# What is gradient operator in geometric algebra with some null basis vectors?

Consider a geometric algebra with the orthogonal basis $\{e_x,e_y\}$ where $e_x\cdot e_x = 1$ and $e_y\cdot e_y = 0$. Then define the vector function $f(r) = r$ where $r = xe_x+ye_y$. Then the directional derivative in the $e_y$ direction is

$$\lim_{h\to 0} \frac{f(r+he_y)-f(r)}{h} = e_y$$

is well defined. In this case (for this metric) can one define a meaningful gradient operator equivalent to $\nabla = e^x\partial_x+e^y\partial_y$ in the case where none of the orthogonal basis vectors are null and $e^x$ and $e^y$ are the reciprocal basis vectors?

• What's a null basis vector? – smcc Jul 26 '16 at 15:12
• @smcc A null vector is a nonzero vector that squares to $0$. In this case $e_y$ is a null vector. – user137731 Jul 26 '16 at 15:14
• @Bye_World Okay, I thought this was about something in $\mathbb{R}^2$. Hence my confusion... – smcc Jul 26 '16 at 15:21
• @user2761381 The problem I see here is how to define $e^i$. Usually I'd define them as $e^x = e_yI^{-1}$ and $e^y = -e_xI^{-1}$, but $I$ is not invertible in this space (and even the equation $e^yI = -e_x$ has no solution). I'm not very familiar with spaces equipped with degenerate bilinear forms, though, so I couldn't tell you if there were some standard definition of the reciprocal basis in this space. – user137731 Jul 26 '16 at 15:23

One of $e_1$ and $e_2$ being null does not imply that $I$ has no inverse. For example, consider a Dirac (relativistic) basis $\{\gamma_0, \gamma_1\}$, where

$$\gamma_0^2 = 1 = -\gamma_1^2,$$

and \begin{aligned}\gamma^0 &= \gamma_0 \\ \gamma^1 &= -\gamma_1.\end{aligned}

Suppose that \begin{aligned}e_x &= \gamma_0 \\ e_y &= \gamma_0 + \gamma_1.\end{aligned}

These vectors can be used as a basis for the space, and have the properties specified $e_x^2 = 1, e_y^2 = 0$. For this basis the pseudoscalar is

\begin{aligned}I &= e_x \wedge e_y \\ &= \gamma_0 \wedge \left( { \gamma_0 + \gamma_1 } \right) \\ &= \gamma_0 \wedge \gamma_1 \\ &= \gamma_0 \gamma_1.\end{aligned}

This is actually its own inverse

\begin{aligned}I^{-1} &= \gamma^1 \gamma^0 \\ &= -\gamma_1 \gamma_0 \\ &= \gamma_0 \gamma_1 \\ &= I.\end{aligned}

The reciprocal basis vectors can now be calculated \begin{aligned}e^x&= e_y I \\ &= \left( { \gamma_0 + \gamma_1 } \right) \gamma^1 \gamma^0 \\ &= \gamma^0 - \gamma^1,\end{aligned}

and \begin{aligned}e^y&= -e_x I \\ &= -\gamma_0 \gamma^1 \gamma^0 \\ &= \gamma^1.\end{aligned}

In this case the representation of the gradient corresponding to the vector parameterization $r = x e_x + y e_y = (x + y) \gamma_0 + y \gamma_1$ is

$$\boldsymbol{\nabla} = \left( { \gamma^0 - \gamma^1 } \right) \frac{\partial {}}{\partial {x}} + \gamma^1 \frac{\partial {}}{\partial {y}}.$$

Aside: An alternative way of looking at this problem is presented in the excellent little text "Vector and Geometric Calculus" by Alan Macdonald. See 5.4 Curvilinear Coordinates, and 5.5 The Vector Derivative. For a complete parameterization of the space, as is the case here, the vector derivative (defined for curvilinear coordinates on a surface) is identical to the gradient, and the text presents the differential operators that can be used to define both the curvilinear basis and its reciprocal basis. In this case neither basis varies with the parameterization, but the toolbox of curvilinear coordinates applies nicely.

• The OP specified that $e_x$ and $e_y$ should be orthogonal. – mr_e_man Sep 14 '20 at 16:50
• @mr_e_man -- I'd only noticed the requirement for one to be normalized and the other to be null. I'll try reworking this with a basis like: $e_x = \gamma_2, e_y = \gamma_0 + \gamma_1$ (orthogonal, but one null.) – Peeter Joot Sep 14 '20 at 20:43