Limits of overdot notation in geometric algebra

Question

In geometric calculus the over dot notation is used to denote the proper way to do the vector differentiation of a multivector product - $$ \nabla (AB) = (\nabla A)B + (\dot{\nabla}A)\dot{B} $$ The question is does make any sense to have more than one over dotted $\dot{\nabla}$ and one over dotted multivector in an expression. The reason for this question is that I am implementing the over dot symbolism in the galgebra symbolic geometric algebra python module.

Note that when it comes to multivectors I consider $\dot{(ABC)}$ to be a single multivector. The code would be (A*B*C).odot(). The program would first evaluate $ABC$ and then apply the over dot (actually just a flag on the result) to the product. Parenthesis are essential to the proper evaluation of expressiong with multivector differential operators and multivectors since in general $$ \nabla(ABC) \ne (\nabla A)BC \ne (\nabla AB)C. $$

Answer 1

Let’s first see what the overdot actually is:

If I had an expression like $\dot f \dot \partial$, that would mean I first replace $\dot \partial$ with $e^i$, then replace $f$ with $\partial_i f$ (With the summation convention)

That makes the overdot unnecessary, but fine. It also means that you shouldn’t have an expression like $\dot f \dot \partial \dot g$ as that would include two $\partial$ for the one $e^i$. If you want to have something like $e^j e^i \partial_i f \partial_j g$, then you’d use a second overdot-like symbol and get

$\dot \partial \ddot \partial \ddot f \dot g$

Answer 2

I would avoid having two instances of $\dot\nabla$ in a term and instead use distinct symbols; I usually use a check $\check\nabla$ if a dot is already in use. So $$ \dot\nabla\check A\dot B\check\nabla $$ is a perfectly sensible expression. We could sensibly interpret $$ \dot\nabla\dot AB\dot\nabla = \dot\nabla\check{\dot A}B\check\nabla = \dot\nabla\dot{\check A}B\check\nabla. $$ The order of the dots and checks above $A$ indicates in which order the derivatives are being applied, so as long as partial derivatives commute (which they pretty much always do in this context) the double dot notation is sensible.

It is entirely sensible to have a single $\dot\nabla$ and multiple other dots. The expression $$ \dot\nabla\dot AB\dot C $$ is differentiating $A$ and $C$ but not $B$. If what this means is unclear, we could instead write $$ \bigl[\nabla_yA(y)B(x)C(y)\bigr]_{y=x} $$ making the functional dependencies explicit; we differentiate with respect to $y$ and then evaluate at $x$. Of course, we can expand this as $$ \dot\nabla\dot AB\dot C = \dot\nabla\dot ABC + \dot\nabla AB\dot C. $$

And please, please do not use $\dot{(ABC)}$ to mean $\dot A\dot B\dot C$. You are just asking for trouble and confusion with that. There plenty of situations where we would want dots inside of parentheses anyway; for instance $$ \dot\nabla A\wedge(\dot B{{\rfloor}} C),\quad \dot\nabla(A + B\dot C) = \dot\nabla B\dot C. $$ Dots even need to come out of parentheses; for instance, if $L_x(v, w)$ is linear in $v$ but depends on $x$ then $$ L_{\dot x}(\dot\nabla, f(x)),\quad L_x(\dot\nabla, f(\dot x)) $$ are different expressions. Suppressing the $x$ dependence we would usually write these more like $$ \dot L(\dot\nabla, f),\quad L(\dot\nabla, \dot f) $$

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As a final note, I think that have $\nabla$ differentiate directly to its right by default is terrible, and is the root a much confusion even in standard vector calculus. Especially in the context of geometric algebra, I think that if anything $\nabla$ should differentiate its entire term, or otherwise its scope should always be specified with overdots or something else. For instance, I would prefer that $$ A\nabla B = \dot A\dot\nabla\dot B,\quad L(\nabla, f) = \dot L(\dot\nabla, \dot f),\quad (A\nabla B(C + D))\wedge X = (\dot A\dot\nabla\dot B(\dot C + \dot D))\wedge X, $$$$ \nabla AB = \dot\nabla\dot A\dot B,\quad (ABC\nabla)D = (\dot A\dot B\dot C\dot\nabla)D. $$ Note that in the $\wedge$ expression, $X$ is not differentiated since it is outside of the $\nabla$ terms, and similarly in the last expression $D$ is not differentiated. Of course, we could explicitly have $\nabla$ differentiate outside its term: $$ \dot A\wedge(\dot\nabla{{\rfloor}} C). $$