# Why does $C_{n+8}\approx C_n\otimes_\mathbb{R}M_{16}(\mathbb{R})$?

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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You should probably tell us at least what documents you looked at, so that we do not repeat arguments that do not satisfy you—much better would be telling us what is unsatisfying about them, for then we might be able to fix them. – Mariano Suárez-Alvarez Apr 30 '12 at 7:53
@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. – Adelaide Dokras Apr 30 '12 at 19:56

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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This is essentially the same as the argument in Atiyah-Bott-Shapiro paper on Clifford modules (or the minimal subset of ABS necessary to prove the 8-periodicity over R). – zyx Apr 30 '12 at 8:12
I learned it (and extracted it) from Husemoller's Fiber Bundles. – Mariano Suárez-Alvarez Apr 30 '12 at 8:13
+1 Just adding my .02 cents: The first step is a special case of Lemma 4.8.5 in Jacobson's Basic Algebra II (and surely also in many other books covering the basics of Clifford algebras). – Jyrki Lahtonen Apr 30 '12 at 9:44
Thank you for this outline, it is more than sufficient for me to fill out. – Adelaide Dokras Apr 30 '12 at 19:56