As explained in the book "Spinors in Hilbert Space" by Plymen and Robinson, if $V$ is a complex (separable) Hilbert space with a real structure, and $\mathrm{Cl}(V)$ the corresponding Clifford algebra, there is a unique completion $\mathrm{Cl}[V]$ of this algebra to a $C^*$-algebra: Any $*$-representation of $\mathrm{Cl}(V)$ on a Hilbert space will induce that same $C^*$-norm on it, and the corresponding closure is the Clifford $C^*$-algebra $\mathrm{Cl}[V]$.

The situation is different for von-Neumann algebras. If $\pi: \mathrm{Cl}(V) \rightarrow B(\mathcal{H})$ is a $*$-representation, then the von-Neumann-closure $M_\pi := \pi(\mathrm{Cl}(V))^{\prime\prime}$ will depend drastically on $\pi$. For example, if $\pi$ is the left regular representation, then $M_\pi$ is the hyperfinite $\mathrm{II}_1$ factor, if $\pi$ is a Fock representation, then it is type $\mathrm{I}_\infty$, and there are other settings where we obtain type $\mathrm{III}$ factors.

However, there is another canonical von-Neumann algebra containing $\mathrm{Cl}(V)$, namely the universal enveloping algebra, coming from the case where $\pi$ is the universal representation of $\mathrm{Cl}[V]$. Alternatively, it is the double dual of $\mathrm{Cl}[V]$.

Q: What can we say about the enveloping von-Neumann-algebra of $\mathrm{Cl}[V]$? Does it happen to be a factor? Is it possibly the hyperfinite $\mathrm{II}_1$ factor?

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    $\begingroup$ No way! Every von Neumann completion is a quotient of the universal enveloping algebra. It's as far from being a factor as you can get. $\endgroup$ – Nik Weaver May 12 at 5:51
  • $\begingroup$ I know that there is a unique map to every von-Neumann completion; but how do you see it is surjective? $\endgroup$ – Matthias Ludewig May 12 at 12:13
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    $\begingroup$ Any weak* closed ideal of a von Neumann algebra $M$ has the form $pM$ for some central projection $p$. The quotient $M/pM$ is isomorphic to $(1-p)M$. Thus the image of $M$ under any normal (i.e., weak* continuous) $*$-homeomorphism is a von Neumann algebra. $\endgroup$ – Nik Weaver May 12 at 15:38

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