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The defining property for the gamma matrices to generate a Clifford algebra is the anticommutation relation

$\displaystyle\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4 $

where $\{ , \}$ is the anticommutator, $\eta^{\mu \nu}$ is the Minkowski metric with signature (+ − − −) and $I_4$ is the 4 × 4 identity matrix.

This makes perfect sense if $\eta^{\mu \nu}$ is a scalar that is equal to zero when $\mu \neq \nu$ and is either +1 or -1 when they are the same. This would make sense because squaring a gamma matrix results in either $I_4$ or $-I_4$

But everything I read online says that $\eta^{\mu \nu}$ is a tensor

$\eta^{\mu \nu} = \pm \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},$

Just for reference the 4 gamma matrices are:

$\begin{align} \gamma^0 &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix},\quad& \gamma^1 &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\ \gamma^2 &= \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{pmatrix},\quad& \gamma^3 &= \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}. \end{align}$

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  • $\begingroup$ Well. $A$ is a matrix but its entries are real numbers. $\endgroup$ Apr 1, 2018 at 6:20
  • $\begingroup$ You can in fact work out that if $\mu \neq \nu$ that the right hand side yields zero. and you can work out that it will yield the diagonal components of $\eta$ for when the indices are the same. Obviously the object on the right depends on the indices of the objects on the left (there are two indices that describe the pieces so we know it is an array at least which is definitely not a scalar) $\endgroup$
    – Triatticus
    Apr 1, 2018 at 6:52
  • $\begingroup$ multiplying 2 gamma matrices results in $I$ or $-I$ not $\eta^{\mu \nu}$ $\endgroup$
    – R. Emery
    Apr 1, 2018 at 6:54
  • $\begingroup$ The thing is you are looking at it component-wise, that is the key thing here, for a given pair of indices the equation is completely true $\endgroup$
    – Triatticus
    Apr 1, 2018 at 6:58
  • $\begingroup$ Take for instance $\mu=1,\nu=2$ and apply $\gamma^1 \gamma^2 + \gamma^2 \gamma^1 = \gamma^1 \gamma^2 - \gamma^1 \gamma^2 = 2 \eta^{12} I = 0I$ $\endgroup$
    – Triatticus
    Apr 1, 2018 at 6:59

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$\eta= \pm \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix},$

but $\eta^{\mu \nu}$ is just one component of $\eta$

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