# 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?

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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. –  Michael Brown Feb 7 '13 at 16:21
Dear Andrea, I was a bit confused by the term 'order'. At this level, people prefer the term 'dimension'. Your question can be rephrased as: does there exist a representation of dimension 6 of the Clifford algebra? I'm not 100% sure of the answer, so I'll let other people think about it. –  Vibert Feb 7 '13 at 17:13
yes, it's definitively possible to have matrices larger than 4x4 satisfying the above relations, see this Wikipedia article and this link –  Andre Holzner 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. –  Michael Brown Feb 8 '13 at 0:17
@Vibert: I was actually going to use the term "dimension", but I just went along with the choice of words in the quoted work. –  Andrea Colonna Feb 8 '13 at 20:34

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 $2^p$.

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# 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.

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