$\newcommand\CAR{\mathit{CAR}}\newcommand\Cl{\mathbb C\mathit l}$This question will be rather long and it will be my attempt to finally clarify many issues concerning CCR, CAR and Clifford algebras together with the Fock spaces and the classification of the corresponding von Neumann algebras appearing in this way. I must say that I'm very confused by all this stuff. Let me summarize what I have learned so far. Let $V$ be a Hilbert space (over $\mathbb{R}$ I suppose—at least for the Clifford algebra):

  1. First of all, let me emphasize the slight difference between the so called CAR (canonical anticommutation relation) algebra over a Hilbert space and the Clifford algebra (in the $C^*$-algebraic version, to be denoted by $\Cl^*(V)$: these two are very closely related. However, note that the Clifford algebra is generated by all $j(v)$'s where $j:V \to \Cl^*(V)$ is the natural embedding (in other words this algebra is the universal $C^*$-algebra satisfying Clifford relations) while the CAR algebra is generated by all creation and anihilation operators (abstractly) so is generated by a priori larger set of relations.

Question 1 Is it true that abstractly $\CAR(V)$ and $\Cl^*(V)$ are isomorphic as $C^*$-algebras?

  1. Now it is shown in Plymen's book „Spinors in Hilbert space‟ that given the von Neumann version of the Clifford algebra (over infinite dimensional $V$) acting via the left regular representation (meaning the GNS representation coming from the trace) one obtains the hyperfinite $II_1$ factor. This $II_1$ hyperfinite factor is unique (up to isomorphism) and admits various incarnations: one of them is an infinite tensor product of $M_2(\mathbb{C})$. This means that $\operatorname{v.N.}\Cl(V)$ acts naturally on the infinite tensor product.
  2. On the other hand, there are plenty of nonequivalent representations of this Clifford algebra, constructed with the help of the Fock space. The construction of this Fock space requires the choice of the so called unitary (or complex) structure $J$ on our real Hilbert space $V$. Different choices of $J$ can lead to inequivalent representations and the condition for equivalence is the content of Shale–Stinespring theorem and is of the form that $J-K$ should be a Hilbert–Schmidt operator.
  3. One shows that Fock representations are irreducible: therefore (the v.N. version of) our Clifford algebra will be type $I_{\infty}$. However, we also stated that in the left regular representation our Clifford algebra is just the infinite tensor product $\bigotimes M_2(\mathbb{C})$ which is of type $II_1$. Therefore a natural question arises:

Question 2 a) Is there a way to find a representation $\pi$ of $\Cl^*(V)$ such that $(\pi(\Cl^*(V)))''$ is of type $III$? Is the Hilbert space of this representation some kind of infinite tensor product (or wedge product)?
b) Is there a some sort of canonical isomorphism between Fock space and the infinite tensor product (possibly depending whether we take full, or symmetric or antisymmetric Fock space) allowing us to see Fock representation as a representation on this infinite symmetric tensor product?

  1. Recall that the Fock representation $\pi$ is defined as $\pi(v)=a^+(v)+a^-(v)$ and turns out to be irreducible: thus the von Neumann algebra generated by $\{\pi(v), v \in V\}$ will be the same as the von Neumann algebra generated by $\{a^+(v),a^-(v): v \in V \}$ both being type $I_{\infty}$. Note that the family $\{\pi(v): v \in V\}$ consists from self adjoint operators but the operators $\pi(v)$ do not commute with each other—unlike in the bosonic case for which I would like to turn now.
  2. So in the bosonic case the analogue of the algebraic version of Clifford algebra is the so called Weyl algebra—however one can show that it is impossible to represent a Weyl algebra as bounded operators in some Hilbert space (unlike for Clifford algebra)—therefore one considers the associated canonical commutation relation in the integrated form. One then considers the universal $C^*$-algebra generated by these relations. Recall that for the Clifford algebra we had a particular way to represent it as a type $II_1$ hyperfinite factor (using the trace).

Question 3 a) Is there a canonical „left regular representation‟ picture for CCR? If the answer is yes, what is the corresponding von Neumann algebra (its type? Is it a factor)?
b) Is there a natural and canonical way to represent CCR algebra as an infinite tensor product?

