# Trace of six gamma matrices

I need to calculate this expression:
$$Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})$$ I know that I can express this as: $$Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})=-4i(g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\mu\sigma}\epsilon^{\nu\rho\alpha\beta}-g^{\mu\alpha}\epsilon^{\nu\rho\sigma\beta}+g^{\mu\beta}\epsilon^{\nu\rho\sigma\alpha}+g^{\nu\rho}\epsilon^{\mu\sigma\alpha\beta}-g^{\nu\sigma}\epsilon^{\mu\rho\alpha\beta}+g^{\nu\alpha}\epsilon^{\mu\rho\sigma\beta}-g^{\nu\beta}\epsilon^{\mu\rho\sigma\alpha}+g^{\rho\sigma}\epsilon^{\mu\nu\alpha\beta}-g^{\rho\alpha}\epsilon^{\mu\nu\sigma\beta}+g^{\rho\beta}\epsilon^{\mu\nu\sigma\alpha}+g^{\sigma\alpha}\epsilon^{\mu\nu\rho\beta}-g^{\sigma\beta}\epsilon^{\mu\nu\rho\alpha}+g^{\alpha\beta}\epsilon^{\mu\nu\rho\sigma})$$ So, some of this terms are the same and some vanish. My question is how to show that.
I know that:
$$Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})=-4i(g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\rho\nu}\epsilon^{\mu\sigma\alpha\beta}-g^{\alpha\beta}\epsilon^{\sigma\mu\nu\rho}+g^{\sigma\beta}\epsilon^{\alpha\mu\nu\rho}-g^{\sigma\alpha}\epsilon^{\beta\mu\nu\rho})$$ So only six terms survive, but how?

EDIT:
The gamma matrices are defined here, the metric tensor is $$g=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ and $\epsilon^{ijkl}$ is Levi Civita symbol in four dimensions

• I think you need to include some explanation of the symbols you are using. Nov 4, 2015 at 9:10

To get from your 15 term expression to your 6 term expression, you can use what is sometimes called the Schouten identity, $g^{\mu [ \nu} \epsilon^{\rho \sigma \alpha \beta]} = 0$. The square brackets denote antisymmetrization, and an antisymmetrization of 5 indices in 4 dimensions must vanish. There are 5 terms when expanded: $g^{\mu \nu} \epsilon^{\rho \sigma \alpha \beta} + g^{\mu \rho} \epsilon^{\sigma \alpha \beta \nu} + g^{\mu \sigma} \epsilon^{\alpha \beta \nu \rho} + g^{\mu \alpha} \epsilon^{\beta \nu \rho \sigma} + g^{\mu \beta} \epsilon^{\nu \rho \sigma \alpha} = 0$. Apply the Schouten identity three times using $g^{\mu [ \nu} \epsilon^{\rho \sigma \alpha \beta]} = 0$, $g^{\nu [ \rho} \epsilon^{\mu \sigma \alpha \beta]} = 0$ and $g^{\rho [ \mu} \epsilon^{\nu \sigma \alpha \beta]} = 0$. The 9 terms you want to cancel will turn into 6 terms, but they are opposite sign pairs of $g^{\mu \nu} \epsilon^{\rho \sigma \alpha \beta}$, $g^{\nu \rho} \epsilon^{\mu \sigma \alpha \beta}$, $g^{\rho \mu} \epsilon^{\nu \sigma \alpha \beta}$, so they cancel.

• Thank you, thanks a million @DavidChow. Just one thing, I can choose different Schouten identities and get different solutions ( I mean different six terms ) that are the same?
– VlS
Nov 4, 2015 at 20:29

I am not sure what your conventions are for the Clifford algebra, but let's assume that $\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu = 2 g_{\mu\nu} 1$. The basic rule for multiplying $\gamma$ matrices is $$\gamma_\mu \gamma_\nu = \gamma_{\mu\nu} + g_{\mu\nu} 1~,$$ where $\gamma_{\mu\nu} = \frac12 [\gamma_\mu, \gamma_\nu]$, which generalises to $$\gamma_\mu \gamma_{\nu_1\cdots \nu_k} = \gamma_{\mu\nu_1\cdots\nu_k} + \sum_{i=1}^k g_{\mu\nu_i} (-1)^{i+1} \gamma_{\nu_1\cdots\widehat{\nu_i}\cdots \nu_k}~,$$ where $\gamma_{\nu_1\cdots \nu_k}$ is the antisymmetrisation (with weight one) of $\gamma_{\nu_1} \cdots \gamma_{\nu_k}$. In other words, there should be $k!$ terms and you divide by $k!$. In addition, you can derive some formulae like $$\gamma_{\mu_1\cdots\mu_4} = \epsilon_{\mu_1\cdots\mu_4} \gamma_5$$ and $$\gamma_{\mu\nu\rho} = \epsilon_{\mu\nu\rho\sigma} \gamma^\sigma\gamma_5$$ where I raise and lower indices with $g$. (There maybe a sign here which I'm too lazy to track.) Finally, you need to use that $\gamma_\mu$, $\gamma_{\mu\nu}$, $\gamma_\mu\gamma_5$ and $\gamma_5$ are traceless.

