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In an earlier answer, rschwieb kindly pointed me in the direction of Bott periodicity. Just out of curiosity I was reading through a paper on periodicity of Clifford algebras. There was a list of isomorphisms, "all of them easy to prove according to the author," but the last one I couldn't really work out at all. I think it's pretty well known, the isomoprhism in question is $ C_{n+8}\approx C_n\otimes_\mathbb{R}M_{16}(\mathbb{R}) $ regardless of whether $C$ is the clifford algebra associated with a positive or negative definite form.

This isomorphism is in a lot of documents that popped up on google, but nowhere a satisfying proof. Does someone have one here?

As for notation, I'm denoting the Clifford algebras $C_n$ associated with the vector space $\mathbb{R}^n$ with negative definite form, and $C'_n$ associated with $\mathbb{R}^n$ with positive definite form, although from what I understand the isomoprhism is true in either case.

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    @MarianoSuárez-Alvarez The paper I was reading was _Modulo (1,1) Periodicity of Clifford algebras_ by J.G. Maks. I wanted to link to it, but, at least for me, it induces an automatic download of the pdf.2012-04-30

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The argument I know, and which satisfies me is:

  • Show first that $C_{k+2}\cong C_k'\otimes C_2$ and $C_{k+2}'\cong C_k\otimes C_2'$ by exhibiting isomorphisms.

  • Notice that this implies that $C_4\cong C_2\otimes C_2'$.

  • Using the first point twice, $C_{k+4}\cong C_k\otimes C_2\otimes C_2'$ and, by the second point, this is $C_k\otimes C_4$.

  • Using this last point twice now, we see that $C_{k+8}\cong C_k\otimes C_4\otimes C_4$.

  • Finally, show by hand that $C_4\cong M_2(\mathbb H)\cong M_2(\mathbb R)\otimes\mathbb H$, and, using this, that $C_4\otimes C_4\cong M_2(\mathbb R)\otimes M_2(\mathbb R)\otimes\mathbb H\otimes\mathbb H\cong M_{16}(\mathbb R)$, because $\mathbb H\otimes\mathbb H\cong M_4(\mathbb R)$.

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    Thank you for this outline, it is more than sufficient for me to fill out.2012-04-30