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.


2 Answers 2


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.


[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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Nikolawn
    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.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    May 9, 2022 at 5:50
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – arnav_de
    May 9, 2022 at 6:09
  • $\begingroup$ Thank you, the paper shows some interesting properties which I haven't seen in other papers or works. I appreciate the contribution! $\endgroup$
    – Nikolawn
    May 15, 2022 at 14:28
  • $\begingroup$ Oh good! Yeh there's some interesting insights aren't there. FYI since posting this I read David Hestenes' 1968 paper 'Multivector Calculus' (it's only 10 pages) where he defines derivatives in a much different way - as the limit of a boundary integral. He also introduces them like this in a 1993 paper titled 'Differential Forms in Geometric Calculus' in which he states that his 1968 paper is the work he's most proud of. I think in both papers he states that this is the only meaningful way to define the derivative. Well worth a read, as it's completely changed my perspective on this. $\endgroup$
    – Matt B.
    May 19, 2022 at 11:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .