# Trace of $2n$ gamma matrices

To proof $$\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2n}}) =\mathrm{Tr}(\gamma_{\mu_{2n}}\cdots\gamma_{\mu_1}),$$

I use $\gamma_\mu^\dagger=\gamma^0\gamma_\mu\gamma^0$ and get

$$\cdots=\mathrm{Tr}(\gamma^0\gamma_{\mu_1}^\dagger\cdots\gamma_{\mu_{2n}}^\dagger\gamma^0) =\mathrm{Tr}(\gamma_{\mu_1}^\dagger\cdots\gamma_{\mu_{2n}}^\dagger) =\mathrm{Tr}[(\gamma_{\mu_{2n}}\cdots\gamma_{\mu_1})^\dagger].$$

The final step is to remove the conjugate transpose. In wikipedia, it says that the trace of a product of gamma matrices is real. How can I proof it?

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Dec 7 '17 at 14:37

Now I have found a proof for $\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2n}})\in\mathbb{R}$ with mathematical induction:

1. For $n=1$. We know that $\mathrm{Tr}(\gamma_\mu\gamma_\nu)=4\eta_{\mu\nu}\in\mathbb{R}$.

2. Assume the conclusion is correct for $n=k$, i.e.

$$\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}})\in\mathbb{R}.$$

Then for $n=k+1$, we have

\begin{align} \mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} \gamma_{\mu_{2k+1}}\gamma_{\mu_{2k+2}}) &=\mathrm{Tr}[\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} (2\eta_{\mu_{2k+1},\,\mu_{2k+2}} - \gamma_{\mu_{2k+2}}\gamma_{\mu_{2k+1}})] \\ &=2\eta_{\mu_{2k+1},\,\mu_{2k+2}} \mathrm{Tr}(\gamma_{\mu_1}\cdots \gamma_{\mu_{2k}}) -\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} \gamma_{\mu_{2k+2}}\gamma_{\mu_{2k+1}}). \end{align}

Note that the first term is real, and the second term can be expanded in the same way:

$$\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} \gamma_{\mu_{2k+2}}\gamma_{\mu_{2k+1}}) =2\eta_{\mu_{2k},\,\mu_{2k+2}}\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k-1}} \gamma_{\mu_{2k+1}}) -\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k-1}} \gamma_{\mu_{2k+2}}\gamma_{\mu_{2k}}\gamma_{\mu_{2k+1}}).$$

The first term has $2n$ gamma matrices and hence is real as well. Do the same process $2n+1$ times, we can move $\gamma_{\mu_{2k+2}}$ to the first position. Each movement will lead to a minus sign, so we have

\begin{align} \mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} \gamma_{\mu_{2k+1}}\gamma_{\mu_{2k+2}}) &=S+(-1)^{2n+1}\mathrm{Tr}(\gamma_{\mu_{2k+2}}\gamma_{\mu_1}\cdots \gamma_{\mu_{2k}}\gamma_{\mu_{2k+1}})\\ &=S-\mathrm{Tr}(\gamma_{\mu_{2k+2}}\gamma_{\mu_1}\cdots \gamma_{\mu_{2k}}\gamma_{\mu_{2k+1}})\\ &=S-\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k}} \gamma_{\mu_{2k+1}}\gamma_{\mu_{2k+2}}) \tag*{\mathrm{Tr}(AB)=\mathrm{Tr}(BA)}, \end{align}

where $S$ contains only $\eta_{\mu\nu}$ and trace of $2n$ gamma matrices, so $S\in\mathbb{R}$.

Now obviously we have

$$\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2k+2}}) =\frac{S}{2}\in\mathbb{R}.$$

Given the above, we now reach the conclusion that

$$\mathrm{Tr}(\gamma_{\mu_1}\cdots\gamma_{\mu_{2n}})\in\mathbb{R}$$

for any $n$.