How do I take the divergence of a multi-vector?

Question

Say I have a multi-vector of $\mathbb{G}(2,\mathbb{R})$:

$$ \mathbf{u}=a+xe_0+ye_1+be_0e_1 $$

How do I take the divergence? How do I even define it for multi-vector in general?

Answer 1

Given a 2D multivector $$X = x^0 + x^1 e_1 + x^2 e_2 + x^{12} e_{12},$$ where $ e_{12} = e_1 \wedge e_2 $, you can show that the multivector gradient has the form $$\partial_X = \frac{\partial {}}{\partial {x^0}} + e^1 \frac{\partial {}}{\partial {x^1}} + e^2 \frac{\partial {}}{\partial {x^2}} + e^{21} \frac{\partial {}}{\partial {x^{12}}}.$$

I think that the only sensible definition of the divergence of $ F = F^0 + F^1 e_1 + F^2 e_2 + F^{12} e_{12} $ would be the scalar product of the gradient with $ F $. That is $$\begin{aligned}\partial_X * F &= \left( {\frac{\partial {}}{\partial {x^0}} + e^1 \frac{\partial {}}{\partial {x^1}} + e^2 \frac{\partial {}}{\partial {x^2}} + e^{21} \frac{\partial {}}{\partial {x^{12}}}} \right) * \left( { F^0 + F^1 e_1 + F^2 e_2 + F^{12} e_{12} } \right) \\ &=\frac{\partial F^0}{\partial {x^0}} + \frac{\partial F^1}{\partial {x^1}} + \frac{\partial F^2}{\partial {x^2}} + \frac{\partial F^{12}}{\partial {x^{12}}}.\end{aligned}$$

Note that the above expression allows for non-Euclidean basis vectors, and utilizes the reciprocal basis vectors $ e^k \cdot e_j = {\delta^{k}}_j $. For the Euclidean case that can be simplified with $ e_1 = e^1, e_2 = e^2, e_1^2 = e_2^2 = 1 $.