Gradient Operator

I think some of the physics that is encoded in the algebra manifests itself mainly in the derivatives. For instance, the true meat of the complex algebra is the analytic functions. Can we come up with some type of space-time version of an analytic function? I've made a few attempts, and I've seen other attempts, but in general I'm just not satisfied. There is no parallel that is as clean and clear as the analytic functions of a complex variable. The troubles with derivatives in the algebraic representation of space-time all boil down to non-commutativity. I'll keep searching.

In the mean time, let's discuss a derivative operator, which I believe is a valid operator, though I wish it would arise in a more clean fashion. Since I am not satisfied with any particular derivation of this operator, I will just introduce it without derivation: the algebraic gradient

∂ = ∂^μe_μ

Here the notation ∂^μ makes reference to contra-variant (raised index) partial derivative with respect to the μ^th coordinate. Remember that a standard partial derivative is defined with the index lowered.

∂_μ = ∂/∂x^μ

If we use the minkowski metric to raise the index, we must change the sign on the spatial terms of the gradient.

∂ = (1/c) ∂/∂t e₀ - ∂/∂x e₁ - ∂/∂y e₂ - ∂/∂z e₂

Now, in tensor algebras, we use index balancing in order to construct relativistically significant quantities. In this algebra we don't deal with the component indexes much, but we still need to be concerned with making relativistically significant quantities.

We can use the fact that the gradient transforms like a para-vector. Thus, if we want the gradient to act on a paravector, it needs to do so like this

∂A

Which results in a biparavector.

If the gradient acts on a biparavector it must do so like this

B∂ or B∂

Which results in a paravector. Note that the gradient acting from the right is a distinct operation due to non-commutativity.

And if the gradient acts on a spinor, it must do so like this

∂S^†

which results in another spinor. I only know that this expression is relativistically significant, but since I haven't yet seen this operation in action, I have no physical intution as to its significance.