Gradient in geometric calculus In geometric calculus I see that we can unify the three fundamental derivatives from vector calculus; the gradient, the curl, and the diverge; into one operator.  However, $\nabla$ is defined for any multivector field, not just scalar and vector fields.  What does the geometric calculus gradient of a multivector field really mean?  For instance, if $B$ is a bivector field on an open set $D\subseteq \Bbb R^n$ then what does $\nabla B$ tell us?  It should be the case that $\nabla B = \nabla \cdot B + \nabla \wedge B$, but then that just leads to the question: what do the divergence and curl of $k$-vector fields, where $k\gt 1$, mean?
As an example of the type of info I'm looking for, just consider the geometric meaning the usual vector calculus derivatives:


*

*The gradient of a scalar field is the direction of steepest ascent.

*The divergence of a vector field is flux per unit volume through an sufficiently small volume around a point

*The curl of a vector field is the circulation per unit area in a sufficiently small area of a plane around a point (alternatively: the circulation around the normal vector in $\Bbb R^3$)


With these intuitive concepts at hand I know roughly what I get from performing any of these operations.  But when it comes to geometric calculus, I know how to take the gradient of a bi(/$k$-)vector field but I don't know what information it should be telling me.
 A: I'm not sure that you'll find a clear cut geometrical interpretation, since the gradient applied to either a vector or bivector field
encodes multiple pieces of information, each in a different logical geometric space.  Consider Maxwell's equations in free space with electric currents and sources, which can be written in GA form as a pair of gradients
$$\begin{aligned}\boldsymbol{\nabla} \mathbf{E} &= \rho/\epsilon_0 - \frac{\partial {I \mathbf{B}}}{\partial {t}}  \\ \boldsymbol{\nabla} I \mathbf{B} &= -\left( {\mu_0 \mathbf{J} + \epsilon_0 \mu_0 \frac{\partial {\mathbf{E}}}{\partial {t}}} \right).\end{aligned}$$
The first equation applies the gradient to the electric field $\mathbf{E}$ (a vector), whereas the second applies the gradient to a bivector $I \mathbf{B}$.  The gradient of the electric field has both scalar and bivector components.  Selecting out those components provides two of Maxwell's equations
$$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= \rho/\epsilon_0 \\ \boldsymbol{\nabla} \wedge \mathbf{E} &= - \frac{\partial {I \mathbf{B}}}{\partial {t}} = I \boldsymbol{\nabla} \times \mathbf{E}.\end{aligned}$$
Similarly, the gradient of the bivector $I \mathbf{B}$ has both vector and pseudoscalar components, and selecting out those components provides the other two Maxwell equations
$$\begin{aligned}\boldsymbol{\nabla} \cdot (I \mathbf{B}) &= -\left( {\mu_0 \mathbf{J} + \epsilon_0 \mu_0 \frac{\partial {\mathbf{E}}}{\partial {t}}} \right) = - \boldsymbol{\nabla} \times \mathbf{B} \\ \boldsymbol{\nabla} \wedge (I \mathbf{B}) &= 0 = I (\boldsymbol{\nabla} \cdot \mathbf{B})\end{aligned}$$
For both the application of the gradient to a vector and to a bivector, the gradient equation encodes a lot of information, providing many constraints on the solutions of the equations.  It's not clear that there is any clear geometric meaning to either without considering the context.
If this question has an answer, perhaps you should start with asking what the geometric interpretation of a vector-bivector product is, and then if you figure that out, see how that interpretation is modified by making that vector the gradient.
A: I realize this is an extremely old question, but it looks like it has yet to be answered well.
First, let's simply consider why the del notation works so well in geometric algebra. Anytime del shows up in $R^3$, you can essentially replace it with the vector
$\begin{pmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \end{pmatrix}$
or, more generally we can say
$\nabla = \sum_ie_i\frac{\partial}{\partial x_i}$ or $\nabla_i = \frac{\partial}{\partial x_i}$. So, when we consider the geometric derivative of a bivector, we can expand it to see the components we get. In n dimensions, for some function $B(x_1, x_2, ...)$, this gives us (if you'll forgive the awkward jk notation):
$$\nabla B = \sum_{i,jk}\nabla_iB_{jk}e_ie_je_k$$
and in 3 dimensions, with $I$ being the pseudoscalar, this expands to:
$$\nabla B = (\nabla_3 B_{31}-\nabla_2B_{12})e_1\\
+(\nabla_1B_{12}-\nabla_3B_{23})e_2\\
+(\nabla_2B_{23}-\nabla_1B_{31})e_3\\
+(\nabla_1B_{23}+\nabla_2B_{31}+\nabla_3B_{12})I$$
I used $\nabla_iB$ instead of $\frac{\partial B}{\partial x_i}$ because it shows the geometric product more naturally, but you can swap it freely.
If you interpret a 3D bivector as a pseudovector, then the equation above shows that the dot product portion (which I've generalized to lower the grade of the multivector) is analogous to the curl of a normal vector. Similarly, the wedge product portion is similar to the divergence of a normal vector.
In general, the del operator represents how something is changing with respect to its location in the input space. So, if you're taking the del operation on a multivector field, the result will tell you how the multivector is changing at that point. For a bivector, this would be how the area and orientation of the plane changes (similar for a 3-vector+). I haven't verified this mathematically, but I suspect the vector portion of the bivector derivative can be thought of as the change in orientation while the pseudoscalar portion can be considered the change in area. This is consistent with curl as the change in orientation of a vector and divergence as the change in magnitude as you follow the flow in the input space. As an interesting side note, the value returned by the del operator on a vector field can be thought of as a quaternion representing both the rotational and non-rotational change in the vector flow.