The above rules are sufficient to arrive at the formula you wrote down.

• Thank you @José. Seems like Schouten identity solves my problem.
– VlS
Nov 4, 2015 at 20:31

Unfortunately, I can not comment on other answers, hence the reply. I do not agree with @Zurab Silagadze. The identity

$$\gamma_\mu \gamma_\nu \gamma_\rho = g_{\mu \nu} \gamma_\rho - g_{\mu \rho} \gamma_\nu + g_{\nu \rho} \gamma_\mu - \mathrm{i} \epsilon_{\mu \nu \rho \tau} \gamma^\tau \gamma^5$$

from Pal - Representation-independent manipulations with Dirac matrices and spinors, eq. (4.15) seems correct to me. This can explicitly be checked for fixed values of the indices. It depends on the convention of the Levi-Civita Tensor, that is $$\epsilon_{0123}=1$$ (implying $$\epsilon^{0123}=-1$$) or $$\epsilon^{0123}=1$$ (implying $$\epsilon_{0123}=-1$$), which sign has to be used. See also Appendix C: Dirac Matrix and Gamma Matrix Traces of Nagashima - Elementary Particle Physics: Quantum Field Theory and Particles.

The other steps that are done then possibly obtain an additional minus sign, as also pointed out by @José Figueroa-O'Farrill in his answer. The procedure stays the same.

I was having problems with this specific formula because of the exact same sign. So hopefully this turns out useful for somebody else as well.

You can use the identity $$\gamma_\mu\gamma_\nu\gamma_\rho=g_{\mu\nu}\gamma_\rho-g_{\mu\rho}\gamma_\nu+g_{\nu\rho}\gamma_\mu+i\epsilon_{\mu\nu\rho\tau}\gamma^\tau\gamma^5$$ which can be found, for example, in Pal - Representation-independent manipulations with Dirac matrices and spinors (identity 4.15, sign error in the last term was corrected — see the identity C2b in the book Y. Nagashima, Elementary Particle Physics. Volume 1: Quantum Field Theory and Particles, p. 818). With the help of this identity, traces involving $$\gamma_5$$ can be calculated recursively. For example, in your case we will have $$\operatorname{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)=g^{\mu\nu}\operatorname{Tr}(\gamma^\rho\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)-g^{\mu\rho}\operatorname{Tr}(\gamma^\nu\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)+ \\ g^{\nu\rho}\operatorname{Tr}(\gamma^\mu\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)+i\epsilon^{\mu\nu\rho\tau} \operatorname{Tr}(\gamma_\tau\gamma^5\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5).$$ Now we can use $$\operatorname{Tr}(\gamma^\rho\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)=-4i\epsilon^{\rho\sigma\alpha\beta},$$ and (because $$\gamma^5$$ anticommutes with all $$\gamma^\alpha$$, and $$\gamma^5\gamma^5=1$$) $$\operatorname{Tr}(\gamma_\tau\gamma^5\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)=-\operatorname{Tr}(\gamma_\tau\gamma^\sigma\gamma^\alpha\gamma^\beta)=-4(g_\tau^\sigma g^{\alpha\beta}-g_\tau^\alpha g^{\sigma\beta}+g_\tau^\beta g^{\sigma\alpha}),$$ to get your desired identity $$\operatorname{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^\alpha\gamma^\beta\gamma^5)=-4i\left (g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\nu\rho}\epsilon^{\mu\sigma\alpha\beta}+g^{\alpha\beta}\epsilon^{\mu\nu\rho\sigma}-g^{\sigma\beta}\epsilon^{\mu\nu\rho\alpha}+g^{\sigma\alpha}\epsilon^{\mu\nu\rho\beta}\right ).$$

Very interesting discussion, and I would like to put there my version of this problem.

First, all identities only have any sense if basic conventional points are postulated. Namely, below I will use the following definition of four-dimensional Levi-Civita symbol: $$\varepsilon^{\alpha\beta\gamma\delta}=-\varepsilon_{\alpha\beta\gamma\delta}= \begin{cases} +1,& \text{if } \alpha\beta\gamma\delta=\text{even permutation of 0123}\\ -1,& \text{if } \alpha\beta\gamma\delta=\text{odd permutation of 0123}. \end{cases}$$ Next, the definition of fifth gamma-matrix is the following: $$\gamma^5=\gamma_5=i\gamma^0\gamma^1\gamma^2\gamma^3=-i\gamma_0\gamma_1\gamma_2\gamma_3.$$ With such definitions $$\mathrm{Tr}\{{\gamma^5\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}}\}=\mathrm{Tr}\{{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\gamma^5}\}=-4i\varepsilon^{\alpha\beta\gamma\delta},\tag{1}\label{1}$$ what is checked for the component $$\alpha=0,\ \beta=1,\ \gamma=2,\ \delta=3$$ and is carried over to the other components which are permutations of 0123.

