= Solution
A <Riemannian metric> is a smooth field of positive-definite <symmetric bilinear forms> $g_x$ on the <tangent spaces>, equivalently a smooth section of $\operatorname{Sym}^2T^*M$ with that positivity. Its <musical isomorphism> is
$$
\boxed{\flat_g:TM\longrightarrow T^*M,\qquad v\longmapsto g(v,\cdot)}.
$$
Nondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse $\sharp_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing <determinant>. Both maps cover the identity on $M$, giving a smooth <vector bundle isomorphism>.
For <local flattening of a Riemannian metric>, choose a <manifold chart> neighborhood $V$ around $p$ with closure contained in $U$, and let $h=\phi^*\delta$ on $V$. Choose a <smooth bump function> $0\leq\chi\leq1$, with <compact support> in $V$ and equal to one on a smaller neighborhood $W$ of $p$. Set
$$
\boxed{\widetilde g=(1-\chi)g+\chi h\text{ on }V,\qquad\widetilde g=g\text{ off }V}.
$$
The formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\phi$ gives an <isometry> to a Euclidean open subset. Outside $U$ the original metric remains unchanged.
<Geodesic completeness> means that every <geodesic> with arbitrary initial point and tangent vector extends for every real affine parameter. On each <connected> component, the <Hopf-Rinow theorem> equates this with completeness of the <Riemannian distance>.
The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in\mathbb R^n:|x|>1\}$ with $n\geq2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The <geodesic> $\gamma(t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.
Here is the intended <compact-core completeness for a Euclidean end>. Write $\Phi:M\setminus X\to\{|x|>1\}$ for the end coordinates and assume additionally that
$$
K_R=X\cup\Phi^{-1}\{1<|x|\leq R\}
$$
is <compact> for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a <geodesic>. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the <compact> set $K_{R+sT}$. Its velocities also stay in a <compact> subset of $TM$, since their metric <norm> is fixed and the base set is <compact>. The smooth <geodesic> ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.
For <upward stability of Riemannian completeness>, lengths of all curves satisfy $L_{\widetilde g}\geq L_g$, hence $d_{\widetilde g}\geq d_g$. A $d_{\widetilde g}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\widetilde g$ has a bounded largest matrix <eigenvalue>, so the coordinate straight segment gives $d_{\widetilde g}(p,q)\leq C|\phi(q)-\phi(p)|$. Thus convergence to $p$ is also in $d_{\widetilde g}$. That distance is complete, and Hopf-Rinow yields
$$
\boxed{\widetilde g\geq g,\quad g\text{ complete}\quad\Longrightarrow\quad\widetilde g\text{ complete}}.
$$
For disconnected $M$, apply this argument on each component.
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