# Bott periodicity homeomorphisms for spaces of Clifford extensions

I am trying to prove the following statement of real Bott periodicity, on the level of actual spaces of Clifford module extensions (i.e., not equivalence classes of modules).

Let $$W = \mathbb{R}^{\infty}$$ (in the direct sum sense), equipped with the standard inner product. Fix in advance a set of skew-symmetric operators $$e_1, e_2, \ldots$$ on $$W$$ for which $$e_i^2 = -I$$ and $$e_ie_j = -e_je_i$$ for $$i \neq j$$. Let $$X(n,W)$$ be the space of operators $$f_n$$ on $$W$$ which anticommute with $$e_1, \ldots, e_{n-1}$$ have $$f_n^2=-I$$, and for which $$\ker(f_n-e_n)^{\perp}$$ is finite dimensional, and equip this space with the topology induced by the operator norm. I wish to show that $$X(n,W) \cong X(n+8,W)$$ where $$\cong$$ denotes homeomorphism, or barring that, homotopy equivalence. Note that $$X(n,W)$$ is the space of orthogonal $$Cl_n$$-module structures on $$W$$ which restrict to the "standard" $$Cl_{n-1}$$-module determined by $$e_1, \ldots, e_{n-1}$$. Here $$Cl_n$$ denotes the real Clifford algebra on $$n$$ generators with negative definite quadratic form.

I hope to make use only of the well-known isomorphism of real Clifford algebras $$Cl_{n+8} \cong Cl_{n}\otimes_{\mathbb{R}} Cl_{8} \cong Cl_{n}\otimes_{\mathbb{R}}\mathbb{R}(16)$$. I'm aware that additional technical assumptions may be necessary, for instance regarding a "complete universe" of representations, but my hope is just to get the basic idea.

I would be content to show that the space of irreducible $$Cl_n$$-module extensions on $$V$$ of appropriate dimension is homeomorphic to the space of $$Cl_{n+8}$$-module extensions on $$V \otimes \mathbb{R}^{16}$$, with the map induced by tensoring with the canonical representation of the $$16 \times 16$$ real matrix algebra $$\mathbb{R}(16)$$.

The main paper I have been following is Behrens - Addendum to "A New Proof of Bott Periodicity".

• A non-mathematical remark: an easy, and I think reasonably standard, way to write "$\mathbb R^\infty$ (in the direct sum sense)" is "$\mathbb R^{\oplus\infty}$". – LSpice Dec 2 '19 at 19:14
• @LSpice And yet $\mathbb{R}^\infty$ is the standard notation in algebraic topology and related fields :). – Denis Nardin Dec 2 '19 at 19:52

We follow the form of this isomorphism: $$Cl_{n+8} \cong Cl_n\otimes_\mathbb{R}Cl_8$$: $$\begin{cases} e_i \mapsto 1\otimes \eta_i, &\text{for } i=1,\ldots,8\\ e_j \mapsto e_{j-8}\otimes \eta_1\eta_2\cdots \eta_8, &\text{for } j=9,\cdots,n+8, \end{cases}$$ where $$\eta_1,\ldots,\eta_8$$ are orthonormal generators of $$Cl_8$$. Taking $$e_i$$ now to denote the operators corresponding to these generators under the "standard" representation of $$Cl_n$$ on $$W$$ and $$\eta_i$$ the operators corresponding to the "standard" representation of $$Cl_8$$ on $$\mathbb{R}^{16}$$, we get a map $$X(n,W) \to X(n+8,W\otimes_{\mathbb{R}}\mathbb{R}^{16})$$, where $$f_n$$ maps to $$f_n\otimes\eta_1\cdots\eta_8$$. This is apparently continuous and injective, but not clearly surjective. So we take a different tack:
Let $$O_{Cl_n}(W)$$ denote the subgroup of $$O(W)$$ which fixes each of $$e_1,\ldots,e_n$$ under conjugation. Then we may identify $$X(n,W)$$, as the homogeneous space $$O_{Cl_{n-1}}(W)/O_{Cl_n}(W)$$ by considering the orbit of $$e_n$$ (actually for $$n\equiv 3(\text{mod }4)$$ there will be multiple homeomorphic orbits corresponding to inequivalent representations but let's ignore this for now).
Similarly $$X(n+8,W\otimes\mathbb{R}^{16}) \cong O_{Cl_{n+7}}(W\otimes\mathbb{R}^{16})/O_{Cl_{n+8}}(W\otimes\mathbb{R}^{16})$$. However any subgroup of $$O(W\otimes\mathbb{R}^{16})$$ which fixes $$1\otimes\eta_1\ldots,1\otimes\eta_8$$ must be of the form $$O(W) \otimes I$$ by the appropriate form of Schur's lemma: $$\eta_1,\ldots,\eta_8$$ give the only irreducible representation of the simple algebra $$Cl_8$$ on $$\mathbb{R}^{16}$$ up to equivalence, so the action of this subgroup must just "permute" these irreps, while acting on them only by $$\pm 1$$. So $$O_{Cl_{n+7}}(W\otimes\mathbb{R}^{16}) \cong O_{Cl_{n-1}}(W)$$ and $$O_{Cl_{n+8}}(W\otimes\mathbb{R}^{16}) \cong O_{Cl_{n}}(W)$$, thus $$X(n+8,W\otimes\mathbb{R}^{16}) \cong X(n,W)$$.