Does the geometric algebra of curved space have a matrix representation? Suppose the geometric algebra defined by
$$
\frac{1}{2}(e_\mu e_\nu +e_\nu e_\mu)=g_{\mu\nu}
$$
where $e_\mu,e_\nu$ are generators of the algebra, and where $g_{\mu\nu}$ are elements of the reals. I am struggling to find a matrix representation of this algebra. 
In the case of $Cl_{3,0}$ it is well-known that the matrix representation is given by the Pauli matrices, and the case of $Cl_{3,1}$, they are the Dirac matrices. However, these two Clifford algebras do not describe curved spaces. Indeed, the generators form an orthogonal basis and are given by this relation: $\frac{1}{2}(\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu)= \eta_{\mu\nu}
$
I am interested in the matrix representation of the geometric algebra of curved space. For simplicity let us assume 2D space. Then, the constraints are:
$$
e_x e_x = g_{xx}\\
e_ye_y = g_{yy}\\
e_xe_y+e_ye_x=g_{xy}=g_{yx}
$$
To closest I was able to get to finding the correct matrix representation gives me the freedom to set $g_{xx}$ and $g_{yy}$ to any value of the reals (but not the cross-term $g_{yx}$):
$$
e_x= \pmatrix{-\sqrt{g_{xx}} & 0 \\ 0 & \sqrt{g_{xx}}}\\
e_y= \pmatrix{0 & \sqrt{g_{yy}} \\ \sqrt{g_{yy}} & 0}
$$
Then, with these matrices I get
$$
e_x e_x=g_{xx}\\
e_ye_y=g_{yy}\\
e_xe_y+e_ye_x=0
$$
What matrices will give me the full set of relations including $e_xe_y+e_ye_x=g_{xy}=g_{yx}$?
 A: In curved space the coordinate basis will not be orthonormal, but an orthonormal basis still exists, and can be used to construct an arbitrary basis.
So...
Start with $\gamma_\mu$ as a matrix representation of an orthonormal basis in curved space. Then you can express any arbitrary basis $e_\mu$ as their linear combination. To get a useful basis this way, you should be able to write an orthonormal basis in terms of the coordinate basis, then invert the transformation. Then you should have $e_\mu \cdot e_\nu = g_{\mu\nu}$ by linearity of the dot.
Although, it's probably easier to use GA in curved space by forgetting the matrix representation altogether.
EDIT: As requested in comments, here's an example. This is in 2d using just $\sigma_x,\sigma_y$.
Let
$$
e_k 
= a_k \, \sigma_x +b_k \, \sigma_y 
= \left(\begin{array}{cc} 0 & a_k - i b_k \\ a_k + i b_k & 0 \end{array}\right) 
= \left(\begin{array}{cc} 0 & c_k^* \\ c_k & 0 \end{array}\right)
$$
Then
$$
(e_1)^2 = |c_1|^2 \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right),
\qquad
(e_2)^2 = |c_2|^2 \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right),
$$
$$
\qquad
\tfrac{1}{2} (e_1 e_2 + e_2 e_1) = \textrm{Re}(c_1^* c_2) \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right).
$$
Thus
$$
\tfrac{1}{2} (e_i e_j + e_i e_j) = g_{ij} \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)
$$
with
$$
g_{ij} = \left(\begin{array}{cc} |c_1|^2 & \textrm{Re}(c_1^* c_2) \\ \textrm{Re}(c_1^* c_2) & |c_2|^2 \end{array}\right).
$$
An arbitrary metric can be explicitly realized by choosing $c_k = \sqrt{g_{kk}} \; e^{i \phi_k}$ with $\Delta \phi = \phi_2 - \phi_1$ obeying
$$ \cos(\Delta \phi) = \frac{g_{12}}{\sqrt{g_{11}}  \sqrt{g_{22}}} $$
where the magnitude of the RHS is less than 1 by the Cauchy-Schwarz inequality. Note that the "arbitrary" metric under consideration still must be positive-definite if $\sigma_i$ are to provide a representation, so the C-S inequality must hold.
Note that when you extend this to 3d by including $\sigma_z$, the matrix representations involved will still be 2x2, but $g_{ij}$ will be 3x3. Good luck!
But again, using matrix representations is not in the spirit of geometric algebra --- for most purposes it's better to just use the formal rules and geometric interpretations.
