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I recently came across this post by Valter Moretti concerning the utility of von Neumann algebras in mathematical physics. In it, he mentions

The closure of von Neumann algebras with respect to the strong operator topology is the feature that makes them more interesting than simple $C^*$-algebra of operators. Simultaneously it makes them very rigid structures not suitable to describe some quantum systems in general situations (e.g., locally covariant QFT in curved spacetime).

What sort of undesirable features do von Neumann algebras have to disconsider them when doing Quantum Field Theory in Curved Spacetime?

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There are several issues.

The modern approach to QFT in CST considers the $*$-algebra of observables ${\cal A}$ as a fundamental notion. This approach permits, in particular, to define a theory in all globally hyperbolic spacetimes simultaneously (using a functorial description) and, from this perspective, to take advantage of locality and covariance in a very broad sense.

For instance, this approach leads to the noticeable result that the values of renormalization constants of a given theory (abstractly formulated) must be the same in every globally hyperbolic spacetime.

In this framework, states are algebraic: positive functionals $\omega : {\cal A}\to \mathbb{C}$. The the Hilbert space (von Neumann) perspective is recovered through the so called GNS construction. Given a state as before, it provides a Hilbert space, a representation of the algebra ${\cal A}$ in terms densely defined operators in the GNS Hilbert space, and a cyclic vector: the "vacuum" of the theory.

The double commutant of the concrete algebra of operators in the GNS Hilbert space, in principle, gives rise to a von Neumann algebra of observables in that Hilbert space.

An immediate problem arises from the fact that this new structure is not invariant under the $C^*$ or $*$-algebra morphisms relevant at the level of abstract algebra ${\cal A}$.

That is because this structure is connected to the uniform norm (the $C^*$ case) or it is not topological ($*$ algebra case), whereas von Neumann algebras are topological objects fundamentally related to the weak and strong operator topologies, which are weaker than the uniform one.

For instance, suppose that an abstract observable (an element $a$ of ${\cal A}$) is represented by a selfadjoint operator $A$ in the GNS Hilbert space associated to an algebraic state. $A$ is a very complex object, in particular it has an associated spectral measure made of orthogonal projectors with an operationistic meaning (the "elementary propositions" which refer to $A$).

There is no way to see this associated structure as a part of ${\cal A}$. In fact, a spectral measure is a topological structure connected with the strong operator topology of the Hilbert space (the spectral projectors belong to the von Neumann algebra generated by $A$) and there is no chance to lift it to the more abstract level of ${\cal A}$.

In this sense a von Neumann algebra is too rigid to be used in algebraic QFT in curved spacetime if one want to deal with the power of the locally covariant formulation. It may make sense However, once a state has been chosen and in a specific spacetime.

  • This makes sense to me from the concrete perspective. However, couldn't one require that $\mathcal{A}$ has a predual, and hence ask that it be a von Neumann/$W^*$-algebra? – Níckolas Alves Jan 27 '24 at 20:32
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    I do not know, maybe someone tried that way. As far as I know the standard status is the one I illustrated above. Personally, even if I spent a large part of my career by dealing with these topics, I find today all this formalism too far from physics. In my view, in spite of an enormous effort this outstanding mathematical machinery did not succeed in solving fundamental problems. First of all why renormalization? – Valter Moretti Jan 27 '24 at 22:46
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    To be completely frank, I do not know if I believe anymore in the algebraic formalism. I cannot see any cogent physical reason why observables should form an algebra. If I know the instruments to measure a pair of incompatible (bounded) observables and , the mere structure of algebra does not suggest any instrument to measure + or +. Hence it is disputable that these combinations are observables as well. Physics should decide, not mathematics. This is my present viewpoint. – Valter Moretti Jan 27 '24 at 22:47