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Let $H$ be a real separable, infinite-dimensional Hilbert space and let $$\mathrm{Cl}(H) = \mathcal{T}(H_{\mathbb{C}}) / \{v\otimes w + w\otimes w - 2\langle v, w\rangle \cdot \mathbf{1} ~|~ v, w \in H_{\mathbb{C}}\}$$ be its algebraic Clifford algebra (where $\mathcal{T}(H_{\mathbb{C}})$ is the algebraic tensor algebra on the complexification $H_{\mathbb{C}}$).

Q: What is known about the automorphism group of this algebra?

There is a lot known for completions of $\mathrm{Cl}(H)$ to various operator algebras.

  • For example, it is a fact that $\mathrm{Cl}(H)$ has a unique $C^*$-norm, and the completion with respect to this norm is the infinite tensor product algebra $M_2(\mathbb{C})^{\otimes \infty}$, which is known to have contractible automorphism group (where "automorphism" means *-automorphism).

  • There are various different completions to von Neumann algebras, the automorphism groups of which have been studied to various degrees.

However, I am not aware of any results regarding the automorphism group of the algebraic Clifford algebra.

  • One more particular question: By the universal property of the Clifford algebra, the orthogonal group $\mathrm{O}(H)$ of $H$ acts on $\mathrm{Cl}(H)$ (via "Boguliubov automorphisms"), and these automorphisms are *-preserving, hence continuous (as observed by Qiaochu Yuan in the comments). But I don't know if these are the only continuous automorphisms, and whether there are non-continuous automorphisms.
  • In fact, I don't even know if the automorphisms of the norm completion of $\mathrm{Cl}(H)$ are all Boguliubov automorphisms (a positive answer would align well with the fact that both $\mathrm{O}(H)$ and $\mathrm{Aut}(M_2(\mathbb{C})^{\otimes \infty}$ are contractible.
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  • $\begingroup$ I think you assume H is separable and infinite-dimensional. Should the 2x2-matrix algebra you mention be over the reals or am I missing something? Also, how is the topology on Cl(H) for which Boguliubov automorphisms are continuous defined, is it the one coming from the C*-norm? $\endgroup$
    – Luuk Stehouwer
    Commented Sep 5, 2022 at 12:51
  • $\begingroup$ Thanks for the suggestion of several edits: Yes, H is infinite-dimensional, separable; the Clifford algebra should be complex, either formed on complexification $H_{\mathbb{C}}$ or, equivalently, tensored with $\mathbb{C}$ in the end. Boguliubov automorphisms are continuous with respect to the C*-norm. $\endgroup$
    – Matthias Ludewig
    Commented Sep 5, 2022 at 20:04
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    $\begingroup$ If the Clifford algebra is known to have a unique $C^{\ast}$-norm then doesn't any automorphism have to preserve that norm? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 5, 2022 at 20:15
  • $\begingroup$ Well, I guess that is true for any $*$-preserving automorphism, I just clarified in the question above. But are there non-$*$-preserving automorphisms? $\endgroup$
    – Matthias Ludewig
    Commented Sep 6, 2022 at 7:10

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