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Subtag of the [oa.operator-algebras] tag for questions about von Neumann algebras, that is, weak operator topology closed, unital, *-subalgebras of bounded operators on a Hilbert space.
1
vote
Accepted
Definition of Radon measure on Takesaki's first volume
Making no claims of originality, one possible proof can be obtained by combining Example 4.60, Lemma 5.11, and Lemma 3.14 in arXiv:2005.05284.
This shows that for any Radon measure its algebra of equi …
2
votes
Accepted
Continuous surjection between spectra of commutative von Neumann algebras
Is it true that π maps clopen sets into clopen sets?
This is true if and only if the inclusion $V_1→V_2$ is a morphism of von Neumann algebras, i.e., its image is closed in the ultraweak topology.
S …
2
votes
Hochschild/cyclic homology of von Neumann algebras: useless?
I am surprised that nobody mentioned the book by Sinclair and Smith “Hochschild Cohomology of von Neumann algebras” so far.
If I remember it correctly, they claim that nobody knows whether there is a …
13
votes
Accepted
Making sense of "every non-commutative algebra has its own internal time evolution (aka a on...
Given any von Neumann algebra $M$, we can define its noncommutative $\def\L{{\cal L}} \L^p$-spaces $\L^p(M)$ for any $\def\C{{\bf C}} p∈\C$ such that $\Re p≥0$.
Here I use the notation $\L^p:={\rm L}^ …
7
votes
Quick derivation of classical probability theory from von Neumann algebraic framework
I am not sure how far you want to go, but some basics are explained in this answer:
Is there an introduction to probability theory from a structuralist/categorical perspective?
In particular, you have …
23
votes
Accepted
Is the opposite category of commutative von Neumann algebras a topos?
The opposite category of commutative von Neumann algebras is not a topos
because categorical products with a fixed object do not always preserve small colimits.
See Theorem 6.4 in Andre Kornell's Quan …
2
votes
Reference for "the algebra of multiplication by all measurable bounded functions acts in Hil...
One possible interpretation of Connes's statement is that
up to an isomorphism, there is a unique faithful indecomposable representation of any commutative von Neumann algebra on a Hilbert space.
Inde …
37
votes
5
answers
4k
views
Reference for the Gelfand duality theorem for commutative von Neumann algebras
The Gelfand duality theorem for commutative von Neumann algebras states that the following three categories are equivalent:
(1) The opposite category of the category of commutative von Neumann algebra …
6
votes
Accepted
Reference for the Gelfand duality theorem for commutative von Neumann algebras
As shown in the paper Gelfand-type duality for commutative von Neumann algebras, the following categories are equivalent.
The category CSLEMS of compact strictly localizable enhanced measurable spac …
5
votes
1
answer
333
views
One-parameter semigroups of bimodules
Suppose M is a von Neumann algebra.
Consider a monoidal category of bimodules over M.
Here a bimodule is a Hilbert space with two normal representations of M.
The monoidal structure is given by Connes …
4
votes
Accepted
On existence of certain operators in von Neumann algebra
This is false.
Consider, for example, the case of M being the von Neumann algebra
of bounded complex-valued functions on an infinite countable set I.
It acts on the Hilbert space of square-summable fu …
2
votes
Regarding Haagerup $L^{P}$ spaces
How the norm on L^{P} space related to weight φ?
The L^p-spaces and their norms are independent of the choice of the weight φ.
See, for instance, the exposition by Yamagami in “Algebraic Aspects …
4
votes
Why are ultraweak *-homomorphisms the `right' morphisms for von Neumann algebras (and say, n...
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $A\to(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^ …
6
votes
Accepted
A non-commutative Radon-Nikodym derivative.
Such t_0 is unique if its support is at most p, where p is the support of ϕ.
Note that we can replace t_0 by pt_0p and the support of pt_0p is at most p.
Without this additional condition t_0 is high …
3
votes
Accepted
$e\precsim f$ and $1-e\precsim 1-f$ imply $e\sim f$?
No. Take e=0 and 0 < f < 1 such that both f and 1−f are infinite, with (1−f)~1.
Then e≾f because 0≾f for any projection f.
Also 1−e≾1−f because 1≾1−f, which holds by definition of f.