Does the geometric algebra of curved space have a matrix representation?

Question

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 $e_xe_y+e_ye_x=g_{xy}=g_{yx}$?

Answer 1

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.