# Divergence of a vector field in geometric calculus

I have been studying geometric algebra from a couple of books, and have just moved into geometric calculus in the book by Doran-Lasenby(DL).

DL shows that the vector derivative can be written as $$\nabla=e^k\partial_k$$ (where $$e^k$$ are basis vectors) and then moves onto differentiation of multivectors of different grades.

In the section about vector fields, DL takes a vector field $$J(x)$$ and mentions the following:

$$\nabla \cdot J(x)=(e^k\partial_k)\cdot J=\partial_k(e_k\cdot J)=\partial_kJ^k$$

As a definition for the divergence, this looks alright.

But then DL drops a line that I cannot make any sense of :

The divergence can also be defined in terms of the geometric product as $$\nabla\cdot J=\frac12(\nabla J+\dot{J}\dot{\nabla})$$

where he uses the overdot notation to quote "define the scope of the vector derivative".

The last term $$\dot{J}\dot{\nabla}$$ looks awfully like an operator to me, which it shouldn't be if I understand things correctly.

As an example, let us take $$J(x)=x$$. If we were to look at things symbolically, we would have:

$$\nabla x=(e^i\partial_i)(e^jx_j)=(e^ke_k)\partial_kx^k+(\text{mixed terms})=p-q+(0)=p-q$$ where $$p-q$$ is the signature, and is indeed the "divergence". But for the other term:

$$\dot x \dot\nabla=(e^jx_j)(e^i\partial_i)=(e^je_i)x^j\partial_i$$ which is an operator.

The only way I can see any way out is if the overdot notation allows me to interchange the partials with the coordinates, in which case the result matches with what one expects from a "divergence", because then the 2 terms are the same and the half cancels one of them in the definition. But I see no reason why one can do this.

I am guessing the culprit is myself not understanding the overdot notation, but I cannot be sure.

Any help is greatly appreciated!

It's not uncommon to allow the gradient to act bidirectionally in geometric algebra. For example if $$Q, R$$ are multivectors, then the bidirectional action of the gradient in a $$Q, R$$ sandwich is \begin{aligned} Q \nabla R &= \partial_k \left( { Q e^k R } \right) \\ &= \left( { \partial_k Q} \right) e^k R + Q e^k \left( { \partial_k R } \right).\end{aligned}
This chain rule expansion is often written using overdots to designate the scope of the partial operators, where the vector basis elements have to stay in place in whatever multivector expression contains the gradient. i.e.: $$Q \nabla R \equiv \dot{Q} \dot{\nabla} R + Q \dot{\nabla} \dot{R},$$ where we mean the coordinate expansion above.
Recall that the dot product of any two vectors can be written as a symmetric vector product $$A \cdot J = \frac{1}{{2}} \left( { A J + J A } \right).$$ This also works for the gradient, provided you allow the gradient to act bidirectionally. So, when you say $$\dot{J} \dot{\nabla}$$ looks like an operator, you have nailed it. The gradient is acting as an operator (as it always does), but this time to the left, using the overdots to specify the scope of it's action.
If you aren't comfortable just substituting $$A = \nabla$$ above, here's an explicit expansion of that symmetric sum that illustrates that this substitution has the desired effect: \begin{aligned} \frac{1}{{2}} \left( { \nabla J + \dot{J} \dot{\nabla} } \right) &= \frac{1}{{2}} \left( { e^k \partial_k J + \partial_k J e^k } \right) \\ &= \partial_k \frac{1}{{2}} \left( { e^k J + J e^k } \right) \\ &= \partial_k e^k \cdot J \\ &= \nabla \cdot J.\end{aligned}
Similarly, for the wedge product, we can use the gradient in an antisymmetric sum, just like any non-operator vector: \begin{aligned} \frac{1}{{2}} \left( { \nabla J - \dot{J} \dot{\nabla} } \right) &= \frac{1}{{2}} \left( { e^k \partial_k J - \partial_k J e^k } \right) \\ &= \partial_k \frac{1}{{2}} \left( { e^k J - J e^k } \right) \\ &= \partial_k e^k \wedge J \\ &= \nabla \wedge J.\end{aligned}
• Awesome! This is exactly what I suspected. Thanks for explaining this in such lucidity. I just have one more clarification to ask for. Thinking symbolically, I want to say that the derivative essentially works sth like this: Replace $\nabla$ (or $\dot\nabla$) with $e^k$ and then replace whatever the derivative operator is acting on (say $J$ or $\dot J$) and replace it with $\partial_k J$. Would I be correct in saying this? Commented Jun 22, 2022 at 20:40