Problem with a pushforward of vector field formula (Michael Taylor, "Partial Differential Equations")

Question

Let $X$ denote a vector field and let $\mathcal F^t_X$ denote its flow. If $X$ and $Y$ are two vector fields we denote by $\mathcal F^t_{X\#}Y$ the vector field satisfying $$ \mathcal F^s_{\mathcal F^t_{X\#}Y} = \mathcal F^{-t}_X \circ \mathcal F^s_Y \circ \mathcal F^t_X, $$ for small $|t|$, $|s|$. Then $\mathcal F^{t+s}_X = \mathcal F^t_X \mathcal F^s_X$ and hence $\mathcal F^{t+s}_{X\#} = \mathcal F^t_{X\#} \mathcal F^s_{X\#} = \mathcal F^s_{X\#}\mathcal F^t_{X\#}$. Taking into account that $\mathcal L_X Y = [X,Y]$ and differentiating the latter inequality we obtain: $$ \frac{d}{dt} \mathcal F^t_{X\#} Y = [X,\mathcal F^t_{X\#}Y], \\ \frac{d}{dt} \mathcal F^t_{X\#} Y = (D\mathcal F^t_{X\#})(Y) \cdot [X,Y]. $$ But at page 41 of M. Taylor's "PDEs I" formula (8.6) states that $$ \frac{d}{dt} \mathcal F^t_{X\#} Y = \mathcal F^t_{X\#} [X,Y]. $$ Please, help me, where is the problem?

Answer 1

Is it possible that on page 41 of Taylor, the function $\cal F$ is known to be linear? In that case, $D\cal F(Y)$ would be exactly $\cal F$, and would resolve the difference between the last two equations you wrote. If, on the other hand, $\cal F$ is not known to be linear, then it does indeed look as if there's a real problem. (I've omitted the various sub- and superscript decorations, because they're not really at the heart of things here, I believe...)

Answer 2

The key is to note that $\mathcal F^t_{X\#}$ commutes with $L_X$.