# Tag Info

## Hot answers tagged oa.operator-algebras

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 $B\otimes_{\max{}} C = B\otimes_{\min{}} C$...

9

Yes to both.$\newcommand{\Cst}{{\rm C}^*}$ The standard example for the first is: take a discrete group $G$ and let $A$ be its full $\Cst$-algebra, $B$ its reduced $\Cst$-algebra. There is a canonical homomorphism $q:A\to B$ which is injective when restricted to $\ell^1(G)$; but $q$ is injective if and only if $G$ is amenable. So any non-amenable discrete ...

8

There are even commutative counterexamples. Let $A = C[0,2]$ and let $A_0$ be the $*$-subalgebra of all polynomials in $x$. Then let $p: C[0,2] \to C[0,1]$ be the restriction map. (My first example took $A = C[0,3]$ and let $A_0$ be the set of all polynomials in $x$ with rational coefficients and $p: A \mapsto \mathbb{C}$ the point evaluation at $x = e$. ...

5

Go to the Fourier transform picture. Then these become the operators that commute with multiplication by exponentials, which means they commute with all multiplication operators, which means you have just described the operators which become multiplication operators when you conjugate by the Fourier transform.

4

It's not true. For simplicity, suppose $A$ is unital, so that its spectrum is compact Hausdorff. If $X$ is a compact Hausdorff space and $C(X)$ is separable, then you can show that $X$ is second countable. (Let $\{f_k\}$ be a countable dense subset of $C(X)$, and $\{U_n\}$ a countable basis of open sets in $\mathbb{C}$; then the countable collection of ...

3

I think there are some subtle points here about what the "right action" even means. For a general $*$-algebra $A_0$ and a functional $\phi:A_0\rightarrow\mathbb C$, we first of all have to decide what "positive" means for $\phi$. We could take this as being $\phi(a^*a)\geq0$ for all $a$. Then Cauchy-Schwarz holds and we can form $L^2(A_0,\phi)$. Why, ...

2

The problem is still ill-posed, because the set of nilpotent bounded operators is not a vector space. (The sum of two nilpotents need not be nilpotent.) Let's interpret the question as: which algebras admit faithful representations as algebras of nilpotent bounded operators? Note that "nilpotent" usually means $T^n = 0$ for some $n$, not $T^2 = 0$. But ...

1

For separable von Neumann algebras the Haagerup property for $N$ is equivalent to the existence of a real closable derivation $\delta$ such that $\delta^*\delta$ has compact resolvents in $\mathcal B(L^2(N))$. This is true even in the non-tracial case. See Theorem 7.7 in: Martijn Caspers, Adam Skalski, The Haagerup Approximation Property for von Neumann ...

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