# Dimension of Dirac $\gamma$ matrices

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

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

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 that we are given $n\times n$ matrices $\gamma_{1}, \ldots, \gamma_{d}$, that satisfy $$ \{\gamma_{\mu}, \gamma_{\nu}\}_+~=~2\delta_{\mu\nu}{\bf 1}, \qquad \mu,\nu~\in~\{1,\ldots, d\}.\tag{2} $$

We may define $$ \gamma_{\mu\nu}~:=~ \frac{1}{2}[\gamma_{\mu}, \gamma_{\nu}]_- ~=~-\gamma_{\nu\mu}, \qquad \mu,\nu~\in~\{1,\ldots, d\}. \tag{3}$$ In particular, define $p$ elements $$ H_1, \ldots, H_p,\tag{4} $$ as $$ H_r ~:=~i\gamma_{r,p+r}, \qquad r~\in~\{1,\ldots, p\}.\tag{5} $$

Note that the elements $H_1,\ldots, H_p$, (and $\gamma_d$ if $d$ is odd), are a set of mutually commuting involutions $$ [H_r,H_s]_- ~=~0, \qquad r,s~\in~\{1,\ldots, p\},\tag{6} $$ $$ H_r^2 ~=~{\bf 1}, \qquad r~\in~\{1,\ldots, p\}.\tag{7} $$

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, $$H_1 v~=~(-1)^{j_1} v, \quad \ldots, \quad H_p v~=~(-1)^{j_p} v,\tag{8} $$ where $$ j_1,\ldots, j_p~\in ~\{0,1\} \tag{9}$$ are either zero or one.

Apply next the $p$ first gamma matrices $$ \gamma^{1}, \gamma^{2}, \ldots, \gamma^{p}, \tag{10} $$ to the common eigenvector $v$, so that $$ v_{(k_1,\ldots, k_p)}~:=~ \gamma_{1}^{k_1}\gamma_{2}^{k_2}\cdots\gamma_{p}^{k_p} v, \tag{11} $$ where the indices $$ k_1,\ldots, k_p~\in ~\{0,1\} \tag{12} $$ are either zero or one.

Next note that $$ [H_r,\gamma_s]_-~=~0 \quad \text{if}\quad r~\neq~ s \mod p \tag{13} $$ and $$ \{H_r,\gamma_r\}_+~=~0. \tag{14} $$ It is straightforward to check that the $2^p$ vectors $v_{(k_1,\ldots, k_p)}$ also are common eigenvectors for $H_1,\ldots, H_p$. In detail, $$ H_r v_{(k_1,\ldots, k_p)}~=~(-1)^{k_r+j_r}v_{(k_1,\ldots, k_p)}.\tag{15}$$

Note that each eigenvector $v_{(k_1,\ldots, k_p)}$ has a unique pattern of eigenvalues for the tuple $(H_1,\ldots, H_p)$, so the $2^p$ vectors $v_{(k_1,\ldots, k_p)}$ must be linearly independent.

Since $$ \gamma_{p+r}~=~ i H_r \gamma_r, \qquad r~\in~\{1,\ldots, p\}, \tag{16} $$ we see that $$ W~:=~{\rm span}_{\mathbb{C}} \left\{ v_{(k_1,\ldots, k_p)} \mid k_1,\ldots, k_p~\in ~\{0,1\} \right\} \tag{17} $$ is an invariant subspace $W\subseteq V$ for $C$.

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 $2^p$. $\Box$

## Intuitive explanation

**Preliminary:** A vector has a many components as elements of the vector space basis.

A Clifford algebra basis is generated by

all(independent) products of the generators (in Dirac's equation case these are the $\gamma$'s).

### The counting

There are as many $\gamma$'s as the dimension of the spacetime, and according to the definition the algebra includes a unit, $$\bigl\{\gamma^a,\gamma^b\bigr\} = 2 \eta^{ab}\mathbf{1}.$$

For any extra element the *new* basis consists of the *previous* basis elements plus the product of each of those by the extra element. This is the *new* basis has twice the elements. Therefore,
$$\dim(\mathcal{C}\ell(n)) = 2^{n}.$$

In order to represent this algebra one needs "matrices" of $2^{n/2}\times 2^{n/2}$, which is not bad for even dimensional spacetimes.

