Differential of the radial distance of a Riemannian manifold

Asked Dec 12 '19 at 12:49

Active Dec 12 '19 at 13:02

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Let $(M,g)$ be a Riemannian manifold with normal coordinate chart $(U,\varphi)$, $\varphi = (x^{1},...,x^{n})$, and centre $p \in M$.

I have shown that the differential of the radial distance function is $dr=\sum_{i=1}^{n} \frac{x^{i}}{r}dx^{i}$.

Define $\partial_{r}=\frac{x^{i}}{r}\frac{\partial}{\partial x_{i}}$. Let $q\in U\setminus\{p\}$.

How can I show that $dr_{q}(\partial_{r}|_{q})=1$?

I don't really understand here how you can take $\partial_{r}|_{q}$ to be the input for $dr_{q}$. I know that I can choose a radial geodesic $\gamma_{v}$ such that $\gamma_{v}(0)=p$ but I'm not sure how to use this.

edited Dec 12 '19 at 13:02

asked Dec 12 '19 at 12:49

user520830

Shouldn't this be true for every function? I mean, $dr_p(\partial_r)$ equals, by definition, $\partial_r(r)$. Of course it is $1$. You have surely seen the formula $dx^j(\partial_{x^i})=\delta_i^j$. – Giuseppe Negro Dec 12 '19 at 13:01
Sorry I may be missing something obvious here, but why does $dr_{p}(\partial_{r})=\partial_{r}(r)$? – user520830 Dec 12 '19 at 13:05
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By definition: $df_p(X):=X(f)(p)$, for every function $f$, vector $X$ and point $p$. – Giuseppe Negro Dec 12 '19 at 13:21
This post may be useful to understand why $\partial_r r =1$. – Giuseppe Negro Dec 16 '19 at 21:09

Differential of the radial distance of a Riemannian manifold

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