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Can matrix representations of clifford algebras of type Cl(0,n) be found? Specifically for even orders

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What do you mean by "matrix representations"? The algebra has faithful modules (itself, for example) and from each you get an algebra of matrices isomorphic to it, if you want... – Mariano Suárez-Alvarez Apr 27 '11 at 3:13
Is your question answered in the following article: ? – S. Carnahan Apr 27 '11 at 3:47
I think matrix generator is a better word. And I am after explicit matrix generators rather than the classification. – user14713 May 3 '11 at 0:56
up vote 6 down vote accepted

The answer is Yes.

It follows from the classification of real Clifford algebras. First of all, write $n = 8p + q$, where $q=0,\dots,7$. Bott periodicity says that $Cl(0,n) \cong Cl(0,q) \otimes_{\mathbb{R}} \mathbb{R}(16p)$, where $\mathbb{F}(N)$ denotes the real associative algebra of $N\times N$ matrices with coefficients in $\mathbb{F}$. Now the classification of real Clifford algebras gives the first 8 $Cl(0,q)$:

$$Cl(0,0) \cong \mathbb{R}$$

$$Cl(0,1) \cong \mathbb{R}\oplus\mathbb{R}$$

$$Cl(0,2) \cong \mathbb{R}(2)$$

$$Cl(0,3) \cong \mathbb{C}(2)$$

$$Cl(0,4) \cong \mathbb{H}(2)$$

$$Cl(0,5) \cong \mathbb{H}(2)\oplus\mathbb{H}(2)$$

$$Cl(0,6) \cong \mathbb{H}(4)$$

$$Cl(0,7) \cong \mathbb{C}(8)$$

Finally, use the fact that the matrix algebras $\mathbb{R}(N)$ and $\mathbb{H}(N)$ have each a unique irreducible representation up to isomorphism (namely, $\mathbb{R}^N$ and $\mathbb{H}^N$) and that $\mathbb{C}(N)$ has two irreducible representations: $\mathbb{C}^N$ and its complex conjugate.

This answers the question as given. Now, you could have also asked (perhaps this is implied) how can go about finding explicit matrix representations. If so, then one way to do this is to use the periodicities: $$ Cl(0,n) \cong Cl(n-2,0) \otimes_{\mathbb{R}} Cl(0,2) $$ and $$ Cl(n,0) \cong Cl(0,n-2) \otimes_{\mathbb{R}} Cl(2,0) $$ together with explicit matrix representations for $Cl(0,2)$, $Cl(2,0)$, $Cl(1,0)$ and $Cl(0,1)$, which are easy to find by hand.

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This is done in "Lectures in Geometry: Lie Groups and Lie Algebras (Semester V)" by Postnikov – Paolo Aceto Apr 27 '11 at 13:33

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