Said that, the *problem* (which I don't intend to demonstrate) comes with odd dimensional spacetimes... however, intuitively again, this algebra can be represented by two copies of the co-dimension one algebra, *i.e.* one dimension less. This reason is why the minimal dimensionality for the representation of the $\gamma$'s is
$$\dim(\gamma) = 2^{\lfloor n/2\rfloor}\times 2^{\lfloor n/2 \rfloor}.$$

If you wonder whether one can find a bigger representation of the $\gamma$'s, the answer is YES, but you will will end up with either a *non-fundamental* or a trivial extension.

Thats a good question. To answer this lets start with Clifford algebra generated by $\gamma$ matrices. \begin{equation} \gamma_{\mu}\gamma_{\nu}+ \gamma_{\mu}\gamma_{\nu}=2\eta_{\mu\nu} \end{equation} with $\mu,\nu=0,1,2,\cdots N$ with the metric signature $\eta_{\mu\nu}=\text{diag}(+,-,-,-,\cdots,-)$. Using $I$ and $\gamma_{\mu}$ we can construct a set of matrices as follow \begin{equation} I, \gamma_{\mu},\gamma_{\mu}\gamma_{\nu}\quad(\mu<\nu), \gamma_{\mu}\gamma_{\nu}\gamma_{\lambda}\quad(\mu<\nu<\lambda),\cdots,\gamma_{1}\gamma_{2}\cdots\gamma_{N}. \end{equation}

There are \begin{equation} \sum_{p=0}^{N}\binom{N}{p} = 2^{N} \end{equation} such matrices. Lets call them $\Gamma_{A}$, where $A$ runs from $0$ to $2^{N}-1$. Now let $\gamma_{\mu}$ are $d\times d$ dimensional irreducible matrices. Our goal is to find a relation between $d$ and $N$. To this end lets define a matrix \begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} \end{equation}. Where $Y$ is some arbitary $d\times d$ matrix. It is follows that \begin{equation} (\Gamma_{B})^{-1}S\Gamma_{B} = \sum_{A=0}^{2^N-1}(\Gamma_{A}\Gamma_{B})^{-1}Y\Gamma_{A}\Gamma_{B} =\sum_{C=0}^{2^N-1}(\Gamma_{C})^{-1}Y\Gamma_{C}=S \end{equation} Where we have used $\Gamma_{A}\Gamma_{B}=\epsilon_{AB}\Gamma_{C}$, with $\epsilon_{AB}^{2}=1$

Hence \begin{equation}S\Gamma_{A}=\Gamma_{A}S\end{equation} Since $S$ commutes with all the matrices in the set, by Schur's lemma we conclude that $S$ must be proportional to the identity matrix so that we can write \begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \lambda I \end{equation}

Taking trace we get \begin{eqnarray} \text{Tr} S & = & \sum_{A=0}^{2^N-1} \text{Tr} Y = \lambda d\\ \Rightarrow \lambda & = & \frac{2^{N}}{d}\text{Tr} Y \end{eqnarray} or \begin{equation} \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \frac{2^{N}}{d}\text{Tr} Y \end{equation}

Taking the $(j; m)$ matrix element of both sides of last equation yield \begin{equation} \sum_{A=0}^{2^N-1}((\Gamma_{A})^{-1})_{jk}(\Gamma_{A})_{km} = \frac{2^{N}}{d}\delta_{jm} \delta_{kl} \end{equation} where $j; k; l; m = 1; 2;\cdots; d$ and we have used the fact that Y is an arbitrary $d \times d$ matrix. If we set $j = k; l = m$ and sum over these two indices, that gives \begin{equation} \sum_{A=0}^{2^N-1} \text{Tr}[(\Gamma_{A})^{-1}] \text{Tr}[\Gamma_{A}] = 2^{N}\end{equation} There are two cases to consider, namely, $N$ even and $N$ odd. For $N = 2M$ (even), $\text{Tr} \Gamma_{A} = 0$ except for $\Gamma_{0} = 1$ for which $\text{Tr} \Gamma_{0} = d$. Which gives \begin{equation} d^2 = 2^N\qquad \text{or} \quad \boxed{d = 2^{N/2}} \end{equation} This is the main result. For four dimensional Minkowski space time $N=4$ cosequntly the dimension of irreducible representation is $d = 2^{4/2} =4$.