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I've had to rewrite the question as i made agrievious errors in the first go-round.

Given a (pseudo)Riemannian metric g, we can identify its components with the symmetric product of gamma matrices:

$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}I$$

The gammas act as basis vectors and their products and linear combinations form a clifford algebra. Under the geometric algebra our basis vectors (matrices now) transform in a two-sided way:

$$\bar{\gamma}^{\mu}=\psi^{\dagger}\gamma^{\mu}\psi$$

for a general non-unitary transformation (i,e not just a change of coordinate basis) $\psi^{\dagger}\neq\psi^{-1}$ how does the metric g and its determinant Det(g) transform??

I had initially thought $g\longrightarrow\psi^{\dagger}g\psi$ however that is silly since we have the non-unitary generalization.

$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\psi\psi^{\dagger}\gamma^{\nu}+\gamma^{\nu}\psi\psi^{\dagger}\gamma^{\mu}\right)\psi$$

For my problem set $\psi\psi^{\dagger}$ should be a scalar (in case you havent guessed the $\psi$ are spinors). this should leave me with:

$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}\right)\psi\left(\psi\psi^{\dagger}\right)$$

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    $\begingroup$ If $\psi$ is nonsquare then how can $g$ and $\tilde{g}$ have the same size? $\endgroup$ – Qiaochu Yuan Jun 23 '18 at 6:44
  • $\begingroup$ @QiaochuYuan I was very sloppy and was trying to avoid too much depth will edit shortly $\endgroup$ – R. Rankin Jun 23 '18 at 7:02

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