Gamma matrices and trace operator I'm trying to show that the trace of the product of the following three Gamma (Dirac) matrices is zero, i.e. $$\text{tr}(\gamma_{\mu} \gamma_{\nu} \gamma_{5})=0 \text{.}$$ I attempted to use the fact that the trace operator is invariant under cyclic permutations and linear, and that $$\gamma_{\mu} \gamma_{5}= -\gamma_{5} \gamma_{\mu}, \text{   } (\gamma_{5})^{2}= I_4 \text{ (4 $\times$ 4 identity matrix)} \text{,}$$ where $\gamma_{5} \equiv i\gamma_{0} \gamma_{1} \gamma_{2} \gamma_{3}$. But whenever I do that, it seems that I keep going in circles. Any idea on how I should proceed?
 A: Start noticing that ${(\gamma^{\alpha})}^2 =1\cdot g^{\alpha \alpha}$ and that
$$
\textrm{tr} (\gamma^{\mu}\gamma^{\nu}\gamma^5)= 
\textrm{tr}\left(\frac{1}{g^{\alpha \alpha}}{(\gamma^{\alpha})}^2\gamma^{\mu}\gamma^{\nu}\gamma^5\right)=\frac{1}{g^{\alpha \alpha}}\textrm{tr} (\gamma^{\alpha}\gamma^{\alpha}\gamma^{\mu}\gamma^{\nu}\gamma^5).
$$
Now choose $\alpha\neq \mu,\nu$ and commute the second $\gamma^{\alpha}$ three times until the end, to obtain three minus signs as
$$
\textrm{tr} (\gamma^{\mu}\gamma^{\nu}\gamma^5)= \frac{1}{g^{\alpha \alpha}}\textrm{tr} (\gamma^{\alpha}\gamma^{\alpha}\gamma^{\mu}\gamma^{\nu}\gamma^5) =-
\frac{1}{g^{\alpha \alpha}}\textrm{tr} (\gamma^{\alpha}\gamma^{\mu}\gamma^{\nu}\gamma^5\gamma^{\alpha}).
$$
Use at this point the ciclicity of the trace bringing back the last $\gamma^{\alpha}$ to the beginning, which together with the coefficient in the denominator gives rise to the identity leaving back only the additional minus sign in front, which proves that the equation is satisfied only if both members vanish.
A: Substitute two values for $\mu$ and $\nu$. 
If $\mu=\nu$, then using $(\gamma^\mu)^2 =\pm1$ and $tr(\gamma ^5) =0$ you have finished.
If $\mu \neq\nu$, then use $tr(\gamma ^\mu\gamma ^\nu) =0$ with the two  remaining $\gamma$ .
