This issue is for logicians and operator algebraists (but also for anyone who is interested).

Let's start by short reminders on von Neumann algebra (for more details, see [J], [T], [W]):

Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators.

Definition: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant:
$M^{*} = M$ and $M'' = M$.

Theorem: The abelian von Neumann algebras are exactly the algebras $L^{\infty}(X)$ with $(X, \mu)$ a standard measure space. They are isomorphic to one of the following:

  • $l^{\infty}(\{1,2,...,n \})$, $n \geq 1$
  • $l^{\infty}(\mathbb{N})$
  • $L^{\infty}([0,1])$
  • $L^{\infty}([0,1]\cup \{1,2,...,n \})$
  • $L^{\infty}([0,1]\cup \mathbb{N})$

Noncommutative philosophy: There are various schools of noncommutative philosophy, here is the school close to operator algebras. This issue is not about philosophy, so I will explain it quickly (for more details see for example the introduction of this book). First an intuitive idea : in the same way as there are classical physics and quantum physics, there are classical mathematics and quantum mathematics. What does it mean in practice ? It means the following : in the classical mathematics there are many different structures, for example, the measurable, topological or Riemannian spaces. The point is to encode each structure by using the framework of commutative operator algebras. For the previous examples, it's the commutative von Neumann algebras, C$^{*}$-algebras and spectral triples. Now if we take these operator algebraic structures and if we remove the commutativity, we obtain what we call noncommutative analogues : noncommutative measurable, topological or Riemannian spaces.
This school explores noncommutative analogues of more and more structured objects, it goes in one direction. My point is to question about the other direction (back to the Source) :
What's the noncommutative analogue of a set (called a noncommutative set) ?

What is a noncommutative set?

The von Neumann algebras of the standard measure space $[0,1]$, $[0,1]\cup \{1,2,...,n \}$ and $[0,1]\cup \mathbb{N}$ are not isomorphic, but as sets, these spaces are isomorphic (i.e., same cardinal).

Is there a natural equivalence relation $\sim$ on the von Neumann algebras, forgetting the measure space but remembering the set space, on abelian von Neumann algebras?

Remark: If $M \sim N$, we could say that they are isomorphic as noncommutative sets.
The equivalence class could be called the quantum cardinal (a link with cyclic subfactor theory?).

Are there noncommutative analogues of the ZFC axioms ?

What I'm looking for seems different of what is called quantum set in the literature...

  • 6
    $\begingroup$ The NC philosophy did not begin with Connes, of course $\endgroup$
    – Yemon Choi
    Jul 23, 2013 at 20:24
  • 5
    $\begingroup$ You're right @YemonChoi, the NC philosophy began with the pioneers of the quantum mechanics, in particular W. Heisenberg (for the NC aspect). And of course, it also began with J. von Neumann. However, A. Connes is one of the greatest modern representative of the NC philosophy. $\endgroup$ Jul 23, 2013 at 21:07
  • 5
    $\begingroup$ I was thinking more of less trendy names such as Segal, Stinespring, Takesaki, Effros... (not to mention the initial pioneers of K-theory for C-star algebras). My impression as a non-expert is that the key aspect of Connes's NCG, at least initially, is that it is NCDG (see the title of his long IHES paper) $\endgroup$
    – Yemon Choi
    Jul 24, 2013 at 0:15
  • 10
    $\begingroup$ It seems rather misguided to talk of "non-commutative $\mathsf{ZFC}$" in this context. $\mathsf{ZFC}$ is not a theory about sets of reals, or sets of size "close" to the size of the reals, while the objects of your question are, by virtue of the extra structure you impose of them. $\endgroup$ Jul 25, 2013 at 1:43
  • 5
    $\begingroup$ @SébastienPalcoux " the cardinal of $B(H)$ is $\aleph_2$(if I'm not mistaken...) " You are. And please do not tell a set theorist what a paradise for set theorists is. $\endgroup$ Jul 25, 2013 at 13:53

4 Answers 4


I suggest that you look into the paper Quantum Collections by Andre Kornell, available here on the arXiv. The abstract reads:

We develop the viewpoint that the opposite of the category of $W^\ast$-algebras and unital normal $\ast$-homomorphisms is analogous to the category of sets and functions. For each pair of $W^\ast$-algebras, we construct their free exponential, which in the context of this analogy corresponds to the collection of functions from one of these $W^\ast$-algebras to the other. We also show that every unital normal completely positive map between $W^\ast$-algebras arises naturally from a normal state on their free exponential.

