Product of an odd number of Dirac $\gamma$ matrices Suppose $n$ is an odd number. Why can we write $a\llap{/}_1 a\llap{/}_2 ... a\llap{/}_n$ as
$$a\llap{/}_1 a\llap{/}_2 ... a\llap{/}_n = V_\mu \gamma^\mu + A_\mu \gamma^\mu \gamma_5$$
for some $V_\mu, A_\mu$?
I know that $\gamma^{\mu} \gamma^\nu=g^{\mu \nu}-i\sigma^{\mu \nu}$ and I tried to write $a\llap{/}_1 a\llap{/}_2 ... a\llap{/}_n$ as $a_{1\mu}a_{2\nu}… a_{n-1\alpha}  \gamma^{\mu} \gamma^\nu…\gamma^\alpha a\llap{/}_n$ and use $\gamma^{\mu} \gamma^\nu=g^{\mu \nu}-i\sigma^{\mu \nu}$. It gives a term with $a_1 \cdotp a_2...a_{n-2} \cdotp a_{n-1} a\llap{/}_n$ but then there are terms with products of $\sigma^{\mu \nu}$ that I don't know how to deal with. I don't know if this is the best approach to the problem.
 A: All possible (complex) $4\times4$ matrices span a 16-dimensional vector space, on which one can define a scalar product via $\langle A,B\rangle=\text{Tr}(A^\dagger B)$. Using the trace identities for the $\gamma$-matrices, it is easy to see that the 16 matrices $1$, $\gamma^\mu$, $\gamma^\mu\gamma^\nu$ ($\mu<\nu$), $\gamma^\mu\gamma_5$, $\gamma_5$ define an orthogonal basis on this space. Denoting this basis abstractly as $\Gamma_i$, $i=1,\dotsc,16$, any $4\times4$ matrix $M$ can then be expanded as
$$
M=\sum_{i=1}^{16}m_i\Gamma_i,
$$
where $m_i$ is proportional to $\langle\Gamma_i,M\rangle$. Whenever $M$ is a product of an odd number of $\gamma$-matrices, the trace of its product with $1$, $\gamma^\mu\gamma^\nu$ and $\gamma_5$ necessarily vanishes, so the expansion in the basis of $\Gamma_i$ contains only the $\gamma^\mu$ and $\gamma^\mu\gamma_5$ contributions.
A: It follows from the basic anticommutation relation for the $\gamma$-matrices, $\{\gamma_\mu,\gamma_\nu\}=2g_{\mu\nu}$, that (i) two different $\gamma$-matrices anticommute, and that (ii) the square of a single $\gamma$-matrix is plus or minus the unit matrix. Then the statement in the question follows from the following claim: the product of an odd number of $\gamma$-matrices equals, up to a numerical factor, a single $\gamma$-matrix or a product of three $\gamma$-matrices.
The proof is by reduction of $n$. Consider a product $\gamma^{\mu_1}\dotsb\gamma^{\mu_n}$ with odd $n\geq 5$. Since the indices on the $\gamma$-matrices take values from the range $0,\dotsc,3$, there must be a pair of indices, $\mu_i$ and $\mu_j$ with $i<j$, such that $\mu_i=\mu_j$ and all $\mu_k$ with $i<k<j$ are different from $\mu_i=\mu_j$. Using the above properties (i) and (ii), the matrices $\gamma^{\mu_i}$ and $\gamma^{\mu_j}$ can be removed from the product, possibly up to changing the overall sign. One then continues to reduce the number of $\gamma$-matrices in the product in the same manner until it becomes $n=3$. If all the remaining $\gamma$-matrices are different, we are done. If there is still one pair of matrices with equal indices, the whole product is equivalent to a single $\gamma$-matrix. Done.
