TITLE: Confusion about slash notation
QUESTION [4 upvotes]: I am confused about the slash notation and especially taking the square of a slashed operator.
Defining $\displaystyle{\not} a \, = \, \gamma^\mu a_\mu$ we have $\,\,$ $\displaystyle{\not} a \displaystyle{\not} a = a^2 $
I tried to prove that, but I can't really doing it without assumption I didn't prove. That's my (I think wrong) procedure:
$\displaystyle{\not} a \displaystyle{\not} a = \gamma^\mu a_\mu \gamma^\nu a_\nu  = \gamma^\mu \gamma^\nu a_\mu  a_\nu $ assuming in the last equality that the they commute, now using the anticommutation relation of the gamma matrices :
$$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu =2\eta^{\mu\nu}\tag{1}$$ I say that probably $\gamma^\mu \gamma^\nu= \eta^{\mu\nu}$ and substituing it in one i obtain 
$$\eta^{\mu\nu}a_\mu  a_\nu\,=\, a_\mu  a^\nu \, = \, a^2$$
I don't think that's really a proof, can someone provide a right proof without the assumptions I've made?

REPLY [4 votes]: This: $$\gamma ^\mu \gamma ^\nu =\eta ^{\mu \nu}$$is wrong. It would give, for example, $$\gamma ^0 \gamma ^0=1=-\gamma ^1 \gamma ^1\\ \gamma ^0 \gamma ^1=0,$$ and you can see that these relations are inconsistent.
The relation comes from the following identities: $${\not} a \, {\not} a = a_\mu a_\nu \gamma ^\mu \gamma ^\nu=a_\mu a_\nu\left(\frac{\lbrace \gamma ^\mu ,\gamma ^\nu \rbrace }{2}+\frac{[ \gamma ^\mu ,\gamma ^\nu  ]}{2}\right)=a_\mu a_\nu \frac{\lbrace \gamma ^\mu ,\gamma ^\nu \rbrace }{2}.$$ Notice: it is assumed that $a_\mu$ and $a_\nu$, whatever they are, commute.
The third equality is proved here.
The last equality is valid because contracting the symmetric tensor $a_\mu a_\nu$ with the antisymmetric indexed matrices $[\gamma ^\mu ,\gamma ^\nu]$ gives zero, that is, the contraction of a symmetric tensor ($a_\mu a_\nu \equiv A_{\mu\nu}$) and an antisymmetric tensor ($[\gamma ^\mu ,\gamma ^\nu] \equiv \Gamma^{\mu\nu}$) is zero. The proof is as follows.
\begin{eqnarray}
A_{\mu\nu} \Gamma^{\mu\nu} &=& A_{\nu\mu} (-\Gamma^{\nu\mu}) \\
                          &=& -A_{\nu\mu}\Gamma^{\nu\mu} \\
                          &=& - A_{\mu\nu} \Gamma^{\mu\nu}, \quad [\because \mu, \nu \,  \text{are dummy indices.}]
\end{eqnarray}
which implies that $A_{\mu\nu} \Gamma^{\mu\nu}$ is zero.
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