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## Hot answers tagged c-star-algebras

25

Yes. One can take $B$ to be the direct sum of all separable C*-algebras. A more interesting answer would be $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$. For an explanation, let me start with another question of you: It is an open problem whether $B = {\mathcal Q}(\ell_2)$, the Calkin algebra, suffices or not. It's written in my textbook with Nate ...

20

The von Neumann algebra $M$ generated by $\mathcal O_\infty$ is all of $B(\mathcal F(H))$. Indeed, if $a$ belongs to its commutant, let me prove that $a$ is a multiple of the identity. First since for all $v \in H$, $s_v^* (a \Omega)= a (s_v^* \Omega)=0$, we have that $a \Omega=\lambda \Omega$ for some $\lambda \in \mathbb C$. Then for every $\xi \in \... 20 Edit: as pointed out in the comments, the following answers the question for unital C*-algebras presented in terms of generators and relations. When I say C*-algebra, I really mean unital C*-algebra. It may depend on what exactly you mean by "concrete", but I highly doubt that there is a general solution to this; finding a concrete realization of a ... 19 The answer is yes, but I don't know where it is written. If$p$is not central, then$pAp^\perp\neq\{0\}$and one can take$x\in pAp^\perp$such that$0<\|x\|<1/2. Then, $$q:=\left(\begin{array}{cc} \frac{1+\sqrt{1-4xx^*}}{2} & x\\ x^*& \frac{1-\sqrt{1-4x^*x}}{2}\end{array}\right)\quad{\rm in}\quad \left(\begin{array}{cc} pAp & pAp^\perp \\... 19 If C^*(G) is isomorphic to C^*_r(G), then C^*_r(G) has a 1-dimensional representation, i.e. G has a 1-dimensional representation \chi weakly contained in the regular representation \lambda_G. Then 1_G=\chi\otimes\overline{\chi} is weakly contained in \lambda_G\otimes\overline{\lambda_G}\simeq \infty.\lambda_G, hence G is amenable. 19 Answering the question in the body of the original post, which seems to be more restricted than the implicit question in the title of the post.... The answer is YES. See L. Terrell Gardner, On isomorphisms of C^\ast-algebras. Amer. J. Math. 87 (1965) 384–396. MathReview Roughly speaking, the proof works by considering the {\rm C}^*-... 19 One way to tell how active a field is is by looking at what's appearing on the arXiv in that area. I think that will show you that operator algebra is a robust subject with a lot of activity. In the comments, MaoWao points out that UC Berkeley and UCLA have very strong operator algebra groups, and Ulrich Pennig mentions groups in Münster, Göttingen, ... 18 Since free C*-algebras don't exist, we can't give a concrete description of all relations that are allowed. Instead, we need to give axioms that determine what collections of n-tuples (a_1,\dots,a_n) in A are allowed, where A varies over all C*-algebras. It is important that the elements not "know about" the ambient C*-algebra, so a is in a ... 17 There are 2^c mutually non-equivalent irreducible representations. Since \ell_\infty(N) has 2^c many pure states (there are 2^c many ultrafilters on N), any C^*-algebra containing \ell_\infty(N) has at least as much pure states. Since R has c many unitary elements, there are 2^c many mutually non-equivalent pure states. This is a very old ... 17 Yes. I will show that any two positive elements of A commute. Since every element is a linear combination of positive elements, this suffices. Say a and b are positive. Then a^{1/2}ba^{1/2} \in A_{sa}, so by hypothesis ba^{1/2}a^{1/2} = ba \in A_{sa}. That is, ba = (ba)^* = a^*b^* = ab. QED 16 C*-algebras don't see the ISP. The operators T\in B(H) and T\oplus T\in B(H\oplus H) generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces. To have both operators in the same Hilbert space, pick isometries v_1,v_2\in B(H) with orthogonal ranges that add up to H. Then$$ T\mapsto v_1Tv_1^*+v_2Tv_2^* is an ... 15 Here is perhaps the simplest example. Let A be the C*-algebra of all sequences of 2 \times 2 matrices converging to a scalar multiple of diag(1,0). Let p be the constant sequence diag(1,0), and q a sequence of rank 1 projections converging to diag(1,0) but never exactly equal. Then p and q have no upper bound at all. This example can be ... 15 In Theorem 4.6 of their paper http://www.univie.ac.at/nuhag-php/bibtex/open_files/deha85_CanniereHaagerup.pdf de Canni`ere and Haagerup construct an explicit sequence of finitely supported functions on the free group \mathbb{F}_N (N\geq 2), defining positive multipliers of the reduced C*-algebra C^*_r(\mathbb{F}_N), and such that the corresponding ... 15 Kaplansky showed that every non-commutative C*-algebra contains a non-zero nilpotent element. I don't have a reference I'm afraid. 15 There is a nonmetrizable space X with that property so the embedding property fails. I don't know if one could build more elaborate examples along these lines to find infinitely many such spaces. The answer to your question may be different if you require X to be metrizable or Hausdorff or otherwise nice. Let A and B denote two copies of \mathbb R ... 