1
$\begingroup$

I'm currently looking at a $8$-dimensional $\mathbb{R}$-algebra denoted either by $C_n$ or $C(a_1, a_2, a_3)$. After some looking around, I think this is called a Clifford algebra? If not, I apologise for the incorrect title!

We consider the basis of $C_n$ to be $e_I$, where $I$ ranges over the subsets of $\{1,2,3\}$. We also have the following rules for multiplication:

1) $e_\emptyset = 1$

2) $e_i e_j = - e_j e_i, i \neq j$

3) $e_i^2 = 1$

I want to show that the centre of $C(a_1, a_2, a_3)$ is isomorphic to a quotient: $C(a_1, a_2, a_3) \cong \mathbb{R}[T]/(T^2 + a_1a_2a_3)$.

Afterwards, I am meant to generalise this for any $n$.

I feel like I'll be able to generalise for any $n$ once I have done the first part (I am aware that I'll have to look at different cases for values of $n$), but I'm unsure how the actually show this isomorphism! Any help would be greatly appreciated!

$\endgroup$

2 Answers 2

1
$\begingroup$

This is indeed a Clifford algebra. A reference for the proof would be e.g. the first chapter of Friedrich's "Dirac operators in Riemannian geometry."

$\endgroup$
0
$\begingroup$

It's also easily computed from the structure theorem for Clifford algebras.

In your particular case, the algebra is isomorphic to $M_2(\Bbb R)\times M_2(\Bbb R)$ whose center is $\Bbb R\times \Bbb R$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .