# Differentiation in Geometric Calculus, Computing Vector Derivatives of Multivector-Valued Functions

I haven't found an explicit formula and way to compute vector derivatives in geometric calculus. For instance, let $$V \simeq \mathbb{R}^3$$ with the usual orthonormal basis $$\{\textbf{e}_i\}_{i=1}^3$$ and $$C \ell(V)$$ its universal Clifford algebra. Consider the multivector-valued function of a vector, that is $$F: P_1(C\ell(V)) \to C \ell (V)$$ (where $$P_1$$ is the projection operator), defined as $$F(x) = x(\textbf{e}_1 - \textbf{e}_2) + \textbf{e}_1\textbf{e}_2 \textbf{e}_3$$ where $$x \in P_1(C \ell (V))$$. Consider that $$x = \textbf{e}_1$$, then $$F(\textbf{e}_1) = \textbf{e}_1(\textbf{e}_1 - \textbf{e}_2) + \textbf{e}_1\textbf{e}_2 \textbf{e}_3 = {\textbf{e}_1}^2 - \textbf{e}_1 \textbf{e}_2 + \textbf{e}_1\textbf{e}_2\textbf{e}_3$$ $$F(\textbf{e}_1) = 1 - \textbf{e}_1 \textbf{e}_2 + \textbf{e}_1\textbf{e}_2\textbf{e}_3$$

What would it mean to take the vector derivative $$\partial_x$$ of the function $$F$$? My line of reasoning is $$\partial_x F(x) = \partial_x (x \textbf{e}_1) - \partial_x (x\textbf{e}_2) + \partial_x (\textbf{e}_1\textbf{e}_2\textbf{e}_3)$$ and, using $$x=\textbf{e}_1$$ for instance, we would have $$\partial_{\textbf{e}_1} F = \partial_{\textbf{e}_1}({\textbf{e}_1}^2) - \partial_{\textbf{e}_1}(\textbf{e}_1)\textbf{e}_2 + \partial_{\textbf{e}_1}(\textbf{e}_1)\textbf{e}_2\textbf{e}_3$$ where $$\partial_{\textbf{e}_1}({\textbf{e}_1}^2) = 2\textbf{e}_1$$, but $${\textbf{e}_1}^2 = 1$$ and $$\partial_{\textbf{e}_1}(1) = 0$$, this reasoning leads to an ambiguity. In the end $$\partial_{\textbf{e}_1} F = 2\textbf{e}_1 -\textbf{e}_2 + \textbf{e}_2\textbf{e}_3$$ or $$\partial_{\textbf{e}_1} F = 0 -\textbf{e}_2 + \textbf{e}_2\textbf{e}_3$$

This most likely isn't correct, i'm having a hard time undestanding how to compute those derivatives in the Clifford algebra. If the question is answered, I would also like to understand how to compute an $$n$$-vector derivative and even a multivector derivative.

In Alan Mcdonald's book, Vector and Geometric Calculus, he treats $$\mathbb{R}^m$$ as a vector space and simply defines the vector derivative as $$\partial_{h} F = h^i \frac{\partial F}{\partial x^i}$$ where $$h = h^i\textbf{e}_i$$ and $$x^i$$ are coordinates on $$\mathbb{R}^m$$. But this makes any function $$F$$ be implicitly defined on $$\mathbb{R}^m$$ and not general subspaces of $$C \ell(\mathbb{R}^m)$$.

In David Hestenes and Garret Sobczyk's book, Clifford Algebra to Geometric Calculus A Unified Language for Mathematics and Physics, they define the vector derivative using the directional directional derivative as $$a \cdot \partial_X F(x) = \left.\frac{\partial}{\partial \tau} F(x+a\tau ) \right\vert_{\tau =0} = \lim_{\tau \to 0} \frac{F(x+a\tau) - F(x)}{\tau}$$ and, duo the generality desired, they never go on to give $$\partial_x$$ an explicit formula, since this would require a choice of basis. They do derive extensively its properties and its "algebra", and derive that $$\partial_x F = \partial_x \cdot F + \partial_x \wedge F$$

In the Wikipedia article on geometric calculus (https://en.wikipedia.org/wiki/Geometric_calculus), the derivative $$\partial_{\textbf{e}_i} = \partial_i$$ is simply stated as the derivative in the direction of $$\textbf{e}_i$$, does this imply to calculate $$\partial/\partial x^i$$ just like Alan does in his book?

