Gamma matrices in higher (even) dimensions

[KennyAckerman97]

Homework Statement

I define the gamma matrices in this following representation:
\begin{align*}
\gamma^{0}=\begin{pmatrix}
\,\,0 & \mathbb{1}_{2}\,\,\\
\,\,\mathbb{1}_{2} & 0\,\,
\end{pmatrix},\qquad \gamma^{i}=\begin{pmatrix}
\,\,0 &\sigma^{i}\,\,\\
\,\,-\sigma^{i} & 0\,\,
\end{pmatrix}
\end{align*}
where k=1,2,3. Consider 2k+2 dimension, I can then group the gamma matrices into k+1 anticommuting 'raising' and 'lowering' operators. If we define:
\begin{align*}
\Gamma^{0\pm} = \frac{i}{2}\left(\pm\gamma^{0}+\gamma^{1}\right),\qquad \Gamma^{a\pm}=\frac{i}{2}\left(\gamma^{2a}\pm i\gamma^{2a+1}\right)
\end{align*}
with a=1,2,...,k. Then
\begin{align*}
\left\{\Gamma^{a\pm},\Gamma^{b\pm}\right\}=0,\qquad \left\{\Gamma^{a\pm},\Gamma^{b\mp}\right\} =\delta^{ab}
\end{align*}
Then how can I construct the gamma matrices in 6 and 8 dimensions using these information?

Homework Equations

The Attempt at a Solution

I expect the matrices will be a 8 x 8 matrices for 6 dimension and 16 x 16 matrices for 8 dimension. But what I don't get is, in 4 dimensions we have 4 x 4 matrices, how can those operators and their anticommutation relations help me to construct new sets of gamma matrices that is basically not in the same size? What I have in mind is some tensor product, but I don't think it is related to the given information.

 

[anathema]

In order to construct the gamma matrices in 6 and 8 dimensions, we can use the information provided about the anticommutation relations of the operators $\Gamma^{a\pm}$. Since we know that $\Gamma^{a\pm}$ are anticommuting operators, we can use them to construct new gamma matrices in higher dimensions by taking tensor products.

For example, in 6 dimensions, we can construct the gamma matrices as follows:
\begin{align*}
\gamma^{0} &= \begin{pmatrix}
\,\,0 & \mathbb{1}_{4}\,\,\\
\,\,\mathbb{1}_{4} & 0\,\,
\end{pmatrix}, \qquad \gamma^{i} = \begin{pmatrix}
\,\,0 & \Gamma^{i+}\,\,\\
\,\,\Gamma^{i-} & 0\,\,
\end{pmatrix}, \qquad \gamma^{4} = \begin{pmatrix}
\,\,0 & \Gamma^{1+}\,\,\\
\,\,\Gamma^{1-} & 0\,\,
\end{pmatrix}, \qquad \gamma^{5} = \begin{pmatrix}
\,\,0 & \Gamma^{2+}\,\,\\
\,\,\Gamma^{2-} & 0\,\,
\end{pmatrix}
\end{align*}
where $i=1,2,3$ and $\Gamma^{a\pm}$ are the operators defined in the original post. These gamma matrices satisfy the correct anticommutation relations:
\begin{align*}
\left\{\gamma^{a},\gamma^{b}\right\} = 2\eta^{ab}\mathbb{1}_{6}
\end{align*}
where $\eta^{ab}$ is the Minkowski metric in 6 dimensions.

Similarly, in 8 dimensions, we can construct the gamma matrices as follows:
\begin{align*}
\gamma^{0} &= \begin{pmatrix}
\,\,0 & \mathbb{1}_{8}\,\,\\
\,\,\mathbb{1}_{8} & 0\,\,
\end{pmatrix}, \qquad \gamma^{i