representation theory - Dimension of Dirac $gamma$ matrices

While studying the Dirac equation, I came across this enigmatic passage on p. 551 in From Classical to Quantum Mechanics by G. Esposito, G. Marmo, G. Sudarshan regarding the $\gamma$ matrices:

$$\tag{16.1.2} (\gamma^0)^2 = I , (\gamma^j)^2 = -I \ (j=1,2,3)$$ $$\tag{16.1.3} \gamma^0\gamma^j + \gamma^j \gamma^0 = 0$$ $$\tag{16.1.4} \gamma^j \gamma^k + \gamma^k \gamma^j = 0, \ j\neq k$$ In looking for solutions of these equations in terms of matrices, one finds that they must have as order a multiple of 4, and that there exists a solution of order 4.

Obviously the word order here means dimension. In my QM classes the lecturer referenced chapter 5 from Advanced Quantum Mechanics by F. Schwabl, especially as regards the dimension of Dirac $\gamma$ matrices. However there it is stated only that, since the number of positive and negative eigenvalues of $\alpha$ and $\beta^k$ must be equal, $n$ is even. Moreover, $n=2$ is not sufficient, so $n=4$ is the smallest possible dimension in which it is possible to realize the desired algebraic structure.

While I got that the smallest dimension is 4, I fail to find any argument to reject the possibility that $n=6$ could be a solution. I also checked this Phys.SE post, but I didn't find it helpful at all.

Can anyone help me?

Answer

Let us generalize from four space-time dimensions to a $d$-dimensional Clifford algebra $C$. Define

$$\tag{1} p~:=~[\frac{d}{2}],$$

where $[\cdot]$ denotes the integer part. OP's question then becomes

Why must the dimension $n$ of a finite dimensional representation $V$ be a multiple of $2^p$?

Proof: If $C\subseteq {\rm End}(V)$ and $V$ are both real, we may complexify, so we may from now on assume that they are both complex. Then the signature of $C$ is irrelevant, and hence we might as well assume positive signature. In other words, we assume we are given $n\times n$ matrices $\gamma^{1}, \ldots, \gamma^{d}$, that satisfy

$$\tag{2} \{\gamma_{\mu}, \gamma_{\nu}\}~=~2\delta_{\mu\nu}{\bf 1}, \qquad \mu,\nu~\in~\{1,\ldots d\}.$$

We may define

$$\tag{3} \Sigma_{\mu\nu}~:=~ \frac{i}{2}[\gamma_{\mu}, \gamma_{\nu}] ~=~-\Sigma_{\nu\mu}, \qquad \mu,\nu~\in~\{1,\ldots d\}.$$

In particular, define $p$ elements

$$\tag{4} H_1, \ldots, H_p,$$

as

$$\tag{5} H_r ~:=~\Sigma_{r,p+r}, \qquad r~\in~\{1,\ldots p\}.$$

Note that the elements $H_1,\ldots, H_p$, (and $\gamma_d$ if $d$ is odd), are a set of mutually commuting involutions. Therefore, according to Lie's Theorem, then $H_1,\ldots, H_p$, (and $\gamma_d$ if $d$ is odd), must have a common eigenvector $v$.

Since $H_1,\ldots, H_p$ are involutions, their eigenvalues are $\pm 1$. In other words,

$$\tag{6}H_1 v~=~(-1)^{j_1} v, \quad \ldots, \quad H_p v~=~(-1)^{j_p} v,$$

where

$$\tag{7} j_1,\ldots, j_p~\in ~\{0,1\}$$

are either zero or one.

Apply next the $p$ first gamma matrices

$$\tag{8} \gamma^{1}, \gamma^{2}, \ldots \gamma^{p},$$

to the common eigenvector $v$, so that

$$\tag{9} v_{(i_1,\ldots, i_p)}~:=~ \gamma_{1}^{i_1}\gamma_{2}^{i_2}\cdots\gamma_{p}^{i_p} v,$$

where the indices

$$\tag{10} i_1,\ldots, i_p~\in ~\{0,1\}$$

are either zero or one.

It is straightforward to check that the $2^p$ vectors $v_{(i_1,\ldots, i_p)}$ also are common eigenvectors for $H_1,\ldots, H_p$. In detail,

$$\tag{11} H_r v_{(i_1,\ldots, i_p)}~=~(-1)^{i_r+j_r}v_{(i_1,\ldots, i_p)}.$$

Note that each eigenvector $v_{(i_1,\ldots, i_p)}$ has a unique pattern of eigenvalues for the tuple $(H_1,\ldots, H_p)$, so the $2^p$ vectors $v_{(i_1,\ldots, i_p)}$ must be linearly independent.

Since

$$\tag{12} \gamma_{p+r}~=~ i H_r \gamma_r, \qquad r~\in~\{1,\ldots p\},$$

we see that

$$\tag{13} W~:=~{\rm span}_{\mathbb{C}} \left\{ v_{(i_1,\ldots, i_p)} \mid i_1,\ldots, i_p~\in ~\{0,1\} \right\}$$

is an invariant subspace $W\subseteq V$.

This shows that that any irreducible complex representation of a complex $d$-dimensional Clifford algebra is $2^p$-dimensional.

Finally, we believe (but did not check) that a finite dimensional representation $V$ of a complex Clifford algebra is always completely reducible, i.e. a finite sum of irreducible representations, and hence the dimension $n$ of $V$ must be a multiple of $2^p$.