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Maple's documentation says that the Lie derivative of a connection $\nabla_Y(Z)$ is the type (1, 2) tensor field
$S = {\mathcal L}_X \nabla$ that is defined (when viewed as mapping from pairs of vector fields to vector fields) by $$ ({\mathcal L}_X \nabla)(Y,Z) {=} {\mathcal L}_X(\nabla_YZ ) - \nabla_{{\mathcal L}_X Y}(Z) - \nabla_X ({\mathcal L}_Y Z). $$

Based on the answer to Details to Lie derivative of Christoffel symbols I would have expected $$ ({\mathcal L}_X \nabla)(Y,Z) = {\mathcal L}_X(\nabla_YZ) - \nabla_{{\mathcal L}_X Y}(Z) - \nabla_Y ({\mathcal L}_X Z). $$ Is the Maple documentation wrong or is there some identity that says the last terms of each expression are are equal?

mike stone
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1 Answers1

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Short answer: the posted link is right, Maple is wrong.

Long answer: there is a way to determine which of the two proposed formulas cannot be the right answer. If $\mathcal{L}_X\nabla$ is a tensor, then it must be $\mathcal{C}^{\infty}$-linear in its entries, and in particular in $Z$. So, let's take $X$, $Y$ and $Z$ three vector fields, and $f$ a smooth function. We compute: \begin{align} \mathcal{L}_X(\nabla_Y(fZ)) &= \mathcal{L}_X((Yf)Z + f\nabla_YZ) \\ &= (X(Yf))Z + (Yf)\mathcal{L}_XZ + (Xf)\nabla_YZ + f\mathcal{L}_X(\nabla_YZ), \label{eq:1} \tag{1} \\ \\ \nabla_{\mathcal{L}_XY}(fZ) &= ((\mathcal{L}_XY)f)Z + f \nabla_{\mathcal{L}_XY}Z, \label{eq:2} \tag{2} \\ \\ \nabla_X(\mathcal{L}_Y(fZ)) &= \nabla_X((Yf)Z + f\mathcal{L}_YZ) \\ &= X(Yf)Z + (Yf)\nabla_XZ + (Xf)\mathcal{L}_YZ + f\nabla_X(\mathcal{L}_YZ), \label{eq:3} \tag{3} \\ \\ \nabla_Y(\mathcal{L}_X(fZ)) &= \nabla_Y((Xf)Z +f \mathcal{L}_XZ) \\ &= (Y(Xf))Z + (Xf)\nabla_YZ + (Yf)\mathcal{L}_XZ + f\nabla_Y(\mathcal{L}_XZ). \label{eq:3'} \tag{3'} \end{align} Since $(\mathcal{L}_XY)f = X(Yf) - Y(Xf)$, then $$ \eqref{eq:1}-\eqref{eq:2}-\eqref{eq:3'} = f\mathcal{L}_X(\nabla_YZ) -f\nabla_{\mathcal{L}_XY}Z - f\nabla_Y(\mathcal{L}_XZ), $$ while \begin{align} \eqref{eq:1} - \eqref{eq:2} - \eqref{eq:3} &= -((\mathcal{L}_XY)f)Z +(Yf)(\mathcal{L}_XZ - \nabla_XZ) + (Xf)(\nabla_YZ-\mathcal{L}_YZ)\\ &\quad + f \mathcal{L}_X(\nabla_YZ) -f\nabla_{\mathcal{L}_XY}Z -f\nabla_X(\mathcal{L}_YZ). \end{align} It is clear that $\eqref{eq:1} - \eqref{eq:2} - \eqref{eq:3'}$ is $\mathcal{C}^{\infty}$-linear while $\eqref{eq:1} - \eqref{eq:2} - \eqref{eq:3}$ is not.

Didier
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  • @Didier Thanks! I wonder if Maple actually implemented wrongly? I don't have access to Maple, but was just looking for info about ${\mathcal L}_X\nabla$. Curiously $({\mathcal L_X}\nabla)(Y,Z)$ seems to be symmetric in $Y$, $Z$ even if $\nabla$ is not torsion free.... – mike stone Nov 16 '23 at 20:19
  • @Didier. That was a typo. I meant $Y$ and $Z$ -- as if $\delta \Gamma^\alpha_{\mu\nu}$ is symmetric in $\mu$ and $\nu$ even when $\Gamma^\alpha_{\mu\nu}$ is not. – mike stone Nov 16 '23 at 20:22
  • @mikestone Hum, not sure about it. Let me think about it, I'll came back with my insight – Didier Nov 16 '23 at 20:23
  • @Didier Thanks! I find $[ ({\mathcal L}X \nabla)(Y, Z)]^\alpha=Z^\mu Y^\nu \Big(X^\rho \partial\rho \Gamma^{\alpha}{\mu\nu} - (\partial\sigma X^\alpha) \Gamma^\sigma_{\mu\nu} + (\partial_\nu X^\sigma) \Gamma^\alpha_{\mu\sigma} + (\partial_\mu X^\sigma) \Gamma^\alpha_{\sigma\nu}+\partial^2_{\mu\nu}X^\alpha\Big)$ which is symmetric in $\mu\nu$ – mike stone Nov 16 '23 at 20:25
  • @Didier. Ah only if the $X^\rho \partial \rho \Gamma^\alpha_{\mu\nu}$ is. So not a mystery. Sorry! – mike stone Nov 16 '23 at 20:28