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I have the following doubt about the identity in this page: $$ {a\!\!\!/}{a\!\!\!/} \equiv a^\mu a_\mu \cdot I_4 = a^2 \cdot I_4 $$ however: $${a\!\!\!/}{a\!\!\!/}=a^{\mu}\gamma_{\mu}a_{\mu}\gamma^{\mu} = a^2 \gamma_{\mu}\gamma^{\mu}=4 a^2 I_{4}$$

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    $\begingroup$ You have repeated the index $\mu$ four times and should only do so twice. The a components are not contracted until later, as in the answer below $\endgroup$ – lux Dec 28 '19 at 0:37
  • $\begingroup$ @lux That seems more like an answer that an comment. $\endgroup$ – rob Dec 28 '19 at 5:03
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Since $a$ is a vector, its entries $a_\mu$ are just numbers and they commute with everything

$$ {a\!\!\!/}{a\!\!\!/} = a_\mu\gamma^\mu a_\nu\gamma^\nu = \gamma^\mu \gamma^\nu a_\mu a_\nu,$$

using $\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2 \eta^{\mu\nu}I_4 $ to get $\gamma^\mu\gamma^\nu = 2 \eta^{\mu\nu}I_4 - \gamma^\nu\gamma^\mu $, the expression above becomes:

$$ \gamma^\mu \gamma^\nu a_\mu a_\nu = (2 \eta^{\mu\nu}I_4 - \gamma^\nu\gamma^\mu) a_\mu a_\nu = 2 I_4 a^\mu a_\mu - \gamma^\nu\gamma^\mu a_\mu a_\nu.$$

The last term, via commutation of $a_\mu$, can be written $\gamma^\nu\gamma^\mu a_\mu a_\nu = a_\nu\gamma^\nu a_\mu\gamma^\mu = {a\!\!\!/}{a\!\!\!/}$. Bring it to the LHS to get:

$$2 {a\!\!\!/}{a\!\!\!/} = 2 I_4 a^\mu a_\mu, $$

$$ \Rightarrow {a\!\!\!/}{a\!\!\!/} = I_4\,a^\mu a_\mu = I_4\, a^2.$$

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OP has an error in the application of the Einstein summation convention. In their expression $a^{\mu}\gamma_{\mu}a_{\mu}\gamma^{\mu}$ the index $\mu$ is repeated 4 times. OP then incorrectly sees the terms $a^{\mu}a_{\mu}$ and $\gamma_{\mu}\gamma^{\mu}$ and applies the summation convention individually to them.

However in the expression $a\!\!\!/a\!\!\!/$ the contractions are between the vector $a$ and a gamma matrix and not between $a$ and itself: $$a\!\!\!/a\!\!\!/= a_{\mu}\gamma^{\mu}\,\, a_{\nu}\gamma^{\nu}$$ which makes it clear that there is not yet an inner product of $a$ with itself. From here we take advantage of symmetry of the product $a_{\mu}a_{\nu}$ to take only the symmetric part of the decomposition $\gamma^{\mu}\gamma^{\nu} = \frac{1}{2}\{\gamma^{\mu}, \gamma^{\nu}\} + \frac{1}{2}[\gamma^{\mu}, \gamma^{\nu}]$. We get $$a_{\mu}\gamma^{\mu}\, a_{\nu}\gamma^{\nu} = a_{\mu}a_{\nu}\frac{1}{2}\{\gamma^{\mu}, \gamma^{\nu}\} = a_{\mu}a_{\nu}\eta^{\mu\nu}I_{4} = a^{2}I_{4}. $$

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