  • 2
    $\begingroup$ I guess I should mention that Andre also has another paper called Quantum Functions that is concerned with similar notions. $\endgroup$
    – MTS
    Jul 23, 2013 at 19:32
  • $\begingroup$ Thank you @MTS for this answer. These papers seem close to what I'm looking for. I will browse them to know if $^{\omicron}M \simeq$ $^{\omicron}N \Leftrightarrow M \sim N$, with $\sim$ the equivalence relation I have defined in my answer (in particular if $^{\omicron}L^{\infty}(X) \simeq$ $^{\omicron}L^{\infty}(Y) \Leftrightarrow card(X) = card(Y)$). Maybe the functor in these papers forget more than mine. What do you think ? Maybe you already have a sufficient knowledge of these papers to enlighten me. $\endgroup$ Jul 23, 2013 at 20:57
  • $\begingroup$ Sebastian, I don't have a good enough understanding of the technical contents to answer your question, sorry. But I hope that the papers will be helpful to you. $\endgroup$
    – MTS
    Jul 23, 2013 at 21:03

You ask if there are noncommutative analogues of the ZFC axioms. I doubt there are any which provide a workable theory distinct from ZFC.

One key piece of evidence is that natural noncommutative axioms for arithmetic yield a theory identical with Peano Arithmetic. This is a theorem from Michael Dunn.


Andre Kornell wrote a new paper entiteld Quantum sets, available on arxiv from yesterday: arXiv:1804.00581. The first sentence of the paper is the following:

This paper concerns the quantum generalization of sets, in the sense of noncommutative geometry.

Robin Giles and Hans Kummer defined (in this paper published in 1971) a quantum set as an atomic von Neumann algebra (i.e. whose projection lattice is an atomic lattice, which means that every projection contains a minimal projection), in other words, a $\ell^{\infty}$-direct sum of type ${\rm I}$ factors.

The above paper of Andre Kornell takes a smaller class: there, a quantum set is a direct sum of type ${\rm I}$ finite factors, i.e. a direct sum of matrix algebras $M_n(\mathbb{C})$.

These notions should be called atomic noncommutative sets (of type ${\rm I}$).


It could first be relevant to notice that a projection in a von Neumann algebra can also be seen as a noncommutative analog of a set, but the answer below go in a different way.

Let's quickly recall the Tomita-Takesaki modular theory and conditional expectation.

Modular theory : Let $M \subset B(H)$ be a von Neumann algebra. Let $\Omega \in H$ be a cyclic and separating vector (i.e., $M.\Omega$ and $M'.\Omega$ are dense in $H$). Let $S : H \to H$ be the closure of the anti-linear map $x\Omega \to x^{*}\Omega$. Then, $S$ admits a polar decomposition $S = J\Delta^{1/2}$, with $J$ anti-linear unitary and $\Delta$ positive. Then, $JMJ = M'$ and $\Delta^{it} M \Delta^{-it} = M$.
Let $\sigma_{\Omega}^{t}(x) = \Delta^{it} x \Delta^{-it}$ the modular action of $\mathbb{R}$ on $M$.

Conditional expectation (Takesaki 1972) : Let $N \subset M$ be an inclusion of von Neumann algebra, then there is a conditional expectation of $M$ onto $N$ with respect to $\Omega$ (cyclic and separating) if $N$ is invariant under the modular action (i.e., $\sigma_{\Omega}^{t}(N) = N)$.
Notation : if $\exists \Omega$ verifying the previous conditions, we note $N \subset_{e} M$.

