# Computation in Petersen's Riemannian geometry

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 $$d(\text{vol}_{n-1}(\nabla r)) = 0$$?

• It’s always useful when first learning stuff like this to write everything in local coordinates. Commented Jun 13 at 17:43
• @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 $\mathbb{R}$. Commented Jun 13 at 17:47

There's some notational sloppiness here — 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 $$L_{\partial_r}(\lambda\,dr\wedge \text{vol}_{n-1}) = (\partial_r\lambda) dr\wedge\text{vol}_{n-1}$$.
• Hello! "On the other hand, 𝐿∂𝑟(vol𝑛−1)=0 because this 𝑛−1 form is independent of 𝑟." 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? Commented Jun 13 at 17:54
• 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 $r$. Commented Jun 13 at 17:59