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

• In curved space, $e_\mu$ are just linear combinations of Clifford algebra $\gamma_a$ vectors as $e_\mu = e^a_\mu(x)\gamma_a$. See here physics.stackexchange.com/q/514592 Commented Nov 21, 2019 at 20:44

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.

• For instance, using the Pauli matrices as the orthonormal basis, I should be able to define three basis elements $e_x = a_x \sigma_x + a_y \sigma_y + a_z\sigma_z$, $e_y = b_x \sigma_x + b_y \sigma_y + b_z\sigma_z$ and $e_z = c_x \sigma_x + c_y \sigma_y + c_z\sigma_z$ and those three elements $\{e_x,e_y,e_z\}$ would have the desired properties? Commented Nov 21, 2019 at 19:03
• Yes, that should work Commented Nov 21, 2019 at 21:49
• I think your confusion comes from conflating the matrix $g_{\mu\nu}$ with the matrix of the representation. I'll edit the answer to show a quick example. Commented Nov 22, 2019 at 3:35
• @JoeSchindler, the degree of freedom of unconstrained parameters is the numbers of rotation generators, due to global rotation/Lorentz symmetry ($g$ and $e$ are not bijection). For 3D it's 3 (in the example), while for 4D it's 6. Commented Nov 22, 2019 at 21:00
• No you can't. By using the $\sigma_i$ you have represented a metric of Euclidean signature, which means that the metric must be positive definite, so all vectors have non-negative norm. Since the metric is a positive definite inner product it must obey the Cauchy-Schwarz inequality as I stated above. In the metric you are trying to give, some vectors have a negative norm (for example $|e_1-e_2|^2 = -397$). You must specify a positive definite matrix (i.e. all positive eigenvalues). Otherwise you can't use $\sigma_i$ as a representation. Commented Nov 23, 2019 at 3:48