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Consider the Lie derivative of the vector field $\bf{Y}$ with respect to the vector field $\bf{X}$ on manifold $M^{n}(x)$ defined as

$$\displaystyle{[\mathcal{L}_{\bf{X}}Y]_{x}:=\lim_{t\rightarrow 0} \frac{[{\bf{Y}}_{\phi_{t}x}-\phi_{t*}{\bf{Y}}_{x}]}{t}}$$

${\bf{Y}}_{\phi_{t}x}$ is the tangent vector of the vector field $\bf{Y}$ at the point $\phi_{t}x$, where the point $\phi_{t}$ is obtained by starting at point $x$ at time $0$ and traversing along the orbit of $x$ to time $t$.

But I don't understand how to interpret $\phi_{t*}{\bf{Y}}_{x}$. Given the map $\phi_{t}$ which maps points $x$ in $M^{n}$ to points $\phi_{t}(x)$ in $M^{n}$ along the orbit of $x$ parameterised by time $t$, we can define the differential $\phi_{t*}$ that maps the tangent vector of the vector field $Y$ at $x$ to some tangent vector at the point $\phi_{t}x$. Now, there is only one tangent vector of the vector field $\bf{Y}$ at the point $\phi_{t}x$, and this tangent vector is the vector ${\bf{Y}}_{\phi_{t}x}$. This seems to suggest that ${\bf{Y}}_{\phi_{t}x}$ and $\phi_{t*}{\bf{Y}}_{x}$.

What am I missing?

nightmarish
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  • $\phi_{t} (Y_x)$ is the image of the vector $Y_x$ under the map $\phi_{t}: TM\to TM$. This is not necessarily the same as $Y_{\phi_t x}$. However, there is something else strange with your definition. It seems that you are taking the limit of vectors that are at different base-points. In my optinion, you should pull them back to one basepoint ($x$ probably). – Peter Franek Oct 27 '16 at 09:08
  • Is there only one map $\phi_{t*}:TM\rightarrow TM$ that can be defined for a given map $\phi:M\rightarrow M$? I ask this question as there is an infinite number of tangent vectors at any given point of a manifold. – nightmarish Oct 27 '16 at 09:19
  • (1) Yes, seems like a small mistake. but the formulas on page 126 are already meaningful. (2) If you have any smooth map $f: M\to M$, it induces a map $f_{}: TM\to TM$; it maps vectors in $TM_x$ to vectors in $TM_{f(x)}$. In our case, for each $t$* (at least for small enough $t$) you have a map $\phi_t: M\to M$ and hence a map $\phi_{t*}: TM\to TM$. – Peter Franek Oct 27 '16 at 09:27
  • But you didn't answer my question. Can we not define an infinite number of maps $f_{*}:TM\rightarrow TM$ for any given map $f:M\rightarrow M$? I say this because there an infinite number of tangent vectors at any one point of a manifold, and it is possible to map any tangent vector at a point $x$ to an infinite number of tangent vectors at a point $f(x)$. – nightmarish Oct 27 '16 at 09:40
  • No, for one $f$, there is only one $f_{*}$. It maps every tangent vector in every point to exactly one image (in other words, it is a function). – Peter Franek Oct 27 '16 at 09:42
  • I know the definition of a function and I know that $f_{}$ is a function. This is what I meant: for a given map $f:M\rightarrow M$, I think there are an infinite number of maps $f_{}:TM\rightarrow TM$, because of the following: consider tangent vector $\bf{v}$ at point $x$ and tangent vectors $\bf{w}$ and $\bf{y}$ at point $y$. Map the tangent vector $\bf{v}$ at $x$ to the tangent vector $\bf{w}$ at $y$ (where $\phi$ maps $x$ to $y$). Now choose the mapping of the remaining tangent vectors in $M$ arbitrarily. – nightmarish Oct 27 '16 at 09:50
  • Another possible map $\phi_{*}$ is where you map the tangent vector $\bf{v}$ at $x$ to the tangent vector $\bf{y}$ at $y$ and choose the mapping of the remaining tangent vectors arbitrarily. – nightmarish Oct 27 '16 at 09:50
  • Seems like there is some confusion going on. $TM$ is the set of all tangent vectors in all points of $M$. A function $TM\to TM$ maps every element of $TM$ to exactly one element of $TM$. $f_$ is not any* map that "covers" $f$, but a particular one, namely, the derivative.. – Peter Franek Oct 27 '16 at 09:53

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