I am reading certain proof ( John Lee's Introduction to riemannian manifold, Theorem 11.14 ; refer to question Q.3-2) in my questions Understanding the Gunther's Volume comparison theorem ( John Lee's Introductino to Riemannian manifold ) ) and some question arises.
Let $(M, g)$ be a connected riemannian manifold of dimension $n$ and $\psi:=\operatorname{exp}_p|_V^{U} : V:=B_{\delta}(0) \subseteq T_p M \to U:=\operatorname{exp}_p(B_{\delta}(0))$ the exponential map to a geodesic ball.
Let $\varphi : U \xrightarrow{\cong} W\subseteq \mathbb{R}^{n}$ be the assoicated normal coordinates, where the $W$ is the open ball in $\mathbb{R}^{n}$ of radius $\delta$ centered at the origin (C.f. Simple question on normal coordinates on geodesic ball ( image of normal coordinate on geodesic ball can be also ball ? ) ) Consider pull back metric $(W, (\varphi^{-1})^{*}g|_U)$. (Well-defined?)
Q. Then by question is, each $(U,g|_U)$ or $(W, (\varphi^{-1})^{*}g|_U)$, as regarded an open submanifolds, has constant sectional curvatures $0$, or more strongly(?), has constant Riemannian curvature $0$ ?
If the exponential map $\psi : \operatorname{exp}_p$ is local isometry, then from the local isometric invariance of Riemann curvature tensor and the surjectivity of $\psi$, we maybe show the zero constant Riemannian curvature (of $(U,g|_U)$). True? But if not ( C.f. Exponential map as a local radial isometry ), how can we show this? My queston is true?
And more general, how about when the $U$ is just arbitrary normal neighborhood ?
Can anyone help?
