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In Atiyah, Shapiro, and Bott's paper on Clifford modules, they prove Proposition 4.2 on page 11 that there are isomorphisms $C_k\otimes_\mathbb{R} C_2^\prime\cong C_{k+2}^\prime$ and $C_k^\prime\otimes\mathbb{R} C_2\cong C_{k+2}$.

Immediately following the proof, they say it is clear that $C_2\cong\mathbb{H}$ and $C_2^\prime\cong\mathbb{R}(2)$. I get that $C_1\cong\mathbb{C}$ and $C_1^\prime\cong\mathbb{R}\oplus\mathbb{R}$, but those don't seem to be of use with the isomorphisms they proved.

Is there a quick explanation of how those isomorphisms for $C_2$ and $C_2^\prime$ are so easily seen? Thanks.

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up vote 2 down vote accepted

Here we let $Q(x) = -x^2 - y^2$ and $Q'(x) = x^2 + y^2$, so that $$C_2 = \mathrm{C}\ell(\mathbb{R}^2, Q)$$ and $$C^\prime_2 = \mathrm{C}\ell(\mathbb{R}^2, Q').$$

Let $\{e_1, e_2\}$ be an orthornomal basis for $(\mathbb{R}^2,Q)$. Then $C_2$ has generators $\{1, e_1, e_2, e_1 e_2\}$ satisfying the relations $$e_1^2 = e_2^2 = -1,$$ $$e_1 e_2 = -e_2 e_1.$$ Then the map $C_2 \longrightarrow \mathbb{H}$ defined on generators by $1 \mapsto 1$, $e_1 \mapsto i$, $e_2 \mapsto j$, and $e_1 e_2 \mapsto k$ is seen to be the desired isomorphism.

Now let $\{e_1^\prime, e_2^\prime\}$ be a basis for $(\mathbb{R}^2, Q')$. Then $C_2^\prime$ has generators $\{1, e_1^\prime, e_2^\prime, e_1^\prime e_2^\prime\}$ satisfying the relations $$(e_1^\prime)^2 = (e_2^\prime)^2 = 1,$$ $$e_1^\prime e_2^\prime = -e_2^\prime e_1^\prime.$$ Our desired isomorphism is given by $$C_2^\prime \longrightarrow \mathbb{R}(2),$$ $$a + be_1^\prime + ce_2^\prime + de_1^\prime e_2^\prime \mapsto \begin{pmatrix} a + b & c + d \\ c - d & a - b \end{pmatrix}.$$

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