# Gamma matrices

1. Nov 27, 2007

### WarnK

1. The problem statement, all variables and given/known data
Show that $$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{5}) = 0$$

2. Relevant equations
(anti-)commutation rules for the gammas, trace is cyclic

3. 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?

2. Nov 27, 2007

### Avodyne

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

3. Nov 27, 2007

### 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 $$tr(\gamma^{5})=0$$

4. Nov 27, 2007

### 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?

5. Nov 27, 2007

### WarnK

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

Last edited: Nov 27, 2007
6. Nov 27, 2007

### 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.