(x)}$, and that doesn't follow: Consider $C([-1,1])$. Define $$f^*(t)=\overline{f(-t).}$$ That's an ...


1

These results follow from some quite general results about surjective homomorphisms between $C^{\ast}$-algebras which I will state without proof. In the following, $A$ and $B$ are unital $C^{\ast}$-algebras and $\varphi :A\to B$ is a unital surjective $\ast$-homomorphism. [Proposition 4.3.14 in Higson & Roe's Analytic K-homology ] If $f: [0,1] \to ...


1

A typo slipped in; a $k$ became $j$ for no reason. Fixing that, you're almost there, re showing it's a Banach algebra: $$\begin{align}\dots=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{(k-j)}(t)}{(k-j)!}\dfrac{g^{(j)}(t)|}{j!}} &=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=j}^{n}\dfrac{|f^{(k-...


1

Somehow, after I post here a question the solution comes to my mind... So, I'll use the following argument:almost unitaries are close to a unitary element Now, it's enough to show that: If $A$ is a unital $C^*$ algebra and $A=\overline{\bigcup_{k\in \mathbb{N}} A_k}$ ,where each $A_k$ is a unital (same unit of $A$) $C^*$ subalgebra, then for any unitary $...


1

Ignoring your original question: Note that $A = \lim_k A_k$ so $K_1(A) = \lim_k K_1(A_k)$. But $K_1(A_k) = 0$ since $K_1(M_n(\mathbb C)) = 0$ for each natural number $n$. This is a general argument that AF algebras have trivial $K_1$-group.


1

There is a typo in the statement: it should say $e\precsim f$. The rest looks ok to me. As the proof says, one uses Theorem 1.8: there exists central $z$ with $ze\precsim zf$ and $(1-z)e\succsim (1-z)f$. The proof shows that $(1-z)e\sim (1-z)f$. Then $$ e=ze+(1-z)e\precsim zf+(1-z)f=f. $$


1

Another similar approach (inspired by Martin Argerami) is the following: If $x \in A$ and $x^*x, xx^*$ are invertible, then $x$ is invertible. To see this, check that $(x^*x)^{-1}x^* = x^*(xx^*)^{-1}$. This is then the inverse of $x$. Now $x = u \sqrt{x^*x}$ for $u$ unitary since $x$ is invertible. Then the last argument of Martin applies, i.e. that $\...


1

Already $\mathscr{B}(H)^*$ or even $L_\infty^*$ are intractable. However, $\mathscr{M}^{**}$ is a von Neumann algebra too, so $\mathscr{M}^*$ is indeed a predual of the second dual. This is the best we can say in that generality. Maybe you should rethink your question.


1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...


1

Some notation is missing, like for example you need to write your C$^*$-algebra as $C(X)$ for a locally compact Hausdorff space. For your converse, if $\phi_\omega$ is the state given by evaluation at $\omega$ and $\phi_\omega=\alpha\varphi+(1-\alpha)\psi$, by Riesz-Markov you get an equality of measures $$\mu_\omega=\alpha\mu_\varphi+(1-\alpha)\mu_\psi.$$ ...


1

No. The von Neumann algebra $B(H)$ is separable (in the sot topology, say); it follows that any von Neumann algebra $R\subset B(H)$ is separable. This is not totally straightforward, but it is simple: A countable dense subset of $B(H)$ is given by the complex-rational linear combinations of matrix units (because its commutant is trivial); Separability can ...


1

(You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since $A^*A$ is positive and compact, it is orthogonally diagonalizable (spectral theorem): $A^*A=U^*D^2U$ for some unitary $U$ and $D$ diagonal with diagonal $s_1(A),s_2(A),\ldots$ Assume $s_1(A)\geq s_2(A)\geq \cdots$ Since $U$ ...


1

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. ...


1

To answer the new request from the OP (expressed in a commentary), let us see that if $R$ is a ring and $e_i\in R$ are nonzero idempotent and commutative such that $e_1+\ldots+e_n=1$ then they are in fact orthogonal, provided that $R$ is free of torsion up to $n-1$: I have thought of a proof by backwards induction. First observe that $$e_1\ldots e_n = e_1\...


1

A unitary operator is a diagonalizable operator whose eigenvalues all have unit norm. If we switch into the eigenvector basis of U, we get a matrix like: \begin{bmatrix}e^{ia}&0&0\\0&e^{ib}&0\\0&0&e^{ic}\\\end{bmatrix} which is obviously the exponential of a diagonal hermitian matrix.



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