# How to prove $\{\gamma^{\mu}, \gamma^{\nu}\}$ ? (notation problem)

I want to prove that $\{\gamma^{\mu}, \gamma^{\nu} \}=2g^{\mu \nu}$ what are the indices $\mu$ and $\nu$ here? because I know the gamma matrices from 0 to 5 and I need to verify the anti commutation relation. but what should I put in here?

I just tried it for $\gamma^{0}$ and $\gamma^{1}$ this is what I got:

$$\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\\ \end{pmatrix} \begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\ -1&0&0&0\\ \end{pmatrix} + \begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\ -1&0&0&0\\ \end{pmatrix} \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\\ \end{pmatrix}$$

I get the zero matrix.

What am I doing wrong? and what $g^{\mu \nu}$ am I supposed to get?

The gamma matrices in $d=4$ dimensions (one spatial, three temporal) in Minkowski space satisfy
$$\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu} \mathbb{I}_4$$
where $\mathbb I_4$ is the $4\times 4$ identity matrix. Recall that $\mu,\nu$ are Lorentz indices running over $0,1,2,3$. There is a fifth gamma matrix, denoted $\gamma^5$, but the notation $\gamma^5$ is simply because it is the fifth; if we followed the numebering based on the Lorentz indices, it should be $\gamma^4$.
The fifth gamma matrix $\gamma^5$ is not taken to be in the set we iterate over to compute commutation or anti-commutation relations involving the gamma matrices. This is because it is $i\gamma^0 \gamma^1 \gamma^2 \gamma^3$.
The gamma matrices may be extended to any dimension though, simply by extending the defining anti-commutation relation to a $d$-dimensional $\eta^{\mu\nu}$ and a $d\times d$ identity matrix.