  1. For CCR one can again speak about Fock representations and this representation in the bosonic case turns out to be again irreducible: therefore the von Neumann algebra in this representation would be again of type $I_{\infty}$. But note the subtle difference here: the family of field operators $\pi(v):=a^+(v)+a^{-}(v)$ while unbounded, still consists from self adjoint operators but this time this family is commutative! Therefore it should produce some measure: I suspect that this family should produce some variant of Gaussian measure but still I would like to know the details:

Question 4 a) If we consider a von Neumann algebra generated by the family $\{\pi(v): v \in V\}$ it should be naturally isomorphic to $L^{\infty}(X,\mu)$. Can one identify the space $(X,\mu)$ concretely?
b) is there a way to arrive at some canonically defined measure starting from CAR instead of CCR? If yes, what is the corresponding measure space? (Some relation with the Ising model?)

  1. Finally there is also a free (sometimes called full) version of the Fock space where one does not use any symmetrization and antisymmetrization. Here the creation and anihilation operators satisfy the following commutation relation: $a^-(u)a^+(v)=\langle u,v \rangle$. Quite surprisingly (at least for me) Voiculescu in his paper „Symmetries of some reduced free product of $C^*$-algebras‟ has proved that the von Neumann algebra generated by $\{a^+(v)+a^-(v): v \in V\}$ in the Fock representation is isomorphic with the von Neumann algebra generated by the free group where the numbers of generators is equal to $\dim V$. So let me ask a final question:

Question 5 What von Neumann algebra we get by taking the von Neumann algebra $\{a^+(v),a^-(v): v \in V\}''$ generated by all creation and anihilation operators in the free Fock space?

  • $\begingroup$ Lecture Notes on Analysis of Operators by Antony Wassermann should be helpful. It is a ps file, you may need to convert it into pdf. $\endgroup$ Aug 26, 2022 at 6:01
  • $\begingroup$ Concerning your question 1, what is your reference for a rigorous definition of $CAR(V)$? $\endgroup$ Aug 26, 2022 at 9:36

1 Answer 1


Wow, that's a lot of questions. For a more in-depth discussion you might look at my book Mathematical Quantization, particularly Section 2.5 and Chapters 4 and 7. But anyway, let me start with the most interesting questions, 2b and 3a.

Question 2b. Are Fock spaces infinite tensor products? Yes. First, remember that full infinite tensor products of Hilbert spaces are nonseparable and horrible. The version of ``tensor product'' you want to look at is, for any family of Hilbert spaces $H_i$ with distinguished unit vectors $u_i$, the closed subspace of their full tensor product generated by the elements $\bigotimes v_i$ with each $v_i \in H_i$ and $v_i = u_i$ for all but finitely many $i$.

Given that definition of tensor product, let $X$ be a countable set and let $\{e_0, e_1\}$ be the standard basis of $\mathbb{C}^2$. Then $\bigotimes_X \mathbb{C}^2$, using the above definition with $u_i = e_0$ for all $i \in X$, has a natural orthonormal basis indexed by all finite subsets of $X$. To each such subset $S$ we associate the vector $\bigotimes v_i$ where $v_i = e_1$ if $i \in S$ and $v_i = e_0$ if $i \not\in S$. These vectors constitute an orthonormal basis of $\bigotimes_X \mathbb{C}^2$.

This answer is already getting a bit long, so I'll leave it to you to check that the span of the basis vectors coming from singleton subsets of $X$ can be identified with $l^2(X)$; and for arbitrary $n$, the span of the basis vectors coming form subsets of $X$ of size $n$ can be identified with the antisymmetric $n$-fold tensor power of $l^2(X)$. That is, $$\bigotimes_X \mathbb{C}^2 \cong H_0 \oplus H_1 \oplus \cdots$$ where $H_n$ is the $n$-fold antisymmetric tensor power of $l^2(X)$. I.e., it is the fermionic Fock space over $l^2(X)$.