If this is indeed the case, that is, the association of points in $$\mathbb{R}^n$$ to vectors in $$P_1(C\ell(V))$$ is "essential" to compute those derivatives, how would this theory come when considering manifolds as the base space, since this impossibilitates the use of points as vectors.

So, a recap. I haven't been able to understand how to compute vector derivatives of multivector-valued functions on $$P_1(C\ell(V))$$. From all I could see, this operation depends on the base space $$\mathbb{R}^n \simeq V$$ to allow for those calculations, but this seems to restrict those functions to just $$\mathbb{R}^n$$ and not really to vectors, $$p$$-vectors and multivectors.

As you noted, there is some notational inconsistency between different authors on this subject. You mentioned [1] who writes the directional derivative as $$\partial_\mathbf{h} F(\mathbf{x}) = \lim_{t\rightarrow 0} \frac{F(\mathbf{x} + t \mathbf{h}) - F(\mathbf{x})}{t},$$ where he makes the identification $$\partial_\mathbf{h} F(\mathbf{x}) = \left( { \mathbf{h} \cdot \boldsymbol{\nabla} } \right) F(\mathbf{x})$$. Similarly [2] writes $$A * \partial_X F(X) = {\left.{{\frac{dF(X + t A)}{dt}}}\right\vert}_{{t = 0}},$$ where $$A * B = \left\langle{{ A B }}\right\rangle$$ is a scalar grade operator. In the first case, the domain of the function $$F$$ was vectors, whereas the second construction is an explicit multivector formulation. Should the domain of $$F$$ be restricted to vectors, we may make the identification $$\partial_X = \boldsymbol{\nabla} = \sum e^i \partial_i$$, however we are interested in the form of the derivative operator for multivectors. To see how that works, let's expand out the direction derivative in coordinates.

The first step is a coordinate expansion of our multivector $$X$$. We may write $$X = \sum_{i < \cdots < j} \left( { X * \left( { e_i \wedge \cdots \wedge e_j } \right) } \right) \left( { e_i \wedge \cdots \wedge e_j } \right)^{-1},$$ or $$X = \sum_{i < \cdots < j} \left( { X * \left( { e^i \wedge \cdots \wedge e^j } \right) } \right) \left( { e^i \wedge \cdots \wedge e^j } \right)^{-1}.$$ In either case, the basis $$\left\{ { e_1, \cdots, e_m } \right\}$$, need not be orthonormal, not Euclidean. In the latter case, we've written the components of the multivector in terms of the reciprocal frame satisfying $$e^i \cdot e_j = {\delta^i}_j$$, where $$e^i \in \text{span} \left\{ { e_1, \cdots, e_m } \right\}$$. Both of these expansions are effectively coordinate expansions. We may make that more explicit, by writing \begin{aligned} X^{i \cdots j} &= X * \left( { e^j \wedge \cdots \wedge e^i } \right) \\ X_{i \cdots j} &= X * \left( { e_j \wedge \cdots \wedge e_i } \right),\end{aligned} so $$X = \sum_{i < \cdots < j} X^{i \cdots j} \left( { e_i \wedge \cdots \wedge e_j } \right) = \sum_{i < \cdots < j} X_{i \cdots j} \left( { e^i \wedge \cdots \wedge e^j } \right).$$

