Suppose we work with Minkowski flat space $M$ (just to make things easy). If $\textbf X$ is a Killing vector field it is possible to define the Lie derivative of an spinor $\alpha^A$ with respect to $\textbf X$, $\mathcal{L}_\textbf{X}\alpha^A$. This means we would need to drag the spinor along a curve, where $\textbf X$ is tangent to, in every point of it.

Question: But why does there not exists (at least I don't know if it does) a Lie derivative of an spinor field respect to an espinor, $\mathcal{L}_{\beta^B}\alpha^A$?.

I mean, it seems reasonable to think that if you have some complex vector field $\textbf{X}$, such that It can be represented in spinor form as $X^{AB}=\gamma^A\beta^B-\beta^A\gamma^B$, then something like this could happen $\mathcal{L}_{\gamma^A\beta^B-\beta^A\gamma^B}=\mathcal{L}_{\gamma^A}\left(\mathcal{L}_{\beta^B}\right)-\mathcal{L}_{\beta^A}\left(\mathcal{L}_{\gamma^B}\right)$, and therefore some kind of Lie derivative $\mathcal{L}_{\gamma^A}$ can be defined.

I know that there is not definition that involves Lie derivatives respect to tensors in general, but this is because a tensor can't define a curve where it is mean to drag the "object" (tensor, vector, spinor, etc), but an spinor is different, it has in formation for us to construct a curve where we can drag the objects... so, therefore my question. Also, I'm aware that the geometric meaning probably would be lost, but... probably emerge any other.

  • $\begingroup$ At the very least you would have $\mathcal L_{\beta^B\bar\beta^{\bar B}} \alpha^A$ as a candidate definition, treating $\beta^B$ as its null four-vector. $\endgroup$
    – CR Drost
    Oct 26, 2015 at 20:07
  • $\begingroup$ @CR Drost. Yes, but why? I mean, everything you want to know from a vector field could be compacted in just one spinor field even the curve to which is tangent. Everyone only says that the definition was no meant to be for spinors, but that is just not a good answer. And in physics we do stuffs like this all the time. I never think that one could find some Lie derivative for spinors, but there was a way. Let me put this in another way, Is impossible that an spinor define a curve by its own (without the need of its conjugated)? even if this curve is complex. $\endgroup$
    – raul
    Oct 26, 2015 at 20:23
  • $\begingroup$ Just because a definition was meant to be for some objects doesn't mean it only works with those objects. $\endgroup$
    – raul
    Oct 26, 2015 at 20:32
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Oct 26, 2015 at 20:33
  • 1
    $\begingroup$ @raul I'm just not seeing how you'd extend the "canonical" $X^a \nabla_a(Y^b) - Y^a\nabla_a(X^b)$ for any torsion-free connection to spinors, since you don't have $\nabla_A$ in any real sense. In the most obvious case we have maybe $\epsilon^{\bar A\bar B}~\alpha^A \nabla_{A\bar A} \beta^B$ generating some (complex?) 4-vector $v^b$, but if that 4-vector does not turn out to be null then I don't know how it would represent just a single canonical spinor. $\endgroup$
    – CR Drost
    Oct 26, 2015 at 20:50


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