I am currently taking a representation theory class (from a physicist), and I am very confused about the Dirac groups' irreducible representations.

First of all, all the Dirac matrices in the representation have trace = 0, so it does not even seem to include a unit matrix. When we talked about representations in class before, we always had a unit matrix in a representation, what happened?

Also, the lecture went into distinguishing the case for 2n-dimension and 2n-1-dimension. While I understand why there is one more conjugacy classes in the odd dimension (thus even dimension having one more irreducible rep than odd dimension), I can't fully appreciate all the difference in the irreducible representations in the cases of even and odd dimensions; in particular, I was asked in a homework to show that if a Dirac matrices {$\gamma^\mu$} form an irreducible rep, then show that {$-\gamma^\mu$} is equivalent irreducible rep in the case of even dimension, and inequivalent irreducible rep in the case of odd dimension. But then again, if I think about the character table to see whether a representation is equivalent or inequivalent to another representation, I feel like matrices in {$\gamma^\mu$} and {-$\gamma^\mu$} will never have same trace, thus they can never be equivalent (unless they are all 0, which is the case, I believe. Then again, how could they be different representations then?).

I would appreciate any good reading materials / answers to my questions!

  • $\begingroup$ Unit matrix and minus unit matrix are part of the Dirac group, see for instance chapter $6.1$ of this ref. Do you really mean representations of the Dirac group, or representations of the real Clifford algebra $Cl_{1,3}(\mathbb R)$, or representations of another group ? $\endgroup$
    – Trimok
    Oct 2, 2014 at 10:51
  • $\begingroup$ I am talking about the representation of the real Clifford algebra (of arbitrary dimension, and not necessarily Lorentzian signature (Euclidean possibly)). $\endgroup$ Oct 2, 2014 at 12:13
  • $\begingroup$ Thank you for pointing out about the unit matrix and minus unit matrix $\endgroup$ Oct 2, 2014 at 12:13
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    $\begingroup$ I think your confusion arises from thinking about representations of an algebra vs. representations of the group. The group rep. must always have the unit matrix. The group rep. $g$ is obtained from a representation of its algebra $X$ via exponentiation $g = e^X$. Existence of a unit matrix in the group rep. implies existence of a "zero" matrix in the algebra. That a zero matrix exists in the algebra is obvious since it is the additive identity (which must exist for any algebra.) $\endgroup$
    – Prahar
    Oct 2, 2014 at 16:00
  • $\begingroup$ Even for your second point - I think your confusion will be alleviated if you realize that physicists often talk about representations of the algebra, whereas mathematicians like to talk about representations of the group (which in general, are not equivalent things). The conjugacy classes and character tables are usually used to discuss representations of groups (as far as I know). The analogue of these things in an algebra are invariant subalgebras. $\endgroup$
    – Prahar
    Oct 2, 2014 at 16:02

2 Answers 2


The defining relation for the Clifford algebra, $Cl(1,d)$ is $$ \{\gamma_\mu,\gamma_\nu\}=2 \eta_{\mu\nu}\ \mathbf{1}\ , $$ For simplicity, I will assume that $\eta_{\mu\nu}=\text{Diag}(1,-1,\ldots,-1)$ with $\mu,\nu=0,1,\ldots,d$. Other signatures can easily be incorporated. It is easy to see that $\gamma_0^2=-\gamma_i^2=\mathbf{1}$ for $i=1,\ldots,d$. Using the defining relation, one has $$ \gamma_0 \gamma_i + \gamma_i \gamma_0 =0 \ . $$ Multiply the above equation by $\gamma_0$ and then take the trace to obtain $$ \text{Tr}(\gamma_i) + \text{Tr}(\gamma_0 \gamma_i \gamma_0)=0\implies \text{Tr}(\gamma_i)=0\ , $$ on using the cyclic property of the trace. Similarly, one can show $\text{Tr}(\gamma_0)=0$. So the defining property proves the tracelessness of the Dirac matrices.

Two representations, $\gamma_\mu$ and $\gamma_\mu'$, of the Clifford algebra are said to be equivalent if $\gamma_\mu' = S \cdot \gamma_\mu S^{-1}$ for some invertible matrix $S$.

Appendix A of the Physics Reports article by Sohnius might be a good starting point for the other properties.

  • $\begingroup$ I don't think you really answered my question. My question wasn't to prove that it is traceless.. But thank you for the article! $\endgroup$ Oct 2, 2014 at 11:12
  • $\begingroup$ Your long-winded question needs a long answer and so the appendix will do that. I noticed that the tracelessness of the Dirac matrices seemed a mystery to you and hence the proof. I can delete my answer if you think it is a waste. $\endgroup$
    – suresh
    Oct 2, 2014 at 11:15
  • $\begingroup$ No, I do not think it is a waste. Sorry if I sounded obnoxious. Also, I am having trouble getting hold of the article! :( $\endgroup$ Oct 2, 2014 at 12:04
  • $\begingroup$ Type "sohnius introducing supersymmetry" in your favourite search engine and you should find a link to a pdf file. $\endgroup$
    – suresh
    Oct 2, 2014 at 12:21

Although the responses so far have been illuminating, I believe they missed your main point of confusion (the first commenter nailed it I think).

One must differentiate between the Clifford algebra $\text{Cℓ}_{t,s}(\mathbb{R})$, where $t+s=d$ is the dimensionality of spacetime, and the Dirac matrices which generate a basis for it via products. More specifically, we have a set of $d$ Dirac matrices $\{\gamma^\mu\}$ which satisfy the Clifford algebra relation $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$. Products of these Dirac matrices, along with the identity, form a basis for the Clifford algebra:

$$\Gamma = \{1,\gamma^\mu,\gamma^5,\gamma^\mu\gamma^5,\gamma^\mu \gamma^\nu\}$$

where $\gamma^5=\prod_\mu \gamma^\mu$, with a potential factor of $i$ to ensure Hermiticity depending on $(t,s)$. This common choice of basis for the Clifford algebra forms a finite group$^\dagger$, and is what people refer to as the "Dirac group" - not the individual Dirac matrices themselves.

Indeed, as you noticed (and as @suresh confirmed), the unit matrix cannot be a Dirac matrix in any representation (since $\text{Tr}(\gamma^\mu )=0$ for all $\mu$). But the unit matrix has to be present in any representation of any finite group!

$^\dagger$ Actually, as I've written it $\Gamma$ does not generally form a group. One must prepend $\pm$ to each entry: $\Gamma_{\text{Dirac}}=\{\pm 1, \pm \gamma^\mu, \pm\gamma^5, \pm\gamma^\mu\gamma^5, \pm\gamma^\mu \gamma^\nu\}$


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