Uniform convergence of $\sqrt { \det g' (\rho_n \omega)} \to 1$ , where $g' := (\varphi^{-1})^{*} g|_U$.

Question

Let $(M,g)$ be a (connected) Riemannian manifold of dimension $n$

Let $\varphi : U:= \exp_p(B_{\delta_0}(0)) \subseteq M \xrightarrow{\cong} B_{\delta_0}(0) \subseteq \mathbb{R}^{n}$ be a normal coordinate chart of a geodesic ball centered at $p \in M$ ( c.f. Simple question on normal coordinates on geodesic ball ( image of normal coordinate on geodesic ball can be also ball ? ) )

(Notational caution : the left $B_{\delta_0}(0)$ in the definition of $U$ is a subset in $T_pM$, not a subset in $\mathbb{R}^{n}$.)

Let $g' := (\varphi^{-1})^{*} g|_U$. Choose $\rho_n \in (0, \delta_0)$ such that $\rho_n \searrow 0$ ( right limit ). For each $\omega \in \mathbb{S}^{n-1}$, $\rho_n \omega \in B_{\delta_0}(0)$ and $(\rho_n \omega)_{n\in\mathbb{N}}$ converges to the origin in $\mathbb{R}^{n}$.

Then my question is, $\sqrt{ \det g'(\rho_n \omega)} \xrightarrow{ n \to \infty} 1$ uniformly on $\omega \in \mathbb{S}^{n-1}$? Note that $$g'_{ij}(0) := g'_{0}(\frac{\partial}{\partial x^i}|_{0}, \frac{\partial}{\partial x^j}|_{0}) = (g|_{U})_{\varphi^{-1}(0)}( (\varphi^{-1})_{*}(\frac{\partial}{\partial x^i}|_{0}), (\varphi^{-1})_{*}(\frac{\partial}{\partial x^j}|_{0})) =: (g|_U)_{p}( \frac{\partial}{\partial x^{i}}|_{p}, \frac{\partial}{\partial x^{j}}|_{p} ) =\delta_{ij},$$

since the $U$ is a normal coordinate chart centered at $p \in M$. Then how can we show the uniform convergence?

Can anyone help?

Answer 1

O.K. I think that for the proof key point is that $\sqrt{ \det g'}$ is continuous on $B_{\delta_0}(0)$. In particular, it is continuous at $0 \in B_{\delta_0}(0)$. So for every $\epsilon > 0$, there exists $\delta >0$ such that $$| \sqrt{\det g'(p)} -1 |= | \sqrt{\det g'(p)} - \sqrt{\det g'_{ij}(0)}| = | \sqrt{\det g'(p)} - \sqrt{ \det g'(0)}| < \epsilon$$ for all $p \in B_{\delta_0}(0)$ such that $|| p -0 || < \delta.$

Now we show that $\sqrt{\det g' (\rho_n \omega)} \xrightarrow{n\to \infty}1$ uniformly. To show this, we show that for each $\epsilon >0$, there exists $n_1 \in \mathbb{N}$ such that $| \sqrt{ \det g'(\rho_n \omega)} - 1 | < \epsilon $ for all $\omega \in \mathbb{S}^{n-1}$ and $n\ge n_1$.

Since $\rho_n \searrow 0$, there exists $n_1 \in \mathbb{N}$ such that $|\rho_n| < \delta$ for all $n \ge n_1$. Then $n_1$ plays such a role : for all $\omega \in \mathbb{S}^{n-1}$ and $n \ge n_1$, note that $|| \rho_n \omega-0|| = || \rho_n \omega || = |\rho_n| < \delta$ since $\omega \in \mathbb{S}^{n-1}$ (distance $1$ ) !

Correct ?