For $\mu=0,1,2,3$, let $\gamma^{\mu}$ the set of Dirac gamma matrices. What does it mean to say that $\{\gamma^{\mu}: \hspace{0.1cm} \mu=0,1,2,3\}$ is irreducible?
From my previous question, I know that the gamma matrices determine a representation $\rho$ of the Clifford algebra $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$ in the algebra of $4\times 4$ complex matrices. Many references state that the gamma matrices are irreducible representations of $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$, which means that for every $x \in \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$, $\rho(x)$ has no non-trivial invariant subspaces.
However, as discussed here, the gamma matrices can also determine a group $\Gamma_{\text{Dirac}}$ with elements $I, \pm \gamma^{\mu}, \pm \gamma^{\mu}\gamma^{\nu}$, $\mu = 0,1,2,3,5$ and $\mu < \nu$. Here $\gamma^{5} := \gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}$. And the notion of irreducible representation also holds for groups.
So, getting back from the original question, when one says $\{\gamma^{\mu}: \hspace{0.1cm} \mu=0,1,2,3\}$, what does it mean exactly? Maybe both notions defined above have some natural connection?
ADD: I think I got it. It can be shown that the representation $\rho : \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi) \to M_{4\times 4}(\mathbb{C})$ is not only faithful, but also a bijection. Thus, because it is surjective, the representation is irreducible. Now, each element of the Dirac group $\Gamma_{\text{Dirac}}$ is, in particular, the image $\rho(x)$ of an element $x \in \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$ under the representation $\rho$. Thus, there is no (non-trivial) subspace of $\mathbb{C}^{4}$ which is invariant under the action of any of these elements. Hence, the representation $\tilde{\rho}: \Gamma_{\text{Dirac}} \to M_{4\times 4}(\mathbb{C})$ given by $\tilde{\rho}(x) = x$ must be irreducible. In summary, because $\rho$ is irreducible, the group representation $\tilde{\rho}$ is irreducible.
Any comments and corrections are welcome.
EDIT: I actually don't know if the surjectivity of $\rho$ implies $\rho$ to be irreducible.
ADD 2: To clarify, the space $M_{4\times 4}(\mathbb{C})$ mentioned before is being considered as a vector space over $\mathbb{R}$, as well as $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$. Surely one can complexify $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$ to get a new Clifford algebra $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)_{\mathbb{C}}$ which now is isomorphic to $M_{4\times 4}(\mathbb{C})$, the latter being considered as an algebra over $\mathbb{C}$.
After posting this question, some ideas and questions have arisen. Let me summarize the important points/questions below:
- It is not clear to me yet if the irreducibility of $\rho: \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)\to M_{4\times 4}(\mathbb{C})$ is equivalent to the irreducibility of $\tilde{\rho}: \Gamma_{\text{Dirac}} \to M_{4\times 4}(\mathbb{C})$ in any way. However, as I tried to prove above, the irreducibility of $\rho$ seems to imply the irreducibility of $\tilde{\rho}$.
- It is also not clear if the proof I give before is correct, that is, I don't know if the bijectivity of $\rho$ is enough to garantee that it is irreducible.
- In the physics literature, it is sometimes not clear if the matrix algebra is being considered as a real or complex algebra, so I don't if the representation from $\rho: \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi) \to M_{4\times 4}(\mathbb{C})$ is irreducible or if we need to complexify $\mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)$ to get a irreducible representation $\hat{\rho}: \mathcal{C}l_{1,3}(\mathbb{R}^{4},\Phi)_{\mathbb{C}}\to M_{4\times 4}(\mathbb{C})$. Some references suggest that the complex case is irreducible, and I wasn't able to find a proof of this fact or some indiciation if the real case is also irreducible.
