I think I have a good grasp of geometric algebra in orthogonal basis:
$$ \gamma_\mu \cdot \gamma_\nu=\delta_{\mu\nu} $$
For instance, in $Cl_4(\mathbb{C})$ I can write a complex number as:
$$ a+b\gamma_0\gamma_1\gamma_2\gamma_3 $$
a bi-vector as
$$ \mathbf{F}=E_x\gamma_0\gamma_1+B_x\gamma_2\gamma_3=(E_x+IB_x)\gamma_0\gamma_1 $$
My problem is that I cannot understand how the transition to curvilinear coordinates is supposed to be done:
$$ \mathbf{e}_\mu \cdot \mathbf{e}_\nu=g_{\mu\nu} $$
Does a general curvilinear basis erase all the nice properties involving a change of basis using the unit pseudoscalar? For example, are these identities destroyed in curvilinear space?
$$ I\gamma_0\gamma_1\gamma_2=\gamma_3\\ I\gamma_0\gamma_1\gamma_3=\gamma_2\\ I\gamma_0=\gamma_1\gamma_2\gamma_3\\ I\gamma_0\gamma_1=\gamma_2\gamma_3 $$
I can see three different approaches some of which kills the pseudoscalar identities and some who don't.
Write $a+b\mathbf{e}_0\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3$. But then this is no longer a complex number. So are complex numbers banned in curvilinear coordinates?
Use the identity $I=\gamma_0\gamma_1\gamma_2\gamma_3$, then I can write $a+bI$ and this remains unchanged in curvilinear coordinates?
For a bivector $F=E_x\gamma_0\gamma_1+B_x\gamma_2\gamma_3$, does it become
$$ F=E_x\mathbf{e}_0\mathbf{e}_1+B_x\mathbf{e}_2\mathbf{e}_3 $$
which kills the identity, or is it:
$$ F=(E_x+IB_x)\mathbf{e}_0\mathbf{e}_1 $$
which preserves the identity but then restricts the extent of the curvatures that are allowed on $B_x$.
Yet another suggestion I have seen is to define $\gamma_0$ and $\gamma_1$ explicitly, say as
$$ \mathbf{e}_0=t_0 \gamma_0+x_0\gamma_1+y_0\gamma_2+z_0\gamma_3\\ \mathbf{e}_1=t_1 \gamma_0+x_1\gamma_1+y_1\gamma_2+z_1\gamma_3\\ \mathbf{e}_2=t_2 \gamma_0+x_2\gamma_1+y_2\gamma_2+z_2\gamma_3\\ \mathbf{e}_3=t_3 \gamma_0+x_3\gamma_1+y_3\gamma_2+z_3\gamma_3\\ \vdots $$
Which one solves for $\gamma_0,\gamma_1,gamma_2,\gamma_3$. One obtains linears functions such as:
$$ \gamma_0=f[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3]\\ \gamma_1=g[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3]\\ \gamma_2=h[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3]\\ \gamma_3=k[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3] $$
Then one replaces this vector
$$ F=(E_x+IB_x)\gamma_0\gamma_1 $$
With the expressions for $\gamma_0$ and $\gamma_1$ obtained and gets:
$$ F=(E_x+IB_x)(f[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3])\wedge g[t_0,t_1,t_2,t_3,x_0,x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3] $$
In this later case I am not sure what I am getting. Intuitively it feels I am cheating by describing a vector that remains orthogonal regardless of background curvature. Nonetheless, maybe that is the way to go?
Which of these options is used in physics for the Faraday tensor in curved spacetime?