Trace of product of six Pauli matrices Using the standard definition of the Pauli matrices with the zeroth included, i.e.
$$ \sigma^{\mu} = (I, \sigma^i) $$
$$  \bar \sigma^{\mu} = (I, -\sigma^i) $$
it's a standard result that
$$ Tr[\sigma^{\mu} \bar \sigma^{\nu}] = 2\eta^{\mu \nu} $$
which is easy enough to verify. Several sources e.g. https://arxiv.org/abs/1702.08246 also give an analagous result for the trace of the product of four,
$$ Tr[\sigma^{\mu_1} \bar \sigma^{\mu_2} \sigma^{\mu_3} \bar \sigma^{\mu_4}] = 2\eta^{\mu_1 \mu_2}\eta^{\mu_3 \mu_4} + 2\eta^{\mu_1 \mu_4}\eta^{\mu_2 \mu_3} - 2\eta^{\mu_1 \mu_3}\eta^{\mu_2 \mu_4} - 2i \varepsilon^{\mu_1\mu_2\mu_3\mu_4}. $$
None of the standard references go as far as six, but I'd like to know whether there is a similarly compact result for the product of six sigma matrices, i.e.
$$Tr[\sigma^{\mu_1} \bar \sigma^{\mu_2} \sigma^{\mu_3} \bar \sigma^{\mu_4} \sigma^{\mu_5} \bar \sigma^{\mu_6}]?$$
 A: Generalize the result for the product of two Pauli matrices to include the identity matrix. You can verify the factors that I just jotted down:
$$\sigma^\mu\sigma^\nu=(\delta^{\mu\nu}-2\delta^{\mu0}\delta^{\nu0})I+(1-\delta^{\mu0})(1-\delta^{\nu0})\epsilon^{\mu\nu}_{\eta}\sigma^\eta+\delta^{\mu0}\sigma^\nu+\delta^{\nu0}\sigma^\mu.$$ Now iterate this process for the product of more matrices, find which terms are proportional to $I$, and those will be the only ones contributing to the trace (with an extra factor of 2 from $\mathrm{Tr}(I)$. You will need to use concatenation properties of Levi-Civita tensors and the like.
There might be some nice induction you can do, but there will regardless be an increasing number of terms, which is probably why most people don't write these expressions down in general.
If you want to use this for the components of $\bar{\sigma}$ you'll need to add minus signs as appropriate. One can do this by writing, without any summation convention, $\bar{\sigma}^\mu=(\delta^{\mu0}-1)\sigma^\mu+\delta^{\mu0}\sigma^\mu$. One can also do this by not caring about upper and lower indices, using Einstein sum conventions, and writing $\bar{\sigma}^\mu=\eta^{\mu\nu}\sigma^\mu$.