This relates to quantum field theory in the following way. In QFT we have a Hilbert space $H_x$ at each point $x$ of space, or relativistically at each point of some maximal spacelike surface. And morally, a state of the field is an element of the tensor product of all these Hilbert spaces. If you take space to be a discrete set $X$, and let the field have two possible states at each point (one ground state and one excited state), then we can take the field Hilbert space to literally be $\bigotimes_X \mathbb{C}^2 \cong \mathcal{F}_a(l^2(X))$. This is a fermionic field. If space is not discrete, $X = \mathbb{R}^3$ for example, then the natural Hilbert space for the field is $\mathcal{F}_a(L^2(\mathbb{R}^3))$, which is morally $\bigotimes_{\mathbb{R}^3}\mathbb{C}^2$ for some notion of ``measurable tensor product''.

Going back to the discrete case, if we give the field infinitely many states $e_n$ ($n \in \mathbb{N}$), spanning a copy of $l^2$, at each point of $X$, then the natural orthonormal basis now corresponds to finite multisubsets of $X$ and we get an isomorphism $\bigotimes_X l^2 \cong \mathcal{F}_s(l^2(X))$. This is the bosonic case.

So yes, Fock spaces of the two types can be identified literally with infinite tensor powers of either $\mathbb{C}^2$ or $l^2$ over a discrete set, or morally with infinite tensor products over a measure space.

Question 3a. Do CCR algebras have canonical ``left regular'' representations? Yes. First, let $H$ be a finite-dimensional real Hilbert space. For each $v \in H$ define unbounded self-adjoint operators $Q_v$ and $P_v$ on $L^2(H)$ by $$Q_vf(w) = \langle w,v\rangle f(w)\qquad{\rm and}\qquad P_vf(w) = -i\hbar\frac{\partial f}{\partial v}(w).$$ Then $CCR(H)$ is the C*-algebra generated by the unitary operators $e^{iQ_v}$ and $e^{iP_v}$ for $v \in H$.

When $H$ is infinite you can't understand $L^2(H)$ as a genuine $L^2$ space, but you can take it to be a direct limit of genuine $L^2$ spaces for all finite subspaces of $H$. I'm just going to refer you to Section 7.2 of my book for details.

I guess I can answer some of the other questions too.

Question 1. Yes, Theorem 3.19 in these notes. This is basically just a calculation.

Question 2a. It's a theorem of Glimm (``Type I C${}^*$-algebras'', Ann. of Math. 73 (1961), 572-612) that for any C${}^*$ algebra $A$, either $\pi(A)''$ is type I for every representation or $\pi(A)''$ can be any type.

Question 3b. Literally yes for $CCR(H)$ when $H$ is finite-dimensional; it's naturally the tensor product of the algebras $CCR(E)$ where $E$ ranges over all the one-dimensional subspaces of $H$ generated by the elements of some fixed orthonormal basis. Morally this is still true when $H$ is infinite-dimensional, but not literally because (in the $L^2(H)$ model mentioned above) the operators $e^{iQ_v}$ and $e^{iP_v}$, for any $v$ not in the span of finitely many basis vectors, cannot be approximated in norm by tensor products.

Question 4a. Yeah, in the $L^2(H)$ model again you'd now be looking at the C${}^*$-algebra generated by just the operators $e^{iQ_v}$. These all commute, and the C${}^*$-algebra they generate is just the algebra of almost periodic functions on $H$ acting as multiplication operators. For finite-dimensional $H$ the von Neumann algebra they generate would just be $L^\infty(H)$, acting by multiplication. For infinite-dimensional $H$ this doesn't really work and I think the von Neumann algebra there is pretty horrible.

Question 4b. I guess the obvious thing to do here is to go back to the $\mathcal{F}(l^2(X))$ model and identify the finite subsets of $X$ with elements of the product space $\{0,1\}^X$, giving $\{0,1\}$ counting measure.

I haven't thought about Question 5.

  • $\begingroup$ Wow, this is a great answer, thank you! I will accept the answer and post the remaining question 5 for another discussion (it has more to do with the so called ,,free probability theory'') $\endgroup$
    – truebaran
    Aug 28, 2022 at 7:57
  • $\begingroup$ Thank you! Glad I could help. $\endgroup$
    – Nik Weaver
    Aug 28, 2022 at 14:18

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