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?
[Physics] Gamma matrices and trace operator
dirac-matriceshomework-and-exercisestrace
Related Solutions
Ok, after posting this question I have been trying to solve this and finally did, so I am posting the answer.
Just one identity is enough and that is:
$$ \gamma^\mu\gamma^\nu\gamma^\lambda = g^{\mu\nu}\gamma^\lambda + g^{\nu\lambda}\gamma^\mu - g^{\mu\lambda}\gamma^\nu + i \epsilon^{\sigma\mu\nu\lambda} \gamma_\sigma\gamma^5 $$
And a few basic properties of Dirac matrices like: $ (\gamma^5)^2 = 1 $, $ \{ \gamma^\mu, \gamma^5 \} = 0 $, $ \{\gamma^\mu, \gamma^\nu \} = 2 g^{\mu\nu} $ etc.
So we had:
$$ T = [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\mu\gamma_\sigma\gamma_\nu (1-\gamma^5) \nu] $$
We implement the above mentioned identity to the right bracket
$$ T \propto [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}( g_{\mu\sigma}\gamma_\nu + g_{\sigma\nu}\gamma_\mu - g_{\mu\nu}\gamma_\sigma + i \epsilon_{\alpha\mu\sigma\nu} \gamma^\alpha\gamma^5 ) (1-\gamma^5) \nu] $$
Now using using the symmetric property of the metric tensor the we simplify the expression a lot:
$$ T \propto (4+4+4) \{ [\bar{s} \gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu}\gamma_\nu(1-\gamma^5) \nu] + i \epsilon_{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$
Now if we again use the identity of the product of three Gamma matrices in the first bracket of the last term and again the metric tensor-terms will drop because of the anti-symmetric property of the Levi Civita and so the last term will become:
$$ T_{last} = i \epsilon^{\alpha\mu\sigma\nu} [\bar{s}\gamma^\mu\gamma^\sigma\gamma^\nu (1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha\gamma^5 (1-\gamma^5) \nu] = i \epsilon_{\mu\sigma\nu\alpha} i\epsilon^{\mu\sigma\nu\delta} [\bar{s}\gamma_\delta \gamma^5 (1-\gamma^5) d] \otimes [\bar{\nu} \gamma^\alpha\gamma^5 (1-\gamma^5) \nu] $$
Now we can apply the identity: $$ \epsilon_{\mu\sigma\nu\alpha} \epsilon^{\mu\sigma\nu\delta} = 4! \delta^\alpha_\delta $$ and also using the properties of $\gamma^5$ finally we have,
$$ T_{last} = - 24 [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] $$
Hence,
$$ T \propto [\bar{s}\gamma^\alpha(1-\gamma^5) d] \otimes [\bar{\nu} \gamma_\alpha(1-\gamma^5) \nu] = (\bar{s} d)_{V-A}(\bar{\nu}\nu)_{V-A} $$
Quick answer:
We can always define the Dirac adjoint of a spinor to be $\bar \psi := \psi^\dagger \gamma^0$. After that definition is set up, we have to additionally tell how it transforms under a Lorentz transformation and that depends on convention. For example, we could choose $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ (note $\gamma'^0$ is no longer Hermitian) and define $\bar \psi := \psi^\dagger \gamma'^0 $, then we will see that $\bar\psi\psi$ no longer transforms as a scalar.
On the other hand, if we insist that $\bar\psi \psi$ should transform as a scalar, we should find a function of the gamma matrices $F(\gamma'^\mu)$ such that for any Lorentz transformation $\Lambda,\ \color{red}{S(\Lambda)^\dagger FS(\Lambda) \equiv F(\gamma'^\mu)}$. Then, defining $\bar \psi := \psi^\dagger F$ ensures that $\bar\psi\psi$ transforms as a scalar. The question is whether such a function always exists. With your example, you can convince yourself that the ansatz $F= c_\mu \gamma'^\mu$ will work if and only if $c_\mu \Lambda^\mu{}_j=0$ for all spatial indices $j$. This is a system of three equations $(j=1,2,3)$ with four unknowns $(c_\mu)$, and therefore possibly has many solutions. For example, if $\Lambda$ describes a boost in the $x$-direction with speed $\beta$, defining $\bar \psi := \psi^\dagger (\gamma'^0 + \beta \gamma'^1)$ makes $\psi^\dagger\psi$ a scalar. Note that $\gamma'^0 + \beta \gamma'^1 \propto \gamma^0$, so there is no contradiction.
