# Lie derivative in geometric algebra/Clifford algebra

What is the form of the Lie derivative in Clifford algebra?

Context:

Consider the Clifford algebra $\mathcal{C}l (p,q)$ with basis $\{e_i \}$. The geometric derivative following Hestenes is defined as

$$\nabla_x := e^j \frac{\partial}{\partial x^j}$$ (summation) where $\{e^i \}$ is the reciprocal basis and $x_i = x \cdot e^i$ for the radius-vector $x= x^i e_i$.

This paper http://math.columbia.edu/~dlitt/exposnotes/poincare_lemma.pdf defines the Lie derivative of a form $\omega$ as

$$\mathcal{L}_x \omega = \iota_x \circ d \omega + d ( \iota_x \omega )$$

Since in GA $$\nabla_x F = \nabla_x \wedge F + \nabla_x \cdot F$$ and there is a correspondence $\nabla_x \wedge \sim d$

I would like to know what is the equivalent expression of the Lie derivative.

• I'm interested in this as well, but I'm not sure that the Lie derivative is useful in GA. One of its most important properties is that it doesn't depend on a metric, or even a connection. I think $\nabla_x$ requires a connection. Still, you can define $\mathcal L$ by that same formula (Cartan's), using the left-contraction $$\iota_x\omega = x\;\lrcorner\;\omega$$ (This is the same as $x\cdot\omega$ except when $\omega$ is a scalar.) – mr_e_man Sep 5 '18 at 6:27
• Also, $d$ is equivalent to the cocurl, the tangential projection of the curl, not the curl itself. – mr_e_man Sep 5 '18 at 6:57
• Indeed $\nabla$ requires a metric. But the outer derivative does not. – user48672 Sep 5 '18 at 14:18
• In GA, the exterior derivative is defined in terms of $\nabla$. – mr_e_man Sep 5 '18 at 14:19
• What are you expecting for an "equivalent expression"? Is this sufficient $$\mathcal L_x\omega = x\,\lrcorner\,\big(P_\parallel(\nabla\wedge\omega)\big) + P_\parallel\big(\nabla\wedge(x\,\lrcorner\,\omega)\big)$$? – mr_e_man Sep 5 '18 at 16:25