# Calculating traces for triangle diagrams with massless fermions

I am following Schwarz Quantum Field Theory textbook. In particular, I am looking at triangle diagrams with massless fermions. On pg. 623 - 624 Schwarz attempts to calculate $$q_\mu^1 M_{5}^{\alpha\mu\nu}$$ which should be zero by the Ward Identity of QED. This gives (Eq. 30.29),

$$\begin{multline} q_\mu^1 M_{5}^{\alpha\mu\nu} = \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{\text{Tr}[ \gamma^\nu(\not\! k + \not\!q_2)\gamma^\alpha\gamma^5(\not\! k - \not\!q_1) ]}{(k - q_1)^2 (k + q_2)^2} - \frac{\text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} \right.\\[0.25cm] \left. + \frac{ \text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] }{k^2(k - q_2)^2} - \frac{\text{Tr}[ \gamma^\nu(\not\! k + \not\! q_1)\gamma^\alpha\gamma^5(\not\! k - \not\! q_2) ]}{(k + q_1)^2(k - q_2)^2} \right]. \end{multline}$$

Schwarz says that after completing the traces, the result is (Eq. 30.30) $$$$q_\mu^1 M_{5}^{\alpha\mu\nu} = - 4i\epsilon^{\alpha\nu\rho\sigma} \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{(k - q_1)^\rho(k + q_2)^\sigma}{(k - q_1)^2(k + q_2)^2} - \frac{(k - q_2)^\rho(k + q_1)^\sigma}{(k - q_2)^2(k + q_1)^2} \right].$$$$

This leads me to believe that $$I = \int \frac{d^4 k}{(2\pi)^2} \left[ - \frac{\text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} + \frac{ \text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] }{k^2(k - q_2)^2} \right] = 0$$ which I am having difficulty proving. So far, I have evaluated the traces as follows \begin{align} \text{Tr}[ \not\!k \gamma^\nu(\not\! k + \not\! q_2)\gamma^\alpha\gamma^5 ] &= \text{Tr}[ \gamma^\alpha\gamma^5 \not\!k \gamma^\nu(\not\! k + \not\! q_2) ]\\[0.25cm] &= -\text{Tr}[\gamma^5\gamma^\alpha\gamma^\rho\gamma^\nu\gamma^\sigma ]k_\rho(k + q_2)_\sigma\\[0.25cm] &= -4i\epsilon^{\alpha\rho\nu\sigma} k_\rho(k + q_2)_\sigma\\[0.25cm] &= 4i\epsilon^{\alpha\nu\rho\sigma} k_\rho(q_2)_\sigma. \end{align}

Similarly, we can show that $$$$\text{Tr}[\gamma^\nu\not\!k \gamma^\alpha\gamma^5(\not\! k - \not\! q_2)] = -4i\epsilon^{\alpha\nu\rho\sigma}(q_2)_\rho k_\rho .$$$$

From here, the integral $$I$$ becomes, \begin{align} I &= -4i\epsilon^{\alpha\nu\rho\sigma}\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2} \left[\frac{k_\rho(q_2)_\sigma}{(k + q_2)^2} + \frac{k_\sigma(q_2)_\rho}{(k - q_2)^2}\right]\\[0.25cm] &= -4i\epsilon^{\alpha\nu\rho\sigma}\int\frac{d^4k}{(2\pi)^4} \frac{k_\rho(q_2)_\sigma}{k^2}\left[\frac{1}{(k + q_2)^2} - \frac{1}{(k - q_2)^2}\right] \overset{?}{=} 0. \end{align}

Here is where I am stuck. Simplifying the term in square brackets doesn't seem to help. My only other thought is that the integrand is odd as there is a $$k_\rho$$ term and may vanish.

Any help would be appreciated!

1. It is potentially inconsistent to shift the integration variable $$k$$ of divergent integrals, cf. e.g. subsections 30.2.2-3. This is particular sensitive in a discussion of quantum anomalies from triangle diagrams!
2. Pull $$(q_2)_\sigma$$ outside of the integral $$I$$ in OP's last expression.
3. The integral now has a lower external Lorentz index $$\rho$$. Since the integral only depends on $$q_2$$, any Lorentz-covariant regularization of the integral must produce a factor $$(q_2)_{\rho}$$.
4. Contraction of $$\epsilon^{\alpha\nu\rho\sigma} (q_2)_\sigma$$ with $$(q_2)_{\rho}$$ then yields that $$I=0$$ as OP wanted to show.