Detailed answer:
A Clifford algebra over a $D$-dimensional spacetime equipped with the metric $g^{\mu\nu}$ is generated by $D$ hypercomplex numbers $\{\gamma^\mu\}, \mu\in \{0,\cdots,D-1\}$ defined by the following algebraic product:
$$ \{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu} \mathbb I_n\,. \tag1 $$
The algebra is $2^D$-dimensional, meaning there is a list of $2^D$ linearly independent elements, closed under multiplication, formed by various products of the $D$ hypercomplex numbers. Moreover, there is always a representation of the algebra in terms of real $n \times n$ matrices where $n=2^{[D/2]}$, the operation $[\cdot]$ taking the integer part of the given number. If the dimension $D$ is even, then this representation is the one and only irreducible representation of the algebra (up to equivalences). Furthermore, if the representation is unitary, the $2^D$ basis elements can be chosen to be Hermitian.
Without loss of generality, take the metric to be of mostly negative signature, so that $(\gamma^0)^2 = +1$ and $(\gamma^i)^2=-1$. You already begin to notice why $\gamma^0$ is special when contrasted with the rest of $\gamma$-matrices. It is exactly the same way as time is special when contrasted with space, given that the metric has a Lorentzian signature. With this in mind, for D=4 a basis for $4\times4$ matrices is given by $\{\Gamma_j\}, j\in\{1,\cdots, 16\}$ where
\begin{align*} \Gamma_j \in \{&\mathbb I_4, \gamma^0,i\gamma^1,i\gamma^2,i\gamma^3,\\ &\gamma^0\gamma^1,\ \gamma^0\gamma^2,\ \gamma^0\gamma^3,\quad i\gamma^1\gamma^2,\ i\gamma^2\gamma^3,\ i\gamma^3\gamma^1,\ \\ &i\gamma^0\gamma^1\gamma^2,\quad i\gamma^0\gamma^1\gamma^3,\quad i\gamma^0\gamma^2\gamma^3,\quad i\gamma^1\gamma^2\gamma^3,\\ &i\gamma^0\gamma^1\gamma^2\gamma^3\}\,. \tag2 \\ \end{align*}
The imaginary number $i$ has been inserted in the above array so that for all $j, (\Gamma_j)^2=+1.$ Notice that this list is closed under multiplication. This allows you to write any $4\times4$ matrix $X$ as a sum over these matrices: $X = \sum_{i=1}^{16} x^i \Gamma_i$ where $x_i = \frac14 \text{Tr}(X\Gamma_i)$. Moreover, for all $j\ne1, \text{Tr}(\Gamma_j)= 0$. Using this information, you can prove the following lemma (Cf. Schwartz Ch. 4).
Lemma:
Given two sets of gamma matrices $\{\gamma^\mu\}$ and $\{\gamma'^\mu\}$, correspondingly basis vectors $\{\Gamma_j\}$ and $\{\Gamma'_j\}$, there exists a non-singular matrix $S$ such that
$$ \boxed{\gamma'^\mu = S\gamma^\mu S^{-1}}\,, \qquad \text{ where } S=\sum_{i=1}^{16} \Gamma'_i F \Gamma_i\,, $$
and $F$ is so chosen that $S$ is non-singular.
Furthermore, $S$ is uniquely fixed (up to a numerical factor).
