Proving the Trace of Gamma Matrices with (Anti-)Communtation Rules

[WarnK]

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?

 

[Avodyne]

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

 

[WarnK]

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]

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?

 

[WarnK]

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

 

[Avodyne]

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.