Trace of a product of gamma matrices

[BVM]

Homework Statement

A proof of equality between two traces of products of gamma matrices.

$$Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = 2Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu)$$

Where no special property of A is given, so we must assume it is just a random 4x4 matrix.
$1_4$ represents the 4x4 unity matrix.

Homework Equations

$$\gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3$$
The usual product and trace identities as found here.

The Attempt at a Solution

I have been able to prove that
$$Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu) + Tr(\gamma^5 \gamma^\mu A (1_4-\gamma^5) \gamma^\nu)$$
using the definition of the trace and the anticommutativity of the $\gamma^\mu$ and $\gamma^5$.

 

[TSny]

I have been able to prove that
$$Tr(\gamma^\mu (1_4-\gamma^5) A (1_4-\gamma^5) \gamma^\nu) = Tr(\gamma^\mu A (1_4-\gamma^5) \gamma^\nu) + Tr(\gamma^5 \gamma^\mu A (1_4-\gamma^5) \gamma^\nu)$$
using the definition of the trace and the anticommutativity of the $\gamma^\mu$ and $\gamma^5$.

For your last term, try using the fact that the trace is invariant under any cyclic permutation of the matrices inside the trace. So, you can move the γ⁵ on the left to the far right. Then simplify (1₄ - γ⁵) γ$^\nu$ γ⁵.

 

[BVM]

For your last term, try using the fact that the trace is invariant under any cyclic permutation of the matrices inside the trace. So, you can move the γ⁵ on the left to the far right. Then simplify (1₄ - γ⁵) γ$^\nu$ γ⁵.

Got it! That hint really helped. Thanks.