Every geodesic segment is globally length minimizing

We construct the smooth quotient circle directly. If $I\subset \mathbb{R}$ is an open interval of length at most $1$, then $p{\mathrm{\mid }}_I$ is injective by [F3]: two distinct points of the open interval differ in absolute value by less than $1$ and hence cannot differ by a nonzero integer. Moreover $p^{-1}(p[I])={\bigcup }_{n\in \mathbb{Z}}(I+n)$ is open, and for every open $U\subset I$ the saturation $p^{-1}(p[U])={\bigcup }_{n\in \mathbb{Z}}(U+n)$ is open; hence $p[I]$ is open and $(p{\mathrm{\mid }}_I{)}^{-1}$ is a quotient chart. Distinct orbits admit disjoint chart intervals: for $[x]\mathrm{\ne }[y]$, the distance from $x-y$ to $\mathbb{Z}$ is positive by taking the smaller of the two positive distances to the adjacent integers, and intervals of less than one-third that size have disjoint quotient images. Images of rational intervals form a countable basis. On each component of an overlap, two lifted coordinates differ by a fixed integer translation, so these charts define a smooth boundaryless circle and the coefficient $g_{11}=1$ glues by [F4]. Therefore [F2] gives ${\mathrm{\Gamma }}^1{}_{11}=0$ in every such chart.

Define $\gamma :[0,1]\to S^1$ by $\gamma (t)=p(3t\mathrm{/}4)$. Its image lies in the quotient chart lifted from $(-1\mathrm{/}8,7\mathrm{/}8)$, where its coordinate is $\theta (t)=3t\mathrm{/}4$. Thus $\ddot{\theta }=0$, and [F2] and step 1.1 show that $\gamma$ is a geodesic segment. Its speed is constantly $3\mathrm{/}4$, so [F4] gives $L(\gamma )=3\mathrm{/}4$.

Define $c:[0,1]\to S^1$ by $c(t)=p(-t\mathrm{/}4)$. It joins the same endpoints because $p(-1\mathrm{/}4)=p(3\mathrm{/}4)$ by [F3]. Its image lies in the chart lifted from $(-3\mathrm{/}8,1\mathrm{/}8)$, where its coordinate is $-t\mathrm{/}4$ and its speed is constantly $1\mathrm{/}4$, so [F4] gives $L(c)=1\mathrm{/}4<3\mathrm{/}4=L(\gamma )$. Hence the geodesic segment $\gamma$ is not globally length minimizing.

More generally, a lifted change $a$ with $1\mathrm{/}2<a<1$ has length $a$, while the complementary lift $a-1$ has length $1-a<a$; equality occurs exactly at $a=1\mathrm{/}2$. This locates the failure at the half-circumference threshold: the local coordinate equation still makes the long arc geodesic, but the quotient supplies another lift of its endpoint. By step 1.1, the geodesic with initial velocity $v$ at $[0]$ is $t\mapsto p(tv)$, so uniqueness in [F1] gives ${\exp }_{[0]}(v)=p(v)$. Hence the initial vector $3\mathrm{/}4$ lies in no symmetric normal ball on which the exponential is injective: any such ball containing $3\mathrm{/}4$ also contains $-1\mathrm{/}4$, and [F3] gives the same image for those vectors. Thus [F1]'s local theorem does not apply to the long radial representative. Empty and zero-dimensional manifolds provide no counterexample, while this nonconstant one-dimensional witness has included endpoints and no degenerate interval. Assumption [A1] is used only for the uniqueness and local-minimality comparison in this step; steps 1.1--3.1 make no choice.