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".

  • 1
    $\begingroup$ A non-mathematical remark: an easy, and I think reasonably standard, way to write "$\mathbb R^\infty$ (in the direct sum sense)" is "$\mathbb R^{\oplus\infty}$". $\endgroup$
    – LSpice
    Dec 2, 2019 at 19:14
  • 3
    $\begingroup$ @LSpice And yet $\mathbb{R}^\infty$ is the standard notation in algebraic topology and related fields :). $\endgroup$ Dec 2, 2019 at 19:52

1 Answer 1


I believe I have the idea of a proof. Apologies for the mess - hopefully it is somewhat intelligible (and somewhat correct).

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 $X(n+8,W\otimes\mathbb{R}^{16}) \cong X(n,W)$.


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