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Algebras of operators on Hilbert space, $C^*-$algebras, von Neumann algebras, non-commutative geometry

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 …
Dmitri Pavlov's user avatar
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 …
Dmitri Pavlov's user avatar
26 votes
8 answers
3k views

Bimodules in geometry

Grothendieck's approach to algebraic geometry in particular tells us to treat all rings as rings of functions on some sort of space. This can also be applied outside of scheme theory (e.g., Gelfand-N …
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 …
LSpice's user avatar
  • 12.9k
21 votes
3 answers
3k views

Noncommutative smooth manifolds

Connes defined a noncommutative analog of a closed oriented Riemannian spin^c manifold using spectral triples. Using his definition it is unclear how to separate the smooth structure from the metric. …
7 votes
Accepted

Questions about Maharam's classification theorem

The spaces $[0,1]$, $[0,1]^2$, and $S^1$ are all isomorphic as measurable spaces, including their sets of measure 0, as required by the Gelfand-type duality for measurable spaces. For instance, the is …
Dmitri Pavlov's user avatar
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}^ …
Dmitri Pavlov's user avatar
5 votes
Accepted

Free extension of algebra for an operad

This result is true and is due to Spitzweck, Berger–Moerdijk, Fresse, and Elmendorf–Mandell. A complete set of references can be found around Proposition 5.7 in the paper https://arxiv.org/abs/1410.56 …
Dmitri Pavlov's user avatar
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 …
Dmitri Pavlov's user avatar
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 …
Dmitri Pavlov's user avatar
20 votes
0 answers
827 views

Can we define spectral triples using the language of rigged Hilbert spaces?

The traditional mathematical approach to quantum mechanics, as developed by von Neumann, is based on Hilbert spaces and unbounded self-adjoint operators. Another approach, which more closely resembles …
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 …
Dmitri Pavlov's user avatar
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^ …
LSpice's user avatar
  • 12.9k

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