Polar form of the metric in normal coordinates

On $D\setminus \{0\}$ the norm $v\mapsto \mathrm{\mid }v{\mathrm{\mid }}_{g_p}$ is smooth, so composing it with the smooth inverse of [F1] proves that $r$ is smooth on $U\setminus \{p\}$. If $q={\exp }_p(v)$, $v\mathrm{\ne }0$, and $z\in T_{v}D\cong T_{p}M$, differentiation gives $d{r}_q(d{\exp }_v(z))=\frac{g_p(v,z)}{\mathrm{\mid }v{\mathrm{\mid }}_{g_p}}\mathrm{.}$

Put $e_r=v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_{g_p}$. By [F2], ${\mathrm{\partial }}_r=d{\exp }_v(e_r)$ has norm one. For every $X=d{\exp }_v(z)\in T_{q}U$, [F2] and step 1.1 give $g_q({\mathrm{\partial }}_r,X)=g_p(e_r,z)=d{r}_q(X)\mathrm{.}$ By the defining identity for the gradient and nondegeneracy of $g$, this proves $\mathrm{\nabla }r={\mathrm{\partial }}_r$ and $\mathrm{\mid }\mathrm{\nabla }r\mathrm{\mid }=1$.

Decompose uniquely $z=a{e}_r+z_{\perp }$, where $a=g_p(z,e_r)$ and $z_{\perp }\perp v$. Step 1.1 gives $a=d{r}_q(X)$, while [F2] makes $d{\exp }_v(z_{\perp })$ orthogonal to ${\mathrm{\partial }}_r$. Thus $X=d{r}_q(X){\mathrm{\partial }}_r+X_{\perp },X_{\perp }\in \ker d{r}_q\mathrm{.}$ For two vectors $X,Y$, bilinearity and the two vanishing cross terms give $g_q(X,Y)=d{r}_q(X)d{r}_q(Y)+g_q(X_{\perp },Y_{\perp })\mathrm{.}$ Since $dr\mathrm{\ne }0$, $\ker d{r}_q$ is the tangent space of the radial level hypersurface through $q$. This is exactly $g=d{r}^2+g_r$.

The centre is excluded because the norm need not be differentiable at zero; no polar formula is asserted there. In dimension zero $U\setminus \{p\}$ is empty. In dimension one the level tangent space is zero and the formula reduces to $g=d{r}^2$. An empty manifold has no centre. Positive and negative radial coordinate endpoints do not occur: $r>0$ on the stated domain, while arbitrary boundaries of the star-shaped set $D$ are not included. Assumption [A1] is inherited exactly through [F1]--[F2]; the unique orthogonal decomposition is a formula and requires no further choice.