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Algebras of operators on Hilbert space, $C^*-$algebras, von Neumann algebras, non-commutative geometry
6
votes
0
answers
104
views
Automorphisms of algebraic Clifford algebra of a Hilbert space
Let $H$ be a real separable, infinite-dimensional Hilbert space and let
$$\mathrm{Cl}(H) = \mathcal{T}(H_{\mathbb{C}}) / \{v\otimes w + w\otimes w - 2\langle v, w\rangle \cdot \mathbf{1} ~|~ v, w \in …
7
votes
3
answers
408
views
Are nearby subalgebras of matrix algebras conjugate?
Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are c …
6
votes
0
answers
124
views
Meagre sets of bounded operators
Let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathbb{B}(H)$ be the algebra of bounded operators on $H$. The norm topolology on $\mathbb{B}(H)$ is stricly finer, hence the identi …
7
votes
0
answers
126
views
Strong contractibility of unitary group of properly infinite von Neumann algebras
In the introduction of their 1993 paper (see reference below), Popa and Takesaki write
As it turns out, in these topologies [the weak and strong topology] $U(\mathscr{H})$ is again contractible (cf.
…
7
votes
0
answers
266
views
Enveloping von Neumann algebra of Clifford algebra
As explained in the book "Spinors in Hilbert Space" by Plymen and Robinson, if $V$ is a complex (separable) Hilbert space with a real structure, and $\mathrm{Cl}(V)$ the corresponding Clifford algebra …
6
votes
0
answers
231
views
Group $C^*$ vs group von-Neumann algebras
Let $\Gamma$ be a countable (discrete) group (in what follows, make additional assumptions as you wish). Let $C^*_r(\Gamma)$ and $W^*_r(\Gamma)$ be the reduced $C^*$-algebra respectively the reduced v …
3
votes
2
answers
3k
views
Weak convergence implies norm convergence for trace class operators?
It is known that weak convergence implies norm convergence in $\ell^1(\mathbb{N})$, see e.g. here.
Because of the typical analogies of the Schatten ideals $C_p \subset B(H)$ (where $H$ is a Hilbert s …
1
vote
Can a self-adjoint operator have a continuous set of eigenvalues?
The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resol …