You have correctly identified that $\gamma'^\mu := \Lambda^\mu{}_\nu \gamma^\nu$ obeys the same commutation relation as $\gamma^\mu$, and it is precisely because of this that the above lemma tells us of the existence of a non-singular $S$ so that
$$ S^{-1} \gamma^\mu S = \Lambda^\mu{}_\nu \gamma^\nu\,, \tag3$$
which is required for the Lorentz invariance of the Dirac equation.
At this point, let us make a conventional choice (which @akhmeteli has been pointing out all along). Let us choose $\gamma^0$ such that it is Hermitian and the $\gamma^i$s such that they are anti-Hermitian. In other words, this means that $\gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$. Notice that we did not need any such assumption in the previous discussion. But this convention will simplify greatly what is about to follow.
Observe that because $\Lambda^\mu{}_\nu$ is real, and because we chose this convention,
\begin{align*} (S^\dagger\gamma^0) \gamma^\mu (S^\dagger \gamma^0)^{-1} = (S^{-1} \gamma^\mu S)^\dagger &\stackrel{(3)}{=} \Lambda^\mu{}_\nu (\gamma^\nu)^\dagger\\ \text{(Multiplying by $\gamma^0$ from both sides)} \Rightarrow (\gamma^0 S^\dagger\gamma^0) \gamma^\mu (\gamma^0 S^\dagger \gamma^0)^{-1} &= \Lambda^\mu{}_\nu \gamma^\nu \stackrel{(3)}{=}S^{-1} \gamma^\mu S\,. \\ \end{align*}
After rearranging we find that,
\begin{align*} &\Rightarrow (S\gamma^0 S^\dagger\gamma^0) \gamma^\mu (S\gamma^0 S^\dagger \gamma^0)^{-1} = \gamma^\mu \\ &\Rightarrow S\gamma^0 S^\dagger\gamma^0 = c\mathbb I_4 \\ &\Rightarrow S^\dagger\gamma^0 = c \gamma^0 S^{-1}\,, \tag4 \\ \end{align*}
where $c$ is some constant which you can convince yourself is real.
Now, if we normalize $S$ such that $\det(S)=1$, then $c^4 = 1$ or $c= \pm 1$. Let us see which situations correspond to $c=+1$ and $c=-1$. Observe the following identity.
\begin{align*} S^\dagger S = (S^\dagger\gamma^0)\gamma^0S &\stackrel{(4)}= c\gamma^0(S^{-1} \gamma^0 S)\\ &\stackrel{(3)}= c\gamma^0 \Lambda^0{}_\nu \gamma^\nu \\ &= c(\Lambda^0{}_0 \mathbb I_4 - \sum_{k=1}^3 \Lambda^0{}_k\gamma^0\gamma^k)\\ \Rightarrow \text{Tr}(S^\dagger S) &= 4c\Lambda^0{}_0 \,. \tag5 \end{align*}
Since $S^\dagger S$ has real eigenvalues (it is Hermitian) and positive-definite (since S is non-singular), its trace must be positive. This means that when $ \Lambda^0{}_0 \le -1, c=-1$ and when $ \Lambda^0{}_0 \ge +1, c=+1$.
Conclusion:
Defining $\bar \psi := \psi^\dagger \gamma^0$, we see that under a Lorentz transformation (taking $\psi \to S\psi$) that does not involve time reversal ($\Lambda^0{}_0 \ge +1$), $\bar \psi \to +\bar\psi S^{-1}$, whereas for Lorentz transformations that do reverse time ($\Lambda^0{}_0 \le -1$), $\bar \psi \to -\bar\psi S^{-1}$.
If we had not normalized $S$, we would write $\bar\psi \to c\bar\psi S^{-1}$. Then this extra factor of $c$ will need taking care of in the Lagrangian through perhaps a field redefinition.
More generally, if we did not assume the hermiticity properties of the gamma matrices, things would be much more complicated, but a definition is a definition. After defining $\bar \psi := \psi^\dagger \gamma^0$, our task would be to find its transformation rules and then implement it accordingly in the Lagrangian. Alternatively, if you wish to preserve the transformation rules, you will have to change the definition accordingly.
Best Answer
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.