Remark : The modular theory is trivial for $M = L(\Gamma) \subset B(H)$, with $\Gamma$ a discrete group and $H = l^{2}(\Gamma)$ (because $\Delta = I$). In particular, it's trivial for the abelian von Neumann algebras.
As a consequence, in this case: $N \subset M$ $\Leftrightarrow$ $N \subset_{e} M$.

Notation : Let $N$ and $M$ be two von Neumann algebras.
If $\exists P \simeq N$ such that $ P \subset_{e} M$, we note $N \hookrightarrow_{e} M$.

Equivalence relation : $M \sim N$ if $N \hookrightarrow_{e} M \hookrightarrow_{e} N$.

Examples :

  • Among $l^{\infty}(\{1,2,...,n \})$, $l^{\infty}(\mathbb{N})$ and $L^{\infty}([0,1])$ none is equivalent to another.
  • $L^{\infty}([0,1])$, $L^{\infty}([0,1]\cup \{1,2,...,n \})$ and $L^{\infty}([0,1]\cup \mathbb{N})$ are pairwise equivalent,
    because $L^{\infty}([0,1]) \subset L^{\infty}([0,1] \cup \{2,3,...,n\}) \subset L^{\infty}([0,1] \cup \mathbb{N}_{\geq 2}) \hookrightarrow L^{\infty}(\mathbb{R})$
    and $L^{\infty}([0,1]) \simeq L^{\infty}(\mathbb{R})$
  • Obviously $L^{\infty}([0,1]) \not\sim B(H)$.
  • Let $R \subset B(H)$ be the hyperfinite $II_{1}$ factor, $R_{\infty} = R \otimes B(H)$ the hyperfinite $II_{\infty}$ factor. $ B(H) \hookrightarrow_{e} R_{\infty} \hookrightarrow_{e} B(H \otimes H)$ and $B(H) \simeq B(H \otimes H)$. So, $R \not\sim B(H) \sim R_{\infty}$.
  • Let $\Gamma$ be a non-amenable ICC discrete group. Then $L(\Gamma) \not\hookrightarrow_{e} B(H)$ and $L_{\infty}(\Gamma) = L(\Gamma) \otimes B(H) \not\hookrightarrow_{e} B(H \otimes H) $ so $L(\Gamma) \not\sim B(H) \not\sim L_{\infty}(\Gamma)$.
  • Let $\mathbb{F}_{2} = \langle a,b \vert \ \rangle $ and $\mathbb{F}_{\infty} = \langle a_{1},a_{2},... \vert \ \rangle $.
    Then $\mathbb{F}_{2} \hookrightarrow \mathbb{F}_{n} \hookrightarrow \mathbb{F}_{\infty} \hookrightarrow\mathbb{F}_{2} $ (the last injection is given by $a_{n} \to b^{-n}ab^{n}$).
    Consequence : $L(\mathbb{F}_{2}) \sim L(\mathbb{F}_{n}) \sim L(\mathbb{F}_{\infty}) $

Question : Is the fundamental group $\mathcal{F}(M)$ of a $II_{1}$ factor $M$, invariant under $\sim$ ?
Remark : an affirmative answer would solve the free group factor isomorphism problem.

  • 4
    $\begingroup$ With regards to your last questions it is equivalent to the free group factor isomorphism problem. Specifically $L(\mathbb{F}_n)\sim L(\mathbb{F}_\infty)\Leftrightarrow L(\mathbb{F}_n)\simeq L(\mathbb{F}_\infty)$. This is because in the non-isomorphic case you need $P\otimes N\simeq P$ and these factors are prime. More generally we have that if $M, N$ are prime $II_1$ factors then $M\sim N\Leftrightarrow M\simeq N$. $\endgroup$ Jul 23, 2013 at 23:43
  • $\begingroup$ You're right @OwenSizemore, thank you. After your comment, I have completely recasted my answer and improved the definition of $\sim$, in order to keep the observation on the abelian case, and to have $L(\mathbb{F}_{2}) \sim L(\mathbb{F}_{\infty})$ independently of the free group factor isomorphism problem ! $\endgroup$ Jul 24, 2013 at 14:03
  • $\begingroup$ The last question has been posted here. $\endgroup$ Jul 26, 2013 at 8:38

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