# Generators of a non-orthonormal Clifford algebra

In the case of orthonormal Clifford Algebra, matrix representations of the generators are easy to find. For example, in 3d-Euclidean space,

$$\frac{1}{2} \left( \sigma_i \sigma_j +\sigma_j \sigma_i \right) = \delta_{i j}$$

The generators are the Pauli matrices.

However, in the case of non-orthonormal algebras, what can and cannot serve as a representation is unclear to me. Suppose the following Clifford Algebra

$$\frac{1}{2} \left( e_\mu e_\nu+e_\nu e_\mu \right) = g_{\mu\nu}$$

Can we find matrix representations for $$e_\nu, e_{\mu}$$? If not, what can we use?

Edit: example

For example, one can think of the interval (metric tensor) of general relativity

$$e_0e_0=g_{00}\\ e_1e_1=g_{11}\\ e_2e_2=g_{22}\\ e_3e_3=g_{33}\\ e_0e_1+e_1e_0=2g_{01}\\ e_0e_2+e_2e_0=2g_{02}\\ e_0e_3+e_3e_0=2g_{03}\\ e_1e_2+e_2e_1=2g_{12}\\ e_1e_3+e_3e_1=2g_{13}\\ e_2e_3+e_3e_2=2g_{23}$$

Is there a valid matrix representation of the generators $$e_0,e_1,e_2,e_3$$?

• What is an “orthonormal algebra”? Commented Jul 21, 2019 at 14:59
• @rschwieb an algebra whose generators are: $\frac{1}{2} \left( \sigma_i \sigma_j +\sigma_j \sigma_i \right) = \delta_{i j}$ Commented Jul 21, 2019 at 17:51
• Who uses this term? I have trouble seeing its usefulness. It looks like a real Clifford algebra is “orthonormal” iff the signature of its metric is positive definite, and that every complex Clifford algebra for a non degenerate metric is “orthonormal”. Commented Jul 21, 2019 at 18:41
• @rschwieb I would think that if the quadratic form contains cross-terms, it would be non-orthonormal. Commented Jul 21, 2019 at 21:19
• if you give me an example, I could consider it with you. Commented Jul 21, 2019 at 22:20

If you let $$V$$ be the underlying vector space of a $$k$$ algebra, then each element $$a\in V$$ acts by left multiplication on $$V$$, creating a linear transformation $$\lambda_a$$ of $$V$$.
The mapping $$a\mapsto \lambda_a\in End_k(V)\cong M_{\dim(V)}(k)$$ turns out to be a homomorphism of $$k$$-algebras, and its image is an algebra of matrices isomorphic to the original algebra.
• @AlexandreH.Tremblay Why would one leave the bilinear form in terms of $g_{ij}$'s? Why wouldn't you choose a basis to diagonalize it? Commented Jul 23, 2019 at 10:50
• It is not mathematically hard, but a bit tedious to write out $16$ $16\times 16$ matrices for the example you requested, at least for the diagonalized case. Doing it with the $g$'s would be very messy. Commented Jul 23, 2019 at 11:00