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I am reading Lee's book of riemannian geometry and he asks to show that Lie derivative of two vector fields on a riemannian manifold is not a connection.

How can I argue that this is true?

He also asks to show that there is a vector field $V$ on $\mathbb{R}^2$ such that $V$ vanishes on the $x$-axis but $\mathcal{L}_{\partial_x}V$ does not.

This was a confusion to me too.

I can take for example: $V = x\partial_x.$

Then $$[\partial_x,x\partial_x]f = \partial_x(x\partial_xf) - x\partial_x(\partial_xf) = \partial_xf + x\partial^2_{xx}f - x\partial_{xx}^2f = \partial_xf.$$

Then $$[\partial_x,x\partial_x] = \partial_x.$$

And that is a possible solution.

Is this right?

L.F. Cavenaghi
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2 Answers2

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Well, isn't it due to the definition of connection and a lie derivative of a vector field? Let $X$ and $Y$ be smooth vector fields and $\lambda$ a smooth function.

Connection: $\nabla_{(\lambda \, X)} Y = \lambda \, \nabla_X Y$

Lie derivative: $\mathcal{L}_{(\lambda \, X)} Y = [{(\lambda \, X)} , Y ] = {(\lambda \, X)} Y - Y{(\lambda \, X)} = \lambda \, X Y - \lambda \, Y X - Y(\lambda) \, X =\lambda [X,Y] - \left(\mathcal{L}_{ Y} \lambda \right) \, X = \lambda \, \mathcal{L}_{ X} Y -\left(\mathcal{L}_{ Y} \lambda \right) \, X $

So basically any function $\lambda$, a vector field $Y$ such that $Y(\lambda) \neq 0$ and a vector field $X$ non-commuting with vector filed $Y$ will produce a counterexample.

Futurologist
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Your example for the second problem would be OK, except that I suppose your $V$ vanishes on the $y$-axis instead of the $x$-axis.

Your example is also the sort of thing you should think about to solve your first problem. The axiom $\nabla_{fX} Y = f \nabla_X Y$ for a connection is actually equivalent to saying that, for fixed $p \in M$, the map $Y \mapsto (\nabla_XY)(p)$ only depends (linearly) on $X(p)$. This equivalence follows from the divisibility properties of smooth functions (if $f : \mathbb{R}^n \to \mathbb{R}$ is smooth and vanishes at the origin, then one can write $f$ as a finite sum $\sum f_i g_i$ where all functions are smooth and the $f_i$ vanish at the origin).

The conclusion is that connections have the following property: $(\nabla_XY)(p)$ must be zero if $X(p)=0$. The Lie derivative does not have this property, as you can discover by thinking about your example.

Mike F
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  • thanks @Mike! I interpreted wrong what was asked. So what does mean in fact that a vector field vanishes along the an axis? – L.F. Cavenaghi Aug 28 '16 at 17:48
  • Reading your post again, I suppose the statement is a little bit ambiguous, but I suspect they want $V(p) = 0$ for all $p$ on the $x$-axis. – Mike F Aug 28 '16 at 17:51
  • yes! Indeed. It was confusing to me. Thank you so much for answering. I will think about it. – L.F. Cavenaghi Aug 28 '16 at 17:52
  • My example answers the first question! Since $V(0,y) = 0$ but $\mathcal{L}{\partial_x}V(0,y) \neq 0.$ But, I cannot produce and example of a vector field that vanishes on $x-$axis but such that $\mathcal{L}{\partial_x}V$ does not. Can you help me? – L.F. Cavenaghi Aug 28 '16 at 18:31
  • Re. the first question: you are close.... for a connection, one has $X(p) = 0 \Rightarrow (\nabla_XY)(p) = 0$, but one does not have $Y(p)=0 \Rightarrow \nabla_XY(p)=0$. So, what you wrote does not give a contradiction. However, the problem is easy to fix by recalling $\mathcal{L}_XY = - \mathcal{L}_YX$. – Mike F Aug 28 '16 at 18:56
  • thanks for the observation that $\mathcal{L}_XY = - \mathcal{L}_YX.$ This solves one. The second does not work since If I am not mistaken $[\partial_x,V] = y\partial_x$ that vanishes if $y = 0.$ – L.F. Cavenaghi Aug 28 '16 at 19:02
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    I don't think it's possible to give an example. Let's say $V = f \partial_x + g\partial_y$. I get that $\mathcal{L}_XV = \frac{\partial f}{\partial x} \partial_x + \frac{\partial g}{\partial x} \partial_y$. If $V$ vanishes on the $x$-axis, then $f=g=0$ along the $x$-axis. In particular, $f$ and $g$ are constant along the $x$-axis, which implies $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial x} =0$ along the $x$-axis, so $\mathcal{L}_X V$ vanishes there too. – Mike F Aug 28 '16 at 19:12
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    you are right. I checked and there is an errata. Thanks. – L.F. Cavenaghi Aug 28 '16 at 20:16