# Why the imaginary constant on the spin-connection?

I have been studying the spin-connection for the Dirac equation. The covariant derivative is defined as:

$$D_\mu \psi = (\partial_\mu - {i \over 4} {\omega_\mu}^{ab} \sigma_{ab}) \psi$$

where $$\sigma_{ab} = i[\gamma^a,\gamma^b]$$ and $${\omega_\mu}^{ab}$$ is the spin-connection.

When working out the consequence I find that this $$i$$ factor before the spin-connection leads to pseudo-vector terms which treat left and right differently. (Perhaps this is good as maybe left and right handed fermions move differently under gravity but this seems to go against the idea that all things moves the same under gravity)

So my very technical question is, is this $$i$$ really necessary. Or would the theory still be self consistent if we removed the $$i$$? What is the justification for this $$i$$? It would be a different theory but would it be self-consistent? Could we even add a factor of $$\gamma^5$$ before the spin-connection and keep it self-consistent? (Since this would be the same except having a factor of -1 for right-handed fermions?) Basically I'm thinking of any ways that this equation can be changed and remain consistent with general covariance.

The $$i$$ is necessary.
In order to derive the Dirac equations in curved spacetime, the approach is the following: We know how Lorentz vectors can be transported in a generic curved spacetime using covariant derivatives. We use the fact that bilinears of $$\psi$$(a spinor) should transform like a vector. This then gives rise to the above formula.