# Gamma matrices

## Homework Statement

Show that $$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = 0$$

## Homework Equations

(anti-)commutation rules for the gammas, trace is cyclic

## The Attempt at a Solution

I can do
$$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = -tr(\gamma^{\mu}\gamma^{5}\gamma^{\nu}) = - tr(\gamma^{\nu}\gamma^{\mu}\gamma^{5})$$
and so on, but I don't see how that helps me. Any suggestions?

## Answers and Replies

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Avodyne
Science Advisor
You've used the first of your "relevant equations", now use the second.

If I continue like this
$$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = - tr(\gamma^{\nu}\gamma^{\mu}\gamma^{5}) = - tr( (2\eta^{\nu \mu} - \gamma^{\mu}\gamma^{\nu} )\gamma^{5}) = tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) - tr(2\eta^{\nu \mu} \gamma^{5})$$
which I guess means that $$tr(\gamma^{5})=0$$

Avodyne
Science Advisor
Oops, I didn't notice that you've already gotten as far as you need to. You've shown that your original expression is equal to minus itself. What is the only number with this property?

But, I have $$\gamma^{\mu}$$ and $$\gamma^{\nu}$$ in a different order, and the entire point is that they don't commute?

Last edited:
Avodyne
Science Advisor
Oops, sorry, I screwed up again.

To do this one, I think you need to insert (gamma^rho)^2 (which is either one or minus one, depending on your conventions and on whether or not rho=0) into the trace,
with rho not equal to mu or nu, and then move one of the gamma^rho's around by commutation.