14 For x=(x_i)_{i=1}^n, y=(y_i)_{i=1}^n \subseteq A define (x,y) = \sum_i x_i y_i^* \in A, and set \|x\| = \|(x,x)\|^{1/2}. Lemma: We have that (x,y)^* (x,y) \leq \|x\|^2 (y,y) the order in the C^*-algebra sense. Proof: (Copied from Lance's Hilbert C^*-module book). Wlog \|x\|=1. For a\in A let a\cdot x = (ax_i). Then \begin{align*... 14 Here is a direct argument which may not differ much in its essence from the one in the book that you mention: Say A is simple and B is a closed hereditary subalgebra of A. This means that if a\in A and b_1,b_2\in B then b_1ab_2\in B. Let x\in B be non-zero and let us show that it generates B as a closed two sided ideal. Since x generates ... 14 Probably the most standard metric is Banach-Mazur distance, and there is indeed a theorem due to Amir which says that if the Banach-Mazur distance between C(K) and C(L) is less than 2 then K and L are homeomorphic. There's also something called the Kadets distance which is basically a linearized Gromov-Hausdorff distance. I don't know what the ... 14 Such examples do exist. In the paper "Stability of C*-algebras is not a stable property, Doc. Math. J. DMV , 2, (1997), 375-386.", Rordam gives examples of simple C*-algebras A such that M_2(A) is a stable C*-algebra but A is not. Now take the pair A and B=M_2(A). These C*-algebras are not isomorphic, since A is not stable, but B is. On the ... 14 Yes: A C^*-algebra satisfies the identity e^{[xy-yx]}=e^xe^ye^{-x}e^{-y} iff it is commutative. This follows from two independent facts (I write [x,y]=xy-yx) 1) A (real/complex) unital Banach algebra satisfies the identity e^{[xy-yx]}=e^xe^ye^{-x}e^{-y} \Leftrightarrow it satisfies the identity [x,[x,y]]=0 \Leftrightarrow it satisfies ... 13 Let me begin with inverse semigroup C^{\ast}-algebras. Inverse semigroups are semigroups S with the property that for all s\in S, there is a unique element s^* with ss^\ast s=s and s^\ast ss^\ast=s^\ast. The key example of an inverse semigroup is the symmetric inverses semigroup of all partial bijections of a set X (also called the rook ... 12 According to what I have seen in the literature so far, the standard procedure consists of two main steps: Prove the existence of a universal C^{*} -algebra A_{\theta} generated by two unitaries u and v that satisfy u v = e^{2 \pi i \theta} v u. $$Note: We are assuming that \theta is irrational. Prove that A_{\theta} is simple, ... 12 Theorem Let A be a unital ring and I_1,\dots,I_n \subset A be 2-sided commutative ideals such that A=I_1+\dots + I_n. Then, A is commutative. Proof: If A=I_1+\dots+I_n, then 1 = x_1+\dots+x_n for x_i \in I_i. But then,$$1 = (x_1+\dots+x_n)^{n+1} \in I_1^2 +\dots+ I_n^2$$and we conclude that A=I_1^2 + \dots + I_n^2. Now, if I \subset A ... 12 I think the canonical connection between C*-algebra and differential operators is Connes' index theorem for foliated manifolds. I don't know if that counts as PDEs but it's certainly related. Every foliated manifold M has an associated C*-algebra A which is noncommutative (except in trivial cases) but in some way embodies the idea of "the continuous ... 11 Just a supplement to the answer by Tobias Fritz: All your examples are obviously commutative, since there is only one generator which is normal. Thus the question is really about finding certain terminal compact Hausdorff spaces. For example 1. comes from the terminal compact Hausdorff space X equipped with a continuous function X \to \mathbb{C} which is ... 11 I may be showing my ignorance of category theory, but don't think this is true. Work on l^2. Let \mathcal{A}_n consist of the operators A satisfying \langle Ae_i, e_j\rangle = 0 if \max(i,j) > n (so \mathcal{A}_n is isomorphic to the n\times n matrices). The direct limit of the \mathcal{A}_n is the compact operators on l^2. Let P be a ... 11 The answer to both questions 1 and 2 is false, due to the following example. Consider (n+1) consecutive intervals I_0,...,I_n of lengths 1,2,4,...,2^{n-1}, 2^n. Let the map T_n cyclically permute them in an affine way. In fact, if you choose I_j=[2^j,2^{j+1}], then$$ T_n(x)=\begin{cases} 2x, & x \not\in I_{{n}}\\ 2^{-n}x, & x\in I_{n}. \... 11 The groupK_0(C_0(\mathbb{C}))$is generated by by the class$[p_{Bott}] - [1]$where$p_{Bott} \in M_2(C_0(\mathbb{C})^\sim)$is the so-called "Bott projection" given by $$p_{Bott}(z) = \frac{1}{1+|z|^2} \begin{pmatrix} |z|^2 & z \\ \overline{z} & 1 \end{pmatrix}.$$ This class comes from the tautological line bundle on$S^2 \simeq \mathbb{C}P^1$... 11 Any infinite discrete group$\Gamma$with Kazhdan's property (T) gives an example. Since it is not amenable, the full and reduced C*-algebras (which are both completions of the group algebra) do not coincide. Moreover, the full C*-algebra contains a projection with non-trivial K-theory class (so-called Kazhdan projection) which is mapped to$0$in the ... 11 Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear$C^\ast$-algebra$A$with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if$B$has WEP and$C$has LLP, then$B\otimes_{\max{}} C = B\otimes_{\min{}} C\$...

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