I'm interested in representations of the elements of a Clifford algebra, $\gamma^a$ with $a\in \{1,...,n\}$, such that $(\gamma^a)^2 =\pm 1$.
If these were numbers, for those $\gamma^a=1$ one have that $(\gamma^a)^\dagger=\gamma^a$, and for those $\gamma^a=-1$ one have that $(\gamma^a)^\dagger=-\gamma^a$.
QUESTION
- Is this statement still valid for every other matrix representation?
Thank you.
More explanations
The dagger notation denotes the Hermitian conjugation, in the Linear algebra sense, i.e., if you have a Matrix representation of the $\gamma$ elements, dagger is the transpose of the complex conjugate, $$(\gamma^a)^\dagger = \left((\gamma^a)^*\right)^t.$$
As an example of representations of Clifford Algebra, take $\mathcal{C}\ell(2,0)$, generated by the elements$$\gamma^1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad\gamma^2=\begin{pmatrix}0&-\imath\\\imath&0\end{pmatrix}.$$Note that $(\gamma^1)^2=1_2$, and $(\gamma^1)^2=1_2$, where $1_2$ is the $2\times2$ identity matrix.
Additionally, they both are Hermitian.
Question again
- Consider a matrix that squares to the identity. Is it necessarily a Hermitian matrix? $$M^\dagger = M?.$$
- Consider a matrix that squares to minus the identity. Is it necessarily a anti-Hermitian matrix?$$M^\dagger = -M?.$$
To construct a real Clifford algebra, you must have some vector space over $\mathbb R$ with a quadratic form defined on it. I take it $\mathcal{Cl}(2,0)$ refers to a Clifford algebra on a 2D vector space with the quadratic form defined by $e_1^2=1$, $e_2^2=1$ on the orthogonal vectors $e_1,e_2$. An element of the algebra can be written as $M=x_0+x_1e_1+x_2e_2+x_3e_{12}$, where $e_{12}=e_1e_2$. The algebra $\mathcal{Cl}(2,0)$ can be represented by $2\times2$ real matrices. Its basis elements are represented by the following matrices
$1 \to\begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad e_1\to \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad e_2\to \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad e_{12}\to \begin{pmatrix}0&-1\\1&0\end{pmatrix},$
so that $M$ is represented by $\begin{pmatrix}x_0+x_2&x_1-x_3\\x_1+x_3&x_0-x_2\end{pmatrix}$. There is no way you could get a complex matrix such as $\gamma^2$ in this representation.
On the other hand, $\mathcal{Cl}(0,2)$ is a Clifford algebra with the quadratic form defined by $e_1^2=-1, e_2^2=-1$. Its elements can be written as $M=a+be_1+ce_2+de_{12}$, but the representation by matrices is completely different from that of $\mathcal{Cl}(2,0)$. The algebra $\mathcal{Cl}(0,2)$ is isomorphic to the algebra of quaternions $\mathbb H$, so you can use the representation of quaternions with complex $2\times2$ matrices to represent elements of the Clifford algebra. The basis elements are represented by
$1 \to\begin{pmatrix}1&0\\0&1\end{pmatrix}, \quad e_1\to \begin{pmatrix}i&0\\0&-i\end{pmatrix}, \quad e_2\to \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \quad e_{12}\to \begin{pmatrix}0&i\\i&0\end{pmatrix},$
so that $M$ is represented by $\begin{pmatrix}a+ib&c+id\\-c+id&a-ib\end{pmatrix}$. There is no way you could get the matrix $\gamma^1=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ in this representation.
In $\mathcal{Cl}(0,2)$, there are only two elements that square to 1. It is the unit 1 itself and $-1$; they are both Hermitian. Any element with $a=0$ and $b^2+c^2+d^2=1$ squares to $-1$ and it is easy to check that it is anti-Hermitian. There are no other elements that square to $-1$.
In $\mathcal{Cl}(2,0)$, the element $e_1+e_2+e_{12}$ squares to 1 but it is not Hermitian.