Geometric differentiation and chain rule for a function of rotor variables

1

I’m reading the Chapter 8 of “Geometric Algebra for Computer Science”, and trying to find the derivative of a function like this f(B) = \parallel \! B \wedge \mathbf{v}_0 \! \parallel^2, with \parallel \! \mathbf{v}_0 \! \parallel^2 = 1.

Intuitively, I would like to know how rotating a plane represented by a bivector B would affect the trivector volume.

And I think I should further write the bivector B as a function of a rotor R (maybe two rotors), e.g. B(R) =RB_0R^{-1}, by following the saying:

“Elements of geometric algebra should only be perturbed by rotors” :
e^{-\delta P/2} X e^{\delta P/2} = X + X \times \delta P + O(\delta^2) \quad\quad (8.4)
(page 217 of “Geometric Algebra for Computer Science”).

  • I’m a little confused in how to apply the chain rule to a function that involves one or two rotor variables.
  • I get stuck and am not sure how to go further after doing the following for \partial_B f(B).

Here comes the part for f(B) and \partial_B f(B):

For any two multivectors X and Y, and a k-blade M, we have
X Y = X \wedge Y + X \rfloor Y + X \times Y \quad and \quad MM = M \rfloor M,
Then we have \parallel \! M \! \parallel^2 = M * \tilde M = M\rfloor \tilde M = (-1)^{\frac{1}{2}k(k-1)} MM = -M^2 (for k=2),
and f(B) = \parallel \! B \wedge \mathbf{v}_0 \! \parallel^2 = -(B \wedge \mathbf{v}_0 )^2 = - \frac{1}{4}(B\mathbf{v}_0 + \mathbf{v}_0 \hat B)^2 = - \frac{1}{4}(B\mathbf{v}_0 + \mathbf{v}_0 B)^2 .
Then for the derivative,
\partial_B f(B) \overset 0 = -\frac{1}{4} ( \overset \backprime \partial_B ( \overset \backprime B\mathbf{v}_0 + \mathbf{v}_0 \overset \backprime B)(B\mathbf{v}_0 + \mathbf{v}_0 B) + \overset \backprime \partial_B (B\mathbf{v}_0 + \mathbf{v}_0 B)( \overset \backprime B\mathbf{v}_0 + \mathbf{v}_0 \overset \backprime B) )
\quad\quad\quad\quad \!\! \overset 1 = -\frac{1}{4} (2\mathbf{v}_0 (B\mathbf{v}_0 + \mathbf{v}_0 B) + 2(B\mathbf{v}_0 + \mathbf{v}_0 B) \mathbf{v}_0)
\quad\quad\quad\quad \!\! \overset 2 = - (B + \mathbf{v}_0 B \mathbf{v}_0)

  • In the step “\overset 1 =”, I treated the part without ^\backprime , i.e., \mathbf{v}_0 and (B\mathbf{v}_0 + \mathbf{v}_0 B), as constants. Is there anything wrong?

I do not feel confident in:

  • when we can treat such parts as “constants” in taking derivative using product rule and “Table 8.1 / 8.2”, and
  • when we need to apply a projection \mathsf{P}[\ ] (though this can become apparent when one take derivative using definition).

Any hints / corrections? Thank you!

1 Like
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Without looking at the computation in detail, it looks like the units are immediately incorrect due to the lack of a differential-sized factor on the RHS of the final expression. When you say you are trying to “find the derivative,” are you referring to a direction derivative, the total derivative, or partial derivative with respect to a coordinate?

I think the chain rule applies as you would expect without needing to reexpress the \wedge in terms of the geometric product also.

3

Thank you!
Yes. There must be something wrong…
f(B) is scalar valued, and I’m trying to take the total derivative.

What do you mean by “differential-sized factor”? (“RHS” is short for “right hand side”, correct?)

4

I mean simply that if we consider \partial_B f(B) to be a “change of f,” I would expect to see a “change” (aka a \partial) on the right hand side (RHS) of the equation. In this case, because only B is varying, I would expect \partial B to show up in the solution as a factor of all terms (i.e. \partial B \wedge \mathbf{v}_0).

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\partial_B f(B) is the “rate of change of f” by treating \partial_B as a differentiation operator.

For notation, I’m following Prof. LeoDorst’s convection in the book GA for CS:

Let F(\mathbf{x}) be an element of geometric algebra dependent on a vector \mathbf{x}.

The directional differentiation is denoted by (\mathbf{a} * \partial_\mathbf{x}), and is defined as
(\mathbf{a} * \partial_\mathbf{x})F(\mathbf{x}) \equiv \lim_{\epsilon \to 0} \frac{F(\mathbf{x} + \epsilon \mathbf{a}) - F(\mathbf{x})}{\epsilon}. \quad (section 8.5)
The vector differentiation operation is defined as
\partial_\mathbf{x} \equiv \sum_{i = 1}^m \mathbf{e}^i (\mathbf{e}_i * \partial_{\mathbf{x}}) = \sum_i \mathbf{e}^i \frac{\partial}{\partial_{x^i}},
where (\mathbf{e}_i * \partial_\mathbf{x}) = \frac{\partial}{\partial{x^i}} = \partial_{x^i}.

Directional multivector differentiation and multivector derivative are defined similarly.

(A*\partial_X)F(X) \equiv \lim_{\epsilon \to 0} \frac{F(X+ \epsilon A) - F(X)}{\epsilon}; \quad (section 8.7.1)
\partial_X = \sum_{I=1}^{2^m} \mathbf{e}^I (e_I * \partial_X).

6

I’ve been reading [Hitzer's Multivector Differential Calculus](https://arxiv.org/abs/1306.2278).
I find his definition of A-derivative that involves projection quite reasonable.