Trace of two Dirac matrices in 4 dimensions

Question

I want to show that tr($gamma^mu gamma^nu$) = 4$eta^{mu nu}$.
I know that  {$gamma^mu , gamma^nu$} = 2$eta^{munu}I_4$ and tr($gamma^mu gamma^nu$) = tr($gamma^nu gamma^mu$),
so tr($gamma^mu gamma^nu$) = tr(2$eta^{munu} I_4$ - $gamma^nu gamma^mu$) $implies$ 2tr($gamma^mu gamma^nu$) = 2tr($eta^{munu} I_4$), and ony way to get the answer is by pulling out $eta^{munu}$ out of the trace.
I want to ask how we can do that? since $eta^{munu}$ is itself a matrix, then why it is taken out of the trace and its diagonal elements are not summed?

SolubleFish · Accepted Answer

The trace is over the matrix indices which represent the internal degrees of freedom of the spinor, not over the space-time indices.
Each $gamma^mu$ is a matrix, so  if we want to write out all the indices explicitely, it would be $(gamma^mu)^{i}_{~~j}$. Then ${gamma^nu,gamma^mu} = 2eta^{munu}$ becomes :
$$(gamma^mu)^i_{~~j}(gamma^{nu})^j_{~~k} + (gamma^nu)^i_{~~j}(gamma^{mu})^j_{~~k} = 2 eta^{munu}delta^i_k$$
and what you want to show is :
$$(gamma^mu)^i_{~~j}(gamma^nu)^j_{~~i} = 2 eta^{munu}$$
So the short answer is : the $eta^{munu}$ goes out of the trace freely.

Trace of two Dirac matrices in 4 dimensions

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