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



On matrix representations of the Clifford algebras of type $Cl(0,n)$
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.