To make things more concrete, assume that the domain of $$F$$ is a two dimensional geometric algebra, where we may represent a multivector with coordinates $$X = x^0 + x^1 e_1 + x^2 e_2 + x^{12} e_{12},$$ where $$e_{12} = e_1 \wedge e_2$$ is a convenient shorthand. We can now expand the directional derivative in coordinates \begin{aligned} {\left.{{\frac{dF(X + t A)}{dt}}}\right\vert}_{{t = 0}} &= {\left.{{ \frac{\partial {F}}{\partial {(x^0 + t a^0)}} \frac{\partial {(x^0 + t a^0)}}{\partial {t}} }}\right\vert}_{{t = 0}} + {\left.{{ \frac{\partial {F}}{\partial {(x^1 + t a^1)}} \frac{\partial {(x^1 + t a^1)}}{\partial {t}} }}\right\vert}_{{t = 0}} \\ &\quad + {\left.{{ \frac{\partial {F}}{\partial {(x^2 + t a^2)}} \frac{\partial {(x^2 + t a^2)}}{\partial {t}} }}\right\vert}_{{t = 0}} + {\left.{{ \frac{\partial {F}}{\partial {(x^{12} + t a^{12})}} \frac{\partial {(x^{12} + t a^{12})}}{\partial {t}} }}\right\vert}_{{t = 0}} \\ &= a^0 \frac{\partial {F}}{\partial {x^0}} + a^1 \frac{\partial {F}}{\partial {x^1}} + a^2 \frac{\partial {F}}{\partial {x^2}} + a^{12} \frac{\partial {F}}{\partial {x^{12}}}.\end{aligned} We may express the $$A$$ dependence above without coordinates by introducing a number of factors of unity \begin{aligned} {\left.{{\frac{dF(X + t A)}{dt}}}\right\vert}_{{t = 0}} &= \left( {a^0 1} \right) 1 \frac{\partial {F}}{\partial {x^0}} + \left( { a^1 e_1 } \right) e^1 \frac{\partial {F}}{\partial {x^1}} + \left( { a^2 e_2 } \right) e^2 \frac{\partial {F}}{\partial {x^2}} + \left( { a^{12} e_{12} } \right) e^{21} \frac{\partial {F}}{\partial {x^{12}}} \\ &= \left( { \left( {a^0 1} \right) 1 \frac{\partial {}}{\partial {x^0}} + \left( { a^1 e_1 } \right) e^1 \frac{\partial {}}{\partial {x^1}} + \left( { a^2 e_2 } \right) e^2 \frac{\partial {}}{\partial {x^2}} + \left( { a^{12} e_{12} } \right) e^{21} \frac{\partial {}}{\partial {x^{12}}} } \right) F \\ &= A * \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) F.\end{aligned} Now we see the form of the multivector derivative, which is $$\partial_X = \frac{\partial {}}{\partial {x^0}} + e^1 \frac{\partial {}}{\partial {x^1}} + e^2 \frac{\partial {}}{\partial {x^2}} + e^{21} \frac{\partial {}}{\partial {x^{12}}},$$ or more generally $$\partial_X = \sum_{i < \cdots < j} e^{j \cdots i} \frac{\partial {}}{\partial {x^{i \cdots j}}}.$$

Let's apply this to your function \begin{aligned} F(X) &= X \left( { e_1 - e_2 } \right) + e_1 e_2 e_3 \\ &= \left( { x^0 + x^1 e_1 + x^2 e_2 + x^3 e_3 + x^{12} e_{12} + x^{23} e_{23} + x^{13} e_{13} + x^{123} e_{123} } \right) \left( { e_1 - e_2 } \right) + e_1 e_2 e_3.\end{aligned} Our multivector gradient is \begin{aligned} \partial_X F(X) &= \left( { 1 + e^1 e_1 + e^2 e_2 + e^3 e_3 + e^{21} e_{12} + e^{32} e_{23} + e^{31} e_{13} + e^{321} e_{123} } \right) \left( { e_1 - e_2 } \right) \\ &= 2^3 \left( { e_1 - e_2 } \right).\end{aligned} We have had to resort to coordinates to compute the multivector gradient, but in the end, we do end up (at least in this case) with a coordinate free result.

# References

[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

• Thank you so much, I see now that we take the variations with respect to the coefficients of the vector and multivector, not with respect to the basis vectors. Apr 9, 2022 at 21:25

I'm leaving this as an 'answer' as I don't have enough rep. to comment but, I found the following paper useful, which brings together a lot of the definitions and propositions of multivector-calculus in one place, along with detailed proofs.

Eckhard Hitzer - Multivector Differential Calculus

The coordinate expansion which Peeter used in his answer is missing from this paper (which is a shame as that approach contains the intuition of why the operator part of the derivative is a scalar) however the paper does provide more stepping stones for things which feel like they're just stated in other texts.