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


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.

  • $\begingroup$ 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.) $\endgroup$ – mr_e_man Sep 5 '18 at 6:27
  • $\begingroup$ Also, $d$ is equivalent to the cocurl, the tangential projection of the curl, not the curl itself. $\endgroup$ – mr_e_man Sep 5 '18 at 6:57
  • $\begingroup$ Indeed $\nabla$ requires a metric. But the outer derivative does not. $\endgroup$ – user48672 Sep 5 '18 at 14:18
  • $\begingroup$ In GA, the exterior derivative is defined in terms of $\nabla$. $\endgroup$ – mr_e_man Sep 5 '18 at 14:19
  • $\begingroup$ What are you expecting for an "equivalent expression"? Is this sufficient $$\mathcal L_x\omega = x\,\lrcorner\,\big(P_\parallel(\nabla\wedge\omega)\big) + P_\parallel\big(\nabla\wedge(x\,\lrcorner\,\omega)\big)$$? $\endgroup$ – mr_e_man Sep 5 '18 at 16:25

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