# Tag Info

3

Let $A$ be a finite-dimensional associative $\mathbb{C}$-algebra. Suppose $B$ is another such algebra and $f : A \to B$ is an injective algebra homomorphism. Now if there exists a submultiplicative norm for $B$, then we get one for $A$ by restriction. Clearly, we can embed $A$ into a finite dimensional associative $\mathbb{C}$-algebra which has a unit and so ...

0

If $G$ is not a subset of $H$, it is not clear how $L(G)$ could be a subset of $L(H)$. I will assume "isomorphic to". In that case, the answer is "no". Let $G=\mathbb Z_2$, $H=\mathbb F_2$. Then $L(G)=\mathbb C^2$. AS $L(H)$ is a II$_1$ factor, it has a nontrivial projection and so $L(G)\simeq\text{span}\,\{p,1-p\}$. As $H$ has no finite-subgroups, $G_1$ ...

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 ...

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 ...

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 $L^1[0,\infty)$. ...

2

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 $f: [0,1] \to ... 2 A typo slipped in; a$k$became$jfor 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-... 2 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 ... 2 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. 0 Yes, that's a very natural argument. Simpler and nicer than the one I gave on the answer you quoted. 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\}}.$$Then p_nAp_n=p_nBp_n for all n, but B is separable while A is non-separable. 2 Yes, this follows from the monotone convergence theorem for nets (see Theorem IV.15 of Reed and Simon's Functional Analysis, for instance) Let \mu be a regular Borel measure on a compact Hausdorff space X and let (f_{\alpha}) be an increasing net of continuous functions converging pointwise to f. If \sup \|f_{\alpha}\|_1 < \infty, then \... 0 The functional calculus for selfadjoint and normal operators requires projections and integrals. The holomoprhic functional calculus deals with general linear operators, but less general functions of those operators. For example, if C is a positively-oriented simple closed rectifiable curve enclosing the spectrum of a bounded operator A on a complex ... 3 Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense. The right notion to ... 4 The kernels of nonzero homomorphisms to \mathbb C are modular ideals, terminology that might help you find more references. Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take \mathbb C with the 0 product, which has maximal ideal \{0\} and no nonzero homomorphisms to \... 1 There is a typo in the statement: it should say e\precsim f. The rest looks ok to me. As the proof says, one uses Theorem 1.8: there exists central z with ze\precsim zf and (1-z)e\succsim (1-z)f. The proof shows that (1-z)e\sim (1-z)f. Then$$ e=ze+(1-z)e\precsim zf+(1-z)f=f. $$1 To answer the new request from the OP (expressed in a commentary), let us see that if R is a ring and e_i\in R are nonzero idempotent and commutative such that e_1+\ldots+e_n=1 then they are in fact orthogonal, provided that R is free of torsion up to n-1: I have thought of a proof by backwards induction. First observe that$$e_1\ldots e_n = e_1\... 2 Yes. What you do is show that ifA$is not simple, then there is a non-faithful irrep. If$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$, so it is not faithful. 1 Already$\mathscr{B}(H)^*$or even$L_\infty^*$are intractable. However,$\mathscr{M}^{**}$is a von Neumann algebra too, so$\mathscr{M}^*$is indeed a predual of the second dual. This is the best we can say in that generality. Maybe you should rethink your question. -1 This is answered in "Topics in Dynamics 1: Flows" by Edward Nelson (Theorem 7, Chapter 8): if$A$and$B$are skew-adjoint operators on a Hilbert space$\mathcal{H}$and the restriction of$[A,B]$to$\mathcal{D}=\mathcal{D}(AB)\cap\mathcal{D}(BA)\cap\mathcal{D}(A^2)\cap\mathcal{D}(B^{2})$is essentially skew-adjoint, then$\lim_{n\rightarrow\infty}[e^{-\...

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 $0$.

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

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 $H^... 1 Another similar approach (inspired by Martin Argerami) is the following: If$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

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.

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 ...

0

I think we were on the right track. Here is a slightly different approach that I believe will work to get a formal contradiction. Start with a countable basis $(V_n)$ for the SOT of $B(H)$ (the set of all bounded operators on the separable space $H$). Then define, for all $\alpha$, $A_\alpha = \{ n : V_n \cap \mathcal{R}_\alpha \neq \emptyset \}.$ Claim: ...

Top 50 recent answers are included