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Let $M^n$ be a Riemannian manifold with metric $g$ and $r : M \rightarrow \mathbb{R}$ be a smooth function such that $|\nabla r| = 1$. In a neighbourhood of $q \in M$ we can write the metric in polar coordinates via the exponential map such that $$g = dr^2 + g_r,$$where $g_r$ is the metric on $S^{n-1}$. The volume form is then $$\text{vol}_g = \lambda(r,\theta)dr \wedge \text{vol}_{n-1},$$with $\text{vol}_{n-1}$ being the volume form on $S^{n-1}$. Let $\partial_r := \nabla_r$. It is clear, by definition, that $$L_{\partial_r}\text{vol}_g = \text{div}(\partial_r)\text{vol}_g = \Delta r \text{vol}.$$On the other hand, I am trying to compute $$L_{\partial_r}(\lambda(r,\theta)dr \wedge \text{vol}_{n-1}),$$as the Lie derivative a $n-$tensor field. This is supposed to give $$\partial_r(\lambda),$$but I am not seeing how we have $$L_{\partial_r}dr = 0, \ \ \ \ \text{and } \ \ \ \ L_{\partial_r}\text{vol}_{n-1} = 0.$$I believe my question ends up being, why is $d(dr(\nabla r)) = 0$ and $d(\text{vol}_{n-1}(\nabla r)) = 0$?

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    $\begingroup$ It’s always useful when first learning stuff like this to write everything in local coordinates. $\endgroup$
    – Deane
    Commented Jun 13 at 17:43
  • $\begingroup$ @Deane My issue is with the distance function. I am not seeing if there is something obvious about how it relates with the coordinates on $S^{n-1}$ and on $\mathbb{R}$. $\endgroup$
    – user57
    Commented Jun 13 at 17:47

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There's some notational sloppiness here β€” perhaps yours, perhaps Petersen's. $g_r$ is the metric on the sphere of radius $r$, not $S^{n-1}$, but I believe $\text{vol}_{n-1}$ is indeed the volume form on the unit sphere, which does not depend on $r$. (The scalar function $\lambda$ gives the necessary factor to relate the spheres.) The Lie derivative of $\lambda\,dr\wedge\text{vol}_{n-1}$ is again an $n$-form, so I don't know how it's "suppposed to give" a scalar function.

But, yes, $L_{\partial_r}dr = 0$, for many reasons. For example, by Cartan's magic formula, $$L_{\partial_r}dr = \iota_{\partial_r}(d(dr)) + d(\iota_{\partial_r}dr) = 0 + d(1) = 0.$$ You can also verify the result directly by differentiating along the flow of $\partial_r$. On the other hand, $L_{\partial_r}(\text{vol}_{n-1}) = 0$ because this $n-1$ form is independent of $r$. This leaves us with $L_{\partial_r}(\lambda\,dr\wedge \text{vol}_{n-1}) = (\partial_r\lambda) dr\wedge\text{vol}_{n-1}$.

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  • $\begingroup$ Hello! "On the other hand, πΏβˆ‚π‘Ÿ(volπ‘›βˆ’1)=0 because this π‘›βˆ’1 form is independent of π‘Ÿ." But why isn't $r$, our distance function, dependent of $\theta$, where $\theta$ is a coordinate on $S^{n-1}$. I see that if we take distance function literally, then this is invariant per rotations. Is this function supposed to be rotationally invariant? $\endgroup$
    – user57
    Commented Jun 13 at 17:54
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    $\begingroup$ I don't know what "rotationally invariant" means in general. Again, I think you need to check Petersen's book to clarify very carefully where these things live and what $\text{vol}_{n-1}$ means. (Having retired, I no longer have a copy of the book at hand.) I'm saying that $\lambda(r,\theta)\text{vol}_{n-1}$ is the volume form on the sphere of radius $r$. $\endgroup$ Commented Jun 13 at 17:59

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