Clifford Algebra $C_4\simeq \mathbb{H}[2]$

Question

I have been stuck trying to prove this isomorphism between algebras $C_4\simeq \mathbb{H}[2]$, where $C_4$ is the Clifford Algebra generated by $1$ and $x_1, ..., x_4$ with the rules $$x_i^2=-1, \quad x_ix_j+x_jx_i=0,\, (i\neq j), $$ and $\mathbb{H}[2]$ is the $(2\times 2)$ matrix algebra over the quaternions.

I tried constructing the isomorphism explicitly but got nowhere, then I saw the Artin-Wedderburn theorem which seems to imply this, but doesn't explicitly give the algebra of the matrix space. I saw the proof for this theorem in this page

http://www.thebookshelf.auckland.ac.nz/docs/NZJMaths/nzjmaths022/nzjmaths022-01-010.pdf

Any help would be appreciated.

Answer 1

Essentially you need matrices $X_1,\ldots,X_4$ over the quaternions satisfying these identities. How about $$X_1=\pmatrix{i&0\\0&i},$$ $$X_2=\pmatrix{j&0\\0&j},$$ $$X_3=\pmatrix{0&1\\-1&0},$$ $$X_4=\pmatrix{0&k\\k&0}?$$