I've been considering the Clifford algebra over $\mathbb{R}$. For notation, I denote 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.

I'm aware of the basic isomorphisms $C_1\approx\mathbb{C}$ and $C'_1\approx\mathbb{R}\times\mathbb{R}$ and the like. I was looking further into one of my reference books though, and there are the more general isomorphisms $$ C_{n+2}\approx C'_n\otimes_\mathbb{R} C_2;\qquad C'_{n+2}\approx C_n\otimes_\mathbb{R} C'_2. $$

No proof is offered though, and I don't see how these isomorphisms are established. Is there a more in depth reference where these are exhibited, or even a proof or sketch here? Cheers.


Let $\{e_1, \dots, e_{n+2}\}$ be an orthonormal basis for $\mathbb{R}^{n+2}$ (with respect to the standard inner product), $\{e^\prime_1, \dots, e^\prime_n\}$ the standard generators for $C^\prime_n$ and $\{e^{\prime\prime}_1, e^{\prime\prime}_2\}$ the standard generators for $C_2$. Now we define a map $$f: \mathbb{R}^{n+2} \longrightarrow C^\prime_n \otimes C_2$$ on generators by $$f(e_i) = \left\{ \begin{array}{l l} e^\prime_i \otimes e^{\prime\prime}_1 e^{\prime\prime}_2, & 1 \leq i \leq n, \\ 1 \otimes e^{\prime\prime}_{i-n}, & i = n + 2, i = n + 2 \end{array} \right.$$ and on all of $\mathbb{R}^{n+2}$ by extending linearly. It is easy to see that $$f(x)f(x) = \|x\|^2 (1 \otimes 1),$$ so $f$ is a Clifford map and hence extends uniquely to an algebra homomorphism $$\tilde{f}: C^\prime_{n+2} \longrightarrow C^\prime_n \otimes C_2.$$ $\tilde{f}$ is surjective since it maps onto a set of generators for $C^\prime_n \otimes C_2$, and since $\dim(C^\prime_{n+2}) = \dim(C^\prime_n \otimes C_2)$, $\tilde{f}$ is in fact an isomorphism.

The isomorphism $C_{n+2} \cong C_n \otimes C^\prime_2$ is constructed analogously.

  • $\begingroup$ Thanks Henry. Forgive me if I'm missing something, be you define a map $f$ into $C'_n\otimes C_2$ and then extend it to $\tilde{f}$ into $C_n\otimes C'_2$. Is it still supposed to be into $C'_n\otimes C_2$, or how did the primes switch places? $\endgroup$ – Adelaide Dokras Apr 30 '12 at 3:03
  • $\begingroup$ @Adeal: Ah, it's a typo. I was trying to stick to your notation but it's more natural for me to think of the Clifford algebra corresponding to the positive definite quardratic form on $\mathbb{R}^n$ as being "unprimed." Is everything else clear? $\endgroup$ – Henry T. Horton Apr 30 '12 at 3:06
  • $\begingroup$ Thanks, I just wanted to make sure. And to construct the second isomorphism, the proof follows the same way, except I would replace the generators for $C'_n$ with those of $C_n$, and the generators of $C_2$ with those of $C'_2$, correct? $\endgroup$ – Adelaide Dokras Apr 30 '12 at 3:10
  • $\begingroup$ @Adeal: Yes, that is correct. $\endgroup$ – Henry T. Horton Apr 30 '12 at 3:11
  • $\begingroup$ Everything else is clear then, thank you. $\endgroup$ – Adelaide Dokras Apr 30 '12 at 3:15

There is pretty good information on this at Wikipedia. The recurrence of smaller Clifford algebras in larger ones is a manifestation of Bott periodicity.


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