# 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?

• Don't know if this is the way - just thinking out loud here - but the projection operators $P_{\pm} = (1\pm\gamma^5)/2$ cut the space in half, so if you can prove that $P_{\pm}$ can't have an odd number of nonzero eigenvalues you're done. You can probably use the representation theory of SU(2) to complete a proof since in four dimensions SO(3,1) ~ SU(2)xSU(2) (a double cover) and the projectors drop you onto one of the factors. Feb 7 '13 at 16:21
• yes, it's definitively possible to have matrices larger than 4x4 satisfying the above relations, see this Wikipedia article and this link Feb 7 '13 at 20:28
• @AndreHolzner That's not what the OP is asking. The question is whether there are higher dimensional representations of the 4D Dirac algebra, i.e., looking for four matrices that satisfy the 4D algebra but which are larger than 4x4. The standard construction that you link to gives matrix dimensions which are powers of two, which doesn't answer the question of whether there are any 6x6 representations. Feb 8 '13 at 0:17

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:

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

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

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

4. Therefore, according to Lie's Theorem, then $$H_1,\ldots, H_p$$, (and $$\gamma_d$$ if $$d$$ is odd), must have a common eigenvector $$v$$.

5. 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.

6. 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.

7. 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, $$H_r v_{(k_1,\ldots, k_p)}~=~(-1)^{k_r+j_r}v_{(k_1,\ldots, k_p)}.\tag{15}$$

8. 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.

9. 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 $$C$$.

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

11. 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.

A rigorous proof of the dimensionalality of $\gamma$ matrices come from group representation theory. Its about finding the irreducible representation of Clifford algebra. A recent book of Ashok Das of group theory discussed that in a great depth. An etair chapter of this book dedicated for finding the representation of Clifford algebra both in even and odd direction. See page no 162 for the prrof.

A nice and cute prove was given by peter West in

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 $$d = 2^{4/2} =4$$.