# Product of an odd number of Dirac $\gamma$ matrices [closed]

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.

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.
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 $$i, such that $$\mu_i=\mu_j$$ and all $$\mu_k$$ with $$i 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.