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The errata for "Riemannian Manifolds: An Introduction to Curvature" by John Lee has for one of the problems the following correction below.

Page 63, problem 4-3(b): Replace the first sentence by “Show that there are vector fields $V$ and $W$ on $R^2$ such that $V = W = \partial_1$ along the $x^1$-axis, but the Lie derivatives $L_V(\partial_2)$ and $L_W(\partial_2)$ are not equal on the $x^1$-axis."

I don't understand how $L_V(\partial_2)$ and $L_W(\partial_2)$ can be different on the $x^1$-axis. Shouldn't the flow at any point on the $x^1$-axis be the same for V and W and therefore $L_V(\partial_2)=L_W(\partial_2)$ on the $x^1$-axis.

user782220
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1 Answers1

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No. The value $(\nabla_X T)_p$ of a directional derivative of a field $T$ with respect to a vector $X$ depends only on the value $X_p$ of $X$ at $p$ and the value of $T$ along some arbitrarily short curve with tangent vector $X$.

On the other hand, the value $(\mathcal{L}_X T)_p$ of a Lie derivative of $T$ with respect to a vector field $X$ depends on the values of $X$ and $T$ in arbitrarily small neighborhoods of $p$. (Strictly, they depend on less, namely the $1$-jets of $X$ and $T$ at $p$.) This exercise illustrates exactly this difference.

Hint You can write any smooth vector field on $\Bbb R^2$ whose restriction to the $x^1$-axis is $\partial_{x^1}$ as $$U := \partial_{x^1} + x^2 [f \partial_{x^1} + g \partial_{x^2}]$$ for some smooth functions $f, g$. Now, compute $\mathcal L_U \partial_{x^2}$, and observe the dependence of that Lie derivative on $f$ and $g$.

Travis Willse
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  • As I understand the integral curve of $X$ depends on a neighborhood. But in this case for a point starting on the $x^1$-axis isn't it obvious that the integral curve will be just $f(t)=(t,0)$? I don't doubt that it is possible to construct examples showing the lie derivative can be different even if $X$ at $p$ stays the same. But I'm not understanding how the problem given can do that. – user782220 Mar 04 '19 at 18:34
  • An integral curve $\gamma$ of $X$ does not "depend on a neighborhood" in the sense that if one picks a function $h$ that vanishes on the image of $\gamma$, then $\gamma$ is also an integral curve of $X + h Y$ for any $Y$. The problem shows this non-independent just by explicit counterexample, i.e., just find suitable vector fields $V, W$ and compute. – Travis Willse Mar 04 '19 at 19:18
  • Right but the value of $L_{X_1}(V)$ depends on the integral curve of $X_1$. And if $X_1$ and $X_2$ both have the same integral curve on the x axis then shouldn't $L_{X_1}(V)$ and $L_{X_2}(V)$ reduce to just taking difference quotients of $V$ on the x axis. – user782220 Mar 04 '19 at 21:46
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    The Lie derivative of a vector field $U$ w.r.t. a vector field $X$ is defined to be $(\mathcal{L}X U)_p:=\lim{t\mapsto 0}\frac{T_{\theta_t(p)}\theta_{-t}\cdot U_{\theta_t(p)}-U_p}{t}$, where $\theta$ is the flow of $X$. If I understand you correctly, your point is that the quantity $U_{\theta_t(p)}$ only depends on ($U$ and) the integral curve of $X$ through $p$. This is true, but the pushforward map $T_{\theta_t(p)}\theta_{-t}:T_{\theta_t(p)}M\to T_pM$ depends not just on the flow map $\theta$ but also its derivative, and this really does depend on the value of $X$ off the integral curve. – Travis Willse Mar 05 '19 at 01:12
  • Ah I see the tangent space mapping changes how the vectors are compared. – user782220 Mar 05 '19 at 01:21
  • It might be an instructive exercise to compute explicitly the pushforward map $T_{\theta_t(0, 0)} \theta_{-t}$ for explicit $V, W$ as in the problem. – Travis Willse Mar 05 '19 at 17:43