Further, there is the identity $$\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}=\eta^{\alpha\beta}\gamma^{\gamma}-\eta^{\alpha\gamma}\gamma^{\beta}+\eta^{\beta\gamma}\gamma^{\alpha}-i\gamma^5\varepsilon^{\alpha\beta\gamma\delta}\gamma_{\delta},\tag{2}\label{2}$$ where $$\eta^{\mu\nu}$$ is the Minkowski metric with the signature $$\{+1, -1, -1, -1\}$$ and, hence, $$\gamma_0=\gamma^0$$, while $$\gamma_\mu=-\gamma^\mu$$ with $$\mu=\{1, 2, 3\}$$. Such defined the identity is in agreement with the version of @J.G. because $$\gamma^5$$ anticommutes with any of $$\gamma_\mu$$ and $$\varepsilon^{\delta\alpha\beta\gamma}=-\varepsilon^{\alpha\beta\gamma\delta}$$.

Finally, we are ready to calculate the trace $$\mathrm{Tr}\{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\rho}\gamma^{\sigma}\gamma^{\tau}\gamma^5\}$$.

First, replacing $$\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}$$ with the use of the identity \ref{2} one obtains $$\mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}= \eta^{\alpha\beta}\mathrm{Tr}\{\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- \eta^{\alpha\gamma}\mathrm{Tr}\{\gamma^\beta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}+ \eta^{\beta\gamma}\mathrm{Tr}\{\gamma^\alpha\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- i\varepsilon^{\alpha\beta\gamma\delta}\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}.\tag{3}\label{3}$$ First three terms are rewritten with the use of the identity \ref{1}. As for the last term, let's consider the multiplier $$\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}$$ separately: $$\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}$$ because of anticommutation of $$\gamma^5$$ with both $$\gamma_\mu$$ and $$\gamma^\mu$$, further use the identity \ref{2} one more time: $$\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= \eta^{\rho\sigma}\mathrm{Tr}\{\gamma_\delta\gamma^\tau\}- \eta^{\rho\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\sigma\}+ \eta^{\sigma\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\rho\}- i\varepsilon^{\rho\sigma\tau\eta}\mathrm{Tr}\{\gamma_\delta\gamma^5\gamma_\eta\}.\tag{4}\label{4}$$ The last term vanishes (trace of product of fifth gamma-matrix with only two gamma-matrices) and every of the other terms contains the multiplier of the form $$\mathrm{Tr}\{\gamma_\lambda\gamma^\nu\}=\eta_{\lambda\mu}\mathrm{Tr}\{\gamma^\mu\gamma^\nu\}=4\eta_{\lambda\mu}\eta^{\mu\nu}.$$ Using this one can rewrite \ref{4} in the form $$\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= 4\eta^{\rho\sigma}\eta_{\delta\lambda}\eta^{\lambda\tau}- 4\eta^{\rho\tau}\eta_{\delta\lambda}\eta^{\lambda\sigma}+ 4\eta^{\sigma\tau}\eta_{\delta\lambda}\eta^{\lambda\rho}$$ and inserting it to \ref{3} one obtains $$\mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=-4i( \eta^{\alpha\beta}\varepsilon^{\gamma\rho\sigma\tau} -\eta^{\alpha\gamma}\varepsilon^{\beta\rho\sigma\tau} +\eta^{\beta\gamma}\varepsilon^{\alpha\rho\sigma\tau} +\eta^{\rho\sigma}\varepsilon^{\alpha\beta\gamma\tau} -\eta^{\rho\tau}\varepsilon^{\alpha\beta\gamma\sigma} +\eta^{\sigma\tau}\varepsilon^{\alpha\beta\gamma\rho}).$$ Note, that we have used the following transformations: $$\varepsilon^{\alpha\beta\gamma\delta}\eta_{\delta\lambda}\eta^{\lambda\tau}= \varepsilon^{\alpha\beta\gamma}_{\phantom{\alpha\beta\gamma}\lambda}\eta^{\lambda\tau}\equiv \varepsilon^{\alpha\beta\gamma\lambda}\delta^\tau_\lambda= \varepsilon^{\alpha\beta\gamma\tau},$$ where $$\delta^\alpha_\beta=1$$ if $$\alpha=\beta$$ and $$\delta^\alpha_\beta=0$$ if $$\alpha\neq\beta$$ and and so on.