Clifford Algebra Examples

Section 6.7 Clifford Algebra Examples

We consider several examples of \(\Cl(V)=\Cl(p,q)\) below with \(n=\dim(V)=p+q\) small. We will make use throughout of the following standard matrices:

\begin{align} s_x = \begin{pmatrix} 0 \amp -i \\ -i \amp 0 \end{pmatrix}, \qquad s_y = \begin{pmatrix}0 \amp -1 \\ 1 \amp 0\end{pmatrix}, \qquad s_z = \begin{pmatrix} -i \amp 0 \\ 0 \amp i \end{pmatrix},\notag\\ \sigma_x = \begin{pmatrix}0\amp1\\1\amp0\end{pmatrix}, \qquad \sigma_y = \begin{pmatrix}0\amp-i\\i\amp0\end{pmatrix}, \qquad \sigma_z = \begin{pmatrix}1\amp0\\0\amp-1\end{pmatrix},\notag\\ \sigma_i = \begin{pmatrix}0\amp-i\\i\amp0\end{pmatrix}, \qquad \sigma_j = \begin{pmatrix}0\amp-j\\j\amp0\end{pmatrix}, \qquad \sigma_k = \begin{pmatrix}0\amp-k\\k\amp0\end{pmatrix}.\tag{6.7.1} \end{align}

Each of the matrices \(\sigma_m\) squares to \(+\one\text{,}\) whereas each \(s_m\) squares to \(-\one\text{.}\)

Subsection 6.7.1 Dimension 1

The Clifford algebra \(\Cl(1,0)\) requires a basis element that squares to \(+\one\text{,}\) so we choose \(\sigma_z\text{.}\) We're done! The Clifford algebra \(\Cl(1,0)\) is generated by \(\{1,\sigma_z\}\text{,}\) which is easily seen to be a copy of the vector space \(\RR\oplus\RR\text{,}\) that is, two diagonal blocks of \(1\times1\) real matrices. ¹

The Clifford algebra \(\Cl(0,1)\) must be handled separately. Now we require a basis element that squares to \(-\one\text{,}\) and the obvious choice is \(s_y\) (because it is real). Again, we're done, but this time the algebra is a copy of \(\CC\text{.}\) In fact, we could have started with the one-dimensional vector space spanned by \(i\text{!}\)

Remark 6.7.1. Alternate description of \(\Cl(1,0)\).

We could of course have chosen \(\sigma_x\) rather than \(\sigma_z\) when analyzing \(\Cl(1,0)\text{,}\) resulting in a description similar to that of \(\Cl(0,1)\) in that it does not involve diagonal matrices. Since \(\sigma_x^2=\one\text{,}\) we can interpret \(\RR\oplus\RR\) in this case as the split complex numbers \(\CC'\text{;}\) see Section 2.8 and Section 4.1.

Subsection 6.7.2 Dimension 2

For \(\Cl(2,0)\text{,}\) we need two generators that square to \(+\one\text{;}\) we choose \(\{\sigma_x,\sigma_z\}\text{.}\) There is now an element of rank 2, namely

\begin{equation} \omega=\sigma_x\sigma_z=-i\sigma_y=s_y\tag{6.7.2} \end{equation}

and we're done. In this case, \(\omega^2=-\one\text{.}\)

For \(\Cl(1,1)\text{,}\) we need do nothing further; simply swap the roles of \(\sigma_x\) and \(s_y\) in the above construction. Thus, start with \(\{s_y,\sigma_z\}\text{,}\) so now

\begin{equation} \omega=s_y\sigma_z=\sigma_x\tag{6.7.3} \end{equation}

which squares to \(+\one\text{.}\) Thus, as vector spaces, \(\Cl(2,0)\) and \(\Cl(1,1)\) are identical, but they can be distinguished by the sign of the square of their respective elements of rank 2.

Finally, for \(\Cl(0,2)\text{,}\) we need two generators that square to \(-\one\text{.}\) Looking at the matrices in (6.7.1), we choose \(\{s_x,s_y\}\text{,}\) which leads to

\begin{equation} \omega = s_x s_y = s_z .\tag{6.7.4} \end{equation}

Recognizing that these matrices anticommute and all square to \(-\one\text{,}\) we can identify them with the imaginary quaternions. Thus, \(\Cl(0,2)\cong\HH\text{.}\) In fact, we could have started with the two-dimensional vector space spanned by \(i,j\in\HH\text{!}\)

Exercises 6.7.3 Exercises

1. Dimension 3.

Construct the Clifford algebras \(\Cl(p,q)\) with \(n=p+q=3\text{.}\)

Hint.

The class of the resulting Clifford algebra can be found online, including on Wikipedia. Using this information as a guide, you should still construct an explicit basis for \(\Cl(p,q)\text{.}\)

2. Dimension 4.

Construct the Clifford algebras \(\Cl(p,q)\) with \(n=p+q=4\text{.}\)

Hint.

The class of the resulting Clifford algebra can be found online, including on Wikipedia. Using this information as a guide, you should still construct an explicit basis for \(\Cl(p,q)\text{.}\)

The direct sum of algebras is discussed further in Section A.2.